If Ṅ is the rate of disintegration the behavior of the system can be diagrammed as in Fig. 1. The amount of material that remains is a result of past disintegra-tion, and the present decay rate is a negative percentage of the amount of remaining material. Stating this mathematically: Ṅ = (-P/100)\textasteriskcenteredN(t)(1) The exact solution of equation (1) is: N(t) = N_0e^-Pt/100(2) Note that, as t \ding{217} \infty , N(t) \ding{217} 0 as expected since the "goal" of this system is complete disintegration of the material. The program uses the variable AMOUNT for N(t). DATA T/0.0/ COMMON P READ,N,TFIN,AOCUR,P,NO PRINT,T,NO REALN = N DT = TFIN/REALN AMOUNT = NO DO 10 I = 1,N T = T + ET CALL MEULER (T,AMOUNT ,ACCUR, DT) EXACT = N0\textasteriskcenteredEXP(-P\textasteriskcenteredT/100.) ERROR = ABS ( (EXACT-AMOUNT)/EXACT) \textasteriskcentered100. PRINT, T, AMOUNT, EXACT, ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON P G = (-P/100)\textasteriskcenteredW RETURN END

Question:

Acyclindercontaining gas at 27\textdegreeC is divided into two parts of equal volume, each of 100 cm^3, and at equal pressure, by a piston of cross-sectional area 15 cm^2 . The gas is one part is raised in temperature to 100\textdegreeC; the other volume is maintained at the original temperature. The piston and walls are perfect insulators. How far will the piston move during the change in temperature?

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Solution:

The heating of one side of the cylinder increases the pressure of the gas in that portion. If the piston were fixed, the volumes on the two sides would stay equal and there would be a pressure difference across the piston. Since the piston is movable, it alters its position until there is no pressure difference between its two sides: the hotter gas expands and thus drops in pressure, and the cooler gas is compressed and thus increases in pressure. When equilibrium has been reached, the two pressures are equal at p_0; the cooler gas now occupies a volume smaller by an amountdVand the hotter gas a volume greater by a corresponding amountdV. The ideal gas equations for both compartments are p_0(V -dV) =nRT(1) p_0(V +dV) =nRT' where n is the number of moles in each compartment, T and T are the temperatures. Dividing (1) by (2), we get (V -dV)/(V +dV) = T/T' VT' -dVT' = VT +dVT V(T' - T) =dV(T + T') or(dV/v) = (T' - T) / (T' + T) Therefore, the change in the volume of each compartment is dV= [{(373 - 300)\textdegreeC}/ {(373 + 300)\textdegreeC}] × 100 cm^3 = 10.85 cm^3. The piston has an area of 15 cm^2. Hence it moves a distance of 10.85 cm^3/15 cm^2 = 0.723 cm.

Question:

A space probe explodes in flight into three equal por-tions. One portion continues along the original line of flight. The other two go off in directions each inclined at 60\textdegree to the original path. The energy released in the explosion is twice as great as the kinetic energy possess-ed by the probe at the time of the explosion. Determine the kinetic energy of each fragment immediately after the explosion.

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Solution:

Take the direction in which the probe is moving immediately prior to the explosion as the positive x-direction and the point at which the explosion takes place as the origin of coordinates. Let M be the mass of the probe and let the quantities applicable to the probe and its fragments be given subscripts as in the diagram. Momentum must be conserved in the x-direction, the y-direction, and the z-direction independently. It follows that V_1^\ding{217}, V_2^\ding{217}, V_3^\ding{217}, and V^\ding{217} must be coplanar and that MV = (1/3) MV1+ (1/3) MV_2 cos 60\textdegree+ (1/3) MV3cos 60\textdegreeand (1/3) MV2sin 60\textdegree= (1/3) MV_3 sin 60\textdegree. From the second of these equation V_3 = V_2, and thus the first equation becomes MV = (1/3) MV_1 + (1/3) MV_2 cos 60\textdegree+ (1/3) MV_2 cos 60\textdegree = (1/3) MV_1 + (2/3) MV_2 cos 60\textdegree = (1/3) MV_1 + (2/3) MV_2(1/2) = (1/3) MV_1 + (1/3) MV2 ButE = (1/2) MV^2 = (M^2V^2)/(2M)orMV = \surd2ME. Similarly, E_1 = (1/2) [(1/3) M]V_1^2 = (M^2V2_1) / [(3)(2M)] = [(3 M^2V2_1) / (2M)](1/9)or (1/3) MV1= \surd[(2/3) ME_1] and(1/3) MV2= \surd[(2/3) ME_2] The first equation thus becomes \surd(2ME) = \surd[(2/3) ME_1] + \surd[(2/3) ME_2] or \surd(3E) = \surd(E_1) + \surd(E_2). \therefore3E = E_1 + E_2 + 2\surd(E_1E_2). The original kinetic energy of the probe plus the energy released by the explosion must equal the sum of the kinetic energies of the fragments, since no energy can be lost in the process. (That is, we assume an elastic "collision" occurs.) Hence E + 2E = 3E = E_1 + E_2 + E_3 = E_1 + 2E_2. \thereforeE1+ 2E_2 = E_1 + E2+ 2\surd(E_1E_3)orE_2 = 2\surd(E_1E_2). \thereforeE2_2 = 4E_1E_2orE_2 = 4E_1. Thus, 3E = E_1 + 2E_2 = E1+ 8E_1orE_1 = (1/3) E. \therefore E_2 = (4/3) E. Thus the fragment that continues in the line of flight has one-third of the original kinetic energy. The other fragments each have four-thirds of the original kinetic energy. The sum of these kinetic energies is three times the original kinetic energy, as required by the conserva-tion principle.

Question:

Determine the weights of CO_2 and H_2 O produced on burning 104 g. of C_2 H_2. Molecular weights are CO_2 = 44 and H_2 O = 18. The equation for the reaction is 2C_2 H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2 O

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Solution:

Two methods can be used to solve this problem. One is called the mole method and the other the proportion method. Mole method :According to the equation, 2molesof C_2 H_2 react with 5 moles of O_2 to produce 4 moles of CO_2 and 2 moles of H_2 O. Thus, in this problem, the first thing one has to determine is how many moles of C_2 H_2 are contained in 104 g. The molecular weight of C_2 H_2 is 26, by dividing the amount of C_2 H_2 present by the molecular weight of the com-pound, the number of moles present is found. number of moles of C_2 H_2 = [(number of grams of C_2 H_2) / (molecular weight of C_2 H_2)] number of moles of C_2 H_2 present = [(104 g) / (26 g/mole)] = 4 moles. In the equation 2 moles of C_2 H_2 are burned to form moles of CO_2 and 2 moles of H_2 O. Here there are 4 moles of C_2 H_2 present, which is twice the amount in the empirical equation. Therefore, twice as much CO_2 and H_2 O will be formed. 8 moles of CO_2 and 4 moles of H_2 O will be formed. 2 × (2C_2 H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2 O) = 4C_2 H_2 + 10O_2 \rightarrow 8CO_2 + 4H_2 O To find the weight of 8 moles of CO_2, the molecular weight of CO_2 is multiplied by 8. weight of CO_2 = number of moles present × molecular weight of CO_2. weight of CO_2 = 8 moles × 44 g/mole = 352 g. The same procedure can be followed for H_2 O. weight of H_2 O = number of moles present × molecular weight of H_2 O weight of H_2 O = 4 moles × 18 g/mole = 72 g. Proportion method: An alternate method of solution to this problem is the proportion method in which molecular weights (multiplied by the proper coefficients) are placed below the formula in the equation and the amounts of substances (given and unknown) are placed above. Here one has for CO_2: 104 gX 2 C_2 H_2 + 5O_2 \rightarrow 4 CO_2 + 2 H_2 O 2 × 26 g4 × 44 g This becomes (104 g) / (52 g) = (X/176 g) Solving for X one has X = [(104 g) (176 g)] / [52 g] = 352 gr gr CO_2 CO_2 A similar method can be applied to the H_2 O 104 gX 2 C_2 H_2 +5O_2 \rightarrow4CO_2 +2 H_2 O 2 × 26 g2 × 18 g (104 g) / (52 g) = (X/36 g) X = [(104 g) (36 g)] / [52 g] = 72 g H_2 O.

Question:

A bicycle and its rider together weigh 200 lb. If the cyclist free-wheels down a slope of 1 in 100, he has a constant speed of 10 mph, and if he free-wheels down a slope of 1 in 40, he has a constant speed of 20 mph. Suppose that he free-wheels on the level while holding on to the back of a moving truck. Find the power ex-pended by the truck in maintaining his speed at 15 mph. Assume that air resistance varies as the square of his speed, while frictional forces remain constant at all times.

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Solution:

Let the frictional force be F^\ding{217} and the force of air resistance F'^\ding{217} with magnitude kv^2, where k is a constant and v is the speed of the bicycle. On a slope, the forces acting on the bicycle and rider are the weight W^\ding{217} acting downward, which can be resolved into components parallel to and perpendicular to the slope, the normal force exerted by the slope N^\ding{217}, and the forces of friction and air resistance act-ing up the slope opposing the motion. The forces per-pendicular to the slope are equal and opposite and are of no further interest. Since the bicycle is moving with constant speed, the forces parallel to the slope must also cancel out. Hence W sin \texttheta = F + F' = F + kv^2. For the two cases given, values can be inserted. Thus 200lb × (1/100) = F + k × 10^2 mi^2 \bullet hr^-2and 200lb × (1/40) = F + k × 20^2 mi^2 \bullet hr^-2. \therefore2 lb = F + 100k mi2\bullet hr^-2and 5 lb = F + 400k mi2\bullet hr^-2. \therefore300k mi^2 \bullet hr^-2 = 3 lbor k = 1/100 lb \bullet(mph)^-2 and F = 1 lb. For the case of the bicycle traveling on a level surface, a force P^\ding{217} must be supplied to overcome the forces of friction and air resistance and keep the bicycle moving with constant speed. Since there is no acceleration, P^\ding{217} must just balance F^\ding{217} and F'^\ding{217} , or P = F + kv^2 = 1lb + (1/100) lb \bullet (mph)^-2 × 225 (mph)^2 = 3.25 lb. The rate of working (the mechanical power) is P^\ding{217} \textbullet v^\ding{217} and v = 15 mph = 22 ft \textbullet s^-1. \therefore P × v = 3.25 lb × 22 ft \textbullet s^-1. But 1 hp = 550 ft \textbullet lb \textbullet s^-1 Therefore: The rate of working = [(3.25 × 22 ft \textbullet lb \textbullet s^-1)/(550 ft \textbullet lb \textbullet s^-1/hp)] = [(3.25 × 22)/550] hp = 0.13 hp.

Question:

Calculate the velocity of a transverse pulse in a string under a tension of 20 1b if the string weighs 0.003 1b/ft.

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Solution:

Consider the section of string shown in figure A, which is under the tension T. This segment can be interpreted as being part of the pulse being trans-mitted along the string. The net force acting in the y direction is F_y = T sin \textthetax + dx\rule{1em}{1pt} T sin \texttheta_x Since \texttheta_x and \texttheta_x _+ _dx are small, we have approximately, sin \texttheta_x \approx \texttheta_x \approx tan \texttheta_x sin \textthetax + dx\approx \textthetax + dx\approx tan \textthetax + dx Therefore, F_y = T(tan \textthetax + dx\rule{1em}{1pt} tan \texttheta_x) buttan \textthetax + dx= (\partialy/\partialx)x + dx tan \texttheta_x = (\partialy/\partialx)_x(See fig. B) \partialy and \partialx have the same meaning as dy and dx, they are small increments of y and of x, respectively. The \partial symbol indicates that more than one variable is under consideration. Hence, F_y = T [(\partialy/\partialx)x + dx\rule{1em}{1pt}(\partialy/\partialx)_x](1) Now, by Taylor's theorem, given a continuous and differentiable function G(x), G(x + dx) = G(x) + (\partialG/\partialx) dx + ... G(x + dx) \rule{1em}{1pt} G(x) \approx (\partialG/\partialx) dx(2) Higher order terms have been neglected because dx is so small that terms involving (dx)^2 or higher must ne-cessarily be negligible. If G(x) = (\partialy/\partialx), then we have from (2) (\partialy/\partialx)x + dx- (\partialy/\partialx)_x = (\partial/\partialx) (\partialy/\partialx) dx = [(\partial^2 y)/(\partialx^2)] dx Hence, (1) become F_y = T [(\partial^2 y)/(\partialx^2)] dx Denote the mass per unit length of the string by \mu (i.e. \mu = dm/dx). The mass of the segment dx is then m = \mudx. By Newton's second law, the vertical accelera-tion of the string segment, a_y, is F_y = ma_y T [(\partial^2y)/(\partial x^2)] dx = (\mudx) [(\partial^2 y)/(\partial t^2)] [(\partial^2 y)/(\partial x^2)] = (\mu/T) [(\partial^2 y)/(\partial t^2)](3) Equation (3) is of the form of the wave equation [(\partial^2 y)/(\partial x^2)] = (1/v^2) [(\partial^2 y)/(\partial t^2)] where y(x, t) is the displacement of the wave at position x and at time t, and v is the velocity of the wave. For the transverse wave, then v = \surd(T/\mu) Returning to our original problem, T = 20 1b, \mu = (0.003/32) {(1b/ft)/(ft/sec^2)}, v = \surd[(20 1b × 32 ft/sec^2)/(0.003 1b/ft)] = 461 (ft/sec).

Due to the fact that the charge is uniformly distributed over the surface of the sphere (and therefore symmetric), we realize that the electric field must have the same strength at any point a distance r away from the center of the charged sphere. Furthermore, we expect the lines of E^\ding{217} to begin on the positive surface charges and emanate radially from the surface of the sphere. Consider a spherical surface of radius R and with the same center as the charged sphere. If the field at this surface is E, then the total electric flux through the surface is \varphi_E = \intE^\ding{217} \textbullet dS^\ding{217} = \int E dS because E^\ding{217} and dS^\ding{217} are parallel. Hence, \varphi_E = 4\piR^2E where 4\piR^2 is the surface area of the spherical surface. Since the charge enclosed by the surface is q, then by Gauss's Law, \varphi_E = q/\epsilon_0 . Therefore, 4\piR^2E = q/\epsilon_0 from which E = [1/(4\pi\epsilon_0)] (q/R^2) = K_-E (q/R^2) Thus the field outside the sphere is just as if all the charge q were located at the center of the sphere.

Question:

A sports car weighing 1200 lb and traveling at 60 mph fails to stop at an intersection and crashes into a 4000-lb de-livery truck traveling at 45 mph in a direction at right angles to it. The wreckage becomes locked and travels 54.7 ft before coming to rest. Find the magnitude and direction of the constant force that has produced this deceleration.

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Solution:

Let the sports car be traveling in the positive x-direction and the truck in the positive y-direction. After the collision at the origin, the combined mass travels in a direction inclined at \texttheta\textdegree to the positive x- axis with a velocityV. Momentum must be conserved in both the x- and y- directions. Therefore, referring to the diagram, m_1v_1 = (m_1 + m_2)V sin \texttheta(1) andm_2v_2 = (m_1 + m_2)V cos \texttheta(2) Dividing equation (1) by (2), [(m_1 + m_2) V sin \texttheta] / [(m1+ m_2) V cos \texttheta] = (m_1v_1)/(m_2v_2) tan \texttheta = (m_1v_1)/(m_2v_2) = (m_1gv_1)/(m_2gv_2) = (4000 Ib × 45 mph)/(1200 Ib × 60mph) = 2.5 \therefore \texttheta = 68.2\textdegree Squaring equations (1) and (2) and then adding them, we get (m_1 + m_2)^2 v^2 sin^2 \texttheta + (m_1 + m_2)^2 V^2 cos2\texttheta = m^2 _1v^2 _1 + m^2 _2v^2 _2 V2( sin^2 \texttheta + cos^2 \texttheta) = V^2 = [(m^2 _1v^2 _1^ + m^2 _2v^2 _2)] / [(m_1 + m_2)^2] = (m^2 _1g^2v^2 _1 + m^2 _2g^2v^2 _2) / (m_1g + m_2g)^2 = [(4000 Ib)^2 × (45 mph)^2 + (1200 Ib)^2 × (60mph)^2] / [(4000 Ib + 1200 Ib)^2] = 1389.9 (mph)^2. \therefore V = 37.3 mph = 54.7 ft/s, which is the velocity of the combined mass immediately after impact. The wreckage comes to rest in 54.7 ft. Apply the equation for constant acceleration, v^2 = v^2 _0 + 2as. Here V_0 is the initial velocity of the locked mass, s is the distance it travels, and a is the applied acceleration. Hence, when v = 0, s = 54.7 ft and 0 = (54.7 ft/s)^2 + 2a × 54.7 ft. \thereforea = - 27.35 ft/s^2. The deceleration due to friction is thus 27.35 ft/s^2, and since the mass affected is (1200 + 4000)/32 slugs, the magnitude of the frictional force is F = [(5200)/(32)] slugs × 27.35 ft/s^2 F = [(5200)/(32)] slugs × 27.35 ft/s^2 = 4443 l = 4443 l b. This decelerating force must act in a direction opposite to that in which the wreckage is traveling in order to bring it to rest. Thus it is a force of 4443 lb acting at an angle of 68.2\textdegree to the negative x-axis. Note that momentum is conserved only during the collision, for, at that time, the collision forces are much greater than the external forces acting (friction), and the latter may be neglected.

Question:

Most aquatic animals have a special sensory system used fordetecting water vibrations. Describe the anatomy of this system.

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Solution:

This system, peculiar to most aquatic vertebrates, consists of sensorycells arranged in a linear fashion. In cyclostomes, amphibians, andsome other forms, the lines of sensory cells are exposed on the body surface. Usually they are embedded in canals, which may remain open as agroove (e.g., in primitive sharks), or more often are closed over, with poresat intervals. In most bony fish, the canals run through the scales. These canals are the canals of the lateral-line system and consist of the mainlateral-line canal and four main branches on the head. The head canalsare supplied by branches accompanying the facial nerve. The lateral-linesystem is found in cyclostomes, aquatic larvae of amphibians, andin most other fishes. The sense organs of the lateral line system consist ofneuromasts, whichare clusters of sensory and supporting cells. The sensory cells have aterminal hair projecting into the lumen of the canal and are innervated by nervesof the system. The sensory cells functionally resemble those of the internalear in terrestrial vertebrates.

Question:

Calculate the frequency of oscillation of the configura-tion shown in the figure. All surfaces are frictionless.

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Solution:

When the mass m moves, one spring is always stretched, and the other is always compressed by the same length. Thus: ∆x_1 = -∆x_2 where ∆x_1 is the distance that the spring with force constant k_1 stretches, and ∆x_2 that of the other spring. Note that a negative stretching distance represents a distance compressed. We denote the distance of the mass to the right of the origin 0 by ∆x. Thus: ∆x_1 = ∆x ∆x_2 = -∆x taking positive displacement as pointing to the right. The force on the mass at any time is therefore: F= -k_1 ∆x_1 - (-k_2 ∆x_2) = -k_1 ∆x + k_2 (-∆x) = -(k_1 + k_2) ∆x = -k' ∆x where we let k' = k_1 + k_2 Since the frequency of an oscillator having force constant k ׳ is: \nu = 1/(2\pi)\surd(k'/m) then:\nu = 1/(2\pi)\surd{(k_1 + k_2)/m}

Under the influence of primarily FSH from the pituitary, a single ovarian follicle containing an ovum reaches maturity in about two weeks. During the second week the follicle cells and other ovarian cell types are stimulated by FSH to increase their secretion of estrogen. Near the middle of the cycle, or about the fourteenth day, the heightened level of estrogen in the blood reaches a critical value which stimulates the hypothalamus to signal the pituitary to release a surge of LH. This surge of LH induces rupture of the ovarian follicle. This releases the mature ovum into the fallopian tube of the uterus, a process called ovulation. Estrogen secretion decreases for several days following ovulation, perhaps because of the negative feedback effect of excess estrogen on FSH, which stimulates its production. The ruptured follicle is rapidly transformed into the corpus luteum, which secretes progesterone and, in lesser amounts, estrogen. The estrogen secreted by the corpus luteum raises its level in the blood, which inhibits further FSH secretion. Progesterone suppresses additional LH production, and thus prevents additional ovulation from occurring. The combined influence of progesterone and estrogen thickens and prepares the uterus for implantation of the embryo (in the event of fertilization). Failure of oocyte fer-tilization leads to the degeneration of the corpus luteum during the last few days of the cycle. The disintegrating corpus luteum is unable to maintain its secretion of estrogen and progesterone, and their blood concentrations drop rapidly. The marked decrease of estrogen and proges-terone lead to degeneration of the uterine wall followed by sloughing off of the uterine surface tissue. It also results in the removal of inhibition of FSH secretion.The blood concentration of FSH begins to rise, follicle and ovum development are stimulated, and the cycle begins anew. The cycle is usually broken up into three general phases. The first phase, encompassing follicle development up to ovulation, is called the follicular phase and lasts about 9 days FSH and estrogen dominate during the phase. The second phase, lasting from ovulation until the beginning of the disintegration of the uterine lining, is called the luteal phase and lasts approximately 14 days. Progesterone dominates during this phase. Following this phase is the flow phase which lasts 5 days, and is characterized by bleeding due to the sloughing off of the uterine lining.

The eye records the direction from which a light ray enters it as though it has traveled in a straight line from the source. It cannot compensate for refraction of the light. Therefore, the observer sees the dot 0.640 cm below the upper surface of the glass when in actuality it is 1.00 cm below the sur-face. Assuming the index of refraction of air to be one (n_air = 1.00029), we have from Snell's law, n_glass sin \texttheta_glass = n_air sin \texttheta_air(1) n_g = (n_g/n_a) = (sin \texttheta_a/sin \texttheta_g)since n_a \approx 1(2) The sine function can be expressed as sin \texttheta = tan \texttheta cos \texttheta(3) We know from that figure that tan \texttheta_a = (x / 0.64 cm) tan \texttheta_g = (x / 1.00 cm) From the trigonometric relation cos \texttheta = \surd(1 - sin^2 \texttheta) we have cos \texttheta_a = \surd(1 - sin^2 \texttheta_a) cos \texttheta_g = \surd(1 - sin^2 \texttheta_g)(4) From equation (1), Sin \texttheta_g = (n_a/n_g) sin \texttheta_a(5) Substituting equation (5) into equation (4), Cos \texttheta_g = \surd[1 - (n_a/n_g)^2 sin^2 \texttheta_a Using the values found for the cosines and tangents of the angles and equation (3), substitute into equation (2) n_g = (sin \texttheta_a/sin \texttheta_g) = (tan \texttheta_a cos \texttheta_a/tan \texttheta_g cos \texttheta_g) = {(x / 0.64 cm) \surd(1 - sin^2 \texttheta_a)} / {(x / 1.00 cm) \surd[1 - (n_a/n_g)^2 sin^2 \texttheta_a]}(6) The observer looks at the dot through a microscope. This means that he looks at the dot from directly above and \texttheta_a must be very small. We can then make the approximation sin \texttheta_a \approx sin 0\textdegree = 0 Substituting this value in equation (6), we find n_g = {(x/0.64 cm)\surd(1 - 0)} / {(x/1.00 cm)\surd[1 - (n_a/n_g)^2 (0)]} = (1.00/0.64) = 1.57

This is a case of motion in a straight line with constant acceleration. In the figure suppose that at the zero of time the body is at the origin and is mov-ing with a velocity v_0^\ding{217}. Suppose that at time t it has moved through a distance x^\ding{217} and its velocity has changed to v^\ding{217}. If a^\ding{217} is the constant acceleration, then we can derive an equation relating v to x. From the equations v = v_0 + atandx = v_0t + (1/2)at^2 we must eliminate t. Since t = (v - v_0)/a ,then x = v_0 [(v - v_0)/a] + (1/2) a [(v - v_0)/a]^2 = (v^2 - v^2_0)/2a The work done by the force F^\ding{217} in moving the body from the origin through a distance x^\ding{217} is Work Done = Fx = [F(v^2 - v^2_0)]/(2a) But from Newton's second law F = Ma. So work done = [Ma(v^2 - v^2_0)] / [2a] = (1/2)M (v^2 - v^2_0) = (1/2) Mv^2 - (1/2)MV^2_0 Work done = Final kinetic energy - initial kinetic energy.

Question:

The formula Z = [(e^ax- e ^-ax)/ 2] sin (x + b) + a log [(b + x) / 2] is to be evaluated for all combinations of x: 1.0(0,1) 2.0 a: 0.10(0.05)0.80 b: 1.0(1.0) 10.0 where x: 1.0(0.1) 2.0 means x = 1.0, 1.1, 1.2,...,2.0 and so on. (There are 11 × 15 × 10 = 1650 combinations.) For each combination a line giving x, a, b, and Z values is to be written. Write a program containing three DO loops to carry out this computation.

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Solution:

We need to obtain integral subscripts for our DO loops. This is no problem for b. For x, though, note that 10{_\ast}X takes on values 10., 11.,...,20. Thus if I = 10, 11,...,20, X= FLOAT(I)/10. Similarly, 100{_\ast}A takes on values 10.0, 15.0, 20.0,...,80.0 in steps of 5. Let J = 10, 15, 20, ,..,80. So A = FLOAT (J)/100. [Remember that the library function FLOAT changes an integer number into a real number.] Hence we obtain the program below. DO 100 I = 10,20 X = FLOAT (I)/10.0 DO 100 J = 10,80,5 A = ELOAT(J)/100.0 DO 100 K = 1,10 B = FLOAT(K) Z = 0.5{_\ast}(EXP(A{_\ast}X) - EXP(-A{_\ast}X)){_\ast}SIN(X+B) 1+ A{_\ast}AL0G ((B+X)/2.) WRITE (6,10) A,B,X,Z 10FORMAT (4X,3(2X,F6.3), E13.7) 100CONTINUE STOP END

Question:

The diagram below is an example of a phase diagram for a pure substance. To what phases do the regions A, B, and C correspond?

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Solution:

Following the 1 atm constant pressure line from left to right, we are proceeding from low values of the temperature to high values. Therefore, we will intersect the three regions in the order solid-liquid-vapor. The regions A, B, C hence correspond to the solid, liquid, and vapor phases, respectively. Point a denotes the normal freezing (melting) point of the substance and the point b denotes the normal boiling point.

The process of digestion is the breakdown of large, ingested molecules into smaller, simple ones that can be absorbed and used by the body. The breakdown of these large molecules is called degradation. During deg-radation, some of the chemical bonds that hold the large molecules together are split. The digestive enzymes cleave molecular bonds by a process called hydrolysis. In hydrolysis a water molecule is added across the bond to cleave it. Within living systems, chemical reactions require specific enzymes to act as catalysts. Enzymes are very specific, acting only on certain substrates. In addition, different enzymes work best under unlike conditions. Digestive enzymes work best outside of the cell, for their optimum pHs lie either on the acidic (e.g., gastric enzymes) or basic side (e.g., intestinal and pancreatic enzymes). The cell interior, however, requires an almost neutral (about 7.4) pH constantly. Digestive enzymes are secreted into the digestive tract by the cells that line or serve it. Digestion begins in the mouth. Most foods contain polysaccharides, such as starch, which are long chains of glucose molecules. Saliva (and the intestinal secretions) contain enzymes that degrade such molecules. Salivary amylase, an enzyme that is also called ptyalin, hydrolyzes starch into maltose. (Compounds whose names end with -"ase" are enzymes, and those with the suffix "-ose" are sugars.) Glucose is eventually absorbed by the epithelial cells lining the small intestine. The saliva has a pH of 6.5 - 6.8. This is the optimal range for salivary enzyme activity. Food spends a relatively short amount of time in the mouth, and eventually enters the stomach. The stomach is very acidic, with a pH of 1.5-2.5. The acid is secreted by special cells in the lining of the stomach called parietal cells. The low pH is required for the activity of the stomach enzyme pepsin. Rennin coagulates milk proteins in the infant's stomach, making them more susceptible to enzyme attack. Pepsin is a proteolytic enzyme (protease): it degrades proteins. Pepsin starts the protein digestion in the stomach by splitting the long proteins into shorter fragments, or peptides, that are further digested in the intestine. Pepsin will split any peptide bond involving the amino acids tyrosine or phenylalanine. There are 20 different kinds of amino acids that can make up a protein and some proteins are thousands of amino acids long. The body needs the amino acids it obtains from digestion to synthesize its own proteins.

The Exclusive-OR function is defined as a function which gives an output 1 whenever either A or B is 1, but gives an output 0 whenever A and B are both 1 or both 0. The equivalence function of two variables A and B is defined as a function which gives an output 1 whenever both A and B are equal (both 1 or both 0), but gives an output 0 whenever A and B are unequal. These two functions can be represented by a Truth Table as follows: If A and B are binary variables, they can be represented in 4 different ways, giving us the 4 rows in the Truth Table. The Truth Table is shown in figure 1. Col. 1 Col. 2 Col. 3 VARIABLES AB EXCLUSIVE - OR A×B EQUIVALENCE A \equiv B 00 01 10 11 0 1 1 0 1 0 0 1 We note from the Truth Table of fig. 1 that the exclusive-OR: and the equivalence functions are complements of each other. Also, from the Truth Table, we can write the equations: a)Expressions for exclusive-OR A × B = A' \bullet B + A \bullet B' . (This is a sum of products form of the expression obtained by considering entries of column 2 which are equal to 1). Also, A × B = (A+B) \textbullet (A '+B') . (This is a product of sums form, obtained by considering entries of column two which are equal to 0.) b)Expressions for Equivalence: A \equiv B = A'\bulletB' + A\bulletB (Sum of products form) A \equiv B = (A+B') \textbullet (A'+B). (Product of sums form) We shall now realize the exclusive-OR and the equivalence functions, using a) NAND gates only, and, b) NOR gates only

Question:

What mass of calcium (Ca, atomic mass = 40.08 g/mole) mustbe combined with 1.00 g of phosphorus (P, atomic mass= 30.97 g/mole) to form the compound Ca_3p_2?

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Solution:

From the formula of the compound we are trying to form, Ca_3p_2, we seethat the ratio of moles of calcium to moles of phosphorus must be 3/2. By calculating the number of moles of P in 1.00 g, we can determine the re-quirednumber of moles of Ca and then convert this to a mass. The number of moles of P in 1.00 g is molesP = (mass P) / (atomic mass P) = 1.00g / 30.97 g/mole = 0.0322 mole. Then, using the ratio of (moles Ca / moles p) = 3/2 or moles Ca = 3/2 molesP, the number of moles of Ca required to combine with 0.0322 mole P in a 3/2 ratio is molesCa =3/2 molesP = 3/2 × 0.0332 mole = 0.0483 mole. To convert this to a mass we multiply by the atomic mass, obtaining massCa = moles Ca × atomic mass Ca = 0.0483 mole × 40.08 g/mole = 1.94 g Ca .

Question:

In aquatic animals, external respiration is carried out by specialized structures called gills. How do these gills operate?

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Solution:

Fish, molluscs and many arthropods (except insects) possess gills for respiratory purposes (see figure). Organisms such as these require a mechanism to keep a fresh supply of water flowing about them. This is necessary because gas exchange occurs between the blood vessels in the gills and the water which contains dissolved oxygen. Continuous bathing of the gills ensures that there will always be enough oxygen available for the organism. It also ensures that the carbon dioxide diffusing from the gills into the water is removed from the organism. This is important in that the CO_2 concentration gradient is maintained, so diffusion of CO_2 will continue to occur from the gills to the water. A fish opens its mouth and receives a quantity of water. It then closes its mouth and forces the water out past its gills by contracting its oral cavity. Gills have thin walls, are moist, and are well supplied with blood capillaries. Oxygen, which is dissolved in the water, diffuses though the gill epithelium and into the blood capillaries. Simultaneously, carbon dioxide diffuses in the opposite direction in accordance with the CO2 concentration gradient. The amount of oxygen dissolved in sea water is relatively constant, but the amount in freshwater ponds may fluctuate greatly. A fish would suffocate in water that is lacking sufficient amounts of dissolved oxygen. It is interesting to note that the direction of blood flow through a gill is opposite to the direction of water flow over the gill. This counter-current system maximizes the amount of gas exchange that can take place. If both the blood flow and water current were in the same direction, the CO_2 and O_2 concentration gradients across the gills would decrease. This would result in a slower rate of diffusion, and the amount of gas exchange would be reduced.

Write a program in the PDP-8 computer Assembler Language to find the factorial of numbers. The program should use Sub-routine jumps The numbers are stored in an array starting at location 300_8. The results, i.e., the values of the factorials are to be stored in another array starting at location 200_8. The end of the numbers array is indicated by the occurrence of a negative number. The program is to start at location 400_8.

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Solution:

Repetition of the procedure used in the previous problem leads to step 6, where the Euler predictor-corrector method will be used again. Transforming the differential equation to a simultaneous system of first-order equations: ẋ = v(t)(1b) ṿ = -[D{_\ast}v(t) + K{_\ast}x(t)]/M(2b) The relation can be diagrammed as shown in FIG.2 The only change in the program needed is substitution of the new expression for acceleration. Since this expression appears in two lines and is fairly long, a statement function is used. This has the additional effect of making the program more general. It can be applied to any second-order differential equation. Only the statement function definition and the I/O statements vary. The program: REAL K,M,N F(Y,Z) = -(D{_\ast}Y + K{_\ast}Z)/M READ K,M,N,T,TFIN,XCOR,VCOR,ACCUR,D DT = (TFIN - T)/N 50X = XCOR V = VCOR A = F(V,X) PRINT T,X T = T + DT IF(T.GT.TFIN) STOP XPRED = X + V {_\ast} DT VPRED = V + A{_\ast} DT 100XDOT = VPRED VDOT = F(V,XPRED) XCOR = X + 0.5 {_\ast}(V + XDOT) {_\ast} DT VCOR = V + 0.5 {_\ast}(A + VDOT) {_\ast} DT XDIF = ABS (XCOR - XPRED) VDIF = ABS(VCOR - VPRED) IF (XDIF .LE .ACCUR. AND. VDIF{_\ast} LE. ACCUR) GO TO 50 XPRED = XCOR VPRED = VCOR GO TO 100 END

Question:

In the popular nursery rhyme "Mary Had a Little Lamb", her lambwould follow Mary wherever she went. Discuss a possi-blebiological explanation.

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Solution:

The bond established between Mary and her lamb might have resultedfrom a form of learning called imprint-ing. The concept of imprintingwas first formulated byKonradLorenz, an Austrianethologist (ethologyis the study of behavior). Lorenz discovered that new-born geesewill follow the first moving object they see, especially if it makes noise. They form a strong and lasting bond to it, adopting the object as theirparent. Usually, the first moving object the bird sees is its mother. Thus imprinting serves to establish a bond between the young and their motherin nature. This type of response has important survival values, suchas keeping the young near their parent and establishing species recognitionand inter-action. Imprinting is characterized by having a short timeperiod during which it can occur. This critical period lasts only about 32 hours in ducklings. Lorenz demonstrated imprinting by allowing goslings (young geese) tosee him as the first moving creature. They then followed Lorenz as if he weretheir parent. Young birds of most species become independent when they attain sexualmaturity. If birds are reared by a mother of a different species, they attemptto mate with birds of the foster mother's species when sexually mature. Although imprinting was first demonstrated in birds, it is known to occurin sheep, goats, deer and other mammals. Orphan lambs reared by humansfollow them and show little interest in other lambs. That is why "every-wherethat Mary went, her lamb was sure to go."

Question:

Describe the interrelations between the Basic Assembler Language (BAL) Cross Reference Table and its Post Execution Storage Dump.

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Solution:

Along with other types of information, the Cross Reference Table provides the location of all the symbols used within the program. By finding this address in the Post Execution Storage Dump, the contents of a symbol can be easily found. In the Cross Reference Table, the address of the symbols are coded in hexadecimal (HEX). Not only are these addresses coded in hexadecimal in the Post Executional Storage Dump, but so is all other data found in this Dump. The leftmost column of the Dump contains addresses that consecutively increase by 32 bytes; here it is impor-tant to remember the dump is coded in HEX. To the right of the addresses are rows of 32 bytes. These bytes represent the contents of 32 consecutive bytes in core. The first byte of each row represents the content of the location listed in the address column to the left of it. Let's represent a location in the address column by the letters 'DCA'. By hexadecimally adding 1 to 'DCA' we can determine the content of the second byte in this specific row. For example so DCACONTAINS90 DCA + 1CONTAINSEC DCA + 2CONTAINSDO DCA + 30CONTAINS5A and DCA + 31CONTAINSCO Thus, by adding the required amount, the contents of any byte in the row can be determined. It should be noted again that the addition is a hexadecimal one. In summation, we can determine the contents of a symbol by obtaining its address and length from the Cross Reference Table and then locating the content of this symbol in the Post Execution Storage Dump. When the address is found in the Dump, the number of bytes to be observed is dictated by the length of the symbol. It should be noted that in the Dump the addresses are generally denoted by six hexadecimal digits, but on most occasions we need only concern ourselves with the last three. Also, to the right of the dump we usually find a stan-dard (English) translation of each hexadecimal row of the Dump. This translation confirms the contents of the Dump and makes it more comprehensible.

The heart is the muscular organ that causes the blood to circulate in the body. The heart of birds and mammals is a pulsatile four-chambered pump composed of an upper left and right atrium (pi., atria) and a lower left and right ventricle. The atria function mainly as entryways to the ventricles, whereas the ventricles supply the main force that propels blood to the lungs and throughout the body. Depending on where the blood is flowing from, it would enter the heart via one of the two veins: the superior vena cava carries blood from the head, neck and arms; the inferior vena cava carries blood from the rest of the body. The blood from these two veins enters the right atrium. When this chamber is filled with blood, the chamber contracts and forces the blood through a valve called the tricuspid valve and into the right ventricle. Since this blood has returned from its cir-culation in the body's tissues, it is deoxygenated and contains much carbon dioxide. It therefore must be transported to the lungs where gas exchange can take place. The right ventricle contracts, forces the blood through the pulmonary semilunar valve into the pulmonary artery. This artery is unlike most arteries in that it carries de-oxygenated blood. The artery splits into two, with one branch lead-ing to each lung. The pulmonary arteries further divide into many arterioles, which divide even further and connect with dense capil-lary networks surrounding the alveoli in the lungs. The alveoli are small sac-like cavities where gas exchange occurs. Carbon dioxide diffuses into the alveoli, where it is expelled, while oxygen is picked up by the hemoglobin of the erythrocytes. The capillaries join to form small venules which further combine to form the four pulmonary veins leading back to the heart. The pulmonary veins are unlike most veins in that they carry oxygenated blood. These veins empty into the left atrium, which contracts to force the blood through the bicuspid (or mitral) valve into the left ventricle. When the left ventricle, filled with blood, contracts, the blood is forced through the aortic semilunar valve into the aorta, the larg-est artery in the body (about 25 millimeters in diameter). The aorta forms an arch and runs posteriorly and inferiorly along the body. Before it completes the arch, the aorta branches into the coronary artery, which carries blood to the muscular walls of the heart itself, the carotid arteries, which carry blood to the head and brain, and the subclavian arteries, which carries blood to the arms. As the aorta runs posteriorly, it branches into arteries which lead various organs such as the liver, kidney, intestines and spleen and also the legs. The arteries divide into arterioles which further divide and become capillaries. It is here that the oxygen and nutrients dif-fuse into the tissues and carbon dioxide and nitrogenous wastes are picked up. The capillaries fuse to form venules which further fuse to become either the superior or inferior vena cava. The entire cycle starts once again. The part of the circulatory system in which deoxygenated blood is pumped to the lungs and oxygenated blood returned to the heart is called the pulmonary circulation. The part in which oxygenated blood is pumped to all parts of the body by the arteries and deoxygenated blood is returned to the heart by the veins is called the systemic circulation.

Question:

To obtain 500 kg of glass composed of equimolar proportions of Na_2SiO_3 and CaSiO_3, what weights of Na_2CO_3, CaCO_3, and SiO_2 should be used?

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Solution:

When a mixture of limestone, sodium carbonate, and sand is melted together, a clear, homogeneous mixture of sodium and calcium silicates is produced. When the fused liquid is cooled, it becomes more and more viscous and finally hardens to a transparent rigid mass called glass. From the stoichiometry of this equation, an equal number of moles of CaCO_3 and Na_2CO_3 plus twice as many moles of SiO_2 are consumed to produce an equal number of moles of CaSiO_3 and Na_2SiO_3. To solve this problem, determine the number of moles of glass substituents produced; this number will be equal to the number of moles of CaCO_3 and Na_2CO3 and half the number of moles of SiO_2. From the number of moles and each constituent's molecular weight, the actual weights can be determined. The molecular weights are MW of Na_2SiO_3 = 122 g / mole;MW of CaSiO_3 = 116 g / mole; MW of CaCO_3 = 100 g / mole;MW of Na_2CO_3 = 106 g / mole; MW of SiO_2 =60 g / mole. Since the number of moles of each product is unknown (but equal), the number of moles is set equal to x and (x) (122 g / mole) + (x) (116 g / mole) = 500,000 g of glass 238 x = 500,000 g the number of moles = 2101 moles. Thus, the number of moles of the reactants needed are CaCO_3 = 2101 moles; Na_2CO_3 = 2101 moles; SiO_2 = 4202 moles; and the weights of each are CaCO_3 : (2101 moles) (100 g / mole)= 210,100 g = 210.1 kg Na_2CO_3 + (2101 moles) (106 g / mole)= 222,706 g = 222.7 kg SiO_2 : (4202 moles) (60 g / mole)= 252,120 g = 252.1 kg.

Question:

Write a procedure which reads a matrix of non-negative integers and stores it in an array of type MATRICES = ARRAY [1. .XSIZE,1 .. YSIZE] of NONNEGINTEGER where XSIZE and YSIZE are appropriately defined con-stants and NONNEGINTEGER is an appropriately defined type.

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Solution:

To guard against the possibility of the input data being incomplete (orthere being too much input data), the matrix elements are assumed to be followed by a negative end- marker. The procedure checks every datum and, if it detects the end-marker prematurely, it writes a warning message and stops immediately. PROCEDURE READMATRIX (VAR MATRIX: MATRICES; VAR VALID: BOOLEAN); LABEL 99; VAR DATUM: INTEGER; X:1.. XSIZE; Y:1.. YSIZE; BEGIN VALID: = FALSE; (\textasteriskcenteredbeing pessimistic!\textasteriskcentered) FOR X: = 1ToXSIZE DO FOR Y: = 1ToYSIZE DO BEGIN READ (DATUM); IF DATUM> = o THEN MATRIX [X, Y] := DATUM ELSE(\textasteriskcenteredtheendmarkerhas been read \textasteriskcentered) BEGIN WRITELN ('DATA INCOMPLETE'); GOTO 99; (\textasteriskcentered leave the procedure immediately \textasteriskcentered) END END; READ (DATUM);(\textasteriskcentered..... this should be theendmarker\textasteriskcentered) IF DATUM>=o THEN WRITELN ('TOO MUCH DATA') ELSE VALID: =TRUE; 99: END. (\textasteriskcentered READMATRIX \textasteriskcentered)

Question:

(a) isobutane and isobutylene (b) sec-butyl alcohol and n-heptane.

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Solution:

To distinguish between two compounds, try to select for a characterization test. Begin by writing the structures of the compounds involved. You see that isobutylene is an alkene, and that isobutane is an alkane. Thus, you should be able to distinguish isobutylene from isobutane by an addition reaction. Addition reactions are characteristic of alkenes and other unsaturated compounds. When Br_2/CCI_4 , a reddish brown solution, is in contact with an alkene, the following reaction occurs: When the alkene consumes the Br_2, the red color disappears. Thus, alkenes can be detected by their ability to decolorize the red solution of Br_2/CCI_4 . You can also use a cold, dilute, neutral permanganate solution, the Baeyer test, instead of Br_2/CCI_4 to detect the presence of an alkene. Permanganate solution is purple colored, upon reaction with an alkene, the purple disappears and is replaced by a brown manganese dioxide precipitate. The reaction can be written Alkenes do give this reaction. Thus, you now have two simple tests to distinguish isobutane from isobutylene. These two can be distinguished by using a chromic anhydride reagent. Secondary alcohols, such as sec-butyl alcohol, are oxidized by chromic anhydride, CrO_3 , in aqueous sulfuric acid. The clear orange solution turns opaque, blue-green when the alcohol is added. In other words, N-heptane is an alkane, and, as such, does not turn the solution blue green. Thus, you can distinguish sec- butyl alcohol from n-heptane by adding chromic anhydride.

All of these compounds are alcohols; they fit into the general formula R-OH, where R is any alkyl group. In the systematic naming of any alcohol, the following rules should be followed: (1) The longest chain that contains the hydroxyl group (OH) is considered the parent compound. (2) The ^-e ending of the name of this carbon chain is replaced by - ol. (3) The locations of the hydroxyl and any other groups are indicated by the smallest possible numbers. Thus, compound (a) has a 6-carbon chain and an hydroxyl group on carbon number 2. The name of this compound is 2-hexanol. Compound (b) has a 5-carbon chain, because this is the longest chain that contains the hydroxyl group. There is a methyl group on carbon number 3 , and an hydroxyl group on carbon number 1. The name of this compound is, therefore, 3-methyl-1- pentanol. Compound (c), has a 6-carbon chain that contains the OH group, two methyl groups on carbons 2 and 3, and an hydroxyl group on carbon number 3. Thus, the name is 2,3-dimethyl-3- hexanol. Compound (d), has a 5-carbon chain, two methyl groups on carbons 2 and 4 , and an hydroxyl group on carbon 1. Thus, the name is 2, 4-dimethyl-1-pentanol.

(a) The mole fraction of a component in a system is defined as the number of moles of that com-ponent divided by the sum of all the moles present in the system. Here, one must first calculate the number of moles present of each component, then the mole fractions can be found. The number of moles of each gas can be found by dividing the number of grams present of the gas by the molecular weight. (MW He = 4, MW N_2 = 28, MWAr= 40.) no. of moles= no. of grams / MW no. of moles of He = 40 g / (4 g/mole) = 10 moles no. of moles of N_2 = 56 g / (28 g/mole) = 2 moles no. of moles ofAr= 40 g / (40 g/mole) = 1 mole. The total number of moles of gas in the system is the sum of the number of moles of the three gases, Thus, there are 13 moles of gas in the system. The mole fraction can now be found for each gas by dividing the number of moles of each gas by 13, the total number of moles mole fraction = no. of moles / total no. of moles in system mole fraction of He = 10 moles / 13 moles = .77 mole fraction of N_2 = 2 moles / 13 moles= .15 mole fraction ofAr= 1 mole / 13 moles= .08 (b) In a system where various gases are present, the partial pressure of each gas is proportional to the mole fraction of the gas. The relationship between the partial pressure of a particular gas and the total pressure is partial pressure = total pressure × mole fraction In this problem, one is given the total pressure of the system and one has found the mole fraction of He. One can now find the partial pressure of He partial pressure of He = 10atm× .77 = 7.7 atm.

Question:

Design a synchronous sequential circuit with one input line and one output line that recognizes the input string x = 1111. The circuit is also required to recognize overlapping sequences such that if the input string is x = 1101111111010 then the output string should be z = 0000001111000

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Solution:

The first step in the design procedure is to draw a state diagram which represents the behavior of the circuit, which in this case happens to be a sequence 1111 detector. First you assume that the machine is in some starting state A and the first input is a 0. Since 0 is not part of our sequence, the machine remains in state A with an output of 0, 'waiting' for a 1. As soon as a 1 occurs at the input, the machine jumps into state B. The output is still 0, because our sequence has not yet been completed. See fig. 1. If an input of 0 occurs while the machine is in state the 1111 sequence is interrupted and the machine must go back to state A to start all over again. See fig. 1. However, if an input of 1 occurs, the machine jumps from state B to state C, meaning that the second '1' of the sequence was detected. See fig. 2. If the next input is a 0, the 1111 sequence is again interrupted, and the machine must go from C back to A to start all over again; also in fig. 2; but if the input is a 1, while in state C, the machine jumps into state D, meaning that it has just detected the third '1' of the 1111 sequence; see fig. 3. If the next input is a 0, the sequence is again interrupted, and the machine must go back to state A; how-ever, if the input is a 1, the machine has just detected the last '1' of the 1111 sequence, and the output finally becomes 1. See fig. 3 again. Note in fig. 3 that when the machine is in state D, the '1/1' loop will continue giving outputs of 1 as long as the inputs remain 1; thus ensuring the detection of overlapping 1111 sequences. Fig. 3 is the complete state diagram. The next step is to set up a state table from the state diagram. See fig. 4. Note, for example, that the top, right-hand box of the state table means that when the present state is A and in-put becomes a 1, the next state becomes B and the output is 0. Now we choose arbitrary state assignments as follows: A = 00 A = 00 B = 01 C = 10 D = 11 This results in a transition table, and an output table (which is simply extracted from the state table). See fig. 5. Note that states D and C have been interchanged to facilitate the use of K-maps. The next step is to pick a memory element. We shall pick the JK flip-flop. Note that we will need two memory elements, since Log_2 (no. of states)\leq No. of memory elements. Y^K Y^K+1 J K NOTE: d MEANS "DON'T CARE" FIG. 6 0 0 0 d 1 1 d 0 0 1 1 d 1 0 d 1 To work with flip-flop logic, an application table is ex-tremely helpful. This table is easily derived from the truth table of the particular flip-flop. See fig. 6. Next we form JK excitation tables using the JK application table, and the table, and the transition table. See fig. 7a, b, c, d. Note that J_1K_1 corresponds to the y_1^K+1 column on the transition table. Now we solve for the logic equations using the excitation tables as K-maps: J_1 = Y_2XK_1 =X J_2 = XK_2 =X+Y_1 Z = XY_1Y_2 From the logic equations we can now build the actual cir-cuit which will perform the desired function; in this case a 1111 sequence detector. See Fig. 8.

The voltage output of a transformer depends on the input voltage and on the turns ratio according to the relationship (See figure), (V_1/V_2) = (N_1/N_2) Substituting, we find the turns N_2 in the secondary coil to be, N_2 = [(N_1 V_2)/V_1] = {[(1 × 10^4 turns) (2.4 × 10^3 V)]/(2.4 × 10^5 V)} = 1 × 10^2 turns If we assume that power is transferred from primary to the secondary of this transformer at 100 per cent efficiency, the power in equals power out because of the principle of conservation of energy. Therefore, V_1 I_1 = V_2 I_2 or I_1 = [(V_2 I_2)/V_1] {[(2.4 × 10^3 V) (5 × 10^2 A)]/(2.4 × 10^5 V)} = 5 A

Question:

The specific cell processes accelerated or decelerated by hormones are numerous and varied, but most of them fit into one of two general categories. One is the alteration of the rate of membrane transport of substances, and the other is the alteration of the rate of enzyme activity. Explain.

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Solution:

The effect of many hormones is to facilitate or inhibit the transport of substances across the membrane into the cell. Insulin, for example, somehow affects cell membranes so as to increase the rate of glucose transport into the cells. Other hormones, such as glucagon (an antagonist of insulin), inhibit glucose transport. A similar pattern involving hormone-mediated inhibition and stimulation also operates for the membrane transport of amino acids and other sugars. The transport of ions and water into and out of kidney cells is also under the influence of hormones. The hormonal regulation of enzymatic activity usually involves catalysis for which there are different enzymes for the forward and reverse reactions. An example of this is the relationship that exists between glucose and glycogen in the liver: In this particular example, the activity of enzyme 2 is increased by the hormone insulin, and that of enzyme 1 is increased by the hormones glucagon (a pancreatic hormone) and epinephrine (an adrenal medullary hormone). The rate of either the forward or reverse reaction can be increased by production of more of the appropriate enzyme. But there is another way which does not change the total number of enzyme molecules? it does not have any effects on enzyme synthesis. Most enzymes exist in both an active and inactive state. The active state is the functional form of the enzyme. It is believed that the amount of active enzyme can be increased by converting inactive enzyme molecules into active ones. Thus, a hormone may function by converting a certain specific inactive enzyme into its active form, thereby augmenting the enzyme activity and thus the reaction rate. The advantage of this mechanism over synthesis of new enzyme molecules is that protein synthesis requires some time, whereas the activation of enzymes within the cell requires only a few minutes. A hormone may also function by inactivating certain enzymes to bring about a specific physiological change. Note again that in this functional mechanism no new enzymes are being formed, although there is a change in enzyme activity and hence protein production.

For this project, you should construct the main program so that it has a case structure, enabling the user to choose whether he would like to insert or delete a name. The main program should consist only of type declarations, a DO-WHILE loop to choose the case the user wants, and statements to call the subroutines. Since this part is perhaps the easiest, we will do it first and then explain the meaning of the variables. Maximum name length is five letters. INTEGER CHOICE, PTR, DATA, AVAIL, N, NAME DIMENSION DATA (10), PTR (10) N = 10 CALL INIT (PTR, N, AVAIL) CDO WHILE CHOICE NOT EQUAL TO 'STOP' 10IF (CHOICE.EQ.'STOP') GOTO 99 WRITE (5, 100) 100FORMAT (5X,'TYPE INSRT, DELET, OR STOP') READ (5,101) CHOICE 101FORMAT (A5) CIF CHOICE IS INSRT, THEN IF (CHOICE.NE.'INSRT') GO TO 20 WRITE (5, 102) 102FORMAT (5X, 'TYPE THE NAME, PLEASE') READ (5, 103) NAME 103FORMAT (A5) CALL INSERT (DATA, PTR, N, AVAIL, NAME) CALL PRINT (DATA, PTR, N) GO TO 10 CIF CHOICE IS DELET, THEN 20IF (CHOICE.NE.'DELET') GO TO 30 WRITE (5, 102) READ (5, 103) NAME CALL DELETE (DATA, PTR, N, AVAIL, NAME) CALL PRINT (DATA, PTR, N) 30GO TO 10 CEND DO-WHILE 99STOP END DATA (N) is the array which contains the names to be put in order. PTR (N) is the array containing pointers which can be manipulated to accommodate insertions and deletions. The mechanics of pointers will be explained as the problem de-velops, but for now remember that pointers permit the access-ing of data in memory without performing an actual search- and-sort routine. Continuing with the main program, the variable CHOICE can take on three possibilities: INSRT if you want to in-sert a name, DELET if you want to delete a name, STOP to end the session. The program will keep looping back to the beginning, allowing you to adjust the list until you type in STOP. Notice also that after each insertion or deletion, the subroutine PRINT allows you to see the elements of the list. Subroutine INIT sets up the linked list. N is the to-tal number of names (in this case, 10). AVAIL is the ad-dress of the first available location in the storage area DATA. PTR (1), known as the list head, is set to zero. This means that DATA (l) is not used, a necessary precaution that avoids complications in the special case of an empty list. PTR (2) through PTR (9) are initialized such that PTR (2) = 3, PTR (3) = 4, and so on until PTR (9) =10. Finally, PTR (10) is also set to zero. Let us now begin to fill DATA with names to illustrate the activity of the pointers. (See Figure 1.) Subroutine INSERT will go down the list, search for the first available storage location in DATA, insert the name, and most important, rearrange the pointers so that they will cause the list to be printed in alphabetical order. The comparison for alphabetizing is done at statement 23, but the ordering of the pointers is accomplished by the state-ments following statement 33. Let us "walk through" these statements to discover what is happening at each step: J becomes 3 AVAIL becomes PTR (3), which is 4 PTR (3) becomes PTR (1), which is 1 DATA (3) becomes 'CANDY' PTR (1) becomes 3 Now, let us turn to the case where we want to delete a name, say BILLY. The list starts off as it is in Figure 1, after the addition of 'CANDY'. (See Figure 2.) The storage location at DATA(2) is saved in the variable AVAIL; only the pointers are rearranged. In subroutine DE-LETE, statement 14 does the table lookup of the name to be deleted. If it is present, control is passed to statement 34, where the pointers are rearranged to eliminate the de-sired name. A "walk through" yields the following: JD becomes 2 PTR (1) becomes PTR (2), which is 3 PTR (2) becomes 4 AVAIL becomes 2 SUBROUTINE INIT (PTR, N, AVAIL) INTEGER I, N, PTR (N), AVAIL CINITIALIZE LIST HEAD AND START OF FREE LIST AVAIL = 2 PTR (1) = 0 CSET POINTERS FOR FREE LIST DO 12 I = 2, N - 1 PTR (I) = 1+1 12CONTINUE CSET NULL POINTER AT END OF FREE LIST PTR (N) = 0 RETURN END SUBROUTINE INSERT (DATA, PTR, N, AVAIL, NAME) INTEGER II, J, N, PTR (N), DATA (N), NAME IF (AVAIL.NE.0) GO TO 13 WRITE (5, 104) 104FORMAT (5X, 'NO STORAGE LEFT') GO TO 53 13II = 1 CSEARCH LIST FOR INSERTION POINT CDO-WHILE PTR (II) NOT EQUAL TO ZERO CAND DATA (PTR (II)) LESS THAN NAME 23IF (PTR(II).EQ.O.OR.DATA(PTR(II)).GE.NAME) 1GO TO 33 II = PTR (II) GO TO 23 33CONTINUE CEND DO-WHILE CALLOCATE SPACE FROM FREE LIST FOR CNEW ENTRY AND INSERT IT FOLLOWING CENTRY II BY SETTING POINTERS J = AVAIL AVAIL = PTR (J) PTR (J) = PTR (II) DATA (J) = NAME PTR (II) = J 53CONTINUE RETURN END SUBROUTINE DELETE (DATA, PTR, N, AVAIL, NAME) INTEGER ID, JD, N, PTR (N), AVAIL, DATA(N) ID = 1 CFIND NAME TO BE DELETED CDO WHILE PTR (ID) NOT EQUAL TO ZERO AND CDATA (PTR (ID)) NOT EQUAL TO NAME 14IF (PTR (ID).EQ.O.OR. DATA (PTR (ID)).EQ. NAME) 1GO TO 24 ID=PTR (ID) GO TO 14 24CONTINUE CEND DO-WHILE IF (PTR (ID).NE.O) GO TO 34 WRITE (5, 105) NAME 105FORMAT (5X, A5, 'NOT PRESENT IN LIST') GO TO 54 CENTRY FOLLOWING ID NOW CONTAINS CNAME. DELETE IT FROM DATA LIST AND CPUT IT ON FREE LIST. 34JD = PTR (ID) PTR (ID) = PTR (JD) PTR (JD) = AVAIL AVAIL = JD 54CONTINUE RETURN END SUBROUTINE PRINT (DATA, PTR, N) INTEGER IP, N, PTR (N), DATA (N) IP = PTR (1) CDO WHILE IP NOT EQUAL TO ZERO 15IF (IP.EQ.O) GO TO 25 WRITE (5, 107) DATA (IP) 107FORMAT (1X, A5) GO TO 15 25CONTINUE CEND DO-WHILE RETURN END On some systems, the DO-WHILE constructs in INSERT and DELETE may cause problems because of a reference to DATA (0). To amend that difficulty, you may substitute the following: (for INSERT) 23IF (PTR (II).EQ.O) GO TO 33 IF (DATA (PTR (II)).GE.NAME) GO TO 53 (for delete) 14IF (PTR (ID).EQ.O) GO TO 24 IF (DATA (PTR (ID)) .EQ.NAME) GO TO 54

It is believed that diabetes mellitus probably results from an insufficiency in insulin produc-tion. This may happen either as a genetic abnormality in which insulin synthesis is impaired, or as an acquired disease in which insulin-secreting cells of the pancreas are hypofunctional or destroyed. In fact, the latter may also have a genetic origin. Thus in many cases of diabetes mellitus, the disease can be inherited. The children of the person with diabetes may tend to exhibit the symptoms of diabetes. The disease, however, may develop slowly or to a varying degree, for diabetes is not an "all-or-none" disease. Moreover the overt symptoms of diabetes mellitus may all but disappear .with appropriate measures such as a change in diet. Thus it would be more valid to say that a tendency toward diabetes mellitus can be inherited. The untreated person with diabetes, suffering from a deficiency of insulin, will have an excess of glucose in the blood. A large amount of glucose will be excreted and lost from the body in the urine. The person with diabetes will therefore fail to acquire enough energy by glucose metabo-lism alone. He will have to obtain energy by way of triglyc-eride catabolism. The breakdown of triglyceride results in an elevation of fatty acid and glycerol levels in the blood. If further energy is needed, the diabetic body will convert his protein reserve into glucose, which would be accompa-nied by an accumulation of ketones. The products of these two metabolic pathways (i.e. fatty acids and ketones) will upset the delicately balanced pH system of the body, and thus may be considered toxic. Because insulin also has the effect on hindering glycogen formation from glucose, the diabetic individual has very little carbohydratere-serves. Most of the glucose consumed in food, unavailable to the cells for use, simply passes out of the body. In short, the untreated person with diabetes will face the paradox of cell starvation in the presence of an elevated plasma glu-cose concentration. Only the brain cells are spared the glu-cose deprivation since their uptake of glucose is not hormonally regulated. The consequence of these catabolic processes is progressive loss of weight despite any increase in food intake, and a lowered pH of the body fluids. Another- well known consequence of untreated diabetes is dehydration of the body tissues. The large amount of glucose excreted into the kidney tubules induces water to passively follow suit (osmosis), and prevents it from being reabsorbed into the body cells. For some complicated reason, sodium reabsorption is also reduced. The net effect is the marked excretion of water and sodium, resulting in a decreased plasma volume and hence blood pressure. The lowered blood pressure may reduce blood flow to the brain, leading to brain damage, coma and possibly death. Associated with diabetes are arteriosclerosis, small- vessel and nerve disease, susceptibility to infection, and a variety of other complications. At one time it was thought that these problems resulted only from insulin deficiency. However, this view is now being questioned in light of the evidence which has shown that these problems can be delayed but not abolished by administration of insulin. Many researchers now believe that excessive hyperglycemia or possibly even hypoglycemia due to inappro-priate insulin administration contribute to the occurrence of complications. In treating persons with diabetes, the aim is to main-tain the plasma glucose at a near normal value. The admi-nistration of insulin is the major therapy, and must be given through injection. This is because insulin, being a protein hormone, would be broken down by gastrointestinal enzymes if given orally. The dosage must be carefully determined since an overdose would drastically lower the plasma glucose concentration, thereby depriving brain cells of needed glucose and ultimately causing death. Non-insulin drugs which stimulate the islets of Langerhans to produce more insulin have been used in some cases, but are not effective therapeutic measures. There are actually two types of diabetes. Juvenile dia-betes (Type I) is the insulin-deficient diabestsjust de-scribed. These patients require insulin and show some of the more severe effects of the disease. Adult-onset diabetes (Type II) is much more common and less threatening. This type of diabetes is seen more frequently in older often obese patients. Their bodies produce and release insulin yet the receptors for insulin are less sensitive. The receptors are "down-regulated." Type II patients are not insulin-de-pendent. Improvement may occur with proper diet and exer-cise.

Question:

What is instinct? What is instinct?

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Solution:

The term "instinct\textquotedblright is difficult to define since its meaning has been so broadly applied. Instinct has been used to designate behavior that is stereotyped; that is, behavior which is rigidly executed and therefore highly predictable when the appropriate stimuli are presented. For example, when the eggs of herring gulls and other ground-nesting birds accidentally fall out of the nest area, the parent gulls use their bills to roll them back into the nest. This is a stereotyped response and can be repeated by purposely displacing the eggs. The eggs serve as releasing stimuli to elicit the egg-rolling behavior. When investigators placed abnormally large wooden models of eggs adjacent to the gulls' own dis-placed eggs, the gulls preferred to roll the larger models back to the nest. Any such stimuli having a quality that is preferred over that of the natural stimulus encoun-tered in the life of an animal - in this case abnormally large size - is referred to as a supernormal stimulus. Instinct is often believed to consist solely of innate responses whichare determined by set neural path-ways of the brain and not altered by experience. How-ever, recent study has shown that learning may play a role in stereotyped instinctive behavior. For example, ducks and other ground-nesting birds either cry out alarm calls or crouch in fear when a bird of prey flies overhead (see Figure). It was discovered that the basic shape of the flying bird is the releasing stimulus for fear behavior; particular anatomical features are not crucial. Thus, when an investigator passed different shapes of cardboard silhouettes over the heads of ducklings, only the bird shape having a head shorter than its tail evoked the fear response (See Figure). When this same silhouette was passed over the ducklings in the reverse direction, so that the figure appeared to have a long head and short tail, no fear response was evoked. In the first instance, the bird's silhouette represented a hawk, while the second silhouette represented another duck or goose. This fear response was initially thought to be a strictly inherited response to a sign stimulus. Further study, however, has indicated elements of learning in this behavior. Ducklings are initially afraid of all overhead flying objects. Eventually they stop reacting to ones they see repeatedly, such as other, ducks and falling leaves. Hawks, however, are generally scarce, and the ducklings never become accustomed to their peculiar shape. Learning therefore plays a role in what seems to be a totally innate stereotyped response. Hence, it is difficult to make a distinction between instinct and other kinds of behavior, since some instincts can be shown to contain learned elements.

Question:

At 1000\textdegreeK, K = 2.37 × 10-^14 for the reaction, N_2 (g) + 3H_2 (g) \rightleftarrows 2NH_3 (g) . If you inject one mole of each N_2 and H_2 in a one-liter box at 1000\textdegreeK, what per cent of the H_2 will be converted to NH_3 at equilibrium?

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Solution:

To answer this question, determine the concentration of H_2 at equilibrium. Once this is known, subtract it from the initial con-centration. The difference yields the amount that reacted to produce NH_3 . By multiplying the quotient of the difference divided by the original amount of H_2 by 100, you obtain the per cent. To find [H_2] at equilibrium, employ the equilibrium constant expression. It states that K, the equilibrium constant, is equal to the concentration ratio of the products to reactants, each raised to the power of its coefficient in the chemical reaction. For this re-action, then, K = 2.37 × 10-^3 = [NH_3]^2 / {[H_2 ]^3 [N_2] } . To find [H_2 ], proceed as follows. Let x = amount of N_2 that dis-sociates. The initial concentrations of all species are 1M, sincemolarity= moles/liter, and 1 mole of each is placed in a one-liter box. If the N_2 's initial concentration is one and x moles/liter dis-sociates, then, at equilibrium, there is (1\Elzbar x)moles/liter left of N_2 . In other words, at equilibrium, [N_2 ] = 1\Elzbar x. Now, from the chemical equation, it is seen that for every mole of N_2 that reacts, 3molesof H_2 are necessary. Thus, when x moles/liter of N_2 dissociate, 3x moles/Iiterof H_2 are required. The initial concentration is 1, so that, at equilibrium [H_2 ] = (1\Elzbar 3x) moles/Iiter. Notice, also, that for every mole/liter of nitrogen that dissociates, 2(mole/liter) of ammonia is obtained. Substituting these values into the equilibrium constant expression, one obtains 2.37 × 10-^3 = [NH_3 ]^2 / { [N_2 ] [H_2 ]^3 } = (2x)^2 / { (1\Elzbarx) (1\Elzbar3x)^3 } . Solving for x, one obtains x = .0217 moles/liter of N_2 that dis-sociate. Thus, [H_2 ] = 1\Elzbar 3x = .935 moles/liter at equilibrium. The initial concentration was 1 mole/liter. The difference is the amount that dissociated, i.e., the H_2 . The difference = 1\Elzbar .935 = .065, Thus, the percent that dissociated equals [(.065M) / (1M)] × 100 = 6.5% .

Question:

An efficient method of organizing an unordered set of re-cords is by means of a HASH (or SCATTER) FILE. A HASH FILE works as follows: A specified set of arithmetic operations are performed on a data element (also called a key) of a record. The result of this operation is then used as the HASH address for that key. Using this HASH address we can store and locate the record. It should be noted that dif-ferent records may have the same HASH address. Our program will analyze how well a HASH function (the arithmetic oper-ations on a key) performs for a sample input file. We will use as our key element the ID number of the record. The arithmetic operation will be to divide the key by 50, and use the remainder as the HASH address. Thus, our addresses will range from 0 to 49 (possible remainders). Finally, we will set up and print the following tables:

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Solution:

1 HASHFILE START 0 2 SAVE (14,12) 3 BALR 3,0 4 ST 13, SAVE+4 5 LA 13, SAVE \textasteriskcentered THE LOADING OF TABLE 2 \textasteriskcentered BY INCREMENTING REGISTER 5 BY 1 AND STORING \textasteriskcentered THE RESULT IN CONTIGUOUS FULLWORDS. 9 LA 6, ASHTABLE 10 L 5, = F '0' FIRST HASH-ADDRESS OF TABLE 11 L 10, = F '50' LOOP CONTROL FOR REMAINDERS 12 LOAD CVD 5, CONVERT 13 PACK 0 (2, 6), 0 (2, 6) 14 ZAP 0 (2, 6), CONVERT + 6 (2) STORING POSSIBLEREMAINDERS 15 PACK 2 (2, 6), 2 (2, 6) 17 LA 6, 4 (6) GOTO NEXT WORD OF ASHTABLE 18 A 5, = F '1' NEXT REMAINDER FOR ASHTABLE 19 CR 5, 10 DO WE HAVE REMAINDERS 0 - 49? 20 BL LOAD \textasteriskcentered PROCESSING THE KEY AFTER READING IT IN 22 GETKEY READ CARD CARDINPA, CRDEOF 23 CLC CRDIDNO, = C '9999' END OF FILE CARD 24 BE PRTHEAD 2 25 PACK WRKKEY, CARDIDNO 26 ZAP CONVERT, WRKKEY 27 CVB 9, CONVERT REGISTER 9 HOLDS ID# PRIOR TO DIVISION 28 SR 8, 8 CLEAR REGISTER 8. 29 D 8, FINDHASH REMAINDER DETERMINES HASH ADDRESS 30 CVD 8, CONVERT CONVERT REMAINDER IN REGISTER 8 TO DECIMAL \textasteriskcentered PRINTING THE ID NUMBER AND ITS HASH - ADDRESS 32 MVC PRTKEY, CRDIDNO 33 ZAP REM, CONVERT STORE REMAINDER 34 MVC PRTHASH1, PATT1 35 ED PRTHASHI, REM 36 PRINTLIN PRTDETL, 133 \textasteriskcenteredINCREMENTING THE COUNT FOR AN ADDRESS 38 LA 6, ASHTABLE 39 LR 9, 8 40 SR 8, 8 41 M 8, POSITION 42 AR 6, 9 GOTO REMAINDER IN ASHTABLE 43 ZAP CONVERT, 2 (2, 6) STORE COUNT IN 'CONVERT' 44 CVB 7, CONVERT PLACE COUNT IN REGISTER 7 45 A 7, = F '1' INCREMENT COUNT 46 CVD 7, CONVERT 47 ZAP 2 (2, 6) , CONVERT + 6 (2) 48 B GETKEY 49 PRTHEAD2 PRINTLIN HEADING2, 133 \textasteriskcenteredPRINTING HAS\rule{1em}{1pt}ADDRESS AND THEIR COUNTS 51 LA 6, ASHTABLE 52 LA 5, 50 53 PRINTHASH MVC PRTHASH2,PATT1 54 ZAP TEMPHOLD, 0 (2, 6) 55 ED PRTHASH2, TEMPHOLD 56 MVC PRTCOUNT,PATT1 57 ZAP TEMPHOLD,2 (2, 6) 58 ED PRTCOUNT, TEMPHOLD 59 PRINTLIN PRTTABLE. 133 60 LA 6, 4 (6) 61 BCT 5 , PRINTASH 62 CRDEOF L 13, SAVE + 4 63 RETURN (14,12) \textasteriskcenteredDATA DEFINITION FOR CARD INPUT AREA 64 CARDINPA DS \varphi CL80 65 CRDIDNO DS CL5 66 DS CL75 \textasteriskcenteredDATA FOR KEY AND ITS ADDRESS 67 HEADING1 DS \varphi CL133 68 PRTCC1 DS C '1' CARRIAGE CONTROL-RESULTS IN PRINTING ON NEW PAGE 69 DC 20C'' 70 DC C 'KEY HASH-ADDRESS' 71 DC 92C' ' DATA DEFINITION FOR PRINT AREA 72 PRTDETL1 DS \varphi CL133 73 DC C '0' CARRIAGE CONTROL 74 DC 19C ' ' 75 PRTKEY DS CL5 76 DC 8C' ' 77 PRTHASH1 DS CL4 78 DC 96C ' ' \textasteriskcenteredDATA DEFINITION FOR HEADING OF HASH-ADDRESS AND COUNT 79 HEADING2 DS \varphi CL133 80 PRTCC2 DC C '1' 81 DC 20 'C' 82 DC C 'HASH-ADDRESS COUNT' 83 DC 92 'C' \textasteriskcenteredDATA DEFINITION FOR ASHTABLE 84 PRTTABLE DS \varphi CL133 85 DC C '0' 86 DC 25C ' ' 87 PRTHASH2 DS CL4 88 DC 10C ' ' 89 PRTCOUNT DS CL4 90 DC 89C ' ' \textasteriskcenteredDATA DEFINITION FOR WORK AREAS 91 SAVE DS 18F 92 ASHTABLE DC 50F '0' 93 WRKKEY DS PL4 94 POSITION DS F '4' 95 CONVERT DS D 96 FINDHASH DC F '50' THE DIVISOR 97 REM DS PL2 98 TEMPHOLD DS PL2 99 PATT1 DC X' 4020212020' 100 END HASHFILE In general, the format of our program is as follows: The table to contain the hash-addresses, and the amount of records hashed to each address (count), is defined as having a length of 50 fullwords. By use of Register 5, the set of all possible remainders (0-49) are placed in the first two bytes of the words, each of the words containing one possible remainder, increasing from one to 50. The latter two bytes of each word will then store the 'count' for each corresponding hash-address. As each address is found, it is printed along with its key. This constitutes the first table of the program. Af-ter printing the key and its address, we then increment the count of this address in the 'address- count' table (ASH-TABLE) . When no more key elements are found (keys 99999), the contents of the second table are printed. It is to be observed that, in the case of 'ASHTABLE', we manipulate 2 bytes of each word at a time and not the complete word. We first load all the remainders in the first two bytes of each word in the 50-word area. This is accomplished by means of: 13.PACK0 (2, 6), 0 (2, 6) and14.ZAP0 (2, 6), CONVERT + 6 (2) The pack instruction of Line 13 simply converts the contents of the location addressed by 0 (2, 6) to packed decimal. The location addressed by 0 (2, 6) is a part of ASHTABLE, and since ASHTABLE is defined as having zeroes, this location also contains zeroes. The contents of this location are then replaced by a possible remainder. This is done by the ZAP instruction of Line 14. Using the LA instruction of Line 17 we can address the next word of ASHTABLE. LA 6, 4 (6): Coded in such a manner, this instruction adds a displacement of 4 bytes to the present address in Register 6. Hence, if Register 6 contains 1080, the new address after LA 6, 4(6) would be 1084. This technique en-ables us to use just the two instructions of Lines 13 & 14 to load the table. Incidentally, Operand-2 of Line 14 is an example of the combination of implicit and explicit ad-dressing. Since the ZAP instruction proceeds from right to left, this coding was not really necessary, but it - indicates how the two types of addresses can be combined. ZAP 0(2,6),CONVERT+6(2) tells the computer to place the last two bytes of the doubleword 'CONVERT' into Operand- 1. The following figure indicates 'CONVERT+6(2)'. As is always the case, the location of the first byte of a field represents the address of the field in main mem-ory. By 'ADDING' integers to the field (symbol), we can reference a particular byte in the field. Furthermore, if we also specify a length factor, we are able to reference a particular set of bytes in the field. So in the combination of both types of addresses, we have SymbolAdditionLengthFactor \downarrow\downarrow\downarrow CONVERT+6(2). CONVERT+6(2). Note, you should be careful not to make references out-side the Note, you should be careful not to make references out-side the field, since this will result in error messages from the computer or field, since this will result in error messages from the computer or unintended executions. unintended executions. Interestingly enough, we can also reference previous bytes in Interestingly enough, we can also reference previous bytes in memory by "subtracting" integers from our symbol. For example, ZAP memory by "subtracting" integers from our symbol. For example, ZAP ANY, SYM-6. If 'SYM' were at location 1848, then the contents of location ANY, SYM-6. If 'SYM' were at location 1848, then the contents of location (1848-6) 1842 would be placed in 'ANY'. Once again, you should check to (1848-6) 1842 would be placed in 'ANY'. Once again, you should check to be certain that your intentions are carried out. be certain that your intentions are carried out. Let us now backtrack our steps a little. The PACK in-struction of Let us now backtrack our steps a little. The PACK in-struction of Line 15 prepares the latter two bytes of each of the 50 words for the count. Line 15 prepares the latter two bytes of each of the 50 words for the count. In Lines 38-47, each count is incremented when it is found. The technique In Lines 38-47, each count is incremented when it is found. The technique involved in the incrementing of the table is one that should be given some involved in the incrementing of the table is one that should be given some thought. If our HASH-address (remainder) is 37 after dividing, we cannot thought. If our HASH-address (remainder) is 37 after dividing, we cannot simply increment the count at the 37th position in the table. The count to simply increment the count at the 37th position in the table. The count to be incremented will not be at this position, but at the location of the 37th be incremented will not be at this position, but at the location of the 37th fullword of the table. So in order to get to the 37th count, we must multiply fullword of the table. So in order to get to the 37th count, we must multiply 37 (HASH-address) by 4, the length of a fullword. By adding this result to 37 (HASH-address) by 4, the length of a fullword. By adding this result to the address of ASHTABLE, we obtain the desired count. This task is the address of ASHTABLE, we obtain the desired count. This task is per-formed by the following instructions: per-formed by the following instructions: 41.M8, POSITION 41.M8, POSITION 42.AR6, 9 42.AR6, 9 Remember that two registers are needed for a multiplication in fixed point Remember that two registers are needed for a multiplication in fixed point binary arithmetic and that the result is us-ually small enough to be stored in binary arithmetic and that the result is us-ually small enough to be stored in one (the odd) register. one (the odd) register. We will now turn our attention to the LA instruction. The general We will now turn our attention to the LA instruction. The general format of this instruction is: format of this instruction is: (a)LAR, SYMBOL (a)LAR, SYMBOL or(b)LAR, D (L,B) or(b)LAR, D (L,B) Let us concentrate on Case (b). Operand-1 is a register, and Operand-2 Let us concentrate on Case (b). Operand-1 is a register, and Operand-2 specifies a displacement D, a length L, and base register B. Coded like specifies a displacement D, a length L, and base register B. Coded like this, the instruction loads the address specified by Operand-2 into the this, the instruction loads the address specified by Operand-2 into the register. Now, if we were to code the instruction with the length (L) and register. Now, if we were to code the instruction with the length (L) and base register (B) absent (Line 52), it takes on a somewhat different base register (B) absent (Line 52), it takes on a somewhat different meaning. Instead of loading an actual address, the instruction loads the meaning. Instead of loading an actual address, the instruction loads the number specified as a displace-ment into the register. For example, Line number specified as a displace-ment into the register. For example, Line 52, 52, LA5, 50 LA5, 50 The number 50 is loaded into Register 5. Such a method provides The number 50 is loaded into Register 5. Such a method provides us with a simple way of loading numbers in-to registers. Many us with a simple way of loading numbers in-to registers. Many programmers prefer this technique since it is easier to code than that of programmers prefer this technique since it is easier to code than that of the L (load) instruction. the L (load) instruction. Our program coded here is a good example of calcula-tions which Our program coded here is a good example of calcula-tions which involve the combination of packed decimal and fixed point binary involve the combination of packed decimal and fixed point binary arithmetic. Incidentally, if you are not clear on the matter, fixed point binary arithmetic. Incidentally, if you are not clear on the matter, fixed point binary arithmetic is the name given to calculations in which any of the sixteen arithmetic is the name given to calculations in which any of the sixteen general purpose registers are used. Many times, the oper-ands involved in general purpose registers are used. Many times, the oper-ands involved in these calculations must be registers or fullwords. As our program mostly these calculations must be registers or fullwords. As our program mostly worked on two bytes at a time, we found it rather difficult to directly worked on two bytes at a time, we found it rather difficult to directly increment the elements of ASHTABLE using fixed point binary arithme-tic. increment the elements of ASHTABLE using fixed point binary arithme-tic. Thus, like most computer features, binary arithmetic can be restrictive at Thus, like most computer features, binary arithmetic can be restrictive at times. times.

Question:

Given that the H-to-H distance in NH_3 is 0.1624 nm and N-H distance is 0.101 nm, calculate the bond angle H-N-H.

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Solution:

Because the NH distances are equal, an isosceles triangle can be drawn as shown. The altitude NX is drawn to form two right triangles. The magnitude of the angle HNX can be found trigonometrically from its sine. The sine of an angle is equal to the length of the opposite side divided by the hypotenuse. Sin HNX = (0.0812 nm / 0.101 nm) = 0.804. Using a table of trigonometric functions, one can find the angle whose sin is 0.804. sine^-1 0.804 = 53.50\textdegree. The angle H-N-H is twice HNX. As such, the measure of angle H-N-H is 2 × 53.50\textdegree or 107\textdegree. It is known that H-N-H is twice HNX because the altitude of an isosceles triangle bisects the vertex angle.

The general procedure used in solving any problem in mechanics is to calculate all the forces acting on the system and then derive the equation of motion of the system. An easier way to do mechanics problems involves the use of conservation principles. These laws are not applicable to all problems, but when they are, they simplify the calculation of the solution tremendously. In this problem, we may use the principle of conservation of energy. We relate the energy of the block before it was released to the block's energy at the point of maximum compression (see figure). At the moment of release; the kinetic energy is zero. At the moment when maximum compression occurs, there is also no kinetic energy. As shown in the figure, the reference level for gravitational potential energy is the surface S. The initial gravitational potential energy of m is mgy_1. At the point of maximum compression, the potential energy of m is mgy^2. However, at this point, the spring is compressed a distance y and also has elastic potential energy 1/2 ky^2. Hence, equating the energy at the point of release to the energy at the point of maximum compression, mgy_1 = mgy_2 + (1/2) ky^2 mg(y_1 - y_2) = (1/2) ky^2 But y_1 - y_2 = h + y and mg(h + y) = (1/2) ky^2 mg(h + y) = (1/2) ky^2 y^2 = [(2 mg)/(k)] (h + y) y^2 - [(2 mg)/(k)] y - [(2 mgh)/(k)]= 0 Therefore, using the quadratic formula to solve for y, y = (1/2) [(2 mg)/(k)\pm \surd{(2 mg/k)^2 + (8 mg/k)} ].

Lactic acid is first dehydrogenated enzymatically by lactate dehydrogenase to form pyruvic acid. [Refer to Fig. 1.] The two hydrogen atoms and two electrons are transferred to a pyridine nucleotide, nicotinamide adenine dinucleotide (NAD+). NAD+ serves as an electron acceptor in dehydrogenation reactions involving substances (in this case, lactate) with an arrangement as follows: (Secondary alcohols, such as in lactic acid, get oxidized to ketones, such as in pyruvic acid). Following this dehydrogenation reaction is the decarboxy-lation (loss of CO_2) of pyruvic acid to form acetaldehyde. Addition of coenzyme A to acetaldehyde forms a compound which undergoes another dehydrogenation to form acetyl coenzyme A. The oxidation of succinic acid is part of the TCA cycle. [Refer to Fig. 2.] Succinic acid is first dehydrogenated enzymatically by succinic dehydrogenase to form fumaric acid. The two hydrogen atoms and their electrons are not transferred to NAD+, as in lactic acid oxidation, but are transferred to flavin adenine dinuc-leotide (FAD). FAD is an electron acceptor in dehydroge-nation reactions involving substances (in this case, acid. In this case, NAD+ accepts the electrons (with a H) to form NADH. Oxaloacetic acid will either combine with acetyl coenzyme A to form citric acid in the TCA cycle, or be decarboxylated to form pyruvic acid. In most cases though, the oxoloacetata will combine with acetyl coen-zyme A and continue the Krebs cycle. The reader should take note that in both the oxidation of lactate and suc-cinate, dehydrogenation reactions occur, but different electron acceptors are used in each case: NAD+ for lac-tate and FAD for the direct oxidation of succinate. The reducing equivalent NADH and FADH_2 are formed.

The human nervous system, composed of the brain, spinal cord, and peripheral nerves, connects the eyes, ears, skin, and other sense organs (the receptors) with the muscles, organs, and glands (the effectors). The nervous system functions in such a way that when a given receptor is stimulated, the proper effector responds appropriately. The chief functions of the nervous system are the conduction of impulses and the integration of the activities of various parts of the body. Integration means a putting together of generally dissimilar things to achieve unity. Other systems involved in similar functions are the endocrine system and the regulatory controls intrinsic in the enzyme systems within each cell. Examples of the latter are inhibition and stimulation of enzymatic activities. The endocrine system utilizes substances, knows as hormones, to regulate metabolic activities within the body.

(a) The truth table is set up by first specifying the pos-sible combinations of values of x_1 and x_2 . There are 2^2 = 4 of them and they represent the numbers 0 to 3 in the binary system. The next step is to evaluate the terms of statement (1): x_1 x_2 \backsimx_1 \backsimx_2 \backsimx_1 + x_2 x_1 + \backsimx_2 0 0 1 1 0 1 0 1 1 1 0 0 1 0 1 0 1 1 0 1 1 0 1 1 Now transfer columns 1, 2, 5, and 6 to a truth table evaluating statement (1): x_1 x_2 \backsimx_1 + x_2 x_1 + \backsimx_2 x1\equiv x2 = (\backsimx_1 + x_2) \bullet ( x_1 + \backsimx_2) 0 0 1 1 0 1 0 1 1 1 0 1 1 0 1 1 1 0 0 1 Note that x_1 \equiv x2= 1 if and only if x1= x_2, which is why it is called the equivalence function . (b) Recall that variables which are operands of the + operation are inputs to an OR gate and a variable that is the operand of the \textasciitilde operation is the input to a NOT gate. Thus, the gate representations of the terns in parentheses are as shown in Fig. 1. Since each term is an operand of the \bullet operation, each of the above gate circuits is an input to an AND gate. Thus, the gating circuit of statement (1) is as shown in Fig. 2. Note that the output of this network is 1 if and only if = x1= x_2. Thus, the network acts as a comparing device. Specifically, it is the circuit for an equality comparator. An equality comparator compares one bit, x_1, with another bit, x_2. If two strings of bits are to be compared, they could be compared serially. That is, compare a bit in the first position of the first string with a bit in the first position of the second string, then compare a bit in the next position of the first string with a bit in the next position if the second string and continue until either a zero is output, indicating the strings are not equal, or all positions have been compared, indicating the strings are equal. Or they could be compared in parallel. That is, a parallel comparing device that inputs two strings of length n would have n equality comparators. An output of a string of ones indicates that the two inputs are equal. The advantage of parallel comparing is its higher speed compared with series comparison. It will always require k time units to compare strings where k is the operation time of the equality comparator. The series comparator requires n\bulletk time units at worst and k time units at best. The advantage of the series comparator is its economy. But since gates can be manufactured inexpensively, no com-mercial computer would use a series comparator, since every instruction must be identified by using the comparator and hence, there would be a serious loss of speed.

Question:

A trait has the inheritance pattern shown below for four different families, a) Determine whether the trait is dominant or recessive b) Based on your answer to a), determine the possible genotypes for each individual in the four families. (Solid blocks or circles indicate individuals who express the trait).

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Solution:

a) Let us look at what the pattern within each individual family tells us about the trait. This does not really tell us anything about the trait, for each a pattern could result from the inheritance of either a dominant or a recessive trait. offspring. This is definite evidence that the trait must be recessive. We know that since the trait is present in the offspring, then one or both of the parents must carry the gene for that trait. Yet neither parent expresses the trait. Therefore, the gene must be present, but masked, and so it must be recessive. We need go no further since we have estab-lished that the trait is recessive. However, let us see what we might have been able to learn about the trait from the remaining pedigrees. 3. In this family, one parent expresses the trait, yet none of the offspring express it. This is inconclusive, because such a pattern could result from either dominant or recessive inheritance. If the trait were dominant, the parent expressing it could be heterozygous, and so carrying some recessive allele. Although it is not highly probable, it is possible that only the recessive allele was given to the offspring, and none received the domi-nant allele. If the trait were recessive, the parent not expressing the trait would carry at least one dominant gene which could have been inherited by all the offspring. This gene would mask any recessive alleles the offspring may have. 4. Again this pattern is compatible with both dominant and recessive inheritance, assuming that one of the parents is homozygous and the other heterozygous. b) Since we know from family two's pedigree chart that the inherited trait is recessive, we can determine the possible genotypes of all the individuals in all the families. Let A represent the dominant trait (whose inheritance is not shown: ,0) and a the recessive trait (whose inheritance 0 is shown: , ) the genotypes are: 1P^1 ( ) = aa2P^1 ( ) = Aa P^2 ( ) = aaP^3 ( ) = Aa F_1 ( ) = aaF^31( , ) = AA or Aa F^1,2_1() = aa 3P^1 ( ) = aa4P^1 ( ) = aa P^2 ( ) = AA or AaP^2 ( ) = aa F1 - 5( , ) = AaF1 ,2(, ) = aa F3 ,4(0,0) = Aa

The clocked JK flip-flop, for which the logic symbol is given in Fig. 1, has five input signals J, K, S, R, and C. J, C, and K are used for synchronous, clocked operation of the device. In this mode the set (S), and reset (R) inputs are held at zero. To initiate a state change in the device a 0 \rightarrow 1 \rightarrow 0 clock pulse must occur, because internally this flip-flop consists of two flip-flops in a master-slave configuration. This property is shown in the circuit realization of the JK flip-flop in Fig. 2. The first clock transition, 0 \rightarrow 1, causes the master flip-flop to operate, while the second, 1 \rightarrow 0, transfers data from the master to the slave. The outputs Q andQcome from the slave flip-flop. The truth table for clocked operation (when S = R = 0) is shown in figure 3. Inputs clock Paresent State J(t) K(t) C(t) Q(t) Q(t + E) 0 0 0\rightarrow1\rightarrow0 0 0 No change 0 0 0\rightarrow1\rightarrow0 1 1 0 1 0\rightarrow1\rightarrow0 0 0 Reset 0 1 0\rightarrow1\rightarrow0 1 0 1 0 0\rightarrow1\rightarrow0 0 1 Set 1 0 0\rightarrow1\rightarrow0 1 1 1 1 0\rightarrow1\rightarrow0 0 1 Complement 1 1 0\rightarrow1\rightarrow0 1 0 Fig. 3 Truth table - synchronous operation (S = R = 0) Note that when either S or R are NOT EQUAL to zero there will be a STATIC OPERATION. The truth table for static operation is shown in figure 4. S(t) R(t) Q(t) Q(t + E) 0 0 0 ?Clocked 0 0 1 ? Operation 0 1 0 0Reset 0 1 1 0 1 0 0 1Set 1 0 1 1 1 1 0 d Not allowed 1 1 1 d Fig. 4 Truth Table - static operation Note: J and K inputs don't matter in static operation. The clocked JK flip-flop timing diagram is shown in Figure 5.

Question:

A thin square metal sheet of side 6 cm is suspended vertically from a balance so that the lower edge of the sheet dips into water in such a way that it is parallel to the surface. If the sheet is clean, the angle of con-tact between water and metal is 0\textdegree, and the sheet appears to weigh 4700 dynes. If the sheet is greasy, the contact angle is 180\textdegree and the weight appears to be 3000 dynes. What is the surface tension of water?

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Solution:

The contact angle \texttheta is a measure of the curvature of the liquid surface adjacent to the metal sheet (see figure). In either case there are three forces acting on the sheet: the tension in the suspension which gives the apparent weight, the actual weight of the sheet, and the total surface- tension force. In the first case the angle of contact is zero and the surface tension force acts downward since it tries to restore the liquid to its original level. Thus since the sheet is in equilibrium F_1 = W + 2T, the factor 2 being necessary since thereare two sides to the sheet. In the second case the angle of contact is 180\textdegree and thus the surface-tension force is acting upward. Hence F_2 = W - 2T. The surface tension \Upsilon is defined as the ratio of the surface force T, to the length, l, along which the force acts. Each force T acts along one side of the sheet, the thickness of the sheet assumed negligible with respect to the length of the sheet. This length is perpendicular to T and is given to be 6 cm. We have T = \Upsilonl Subtracting F_2 from F_1, we get F_1 - F_2 = (W + 2T) - (W - 2T) = 4T F_1 - F_2 = 4\Upsilonl Therefore \Upsilon = (F_1 - F_2)/(4l) = [(4700 - 3000)dynes]/[4 × 6 cm] = 70.8 dynes \bulletcm^-1.

Question:

In what respects are viruses living things? How are they unlike living things?

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Solution:

All living things possess organization. The fundamental unit of organization is the cell. Living cells are variably bounded by a membrane, which regulates the movement of substances into and out of the cell. Viruses do not have any membranes because they have no need to take in or expel material. Viruses lack all metabolic machinery while cells possess this machinery in order to extract energy from theenviromentto synthesize their components. All cells produce ATP but viruses do not. Cells utilize ATP to build complex materials and to sustain active in- teractions with their environment. Viruses do not perform energy-requiring processes . Cells are capable of growth in size but viruses do not have this capacity . Most living things respond in complex ways to physical or chemical changes in their environment. Viruses do not. Most importantly, livingthinqspossess the cellular machinery necessary for reproduction. They have a complete system for transcribing and translating the messages coded in their DNA. Viruses do not have this system . Although they do possess either RNA or DNA (cells possess both ) they cannot independently reproduce, but must rely on host cells for reproductive machinery and components. Viruses use their host's ribosomes , enzymes, nucleotides, and amino acids to produce the nucleic acids and proteins needed to make new viral particles. They cannot reproduce on their own. However, viruses are unlike non-living things in that they possess the potentiality for reproduction. They need special conditions , such as the presence of a host, but they are able to duplicate themselves . Since non-living things do not contain any nucleic acids, they cannot duplicate themselves. This is the critical argument of those who propose viruses to be living things. The difficulty in deciding whether viruses should be considered living or non-living reflects the basic difficulty in defining life itself.

Question:

What are the expected types of offspring produced by a cross between a heterozygous black, short-haired guinea pig and a homozygous white, long-haired guinea pig? Assume black color and short hair are dominant charac-teristics.

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Solution:

This is an example of a dihybrid cross. A dihybrid cross is one involving parents who differ with respect to two different traits. The principles of the cross are the same as those of monohybrid crosses. Let B be the gene for black color, b the gene for white color, S the gene for short hair and s the gene for long hair. The parental genotypes are Bb Ss (hete-rozygous black, short-haired) and bbss (homozygous white, long-haired). The gametes produced are obtained, as they are in monohybrid crosses, by Mendel's Second Law, such that the genes segregate independently. Note that each gamete formed can contain only one of the allele from each allelic pair. Thus there are only four possible gametes from the heterozygous parent, as shown below: B bS s BSBsbSbs The homozygous parent can produce only one gamete type, bbss. Doing the cross: Thus the offspring are: 1/4 black, short-haired (BbSs), 1/4 black, long-haired(Bbss), 1/4 white, short-haired (bbSs),and 1/4 white, long-haired(bbss).

Paramecia are ciliates (organisms covered with cilia over their entire surface) . On one side of the paramecium there is a slight depression, the oral groove, which serves as a mouth. A paramecium feeds by creating a current (by the beating of the cilia around the oral groove) which sweeps the food particles into the oral groove. In this case, the food particles are the yeast cells. Under pressure from the current, the yeasts and fluids are pushed into the cytoplasm at the base of the groove, called the cytopharynx, and a food vacuole is formed. The vacuole breaks away from the groove and circulates around the cell, diminishing in size as it moves. As the vacuole circulates enzymes made in the cytoplasm are secreted into it. These enzymes break down the yeasts into simple biomolecules (sugars, amino acids, etc.) which diffuse out of the vacuole into the cytoplasm to be used later for the organism's metabolic processes. When the vacuole reaches a tiny specialized region of the cell surface called the anal pore, it becomes attached to the pore and then ruptures, expelling to the outside any remaining bits of indigestible material. The sequence of events is reviewed in the figure below. Since the enzymatic hydrolysis of foodstuff occurs inside the cell of the paramecium, digestion in this organism is of the intracellular type, known as intracellular diges-tion.

The term muscle, as it is commonly used, refers to a number of muscle fibers bound together by connective tissue. Skeletal muscle fibers are multinuc-leated cylindrical cells, 10 to 100\mum in diameter, and may be up to 1 ft. long. Generally each end of an entire muscle is attached to a bone by bundles of collagen fibers known as tendons. Some tendons are very long, and the site of attachment of the tendon to the bone is far removed from the muscle. For example, some of the muscles which move the fingers are found in the lower portion of the arm, between the elbow and the wrist. The most striking feature of muscle fibers is the series of transverse light and dark bands forming a regular pattern along the fiber. Both skeletal and cardiac muscle exhibit such banding patterns; smooth muscle does not. Though the pattern appears continuous across a fiber, the fiber is actually composed of a number of independent cylindrical elements in the cytoplasm of the fiber known as myofibrils. Bundles of these fibrils are enclosed by the muscle cell membrane or sarcolemma. Each myofibril is about a micron in diameter. Between the myofibrils are large numbers of mitochondria, which are to be expected in cells that have such a high energy requirement. The myofibrils show the same pattern of cross striations as the fibers of which they are a part. When viewed with the electron microscope, the structures responsible for the banding patterns become evident. The myofibrils consist of smaller myofilaments which form a regular repeating pattern along the length of the fibril. One unit of this repeating pattern is known as a sarcomere, and is the functional unit of the contractile system of the muscle. Each sarcomere (Figure 1) contains two types of myofilaments: thick and thin. In the central region of the sarcomere are the thick myofilaments, and they appear as a dark band. This band is termed the A band. These thick filaments contain the protein myosin. The thin myofilaments contain the protein actin and are attached at either end of the sarcomere at a structure known as the Z line. The limits of a sarcomere are defined by two successive Z lines. The thin filaments extend from the Z lines toward the center of the sarcomere where they overlap with the thick filaments. Two more bands are distinguishable in the muscle. The I band represents the region between the ends of the A bands of two adjoining sarcomeres. In this region only thin filaments are present; there is no overlap of the thick and thin filaments. Because it contains only thin filaments, the I band usually appears light. The H Zone corresponds to the space between the ends of the thin filaments; only thick filaments are present. The thick and thin filaments are arranged hexagonally with respect to one another. Each thick filament is surrounded by six thin filaments, and each thin filament is surrounded by three thick ones. (See Figure 2) An average muscle fiber contains about 15 billion thick and 65 billion thin filaments. Under extremely high magni-fication, the gap between thick and thin filaments in the region of the A band appears to be bridged by projections at intervals along the filaments. It appears that these projections, or cross bridges, are arranged in a spiral around the thick filament. These projections are thought to play an active role in the contraction of the muscle. (See following question.) An important muscle cell organelle is the sarcoplasmic reticulum, a continuous system of tubules extending throughout the cytoplasm, forming a closely meshed canal network around each myofibril. This organelle corresponds to the endoplasmic reticulum of other cell types, but in muscle it is largely devoid of ribosomes and exhibits a highly specialized repeating pattern. The tubules of the reticulum overlying the A bands have a prevailing longitudinal orientation but anastomose freely in the region of the H band. At regular intervals along the length of the myofibrils the longitudinal tubules of the sarco-plasmic reticulum, called sarcotubules, come together with transversely oriented channels of larger caliber called terminal cisternae. (See Figure 3) Pairs of parallel terminal cisternae run transversely across the myofibrils in close apposition to a slender intermediate element, the transverse tubule, commonly called the T tubule. These three transverse structures, namely the two parallel terminal cisternae and one T tubule, constitute the triads of skeletal muscle. In man, there are two triads to each sarcomere, situated at the junctions of each A band with the adjacent I bands. The cavity of the T tubule does not open into the adjacent cisternae and, strictly speaking, is not part of the sarcoplasmic reticulum. It is merely a slender invagination of the sarcolemma. The lumen of the T tubule is thus continuous with the extracellular medium surroun-ding the muscle fiber. The T tubules are involved in transmitting electrical signals from neurons on the cell surface to deep within the muscle fiber.

The formula for the total impedance is Z = \surd{R^2 + (X_L - X_C)^2} (a) At 400 cycles/sec, X_L = 2\pi × 400 × 0.5 = 1256 ohms, X_C = {1/(2\pi × 400 × 0.2 × 10^\rule{1em}{1pt}6)} = 1990 ohms, X = X_L - X_C = 1256 - 1990 = \rule{1em}{1pt} 734 ohms, Z = \surd{(600)^2 + (\rule{1em}{1pt} 734)^2} = 949 ohms. {see figure, part (a).} (b) At 600 cycles/sec, X_L = 2\pi × 600 × 0.5 = 1885 ohms, X_C = {1/(2\pi × 600 × 0.2 × 10^\rule{1em}{1pt}6)} = 1328 ohms, X = X_L - X_C = 1885 - 1328 = 557 ohms, Z = \surd{(600)^2 + (557)^2 = 818} ohms. {see figure, part (b).}

Question:

29.7 % copper, 15% sulfur, 2.8 % hydrogen, and 52.5 % oxygen. Determine the empirical formula of this hydrate from these percentages.

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Solution:

A hydrate is the chemical combination of water with another compound. For example, copper sulfate combines with water to form the hydrate of the composition CuSO----_4 . 5H_2O CuSO-_4 + 5H_20 \ding{217} CuS0_4 \textbullet 5H_2O One can find the empirical formula for a compound, when given the weight percents of the various elements making up the compound. This is done by finding the number of moles of each element in 100 g of the compound. The weight of each element is equal to the percent weight. The number of moles is equal to the weight divided by the molecular weight of the element. In 100 g of this compound, there are 29.7 g Cu, 15.0 g S, 2.8 g H, and 52.5 g 0.One can now determine the number of moles of each element present. number of moles = (number of grams) / MW MW of Cu = 63.5, MW of S = 32, MW of H = 1, MW of O = 16. number of moles of Cu =(29.7 g) / (63.5 g/mole) = 0.47moles number of moles of S = (15.0 g) / (32.0 g/mole) = 0.47 mole number of moles of H = (2.8 g) / (1 g /mole) = 2.8 moles number of moles of O = (52.5 g) / (16 g/mole) = 3.28 moles. To determine the empirical formula for this hydrate one must look at the ratio of Cu : S : H : O. The ratio of the number of moles of these elements is Cu : S : H : O .47 : .47 : 2.8 : 3.28 To find the empirical formula, these numbers should be made into integers. This is done by making the lowest number equal to 1 and solving for the other three. .47 / 1 = .47 / xx = .47 / .47 = 1 .47 / 1 = 2.8 / xx = 2.8 / .47 = 6 .47 / 1 = 3.28 / xx = 3.28 / .47 = 7 The ratio now becomes Cu : S : H : O 1 : 1: 6 : 7 The empirical formula for the compound is CuSH_6-O_7 or CuSO_4 \textbullet 3H_2O .

Question:

Deuterons with a mass of 3.3 × 10-27kg may have a velocity of 5 × 107(m/s) and an orbit radius of 0.8 m in cyclotron . (a) Find the frequency at which the accelerating fieldmust change. (b) What is the energy of the deuterons in MeV .

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Solution:

As shown in the figure , the particles move in the circular orbit in the dees under the influence ofthe magneticfield and are accelerated across the gap by the electric field . Therefore , the polarity of the electric field varies with time in such a way that each time the particle enters the gap , it is accelerated rather than decelerated . Deuteronstravel during each cycle a distance D = 2 \pi r = 2 × \pi × ( 8 × 10^-1 m ) = 5.0 m in the cyclotron . The frequency of the electric field should be equal to the frequency of the deuteron's revolutions of the dee . f = (1/period) = [(particle speed)/(distance travelled during one revolution)] = [{5 × 10^7 (m/s)}/5m] = 1 × 107 Hz = 10 MHz. b)Since the deuteron speed is much less than the speed of light we can use the non-relativistic expression for kinetic energy : K = (1/2)mv^2 = (1/2) (3.3 × 10^-27 kg) (5 × 10^7 m/s)2 = 4.13 × 10^-12 J Since 1 ev equals 1.6 × 10^-19 J, the kinetic energy may be written K = [(4.13 × 10^-12J)/{1.6 × 10^-19 (J/eV)}] = 2.58 × 10^7 eV = 25.8 MeV

Inheritance of blood groups is an example of a trait controlled by multiple alleles. The term multiple alleles is applied to three or more genes that can control a single trait; different combinations of any two genes may be present in a gene pair determining that trait in an individual. Any individual in the population may have any two of the possible alleles, but never more than two, because only two genes for a particular trait can be carried by an individual. Any gamete may have only one of the possible alleles. However, in the population as a whole, three or more different alleles will occur. The blood types of man, 0, A, B and AB, are coded for by multiple alleles. Gene I^A provides the code for the synthesis of a specific protein, agglutinogen A, in the red cells. Gene I^B leads to the production of a different protein, agglutinogen B. Gene i produces no agglutinogen. Gene i is recessive to the other two genes, but neither gene I^A nor I^B is dominant to the other. The symbols I^A, I^B and i are used to emphasize that all three alleles are alleles at the same locus. Individuals with genotypes I^A I^A and I^A i make up blood group A, and produce agglutinogen A. Those with genotypes I^B I^B and I^B i compose blood group B, and produce agglutinogen B. Blood group 0 individuals have genotype ii, and produce no agglutinoqens. When an individual has the genetic make-up of I^A I^B, he has both agglutinogens A and B and he belongs to blood group AB. These blood types are genetically determined and do not change during an individ-ual's lifetime. In reference to the problem given, it is possible to determine if an exchange occurred by comparing the blood-types of the babies with the possible bloodtypes that could be found in any offspring of each set of parents. Inother words, we can cross the genotypes of each set of other words, we can cross the genotypes of each set of parents and determine what genotypes are possible for their offspring. Then we can see if the bloodtypes of the babies that each family brought home is compatable with the possibilities, and if the parents could then have indeed produced a child with that given bloodtype. We are told that Mrs. Doe is type AB. Therefore, her genotype must be I^A I^B. Mr. Doe is type 0, so his genotype is ii. The possible bloodtypes of any Doe family offspring are obtained as follows: Thus, Mr. and Mrs. Doe can only produce offspring having bloodtypes A or B. In the Roe family, we are told that Mr. and Mrs. Roe are both type Roe are both type B. Therefore, each parent has one of two possible genotypes,. I^B I^B or I^B i. Taking each possiblility: Thus, the Roes can produce only offspring having bloodtype B or 0. Now let us look at the bloodtypes of the babies that each family brought home. The Does brought home Richard, who is type 0. But, as we have seen, the Does can produce only offspring of type A or B. The Roes brought home Nancy, who is type A. But it would be impossible for the Roes to have a child with bloodtype A, for the only possible bloodtypes for their offspring are B and 0. Therefore Richard, who is type 0, must be their child, and Nancy who is type A, must be the Doe's daughter. We see that an exchange did indeed take place.

In the program, the dimension of the search area is left variable, as are the number of trials. The direc-tions are given according to the orientation. (0,0,0) is the origin of the grid and all coordinates are positive. 10PRINT "QUIRKY QUARKS" 20INPUT "DIMENSION OF SEARCH AREA"; S: PRINT 30N = INT (LOG(S)/LOG(2)) + 1 40PRINT "YOU ARE AN EXPERIMENTAL PHYSICIST,"; 50PRINT "TRYING TO LOCATE A SUB-ATOMIC PARTICLE." 60PRINT "YOU HAVE";N;"TRIES. SPECIFY GUESS OF 70PRINT "LOCATION WITH A TRIO OF INTEGERS." 80A = INT (S\textasteriskcenteredRND (1) ) : B = INT (S\textasteriskcenteredRND(1)): C = INT (S\textasteriskcenteredRND (1)) 90FOR D = 1 TO N: PRINT: PRINT "TRIAL #";D;: INPUT X1,X2, X3 100IF ABS (X1 - A) + ABS (X2 - B) + ABS (X3 - C) = 0 THEN 150 110GOSUB 190: PRINT: NEXT D 120PRINT: PRINT "PARTICLE HAS DISINTEGRATED.DON RADIATION"; 130PRINT "SUIT" 140PRINT "THE PARTICLE WAS AT" A;",\textquotedblrightB;",";C 150PRINT: PRINT "YOU FOUND THE PARTICLE IN" ;D; "TRIES" 160PRINT: PRINT: INPUT "ANOTHER GAME(Y OR N)" ;A$ 170IF A$ = "Y" THEN 80 180PRINT "OK.SEE YOU IN OSLO": GO TO 290 190PRINT "CALCULATIONS SHOW, THAT YOUR TRY IS"; 200IF X2 > B THEN PRINT "NORTH"; 210IF X2 A THEN PRINT "EAST"; 230IF X1 < A THEN PRINT "WEST"; 240IF X2 <> B OR X1 <> A THEN PRINT "AND"; 250IF X3 > C THEN PRINT "TOO LOW." 260IF X3 < C THEN PRINT "TOO HIGH." 270IF X3 = C THEN PRINT "HEIGHT OK" 280RETURN 290END

The magnetic field due to each wire in the diagram at the point midway between them will be into the paper. This may be seen by use of the right hand rule. If the thumb of the right hand points in the direction of current through the wire, then the fingers will curl in the direction of the magnetic field (or magnetic flux density) created by the current. Applica-tion of this rule to both current carrying wires in-dicates that the field of each is into the page (see figures A and B). The effects due to the wires are therefore additive at that point and the total effect is twice the effect of either alone. Hence, midway between the wires the magnetic field due to one wire is B = (\mu_0 /2\pi) (I/r) where the permeability \mu_0 = 4\pi × 10^-7 N - A^-2 , I is the current through the wire, and r is the distance from the point being considered to the wire . Thus B = 2 × 10^-7 N - A^-2 × [(100 A)/(0.05 m)] = 4 × 10^-4 × - m^-2 The magnetic field due to both wires is then B_T = 2B = 8 × 10^-4 Wb - m^-2 .

A seed is actually a matured ovule, and consistsof a seed coat surrounding a core of nutritive tissue in which the embryo is embedded. The seed is an interesting structure in that it is composed of tissues from three generations. The embryo consisting of 2n cells derived from the fusion of egg and sperm, is the new sporophyte generation and functions in the continuation of the species by developing into a new reproducing plant. The nutritive tissue, or endosperm, is derived from the female gametophyte. In gymnosperms it is haploid, but in angiosperms it is triploid, resulting from the fusion of both polar bodies with the sperm. Its high starch content provides a source of nourishment for the growing embryo. The seed coat differentiates from the outer layer of the ovule, known as the integument, and as such is 2n and belongs to the old sporophyte generation. The seed coat encloses the endosperm and embryo, and owing to its tough, resistant properties, protects the seed from heat, cold, desiccation, and parasites.

The force on the electron is constant, there-fore the electron moves with a constant acceleration (see the figure). By Newton's Second Law a = F/m = eE/m =1.60 x 10^\rule{1em}{1pt}15 n/9.1 × 10^\rule{1em}{1pt}31 kg = 1.8 × 10^15(m/sec^2). Its velocity after traveling 1 cm, or 10^-2 m, is v^2 = v0^2 + 2a(x - x_0) where v_0 and x_0 are the initial velocity and position of the electron. Both of these are zero in this problem. Hence, v = \surd2ax = \surd[(2)(1.8 x 10^15 m/s^2)(10^\rule{1em}{1pt}2 m)] = 6.0 x 106m/sec. Its kinetic energy is 1/2 mv^2 = (1/2) (9.11 x 10^\rule{1em}{1pt}31 kg) (36 x 10^12 m^2/ s^2) = 1.6 x 10^\rule{1em}{1pt}17 joules. and it represents the work done by the electric force F^\ding{217} on the electron. The time is found from v = v_0 + at . But v_0 = 0 hence t = v/a = 3.3 × 10^\rule{1em}{1pt}9 sec.

The respiratory and circulatory systems have evolved to provide the cells with a continuous and adequate supply of oxygen. There are limitations however: Because oxygen is highly insoluble in water, it has a very low solubility in plasma, since plasma is 90% water. A major adaptation is the evolution of oxygen-carrying molecules. The oxygen carrying molecule found in the blood is hemoglobin. Hemoglobin is contained in the erythrocytes of vertebrates. Human erythrocytes, or red blood cells, are small, biconcave, disc-shaped cells (See Figure 1). During the maturation of red blood cells in mammals, they lose their nuclei, mitochondria, Golgi apparatus and other organelles. It is believed that the evolutionary loss of the nuclei in mammals has the adaptive advantage of leaving more space for hemoglobin. Similarly, since the red blood cell functions to transport oxygen, it would be counterproductive to have mitochondria which utilize oxygen. The mature ery-throcyte does not need these organelles since its main function is to carry hemoglobin, which in turn transports oxygen from the lungs to the tissues. Each human erythrocyte contains about 280 million hemoglobin molecules. A hemoglobin molecule consists of four polypeptide chains, two \alpha chains and two \beta chains. The capacity of hemoglobin to bind oxygen depends on the presence of a non--protein unit called a heme group. The heme group consists of an organic portion, called protoporphyrin, and an inor-ganic iron atom in the center (See Figure 2). The iron atom is capable of binding reversibly to a molecule of oxygen. Since each polypeptide chain contains a heme group, each hemoglobin molecule can combine with a maximum of four oxygen molecules. When all four binding sites contain oxygen molecules (Hb + 4O_2), the resulting compound is called oxyhemoglobin, as opposed to deoxyhemoglobin. Oxy and deoxyhemoglobin differ in quaternary structure- that is, oxygenation changes the three-dimensional configuration of the four polypeptide chains, it is this change in struc-ture which causes oxyhemoglobin to appear a brighter red than deoxyhemoglobin. This, in turn, accounts for the slight change in color between oxygen-rich blood and oxygen-poor blood. While some invertebrates also utilize hemoglobin in oxygen transport, others have different oxygen-carrying molecules. Hemocyanin is a protein in which a molecule of oxygen is carried between two copper atoms. When this mo-lecule is oxygenated, it is a pale blue color (deoxyhemocyanin is colorless). Hemocyanin is the oxygen carrying pig-ment in molluscs, crustaceae and arthropods. It was mentioned earlier that most of the hemoglobin molecules are contained within the red blood cells and not freely flowing in the plasma. By concentrating the hemoglobin molecules, the affinity for oxygen is increased, and the entry of oxygen from the outside to the lungs and from the lung tissues to the circulatory system is made faster and more efficient. Also, if the hemoglobin mo-lecules were found in the plasma, the increased concentration of protein would thicken the blood. This increased vis-cosity would require that the heart perform more work in pumping the blood to peripheral organs and particularly to the brain. Erythrocytes thus act as hemoglobin containers for more efficient physiological functioning.

Question:

Differentiate between homologous and analogous structures. Give examples of each.

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Question:

Ecologists refer to the ocean as the marine ecosystem. How are the major marine communities classified and where are the regions of their habitation?

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The ocean is a very complex ecosystem with many different kinds of communities. Temperature, salinity, availability of light, and depth all affect the nature of these communities and present major barriers to the free movement of organisms that have specific adaptability to certain environmental conditions. The major currents, which keep the waters of the ocean in continous circulation, also affect the nature of marine communities, as do waves and tides. Another important phenomenon, called upwelling, occurs where the winds move surface waters away from the steep coastal slopes, bringing cold water to the surface from the depths which is rich in such nutrients as phosphates and nitrates; the regions of such upswellings are often among the most productive regions of the ocean. The major marine habitats are shown in the accom-panying diagram. Marine organisms in general are divided into two groups, depending upon whether they live in the open water or upon or beneath the bottom. Those organisms that live suspended or swimming in the open waters are said to be pelagic. Bottom dwellers, called benthic organisms, may live on the surface or beneath the surface of the ocean floor. The benthic organisms include both attached organisms and those organisms that move in or on the substratum. Bottom dwellers are generally distinct for each of the neritic, or coastal, regions and the type found in each region is largely dependent on the type of ocean floor, whether it is sand, rock, or mud. Many animals are adapted for living in the sand and this group includes representatives of virtually every phylum of animals. Pelagic organisms are of two types. The microscopic plants and animals that are unable to move against the current are collectively called the plankton. The larger, free swimming animals that are able to move at will are called nekton. The planktonic plants constitute the phytoplankton. The phytoplanktonic algae are the most important producers throughout the ocean. The phytoplank-ton are eaten by a great array of invertebrate animals, ranging in size from protozoans to large medusae. These latter are the primary consumers. These, in turn, are consumed by other, larger invertebrates and vertebrates. As would be expected, plankton attains its greatest density in the upper lighted zone where there is sufficient light for photosynthesis to occur. In productive waters, planktonic organisms may occur in such enormous numbers that the water appears turbid. The animals that are permanent inhabitants of the lower dark zones are carnivorous, suspension, or detritus (dead organic matter) feeders and depend ultimately on the photosynthetic activity of the microscopic algae in the upper, lighted regions. Therefore, vertical distribution of marine organisms is largely controlled by the depths of light penetration. Light sufficient for photosynthesis to exceed respiration penetrates only a short distance below the surface, with this distance being dependent upon the turbidity of the water. Below this is a region where some photosynthesis can occur, but here the production of organic matter is less than the loss through respiration.

Question:

Explain the need for STOP JOB CARDS, and list the differentmethods used in computers to bring about a proper endof the program.

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The computer is just a machine without any intel-ligence of its own. It only does exactly what is has been programmed to do, i.e., told, to do. When an instruction is given to the computer to get some kind of data, say, from a punched card, the computer keeps on getting the data till itis explicitly told again to stop getting any more data. When a computer is taking in data as part of a program, there must besome means of telling the computer how many data items to expect, especiallywhen the appropriation of data is part of an iterative loop. If the computertries tocontinouslyread in data it will eventually not find any moredata. The computer now assumes there is an error, and either terminatesthe program, or gives out an error message. To eliminate this problemwe must have some ways of telling the computer just how much datato expect or what to do (other than bombing the program!) when the dataruns out. There are normally four ways to do this: 1) Header Card method,2) Trailer Card method, 3) End-File condition method,4) Count and compare method

Question:

Describe sexual reproduction in mosses.

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The moss plant has a life cycle characterized by a marked alternation between the sexual and asexual generations. The sexual gametophyte generation is the familiar, small, green, leafy plant with an erect stem held to the ground by numerous rhizoids. When the gametophyte has attained full growth, sex organs develop at the tip of the stem, in the middle of a circle of leaves and sterile hairs called paraphyses. The male organs are sausage-shaped structures, called the antheridia. Each antheridium produces a large number of slender, spirally-coiled swimming sperm, each equipped with two flagellae. After a rain or in heavy dew, the sperm are released and swim through a film of moisture to a neighboring female organ, either on the same plant or on a different one. The female organ, the archegonium, is shaped like a flask and has one large egg at its broad base, the ventor. The archegonium releases a chemical substance that attracts the sperm and guides them in swimming down the archegonium to the ventor. Here one sperm fertilizes the egg. The resulting zygote is the beginning of the diploid, asexual sporophyte generation. The mature sporophyte is composed of a foot embedded in the archegonium and a leafless, spindle-like stalk or seta which rises above the gametophyte. The sporophyte is nutritionally dependent on the gametophyte, absorbing water and nutrients from the archegonium via the tissues of the foot. A sporangium, called the capsule, forms at the upper end of the stalk. Within the capsule each diploid spore mother cell undergoes meiotic division to form four haploid spores. These spores are the beginning of the next gametophyte generation. When the capsule matures, it opens and releases spores under favorable conditions in a mechanism specific to its kind. In the Sphagnum, for example, the mature capsule shrinks, bursts, pushes away the lid, and exposes the spores to the wind. If a spore drops in a suitable place, it germinates and develops into a protonema, a green, creeping, filamentous structure. The protonema buds and produces several leafy gametophytes, thereby completing the moss life cycle.

The two blocks will start sliding radially when the centrifugal force on B is sufficient to overcome (a) the force of friction between block B and the table and (b) the tension in the cord which is equal to the centrifugal force on block A plus the force of friction between block A and the table. Using the formula for centrifugal force: F_C = m v^2 / r = W v^2 / gr = W\omega^2 r^2 / gr = W\omega^2 r / g and for friction:F_f = \mu W, where W is the weight of a block, \omega is the angular velocity, r the radius, g = 32 ft / sec ^2, the acceleration due to gravity, and \mu the coefficient of static friction, we can write the desired equation as: W_B\omega^2 r_B / g = \muWB+ \muW_A + W_A\omega^2 r_A / g where the subscripts A and B refer to the two blocks. (W_B\omega^2 r_B / g) \rule{1em}{1pt} (W_A\omega^2 r_A / g) = \mu(W_A + W_B) [\omega^2(W_B r_B \rule{1em}{1pt} W_A r_A)] / g = \mu(W_A +W_B) \omega^2 = [\mug(W_A +W_B)] / (W_B r_B \rule{1em}{1pt} WAr_A) \omega^2 = \surd[{(\mug)(W_A + W_B)} / (W_B r_B \rule{1em}{1pt} WAr_A)] . Substituting the numerical values, yields \omega = \surd[{.3 × 32 (10 + 20)} / {(20 × 3) \rule{1em}{1pt} (10 × 2)} ] = \surd(7.2) =2.68 radians/sec.

American roulette is played on a wheel with 38 compartments numbered 1 to 36, 0 and 00. Odd and even num-bers alternate around the wheel as do red and black. In the simulation bets can range from $5 to $500 and you may bet on red or black, odd or even, first or second 18 numbers, a column or a single number. Place bets according to the following payoffs: 1) The numbers 1 to 36 signify a straight bet on that num-ber and the payoff is 35:1 (i.e. 35 dollars return for a $1 bet). 2) If you bet that the ball will land in the column or on groups of numbers 1 - 12 or 13 - 24 or 25 - 36, the bets are 2:1. In the program, the bet numbers are 1 - 36 for single numbers and 37 for 1 - 12, 38 for 13 - 24, 39 for 25 - 36. 3) If you bet that the ball will land on i) a number from groups 1 - 18 or 19 - 36 ii) even or odd iii) red or black the payoff is 1:1 ($1 additional for every $1 bet). The even money bet numbers are 43 for 1 - 18, 44 for 19 - 36, 45 for even, 46 for odd, 47 for red and 48 for black. 4) 0 and 00 pay off 35:1 and count only when they are expli-citly betted upon: The bet number for 0 is 49 and 50 for 00. 10PRINT "ROULETTE" 20PRINT "MINIMUM BET IS $5, MAXIMUM IS 500" 30PRINT "WHEN I ASK FOR EACH BET, TYPE"; 40PRINT "THE NUMBER AND THE AMOUNT, SEPARATED BY A"; 50PRINT "COMMA. FOR EXAMPLE: TO BET $500 ON BLACK,"; 60PRINT "TYPE 48,500" 70DIM B(100), C(100), T(100), X(38) 80DIM A(50) 90FOR I = 1 TO 38: X(I) = 0 : NEXT I 100REM MAT X = ZER 110P = 1000 120D = 100000 130PRINT "HOW MANY BETS"; 140INPUT Y 150IF Y < 1 OR Y <> INT(Y) THEN 130 160FOR I = 1 TO 50: A(I) = 0 : NEXT I 170FOR C' = 1 TO Y 180PRINT "NUMBER"; C; 190INPUT X,Z 200B(C) = Z: T(C) = X 220IF X < 1 OR X > 50 OR X <> INT(X) THEN 180 230IF Z < 1 OR Z <> INT(Z) THEN 180 240IF Z < 5 OR Z > 500 THEN 180 250IF A(X) = 0 THEN 280 260PRINT "YOU ALREADY MADE THAT BET" 270GOTO 180 280A(X) = 1: NEXT C 300PRINT "SPINNING" 310PRINT: PRINT 330S = INT(100\textasteriskcenteredRND(1)) 340IF S = 0 OR S > 38 THEN 330 350X(S) = X(S) + 1 360IF S < 37 THEN 420 370IF S = 37 THEN 400 380PRINT "00" 390GOTO 520 400PRINT "0" 410GOTO 520 420RESTORE 430FOR I = 1 TO 18 : READ R: IF R = S THEN 500: NEXT I 470A$ = "BLACK" 480PRINT S; A$ 490GOTO 520 500A$ = "RED" 510GOTO 480 520PRINT 530FOR C = 1 TO Y: IF T(C) < 37 THEN 1210 540ON T(C) - 36 GOTO 590, 690, 720, 750, 800, 850, 900, 970, 1000 560ON T(C) - 45 GOTO 1030, 1060, 1130 570GOTO 1210 580STOP 590REM 1 - 12(37) 2 : 1 600IF S < = 12 THEN 650 610PRINT "YOU LOSE"; B(C); "DOLLARS ON BET"; C 620D = D + B(C) 630P = P - B(C) 640GOTO 680 650PRINT "YOU WIN"; B(C)\textasteriskcentered2; "DOLLARS ON BET"; C 660D = D - B(C)\textasteriskcentered2: P = P + B(C)\textasteriskcentered2 680GOTO 1310 690REM 13-24(38) 2 : 1 700IF S > 12 AND S < 25 THEN 650 710GOTO 610 720REM 25 - 36(39) 2 : 1 730IF S > 24 AND S < 37 THEN 650 740GOTO 610 750REM FIRST COLUMN (40) 2 : 1 760FOR I = 1 TO 34 STEP 3 : IF S = I THEN 650 : NEXT I 790GOTO 610 800REM SECOND COLUMN (41) 2 : 1 810FOR I = 2 TO 35 STEP 3: IF S = I THEN 650 : NEXT I 840GOTO 610 850REM THIRD COLUMN (42) 2 : 1 860FOR I = 3 TO 36 STEP 3 : IF S = I THEN 650: NEXT I 890GOTO 610 900REM 1 - 18 (43) 1 : 1 910IF S < 19 THEN 930 920GOTO 610 930PRINT "YOU WIN"; B(C); "DOLLARS ON BET"; C 940D = D - B(C) 950P = P + B(C) 960GOTO 1310 970REM 19 - 36(44) 1:1 980IF S < 37 AND S > 18 THEN 930 990GOTO 610 1000REM EVEN (45) 1 : 1 1010IF S/2 = INT(S/2) AND S < 37 THEN 930 1020GOTO 610 1030REM ODD (46) 1 : 1 1040IF S/2 <> INT(S/2) AND S < 37 THEN 930 1050GOTO 610 1060REM RED (47) 1:1 1070RESTORE 1080FOR I = 1 TO 18 1090READ R: IF S = R THEN 930: NEXT I 1120GOTO 610 1130REM BLACK (48) 1 : 1 1140RESTORE 1150FOR I = 1 TO 18: READ R: IF S = R THEN 610: NEXT I 1190IF S > 36 THEN 610 1200GOTO 930 1210REM 1 TO 36, 0,00 (1 - 36,49,50) 35 : 1 1220IF T(C) < 49 THEN 1260 1230IF T(C) = 49 AND S = 37 THEN 1280 1240IF T(C) = 50 AND S = 38 THEN 1280 1250GOTO 610 1260IF T(C) = S THEN 1280 1270GOTO 610 1280PRINT "YOU WIN"; B(C)\textasteriskcentered35; "DOLLARS ON BET"; C 1290D = D - B(C)\textasteriskcentered35: P = P + B(C)\textasteriskcentered35 1310NEXT C 1330PRINT "TOTALS:", "ME", "YOU" 1340PRINT " ", D, P 1350IF P > 0 THEN 1380 1360PRINT "YOU JUST SPENT YOUR LAST DOLLAR" 1370GOTO 1470 1380IF D > 0 THEN 1420 1390PRINT "YOU BROKE THE HOUSES" 1410GOTO 1480 1420PRINT "AGAIN"; 1430INPUT Y$ 1440IF LEFT $ (Y$,1) = "Y" THEN 130 1450DATA 1,3,5,7,9,12,14,16,18,19,21,23,25,27,30,32,34,36 1470PRINT "THANKS FOR YOUR MONEY" 1480END.

Question:

Discuss and illustrate the meaning of absolute and relative addressing.

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This example illustrates a problem that is frequently encountered, in which it is necessary to find a desired force by first computing the force that is the reaction to the one desired, and then using Newton's third law. That is, we first calculate the force with which the elevator floor pushes upward on the passenger P; the force desired is the reaction to this. The figure shows the forces acting on the passenger. We may use Newton's Second Law, F = ma, to relate the net force on the man to his acceleration. The resultant force is P - W. The mass m of the passenger is his weight, mg, divided by 32 f/s^2, or 5 slugs, and his acceleration is the same as that of the elevator, \sumF = ma P - 160 lb = 5 slugs × 4 (ft/sec^2) = 20 lb, .P = 180 lb. The passenger exerts an equal and opposite force downward on the elevator floor.

The diversity of organic compounds is so vast that these organic compounds have been divided into families, such as alkanes, alkynes and aromatic compounds, which have no counterparts among inorganic compounds. The tremendous variety of these compounds is made possible by the unique properties of carbon. To be able to understand how carbon can form such a huge number of compounds with a great variety of properties, the way in which the atoms in these molecules are bonded together must be looked at. Carbon has a valence number (the number of bonds an atom of an element can form) of four. This means that each carbon atom always has four bonds which can be either bonded to four other atoms, as in methane; to three other atoms, as in formaldehyde; or to 2 other atoms as in hydrogen cyanide. In other words carbon is capable of forming single, double and triple bonds: But the diversity this element possesses does not stop here. It is capable of bonding to other carbon atoms in an almost unique variety of chain and ring structures. This property is called catenation and accounts for the tremendous variety seen in the following compounds: These examples show the use of a minimum number of carbon atoms. The diversity possible is apparent when one considers the fact that these different types of bond can be joined together in infinite numbers. To exemplify this, look at the compound: As it can be seen, a multitude of different variations could be made with the only constraining factor being that carbon must have four bonds. The freedom allowed carbon is apparent when contrasted with the halogens (iodine, chlorine, bromine, and flourine) which are monovalent and thus can only form one bond, for example: H-ClBr-Br hydrochloric acidbromine Another feature of the compounds of carbon that contributes to their variety is the existence of isomers (compounds composed of the same number and kind of atoms, but with the atoms arranged differently in space). Look, for example, at the compound C_4H_10. This can exist as: Although these compounds have the same chemical formulas they have different chemical, physical and biological properties due to their arrangement in space.

Question:

The "pill" is a birth control device available to women, and is one of the most effective and widely used contraceptives in today's society. How does it function? Are there any side effects when taking the pill?

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The pill contains a small dose of synthetically produced female sex hormones, primarily estrogen and progesterone. These two substances act together to suppress ovulation, thus disrupting the normal reproductive process. The progesterone-estrogen pill acts on the cervix of the uterus to increase mucus secretion which impairs the motility of sperms. It acts upon the uterine lining, preventing implantation because the fertilization of the egg is not in synchrony with the development of the uterus. But most importantly, the pill prevents FSH and LH secretions by direct effect on the pituitary, causing the uterus to be sustained in its post-ovulatory state. It achieves this by acting as a feedback inhibitor on the anterior pituitary, so that FSH and LH secretions are suppressed. With mi-nimal amounts of FSH and LH, follicular maturation and ovulation cannot occur. The pill must be taken daily for 20 or 21 days of the 28-day menstrual cycle, starting with the fifth day after the beginning of menstruation. When taken each day without fail, it is virtually 100% effective. There are, however, some side effects when using the pill. Up to 25% of the women taking the pill develop some of the outward symptoms of pregnancy: swelling and tenderness of the breasts, nausea, irritability, etc. Some users of the pill also develop increased susceptibility to inflam-mation of the veins combined with problems ,of blood clotting. The principle concern over the use of the pill is that the levels of the synthetic estrogens in certain pills are dangerous, since these synthetic estrogens had been shown to be carcinogenic in experimental animals. There are two major points that should be considered in the use of the pill. In the first place, do oral contra-ceptives significantly threaten the users with blood clot problems? Secondly, is prolonged use of the pill by humans going to result in a marked rise in breast cancer? It is the view of many endocrinologists that the evidence in experimental animals suggests that such a threat may exist among women presently planning to use oral contra-ceptives .

Question:

What are the types of simple conditional expressions and compound conditional expressions? Give their formats and any conditions if necessary.

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Solution:

The Simple Conditional Expression: A simple condition reducing to the value true or false can be ex-pressed in the following types of simple conditional ex-pression: i) a relation ii) a condition - name iii) a sign condition iv) a class condition v) a switch status condition. Any of the above are used in a decision-making opera-tion to select different paths of control in a program. 1) The Relational Condition: This condition causes a comparison of magnitude between two quantities. The general form is: Note: means choose one. The first quantity is called the subject and the second is referred to as the object of the condition. A relational operator must be preceded and followed by a space. The list of relational operators are GREATER, LESS, EQUAL. The word, NOT, before an operator specifies the opposite of what the operator normally specifies. e.g.,AGE IS GREATER THAN 21 SUM IS NOT LESS THAN 100 A+B IS EQUAL TO A+M The words IS and THAN are optional and do not alter the meaning of the expression. e.g.,AGE GREATER 21 is equivalent to AGE IS GREATER THAN 21. 2) The Condition-Name condition: A condition name is a name, assigned in the Data Division, to one of the values a conditional variable assumes. The variable is tested to de-termine whether or not its value is equal to a value asso-ciated with a particular condition name. Using an IF statement, the test to determine marital status may appear as follows: IF MARRIED THEN......... which is equivalent to IF MARITAL-STATUS IS EQUAL TO 2 THEN...... NOT is used to specify the opposite of what the condition- name test would normally specify. e.g.,IF NOT SINGLE ................ is equivalent to IF MARITAL-STATUS IS NOT EQUAL TO 1............. (3) Sign Condition: This condition determines whether a numeric quantity is less than, equal to, or greater than zero. The general format is: Note: [ ] means optional e.g., A + B {_\ast} CIS POSITIVE The identifier in a numeric status test must always represent a numeric value. 4) Class Condition: The class condition test deter-mines whether a quantity is purely numeric or alphabetic. The general format of this conditional expression is as follows: The test must be consistent with data description of the item being tested. That is, only the NUMERIC test may be used for data which has been described as numeric and only the ALPHABETIC test for alphabetic data either may be used for alpha numeric data. The usage of the identifier must be defined, either explicitly or implicitly, as DISPLAY. 5) The Switch Status Condition: This condition is used to determine whether a particular hardware switch is off or on. The Compound Conditional Expression A conditional expression having a single condition is referred to as a simple conditional expression. One con-taining more than one condition is referred to as a compound conditional expression. The various conditions are connected by the logical operators AND and OR and the format is: For example: A IS EQUAL TO B OR C IS LESS THAN D AND E IS GREATER THAN F Implied Subject: In a compound conditional expression it is possible for several consecutive conditions to have the same subject. e.g., SUM IS GREATER THAN 100 OR SUM IS LESS THAN 1000 ie equivalentto SUM IS GREATER THAN 100 OR LESS THAN 1000. Implied Operator:Operators, as well as subjects may be im-plied in a compound condition consisting of a consecutive series of relational expressions. e.g., SALES IS EQUAL TO 25 OR EQUAL TO 66 OR EQUAL TO 85 is equivalent to SALES IS EQUAL TO 25 OR 66 OR 85 Implied operators must only be used in expressions where the subjects are also implied.

Consider any point X at a perpendicular distance h from the plane sheet of charge density \rho. See figure A. Drop the perpendicular from the point to the sheet, cutting the latter at 0, and draw two circles, centered at 0, at radii of y and y + dy. Take a small portion of the annulus so formed, of length dl, and consider the electric intensity dE' at the point X due to the small (almost point-like) element of charge. Then the electric field due to a point charge is dE' = dq/4\pi\epsilon_0r^2 Since the charge density \rho = dq/dA = dq/dldy then, dE' = (\rho dl dy)/(4\pi\epsilon_0 r^2). The direction of dE' is the same as that of r, and dE' may therefore be resolved into components along OX and at right angles to it. The component along OX has the same value; no matter what position on the annulus dl occupies. But the element of the annulus diametrical-ly opposite dl produces a component perpendicular to OX equal but opposite to that produced by dl. (See figure B). These two components thus cancel out, as do all components from diametrically opposite elements. The electric intensity from the whole annulus is thus perpendicular to the sheet and has magnitude dE = \oint dE' sin \texttheta = [(\rho dy)/(4\pi\epsilon_0r^2)] sin \texttheta \oint dl = (\rho dy)/(4\pi\epsilon_0r^2) \bullet (h/r)2\piy = [h\rhoy dy] / [2\epsilon_0(h^2 + y^2)^3/2 ]. We used the fact that sin \texttheta = h/r (geometric considerations in figure A) and that the sum of all the infinitesimal elements of length dl about the whole ring is equal to the circumference of the ring. Also used was the fact that r = (h^2 + y^2)^1/2. For the whole sheet of charge the electric intensity is the sum of contributions due to all the annulii of radius y = 0 to y = \infty (for the infinite plane sheet). Or E = \int dE = ^\infty\int_0 [h\rhoy dy] / [2\epsilon_0(h^2 + y^2)^3/2 ] = (h\rho/2\epsilon0) ^\infty\int_0[y dy] / [(h^2 + y^2)^3/2] = \rule{1em}{1pt} [h\rho/2\epsilon_0] [(h^2 + y^2)-1/2]^\infty_0 = \rule{1em}{1pt} [h\rho/2\epsilon_0] [0 - (1/h)] = \rho/2\epsilon_0. To apply Gauss's law to the same problem, construct a cylinder of small and uniform cross-sectional area dA at right angles to the sheet and bisected by the sheet (see figure C). Since the sheet is infinite and the charge uniformly distributed, the electric intensity must be the same at all points equidistant from the sheet, and thus by symmetry must be everywhere per-pendicular to the sheet. Hence E is everywhere parallel to the sides of the cylinder and thus the flux of E from the cylinder through its sides is zero. The magni-tude of E at each end of the cylinder will be the same if the cylinder is bisected by the sheet, and E will be perpendicular to each end. Hence, applying Gauss's law, we obtain \int E^\ding{217} \bullet dA^\ding{217}= \int[(E^\ding{217} \bullet dA^\ding{217})_top + (E^\ding{217} \bullet dA^\ding{217})_side + (E^\ding{217} \bullet dA^\ding{217})_bottom] Since dA^\ding{217} is small, \intE^\ding{217} \bullet dA^\ding{217}\approx E^\ding{217} \bullet dA^\ding{217} and \intE^\ding{217} \bullet dA^\ding{217}= [EdA + 0 + EdA = 2EdA]= dq/\epsilon_0 ; but \rho = dq/dA and 2EdA = (\rhodA/\epsilon_0). Therefore E = \rho/2\epsilon_0. Thus E is everywhere perpendic-ular to the sheet and has the same value \rho/2\epsilon_0 at all points.

Since the acceleration is constant, a_c= (v_Final - v_Initial)/∆t = [(27){m/(sec-0)}]/(30 sec) = (0.9)(m/sec^2 ) The angular velocity vector \omega, which is constant for uniform motion, is The angular velocity vector \omega, which is constant for uniform motion, is perpendicular to the plane of the circle. This will be the linear acceleration of the axle on which the wheel is supported. Imagine that you are sitting on the axle. You will observe the road moving by with velocity V and the tire spinning with a vel-ocity such that, at the maximum speed, the angular velocity \omega= V/R = [27(m/sec)]/(0.36 m) = 75 sec^-1 when the acceleration is constant, the angular acceleration is given by \alpha= (\omega_Final - \omega_initial)/(∆t) = (75 sec^-1 - 0)/(30 sec) = 2.5 sec^-2 and we can see that a_c = \alphaR . A point on the circumference of the wheel experiences an acceleration, directed toward the axle, of magnitude a= V^2/R =\omega^2R = (75 sec^-1)^2(0.36 m) = 2.0 × 10^3 (m/sec^2) since this point is travelling in a circular path.

Since there is a fixed mass of air at all times in the top of the barometer tube, the gas law (pV/T) = const may be applied to it directly. In the first case when the temperature is 25\textdegreeC = 25 + 273 = 298\textdegreeK, the barometric height is 76 cm of mercury. This is the pressure exerted at the free surface of mercury and thus, by the laws of hydrostatic pressure, at the same horizontal level inside the barometer tube. At points in the tube higher than this, the pressure drops off with height. If there were true vacuum instead of air above the mercury column in the tube, the mercury would rise to a height of 76 cm, which is the atmospheric pressure. How-ever, the rise of the mercury is resisted by the pressure of the air trapped in the tube (see the figure) , therefore the pressure of the trapped air must be equal to that ex-erted by (76 - 74.5) = 1.5 cm of mercury. The volume of the trapped air is (89.4 - 74.5)A cm^3 = 14.9A cm^3, where A cm^2 is the cross-sectional area of the tube. In the second case, when the temperature is 11\textdegreeC = 284\textdegreeK, the pressure of the trapped air is (p_0 - 75.2 cm of mercury), where p_0 is the atmospheric pressure on that day. The volume of trapped air is (89.4 - 75.2)A cm^3 = 14.2A cm^3 . Hence (1.5 cm × 14.9A cm^3)/(298\textdegreeC) = [(p^0 - 75.2cm)14.2A cm^3] / [284\textdegreeC] \therefore p_0 - 75.2 cm = 1.5 cm or p_0 = 76.7 cm of mercury.

The collection area A of the mirror is, A = \pir^2 = \pi(5 × 10^-1 m)^2 = 7.9 × 10^-1 m^2 The total power P incident on the reflector is, P = (5.5 × 10^2W/m^2)(7.9 × 10^-1m^2) = 4.345 × 10^2 W Since the conversion efficiency is only 70 percent, the power converted to thermal power H is, H = P(7.0 × 10^-1) = (4.34 × 10^2W) (7 × 10^-1) = 3.04 × 10^2 J/s The temperature must be increased by 80\textdegree K (from 293 to 373\textdegree K). The total thermal energy Q required is then, Q = (4.19 × 10^3 J/\textdegreeK)(8 × 10^1 \textdegreeK) = 3.35 × 10^5 The time would be the amount of thermal energy needed divided by the rate at which thermal energy is produced. Hence, t = Q/H = (3.35 × 10^5J)/(3.04 × 10^2J/s) = 11.0 × 10^2 s = (11.0 × 10^2s)/(60s/min) = 1.84 × 10^1 min.

Question:

What are the functions of the nucleus? What is the evidence that indicates the role of the nucleus in cell metabolism and heredity?

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If the cell is thought of as a miniature chemical plant designed to carry out all the processes of life, then the nucleus can be compared to a central computer that controls a network of sophisticated and highly complicated biochemical machinery. This is because the nucleus contains the chromosomes, which bear the genes, the ultimate regulator of life. The genes comprise a library of programs stored in the nucleus: programs that specify the precise nature of each protein synthesized by the cell. The nucleus monitors changing conditions both inside the cell and in the ex-ternal environment, and responds to input of information by either activating or inhibiting the appropriate genetic programs. Control of protein synthesis is the key to control-ling the activities and responses of the cell, since a tremendeous array of important biological and biochemical processes are regulated by enzymatic proteins. By switch-ing particular genes on and off, the cell controls not only the kinds of enzymes that it produces, but also the amounts. Both qualitative and quantitative control of enzyme synthesis are crucial to the proper functioning of the cell and the whole organism. Experiments with the large single-celled alga Acetabularia. Cells without nuclei have a very limited range of function. The mammalian red blood cell does not contain a nucleus and is unable to reproduce; it is functional only for a relatively short period. Egg cells, from which nuclei have been experimentally removed, may divide for a while, but the products of division never differentiate into specialized cell types, and eventually die. Fragments without a nucleus, cut from such large unicellular organisms as the amoeba or the unicellular algae Acetabularia, survive temporarily; but ultimately they die, unless nuclei from other cells are transplanted into them. These experiments demonstrate that the nucleus is essential to long term continuation of life processes, and to structural and functional differentiation of cells. The nucleus is of central importance also in the transmission of hereditary information. The nucleus carries the information for all the characteristics of the cell. This information is found in the chromosomes, which consist of protein and deoxyribonucleic acid (DNA). DNA is the hereditary material which contains all the information for the growth and reproduction of the cell. When the cell divides, the nuclear information is trans-mitted in an orderly fashion to the daughter cells by replication of the chromosomes and division of thenucleus. Thus, in this type of division, the daughter cells each possess a single complement of genetic information iden-tical to that of the mother cell. The hereditary importance of the nucleus can be demonstrated experimentally (see Figure). If a fragment containing the nucleus is cut from an Acetabularium of one species, characterized by a given appearance, the fragment will regenerate a whole cell of that species. This regenerative ability permits experiments of the type shown in the accompanying figure, in which nuclei of one species are grafted onto the cytoplasm of a different species. The observation from these experiments shows that the appearance of the regenerated cell eventually resembles that of the organism from which the nucleus is taken. Hence, we conclude that the nucleus, as a con-trolling center, produces information that controls the cytoplasm and participates in the regulation of cell growth and structure. Since the crucial information comes from the nucleus, the regenerated cell will resemble the organism from which the nucleus is taken.

We proceed by first defining ''total spin" and "multiplicity". Let the number of electrons having "spin up" be denoted by n_\uparrow and the total number of electrons having "spin down" be denoted by n_\downarrow. Assigning a value of (+ 1/2) to spin up and (- 1/2) to spin down, the total spin s is given by the absolute value of n_\uparrow (+ 1/2) + n_\downarrow (- 1/2) = n_\uparrow/2 + n_\downarrow/2. The multiplicity is then defined as m = 2s + 1. The configuration with the highest spin (highest multi-plicity) will have the greatest paramagnetism. (.Substances that are weakly attracted to magnets are paramagnetic.) The configuration with the lowest spin (lowest multiplicity), will be the least paramagnetic. The p orbital can contain six electrons and the d orbital can contain 10 electrons. When these orbitals are filled, electrons are sequentially placed in each sub-orbital (e.g. the p_x or the d_xy) until there is one electron in each suborbital, all electrons having the same spin. Only when there is one electron in each suborbital does another electron enter a suborbital already containing an electron, this time with a spin opposite to that of the electron already in the suborbital (the spins are then "paired") in accordance with the Pauli Exclusion Principle. In the accompanying table we show the orbital filling corresponding to each configuration. Counting n_\uparrow and n_\downarrow, we then calculate s and m. The configurations p^1, p^5, and d^1, have the same electron paramagnetism [all have spin (1/2)]. The next most paramagnetic configurations are p^3 and d^3 (spin 3/2) and the most paramagnetic configuration is d^5 (spin 5/2).

Question:

An axon can conduct impulses in both directions simultaneously. However, an impulse normally propagates down an axon in one direction only, either axon to dendrite or dendrite to axon. Explain how this unidirectional con-duction of impulses along the neuron is possible.

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Normally, an impulse travels from the dendrites to the cell body, then along the axon to the synaptic junction. However, impulses can be made to travel along an axon to the dendrites. But these impulses are not relayed to the next neuron since dendrites do not con-tain the chemical substance (neurotransmitter) required in the transmission of an impulse across a synapse. Although two directions of transmission along an axon are potentially possible, once an impulse is started at either end of a neuron, it will pass on through the neuron toward the opposite end. This phenomenon is due to the existence of a recovery period, known as the refractory period in the neuronal membrane. After a region of the axon is stimulated, a refractory period must elapse before it can be excited again. To understand the relation between the refractory period and the one-way conduction of impulses, one must recall that a region undergoing an action potential is positive with respect to other points inside the axon. Since that region is of higher potential than the surrounding region, a flow of Na^+ ions spreads from that local region to the adjacent regions on both sides of it. This is confirmed by observations which indicate that within a neuron, an impulse can travel in either direction from its point of origin. However, because the region that has just been stimulated cannot be made to depolarize until it has recovered, the impulse moves in one direction only (see Figure). By the time that region has regained the ability to be depolarized, the impulse has traveled too far away from it, and hence cannot depolarize it. It should be remembered that the unidirectional propagation of impulses is determined by the stimulus location (i.e., at which end the stimulus is given) rather than an intrinsic inability of the axon to conduct in the opposite direction. If action potentials are initiated near the middle of a neuron, impulses propagate from this region toward the two ends. In most nerves, however, action potentials are initiated at one end of the cell and impulses therefore propagate in only one direction toward the other end of the cell.

Pressure is defined as force per unit area. Atmospheric is measured by using a barometer (usually a mercury barometer; see figure). It is constructed by inverting a tube longer than 76 cm filled with mercury into a dish of mercury. The atmosphere will support only that height of mercury which exerts an equivalent pressure; any excess mercury will fall into the reservoir and leave a space with zero air pressure above it. The pressure exerted on the diver from above is equal to the sum of the pressure exerted by the sea water and the pressure exerted by the atmosphere. One standard atmosphere equals the pressure exerted by exactly 76 cm (= exactly 760 mm) of mercury at 0\textdegreeC (density Hg = 13.5951 g/cm^3) and at standard gravity, 980.665 cm/s^2. Thus, 1 standard atm equals 13.5951 g/cm^3 × 76 cm (exactly) × 980.665 cm/s^2 = 1.01325 × 10^6 dynes/cm^2. The pressure exerted by the water is found similarly: pressure of water = density × height × standard gravity = 1.00 g/cm^3 × 1.5 × 10^3 cm × 980.665 cm/s^2 = 1.47 × 10^6 dynes/cm^2. However, the problem states that atmospheric press-ure is 1.02 standard atmospheres and, therefore, equals (1.02) (1.01325 x 10^6 dynes/cm^2) = 1.03 × 10^6 dynes/cm^2. To this atmospheric pressure, the pressure of the water is added to yield a total pressure of (1.03 × 10^6 dynes/cm^2) + (1.47 × 10^6 dynes/cm^2) = 2.50 × 10^6 dynes/cm^2. This answer expressed in atmospheres gives [(2.50 × 10^6 dynes/cm^2 ) / (1.01 × 10^6 dynes/cm^2 /atm)] = 2.48 atm. (b)This part is very similar to part (a). The total pressure exerted is the pressure of the water plus the pressure of the atmosphere. The pressure of the atmosphere from part (a) is 1.03 × 10^6 dynes/cm^2. The pressure of the water must be calculated: pressure of water = density × height × standard gravity = 1.20 g/cm^3 × 1.5 × 10^3 cm × 980.665 cm/s^2 = 1.77 × 10^6 dynes/cm^2. The total pressure is 1.77 × 10^6 dynes/cm^2 + 1.03 × 10^6 dynes/cm^2 equals 2.80 × 10^6 dynes/cm^2 . This answer expressed in atmospheres is [(2.80 × 10^6 dynes/cm^2 )/{1.01 × 10^6 (dynes/cm^2 )/atm}] = 2.77 atm.

Viruses such as the T_2 bacteriophage always multiply when they enter a host bacterial cell. Lysis eventually occurs and the newly synthesized progeny are released from the cell. Some viruses such as the \lambda (lambda) phage do not always multiply and lyse their host. They are referred to as the temperate phages, in contrast to the virulent phages which always kill their host. The temperate virus must be present in some in-active state within the bacterial cell. In this special relationship between the virus and the bacterium, called lysogeny, the viral DNA is incorporated into a specific section of the host's chromosome. The viral chromosome becomes an integral part of the host chromosome, and is duplicated along with it during each cell division. The phage DNA can be transferred from one cell to another during bacterial conjugation. When the viral chromosomeis integrated within the host chromosome it is called the prophage. Bacteria containing prophages are called lysogenic bacteria and viruses whose chromosomes can be-come prophages are called lysogenic viruses (in contrast to lytic viruses). The integration of the viral chromosome occurs by a crossing over between the host chromosome and a circular form of the viral chromosome. Both chromosomes break and rejoin with the viral chromosome attaching to the broken ends of the bacterial chromosome. The point of attachment and crossing over occurs at a specific region of the chromosome. The insertion can be shown as follows: The prophage can remain integrated within the bacterial chromosome for many generations. Under certain natural conditions, the viral DNA becomes active and is removed from the host's chromosome. Once independent of the host chromo-some, the lysogenic virus replicates and eventually lyses the cell. This entrance into the lytic cycle can be arti- ficially induced by ultraviolet light or certain chemicals. It is thought that while the viral DNA is inserted as a prophage, the transcription (the process where DNA is used to code for a complementary sequence of bases in an RNA chain, i.e. production of mRNA) of almost all phage genes is blocked. The only protein made by the \lambda phage DNA is the \lambda repressor, which causes the blockage of viral- specific mRNA transcription. Since no viral components can be made, the viral DNA remains inactive within the host chromosome. However, when the \lambda repressor is inactivated by an agent such as UV light, mRNA transcription occurs and the virus begins replication. Lysis eventually results with the re-lease of new \lambda phages. When a lysogenic virus is in the prophage state, its genes can produce phenotypic effects on the host bacterium. It can modify the cell wall, affect the production of en-zymes and antigens, and possibly confer toxigenicity upon the lysogenic bacteria. Certain lysogenic phages enable the bacteriumCorynebacteriumdiphtheriaeto produce the toxin that causes the disease diphtheria. The lytic and lysogenic cycles can be summarized in the following diagram:

The figure shows the forces exerted on the log by the horses at the moment they bolt. These forces can be resolved into components along the path and at right angles to the path. Thus the total forces in the x- and y-directions are \sumF_x\ding{217} = F cos 45\textdegree \^{\i} + F cos 30\textdegree \^{\i} and\sumF_y = F sin 45\textdegree \^{J} - F sin 30\textdegree \^{\i} where F is the magnitude of the force that each horse exerts. To keep the log on the path the man must counteract the unbalanced force in the y-direction, \sumF_y\ding{217}, by an equal and opposite force -\sumF_y\ding{217}. We can see that any force he may have exerted to keep the log moving along the path, exerted in other than the y-direction of the figure, would not have been the minimum force possible. Any otherwise directed force would have an x-component as well as -\sumF_y\ding{217}. But the latter alone could keep the log moving along the path. The magnitude of the result-ant force would then have a greater magnitude than -\sumF_y\ding{217}. Hence the minimum force he must exert has magnitude P_y = F(sin 45\textdegree - sin 30\textdegree) = F [(1/\surd2) - (1/2)] = 0.207 F, and must be directed in the negative y-direction, i.e., at right angles to the path. In the above analysis, frictional forces have been ignored. The frictional force acting along the path does not affect the solution. The frictional force trying to prevent motion at right angles to the line of the path reduces the magnitude of the force the man need apply. It is, however, assumed that on an icy path this frictional force is small in comparison with F, and its effect is therefore ignored.

Question:

Calculate the standard enthalpy change, ∆H\textdegree, for the combustion of ammonia, NH_3(g), to give nitric oxide, NO(g), and water H_2O(l). The enthalpies of formation, ∆H\textdegree_f, are - 68.32 Kcal/mole for H_2O(l) , - 11.02 Kcal/mole for NH_3 (g) , and 21.57 Kcal/mole for NO(g).

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The enthalpy change, ∆H, refers to the change in heat content between the products and the reactants of a chemical reaction. The symbol ∆H\textdegree is the enthalpy change for a reaction in which each reactant and product is in its standard state at a specified reference temperature (the common reference temperature is 25\textdegreeC) . The standard enthalpy of formation, ∆H\textdegree_f, of a sub-stance is defined as the change in enthalpy for the re-action in which one mole of the compound is formed from its elements at standard conditions. For any chemical reaction, the change in enthalpy, ∆H, may be expressed as ∆H ∆H reaction = \sumH_products - \sumH_reactants (where the symbol \sumH means the summation of the enthalpies of each substance). To solve this problem, one must first write and balance the equation for the combustion of NH_3(g): 4 NH_3 (g) + 5O_2 (g) \rightarrow 4NO(g) + 6 H_2O(l) From the balanced equation, one knows that 4 moles of NH_3(g) reacts with 5 moles of O_2 (g) to form 4 moles of NO(g) and 6 moles of H_2O(l). For the combustion of 4 moles of NH_3(g): 4 \bullet ∆H\textdegree_f, _NH3(g)= 4 [- 11.02 (Kcal/mole)]= - 44.08 Kcal 5 \bullet ∆H\textdegree_f, _O2(g)= 5 (0)= 0 4 \bullet ∆H\textdegree_f, _NO(g)= 4 [21.57(Kcal/mole)]= 86.28 Kcal 6 \bullet ∆H\textdegree_f, _H2O(l)= 6 [- 68.32 (Kcal/mole)]= - 409.92 Kcal Note: The ∆H\textdegree_f of all elements is zero. As such ∆H\textdegree_f,O2 (g)= 0. Thus , ∆H\textdegree = [4\bullet∆H\textdegree_f, _NO(g) + 6\bullet∆H\textdegree_f, _H2O(l)] - [4\bullet∆H\textdegree_f,NH3 (g)+ 5\bullet∆H\textdegree_f,O2 (g)] = [(86.28 Kcal) + (-409.92 Kcal)] - [(-44.08 Kcal) + 0] = - 279.56 Kcal. However, since this result is the standard enthalpy change for the combustion of 4 moles of NH_3 (g) , one must divide - 279.56 Kcal/mole by 4 to give the combustion of 1 mole of NH_3, ∆H\textdegree = (- 279.56 Kcal) / (4 moles) = - 69.89 Kcal/mole. = - 69.89 Kcal/mole.

Question:

(a)Water flows into a tank at a constant rate Q. Q is determined by the position of a valve connected to the inflow pipe. Write a FORTRAN program to simulate this system, using Euler'smethod, from time t = 0 to t = t_f if V(0) = V_0 and the capacity of the tank is C units of volume. Compare the Eulerapproximations with the exact values. (b)A hole is made in the bottom of the tank allowing water to flow out at a constant rate, Q_1. Modify the program of part (a) to simulate this system.

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(a)Changes in this system's state can be viewedthrough changes In the volume of water in the tank, V. Note that once the valve position is set, Q is determined and remains unchanged for the length of the simulation. Therefore the valve is external to the system and does not effect the system during the time we are observing it in any given simulation run. The behavior of the system can be diagrammed as shown in Fig. 3 meaning V is the accumulation of the rate of change, Q(when Q has units of volume/time). If V = dV/dt = constant, Q, then the complete mathematical model looks as shown in Fig. 4, meaning V̇ = Q or dV/dt = Q(1) Using Euler's method to simulate this system means observing changes in V for small, discrete changes in t: V_new = V_old + V_added = V_old + Q \textasteriskcentered(t_new - t_old)(2) The exact solution of equation (1) is : V(t) = Q\textasteriskcenteredt + V_0(3) Note that substituting zero for t_old and V(0) = V_0 for V_old inequation (2) yields equation (3). The following program uses TFIN for t_F and DT for \Deltat: CPROGRAM WATERFLOW DATA T/0.0/ READ,N, TFIN,Q,VO,C REALN = N DT = TFIN/REALN VOL = V0 VEXACT = V0 DO 10 I = 1,N IF(VEXACT.NE.O.) ERROR = ABS((VEXACT-VOL)/ VEXACT)\textasteriskcentered 100. PRINT, T,VOL,VEXACT,ERROR T = T + DT VOL = VOL + Q\textasteriskcenteredDT IF(VGL.GT.C) GO TO 999 VEXACT = VO + Q\textasteriskcenteredT 10CONTINUE ERROR - ABS( (VEXACT-VOL)/VEXACT)\textasteriskcentered 100. PRINT,T, VOL, VEXACT,ERROR GO TO 99 999PRINT,'OVERFLOW AT TIME', T 99STOP END (b)Since Q1is the rate of flow out of the tank, it tends todecrease V, decrease V, and so it has a negative sign. V is the accumulation of the net flow, Q + (-Q_1) (see Fig. 5) From this, the following equation can be written: v̇ = Q - Q_1(1) Since both Q and are constant, Q - Q_1 is constant and equa-tion (1) can be written: V̇ = Q_2 where Q_2 = Q - Q1 Hence, the programmer needs only to substitute the value Q_2 of for Q in hisinput data for the program of part (a).

Question:

Classify the following graphs according to the following criteria: a) directed or undirected b) connected or unconnected c) planar or nonplanar

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Solution:

Figure 1 is an example of an undirected, con-nected, planar graph. Undirected graphs contain no arrows: there is no particular direction associated with any of the branches. This graph is connected, since each node has at least one branch leading to it. Also, the graph is planar, because we can draw it on a plane so that none of the branches overlaps. Figure 2 is an undirected, unconnected, nonplanar graph. It is unconnected because node F is not attached to any branch. In addition, it is a nonplanar graph because of the overlapping of the various branches. There is no way to redraw this graph on a plane without running into over-laps. Finally, Figure 3 is a connected, directed, and planar graph. Notice the arrow along the branches, giving direc-tionality to it. We say it is a digraph, an abbreviated form of directed graph.

First, note that a typical thermostat consists of a bimetallic strip, whose elastic reaction to transient heating or cooling is used to control a triggering mechanism controlling the heating system. Thus, the internal components of the thermostat make it a dynamic element, and it can be considered a subsystem. The internal workings of the thermostat could be included in the simulation. Here we will ignore the internal operations that determine the action of the thermostat and take a "black box approach". That is, we know that information fed into the thermostat will determine whether or not it will switch, but we do not carehowthat mechanism operates: Note that the thermostat's behavior will be nonlinear. An initial picture of the system is shown in figure #1. Since the rate of heat loss, L, acts as a valve, and the thermostat acts as a switch, we can draw them as shown in figure #2. Now L = C(T-T_out) where C is a constant of proportionality which out represents change in heat per degree per time unit. Note that Tout> T \Rightarrow L < 0 and T_out < T \Rightarrow L > 0 . Let U(T) be the heating function: U(T) =\midQ , if switch is on \midO , if switch is off Then the net rate of heat flow is H-L, and the net rate of room temperature change is given by: T = KV(U-L) = KV(U-C(T-T_out)) Arbitrarily, the switch is considered off initially, U changes when T(t) deviates from T_s by h degrees. Otherwise, U remains the same. Also, the programs use TEMP for T(t), TMPOUT for T0ut, TMPSET for T_s and TEMPO for T_0 : Case (1) : REAL T/0.0/,K,V,H,C,TEMP,TMPOUT,TEMPO, ACCUR,TFIN,DT,TMPSET,Q,U/0.0/ INTEGER I,N COMMON U, TMPOUT,K,V,C READ,N,TFIN,ACCUR,TEMPO,K,V,H,C,TMPOUT, TMPSET,Q PRINT T, TEMPO REALN = N DT = TFIN/REALN TEMP = TEMPO DO 10 I = 1,N T = T + DT IF (TEMP.LE.(TMPSET-H))U = Q IF (TEMP.GE.(TMPSET+H))U = 0.0 CALL MEULER (T, TEMP, ACC UR, DT) PRINT T, TEMP 10CONTINUE STOP END FUNCTION G(W) COMMON U, TMPOUT,K,V,C REAL K,V,C, U,TMPOUT, W,G G = K\textasteriskcenteredV\textasteriskcentered(U-C\textasteriskcentered(W-TMPOUT)) RETURN END Case (2) :Using OMEGA for\omega, and PHI for \varphi : \omega REAL T/0.0/,K,V,H,C,TEMP,TMPOUT,TEMPO,ACCUR, TFIN,DT,TMPSET,Q,A,OMEGA,PHI, U/0,0/ INTEGER I,N COMMON U,TMP0UT,K,V,C READ,N,TFIN,ACCUR,TEMPO,K,V,H,C,TMPOUT,TMPSET, Q,A, OMEGA,PHI PRINT,T,TEMPO REALN = N DT = TFIN/REALN TEMP = TEMPO DO 10 I = 1,N T = T + DT IF (TEMP.LE. (TMPSET-H))U = Q IF (TEMP.GE.(TMPSET+H))U = 0.0 TMPOUT = A\textasteriskcenteredS IN ( OMEGA. \textasteriskcenteredT+PHI) CALL MEULER (T,TEMP,ACCUR,DT) PRINT T,TEMP 10CONTINUE STOP END FUNCTION G(W) COMMON U, TMPOUT,K,V,C REAL K,V,C,U,TMPOUT,W,G G = K\textasteriskcenteredV\textasteriskcentered(U-C\textasteriskcentered(W-TMPOUT)) RETURN END

Question:

An economist suspects that a leading indicator in the business cycle has the form xcosx. He wishes to know when the indicator will peak, i.e. achieve a maximum, in different time intervals. Write a FORTRAN program to find the maximum of xcosx in interval [a,b].

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Solution:

There are many methods of computing the maximum of a function over its domain or a subset of its domain (local maximum). A method particularly suited for computer applications is the elimination scheme. To illustrate the method, let a = o, b = \pi and suppose we wish to Maxxcosxx\epsilon [o,\pi].(1) 1) Place two search points close together at the centre of the interval. Let the distance between the points be \epsilon > o. 2) Evaluate f(x) at X_L and X_R and call the results f(X_L), f(X_R). If f(X_L) \geq f(X_R) Max f(X) lies between o and X_R and the segment [X_R \pi] can be discarded. 3) Place two search points close together at the center of the remaining interval and repeat step (2). 4) Suppose f(X_(R)1 ) \geq f (X_(L)1). Then Max f(X) lies in the interval [X_(L)1 ,X_R] and the segment [0, X_(L)1 ] can be discarded. 5) The process continues until an interval less than 2\epsilon is obtained. Since the search cannot continue, the maximum is assumed to occur at the center of this interval. In the program, a statement function is used to define xcosx so that it can be referred to at any further point in the program. Also, if the search points pass a toler-ance limit, the program halts. CDEFINE THE FUNCTION Y(X) Y(X) = X\textasteriskcenteredCOX(X) READ XL, XR, EPSI I = 1 CCALCULATE INTERIOR POINTS 10XL1 = XL + . 5\textasteriskcentered (XR - XL - EPSI) XR1 = XL1 + EPSI YL1 = Y(XL1) YR1 = Y(XR1) PRINT YL1, YR1, XL, XL1, XR1, XR IF (YL1 - YR1) 20, 50, 30 20XL = XL1 GO TO 40 30XR = XR1 CTEST FOR END OF SEARCH 40IF (I.GE.100) GO TO 60 I = I + 1 IF (XR - XL.GT.3\textasteriskcenteredEPSI) GO TO 10 50XMAX = .5\textasteriskcentered(XL1 + XR1) YMAX = . 5\textasteriskcentered (YL1 + YR1) PRINT YMAX, XMAX GO TO 70 CPRINT OUTPUT TERMINATED BECAUSE OF MAX- CIMUM ITERATION COUNT 60WRITE (6, 300) 300FORMAT (54HO THE SOLUTION HAS NOT CONVERGED AFTER 100 ITERATIONS. TERMINATE PROGRAM 70STOP END

Question:

Describe some of the method which insure cross-pollination.

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Solution:

Question:

The sensitivity of the silver halides to light is the basis of photography. The change from white silver chloride to a gray purple solid is the result of photochemical decomposition, The ∆H of this reaction is 30.362 Kcal/mole. From this, determine the frequency of light needed to decompose 1 molecule of AgCl.

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Solution:

To solve this problem (1) determine the energy consumed per molecule AgCl by dividing the ∆H given by Avogadro's number. (2) Convert the energy per molecule of AgCl from units of Kcal/molecule to units of J/molecule using the factors 1 Kcal = 1000 cal and 1 cal = 4.18 J. (3) Use the expression E = hѵ where E is the energy in J/molecule, h is Planck's constant, and ѵ is the frequency of light. Thus, (1)E = [{30.362 (Kcal/mole)} / {6.02 × 10^23 (molecules / mole)}] = 5.04 × 10-^23 (Kcal/molecule AgCl) (2)E = [5.04 × 10-^23 (Kcal/molecule )] [1000(cal/Kcal)] [4.18 (J/cal)] = 2.11 × 10-^19 J/molecule AgCl (3)E = hvorѵ = E/h ѵ = (2.11 × 10-^19 J/molecule) / (6.63 × 10-^34 J-s) = 3.18 × 10^14 cycles/sec per molecule AgCl.

Lichens are composite organisms consisting of algae and fungi. They grow on tree bark, rocks and other substrates not suitable for the growth of plants. Lichens may be found in low-temperature environments characteristic of polar regions and very high altitudes. Structurally, a lichen can be likened to a fungal "sandwich" whose hyphae entwine a layer of algal cells (See figure). Structures known as rhizoids, which are short twisted strands of fungal hyphae, serve to attach the bottom layer to the substrate. Not all species of algae or all species of fungi can enter into a lichenlike relationship. Most of the fungi found in lichens are Ascomycetes, although there are a few of the Basidiomycetes. Algae of the lichens are from the Cyanophyta (blue-green algae) or the Chlorophyta (green algae) phylum. Each lichen thallus consists of a single species of fungus associated with a single species of alga. There are two types of lichens: fruticose and crustaceous. The fruticose have an erect, shrublike morphology. The crustaceous are very closely attached to the substrate or may even grow within its surface. Lichens are the product of a relationship called symbiosis, in which each partner of the association derives something useful from the other for its survival. The algae provides the fungus with food, especially carbohydrates produced by photosynthesis, and possibly vitamins as well. The fungus absorbs, stores and supplies water and minerals required by the alga, as well as providing protection and a supporting framework for the alga. Lichens appear to reproduce in a variety of ways. Fragmentation may occur, in which bits of the thallus are broken off from the parent plant and produce new lichens when they fall on a suitable substrate. Lichens may produce "reproductive bodies" called soredia which are knots of hyphae containing a few algal cells. In addition, the algal and fungal components of a lichen may reproduce independently of each other. The fungal com-ponent produces ascospores and the algal component re produces by cell division or infrequently via sporulation. Some species of lichens in the Arctic have been alive for 4,500 years which suggests a very well balanced association between the symbionts. Lichen grow very slowly because of their low metabolic rate. They are also very resistant to heat and desiccation. Lichens produce many interesting organic products. Unusual fats and phenolic compounds make up from two to twenty percent of the dry weight of the lichen body. Litmus, the pigment indicator, and essential oils used in perfumes are obtained from lichens.

Question:

The ballistic pendulum (see figure) is a device for mea-suring the velocity of a bullet. The bullet is allowed to make a completely inelastic collision with a body of much greater mass. Find the bullet's velocity before the collision.

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Solution:

The momentum of the bullet-block system immediately after the collision equals the original momentum of the bullet since the block is initially at rest. Although the ballistic pendulum has now been super-seded by other devices, it is still an important labora-tory experiment for illustrating the concepts of momentum and energy. In the figure, the pendulum, consisting perhaps of a large wooden block of mass m', hangs vertically by two cords. A bullet of mass m traveling with a velocity v, strikes the pendulum and remains embedded in it. If the collision time is very small compared with the time of swing of the pendulum, the supporting cords remain practically vertical during this time. Hence no external horizontal forces act on the system during the collision, and the horizontal momentum is conserved. Then, if V represents the velocity of bullet and block immediately after the collision, mv = (m + m')V, by the principle of conservation of momentum. Hence V = mv/(m + m')(1) The kinetic energy of the system, immediately after the collision, is E_k = (1/2) (m + n')V^2. The pendulum now swings to the right and upward until all of its kinetic energy is converted to gravitational potential energy. (Small frictional effects can be neglected.) Hence (1/2) (m +m') V^2 = (m + m')gy' V^2 = 2gy By (1) (m^2v^2)/(m+m')^2 = 2gy v^2 = [2gy(m + m')^2] / [m^2] v = [(m + m')/m] \surd(2gy). By measuring m, m', and y, the original velocity v of the bullet can be computed. It is important to remember that kinetic energy is not conserved in the collision. The ratio of the kinetic energy of bullet and pendulum, after the collision, to the original kinetic energy of the bullet, is [(1/2) (m + m')V^2] / [(1/2) mv^2] = m/(m + m'). Thus if m' = 1000 gm and m = 1 gm, only about one- tenth of one percent of the original energy remains as kinetic energy; 99.9% is converted to heat.

According to quantum theory, light may be viewed as having wave-like characteristics as well as particle characteristics. In this problem, it is the particle nature of light which determines the be-havior of the space vehicle. Since the emitted light carries away momentum and because the latter is conserved, the rocket will be propelled to the left in the figure. We will suppose that N photons/sec are emitted, with each photon energy hf. Then the power of the lamp is p = [{(energy of one photon) (# of photons)}/time] p = Nhf By definition, f = (c/\lambda), Hence p = (Nhc/\lambda) Thus N = (p\lambda/hc) = [(100 watt × 4.70 × 10^-7 m)/(6.63 × 10^-34 j-sec × 3 × 10^8m/sec)] = 2.4 × 10^20 Each photon will have momentum, by the de Broglie relations p = (h/\lambda) = [(6.63 × 10^-34 j.s) / (4700 \AA)] = [(6.63 × 10^-34 j.s) / (4.7 × 10^-7 m)] = 1.4 × 10^-27 N-sec The total force on the vehicle will be, by Newton's second law, F = [d(np)/dt] Where n is the number of photons emitted and p is the momentum of each photon. Since p is constant, we then have, photon. Since p is constant, we then have, F = p(dn/dt) = pN for N is just the number of photons emitted per unit time . F = 1.4 × 10^-27 N-sec 2.4 × 10^20 sec^-1 = 3.4 × 10^-7N Thus the acceleration has the very small value a = (F/m) = [(3.4 × 10^-7 N)/50 Kg] = 6.8 × 10^-9 m/sec^2

Just prior to inspiration, at the conclusion of the previous expiration, the respiratory muscles are relaxed and no air is flowing into or out of the lungs. Inspiration is initiated by the contraction of the dome-shaped diaphragm and the intercostal muscles. When the diaphragm contracts, it moves downward into the abdomen. Simultaneously, the intercostals muscles which insert on the ribs contract, leading to an upward and outward move-ment of the ribs. As a result of these two physical changes, the volume of the chest cavity increases and hence the pressure within the chest decreases. Then, the atmospheric pressure, which is now greater than the intrathoracic pressure, forces air to enter the lungs, and causes them to inflate or expand. During exhalation, the intercostals muscles relax and the ribs move downward and inward. At the same time, the diaphragm relaxes and resumes its original dome shape. Consequently, the thoracic volume returns to its pre-inhalation state, and the pressure within the chest increases. This increase in pressure, together with the elastic recoil of the lungs, forces air out of the lungs causing them to deflate. The role of the diaphragm in breathing can be demonstrated by the figure below.

Question:

In the 1930s F.T. Bacon designed hydrogen fuel cells with extremely good performance. Using his concepts Pratt and Whitney has constructed hydrogen fuel cells that produce a potential of 0.9 V and generate a current of about 2.3 × 10^3 A for each square meter of anode area. If a Bacon battery is to deliver 100 W of power, how large must the anode be?

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Solution:

Recall that the power dissipated (or pro-duced) by a device is, P = IV Where I is the current going through a device, and V is the voltage across its terminals. In this example P = 1 × 10^2 W and V = 0.9V. Therefore, the current, required to deliver 100W of energy is, I = (P/V) = {(1 × 10^2 W)/(9 × 10^\rule{1em}{1pt}1 V) = 1.1 × 10^2 A The surface area A required is, A = {(1.1 × 10^2A)/(2.3 × 10^3A/m^2)} = 4.8 × 10^\rule{1em}{1pt}2 m^2 If the anode is designed to have a square shape, then the area A is equal to d^2 , where d is the length of one edge. We find that, d^2 = 4.8 × 10^\rule{1em}{1pt}2 m^2 d = 2.2 × 10^\rule{1em}{1pt}1 m A very compact energy supply may be constructed using a hydrogen fuel cell.

Question:

Differentiate between adaptive radiation and convergent evolution . Diagram illustrating the difference between divergent and convergent evolution. In adaptive radiation, a single stock may branch to give many diverging stocks (in the given diagram,1 stockdiverges \rightarrow 2 stocks). In convergent evolution, many stocks that are originally quite different can come to resemble each other more and more; as time passes, they converge. In the given diagram, 2 stocks are shown converging.

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Solution:

Because of the constant competition for food and living space, a group of organisms will tend to spread out and occupy as many different habitats as pos-sible. This process, by which a single ancestral species evolves to a variety of forms that occupy somewhat different habitats, is termed adaptive radiation. Adaptive radiation is clearly an advantageous process in evolution in that it enables the organisms to tap new sources of food or to escape from predators. A classical illustration of adaptive radiation is the great variety of finches found on the Galapagos Islands west of Ecuador. These finches, derived from a single common ancestor, exhibit diversity in beak size and structures, as well as in feeding habits, all of which are mutually related. Some of these birds feed on seeds, others feed mainly on cacti, and still others live in trees and eat insects. The diversity of food sources on the island has allowed each of the many forms of finches to survive in its particular habitat, and thus, prevent intra specific competition for food and space. Such adaptive radiation is often called divergent evolution, since its result is a diversity of adaptive forms evolving from a common ancestor. An opposite process is convergent evolution, which also occurs quite frequently. Convergent evolution is the process by which two or more unrelated groups become adapted to a similar environment, and in doing so, develop characteristics that are more or less similar. For example, wings have evolved hot only in birds but also in mammals (bats), reptiles (pterosaurs), and insects (grasshoppers). Likewise, the shark (a cartilaginous fish), the dolphin, and the porpoise (both mammals) have developed marked superficial similarity because of their adaptation to a similar environment.

The ground state of an atom is that in which the electrons are in the lowest possible energy level. Each level may contain two electrons of opposite spin. When there are several equivalent orbitals of the same energy, Hund's rules are used to decide how the electrons are to be distributed between the orbitals: (1)If the number of electrons is equal to or less than the number of equivalent orbitals, then the electrons are assigned to different orbitals. (2)If two electrons occupy two different orbitals, their spins will be parallel in the ground state. Hund's rules states that the electrons attain positions as far apart as possible which minimizes the repulsion obtained from interelectronic forces. To solve this problem one must: (1)Find the total number of electrons within the atom (2)Determine the number of valence electrons (3)Find the number of electrons in the highest equivalent energy orbital. The total number of electrons in an atom is equal to that atom's atomic number. Thus,Sihas14electrons Phas15electrons Shas16electrons Clhas17electrons Arhas18electrons Next, from the orbital configuration: 14 Si1s_2 2s^2 2p^6 3s^2 \mid 3p^2 15 P1s^2 2s^2 2p^6 3s^2 \mid 3p^3 16 S1s^2 2s^2 2p^6 3s^2 \mid 3p^4 17 Cl1s^2 2s^2 2p^6 3s^2 \mid 3p^5 18 Ar1s^2 2s^2 2p^6 3s^2 \mid 3p^6 One knows that the highest equivalent energy orbital is the 3p orbital. The number of electrons in this orbital increases by 1 starting with 2 for Si, then, 3 for P, 4 for S, 5 for Cl, and 6 Ar, Thus, Ar closes this orbital. Using Hund's rules, the electron configurations can be written as shown in the accompanying figure.

Question:

Explain excretion in the earthworm.

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Solution:

The earthworm's body is composed of a series of segments internally partitioned from each other by membranes. In each segment of its body, there are a pair of specialized excretory organs, called nephridia. They open independently from the body cavity to the outside. The various nephridia are not connected to each other. A nephridium consists of an open ciliated funnel or nephrostome (corresponding to the flame cell in planaria) which opens into the next anterior coelomic cavity. A coiled tubule running from the nephrostome empties into a large bladder, which in turn empties to the outside by way of the nephridiopore (see Figure). Around the coiled tube is a network of blood capillaries. Materials from the coelomic cavity move into the nephridium through the open nephrostome partly by the beating of the cilia of the nephrostome, and partly by currents created by the contraction of muscles in the body wall. Some materials are also picked up by the coiled tubule directly from the blood capillaries. Substances such as glucose and water are reabsorbed from the tubule into the blood capillaries, while the wastes are concentrated and passed out of the body through the nephridiopore. The earthworm daily excretes a very dilute, copious urine, which amounts to 60 percent of its total body weight. The principal advantage of this type of excretory system over the flame cell is the association of blood vessels with the coiled tubule, where absorption and reabsorption of materials can occur.

The placenta is an organ composed of both maternal and fetal tissues that supplies the developing embryo with nutrients and oxygen and enables the embryo to excrete carbon dioxide and other metabolic wastes. It also functions as an endrocrine gland, secreting estrogens, progesterone, human chorionic gonadotropin (HCG), and human placental lactogen (HPL). The former two are steroids, while the latter two are protein hormones. HCG functions to preven regression of the corpus luteum, while HPL stimulates secre-tion of the steroid hormones from the corpus luteum. The placenta forms following the implantation of the embryo within the uterine wall. Lytic enzymes secreted by the trophoblasts break down the capillaries of the uterine endometrium causing the maternal blood to pool in that area. Subsequently, the trophoblast layer surrounds and sends out villi into this region (see Figure). These villi become integrated by the umbilical arteries and veins which are extensions of the fetal circulatory system. The fetal blood in the capillaries of the villi come in close contact with the maternal blood in the tissue sinuses between the villi. Nutrients diffuse from the maternal blood through the trophoblast layer into the fetal capillaries, and waste products from the fetus is eliminated into the maternal blood. Note, how-ever, that at no time or place does the blood of the fetus mix with that of the mother. Blood from the mother enters the placenta via the uterine artery and leaves the region via the uterine veins. Similarly, the fetal blood never leaves the fetal vessels, which consist of umbilical arteries, the capillary network in the placentalvilli, and the umbilical vein.

This is an ancient puzzle, in which there are three pegs and 64 disks with holes through the center. The object is to move the entire tower of disks from one peg to another in the fewest possible numbers of moves. Disks must always be stacked so that a larger disk is never on top of a smaller disk. We have chosen a stack of eight disks, which requires 255 moves, because to move 64 disks, you have to make 18, 446, 744, 073, 709, 551, 615 moves - a formidable task even for a computer. In general, 2^n - 1 moves are requir-ed to transfer the tower to another peg (where n is the num-ber of disks in the tower). The recursively defined solution in PL/I follows: /\textasteriskcentered MOVE N DISKS FROM POLE X TO POLE Y. USE POLE Z TO \textasteriskcentered/ /\textasteriskcentered STORE DISKS IF NECESSARY \textasteriskcentered/ HANOI: PROCEDURE (X, Y, Z, N) RECURSIVE; DECLARE (X, Y, Z, N) FIXED; IF N = 1 THEN PUT SKIP LIST ('MOVE DISK FROM',X,'TO',Y); ELSE DO; CALL HANOI(X, Z, Y, N - 1) ; PUT SKIP LIST ('MOVE DISK FROM', X, 'TO', Y); CALL HANOI(Z, Y, X, N - 1); END; END HANOI;

Question:

Why does milk \textquotedblleftspoil\textquotedblright when kept in a refrigerator?

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Planarians belong to the group of fresh water flatworms and have a digestive system which lacks an anal opening. The mouth of the planarian opens into a cavity that contains the muscular pharynx, or proboscus, which can be protruded through the mouth directly onto the prey (see Figure) . In feeding, the planarian moves over its food object, which may be a small worm, crustacean or insect larva, and traps it with its body. The pharynx is then extended and attached to the food material, and by sucking movements produced by muscles in the pharynx, the food is torn into bits and ingested. Digestive enzymes are released in order to assist the planarian in breaking down the food prior to ingestion. The pharynx delivers the food into the three-branched gastrovascular cavity. The branching of the planaria's gastrovascular cavity into one anterior and two posterior branches provides for the distribution of the end products of digestion to all parts of the body. Flatworms with three branched gut cavities are called triclads, in contrast to those with many branches, called polyclads. Most of the digestion in planaria is intracellular, which means that it occurs in food vacuoles in cells lining the digestive cavity. The end products of digestion diffuse from these cells throughout the tissues of the body. Undigested materials are eliminated by the planaria through its mouth. The mouth, then serves as both the point of ingestion and the point of egestion. It is interesting to note that planarians can sur-vive without food for It is interesting to note that planarians can sur-vive without food for months, gradually digesting their own tissues, and growing smaller as time passes.

In approaching this problem, a careful plan of attack must be laid out. We are asked to find the force of friction in an inclined plane, given the di-mensions and efficiency of the plane. Reasoning back-wards, we begin by noticing that the effort consists of the force gravity exerts down the plane, plus the friction. The force exerted down the plane by gravity is calculable from the dimensions of the inclined plane and the weight of the box by constructing a proportion between the force triangle and the inclined plane (see diagram). 4'/10' = x/100 lb x = 40 lb Efficiency is calculated from AMA over IMA. IMA is calculated from the dimensions of the inclined plane (length over height, in this case 10 ft/4 ft = 2.5). The AMA is resistance over effort. The resistance is simply the weight of the box. We are only left with one unknown - the frictional force, which we now solve for efficiency= AMA/IMA = (resistance/effort) / (2.5) = [resistance/(x + F_friction)] / 2.5 = [100 lb/(40 lb + F_friction)] / 2.5 = 80 % = .80 Hence,F_friction= 10 lb.

Question:

You have two half-cells coupled together to form a concentration cell: H_2(1 atm) \rightarrow 2H^+(.1M) + 2e^- and H_2 (1 atm) \rightarrow 2H^+ (1.0M) + 2e^-. Calculate the free energy and electrical work obtainable. F = 96,500 coulombs.

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Solution:

Question:

What is the maximum number of emission lines one would expect to see in a molecule containing only the six electronic energy levels depicted in diagram A?

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Solution:

Emission is observed when an electron falls from anelectronic energy level of higher energy to one of lower energy. From the energy level n = 6, electrons can fall to energy levels of n = 5, 4, 3, 2, and 1. Thus, five emission lines are drawn from the energy level n = 6. By using a similar process, one determines that the maximum number of emission lines from energy levels n = 5, 4, 3, 2, and 1 are four, three, two, one, and zero, respectively. Because there are no energy levels lower than n = 1, no emission lines are obtained from it. The total number of emission lines is found by taking the sum of all the emission lines from the various energy levels. Thus, 5 + 4 + 3 + 2 + 1 + 0 = 15 mission lines can be expected. The possible emission processes are depicted in diagram B.

Question:

A triangular plate is immersed in water (of density \rho) with one vertex at the surface and the others at depths of 6 in. and 12 in. What is the thrust on the plate due to the pressure of the water only? Its areas is 63 in^2 (See figure (a))

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Solution:

At depth z from the water's surface, the press-ure is g\rhoz, the symbols having their usual significance. The total thrust on the small element, parallel to the surface, shown in figure (a) is dF = g\rhozy dz. The total thrust on the plate is found by summing this differential element of force over the entire tri-angle. F = ^z0\int_0 g\rhozy dz = g\rho[( ^z0\int_0 \rho'zy dz)/(\int^z0_0 \rho'y dz)] × \int^z0_0 y dz ' ' wherep' is the density of the plate material. p' is the density of the plate material. Here,^z0\int_0 \rho'y z dz/^z0\int_0 \rho'y dzis the location of the plate'scenter of mass relative to the water's surface. ^z0\int_0 y dz is the area of the plate. The total thrust on the plate is thus seen to be the pressure at the center of mass of the plate multi-plied by the area of the plate. This is a general result for all plates, as can be seen from the general nature of the derivation. In the particular case of the triangular plate, the area of 63 in^2 = 63/144 ft^2 is given. The center of mass of a triangular plate is two- thirds of the way from a vertex to the middle of the opposite side. From figure (b), we see that triangles BEC and DFC are similar. Hence CF/CE = CD/CB CF = [CD/CB] CE Since D is the mid-point of BC CF = (1/2)CE = (1/2) (6") = 3" Hence, D is 9" below the water's surface. But the center of mass is two- thirds of AD from A and must be 6 in. = (1/2) ft from the surface. In this case, therefore, F = g\rho {location of C.M.} × {area of plate) = 32 ft/s^2 × 1.94 sl/ft^3 × (1/2) ft × 63/144 ft^2 = 15.58 lb where we used the fact that \rho = 1.94 sl/ft^3 for water.

You want to represent the equilibrium constant expression for the reaction, which necessitates a balanced equation. After writing the expression, you want to express the concentrations in terms of the same variables and solve for it. Begin by writing the balanced equation for the reaction of acetic acid in water. The acid will donate a proton (H^+) to the only available base, H_2O. Thus, HOAC + H_2O \rightarrow H_3O^+ + OAC^-. [H_3O^+], the hydroniumconcentration, is the quantity you are looking for. The equilibrium constant expression measures the ratio of the concentrations of the products to the reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, the constant, Ka, = {[OAC-] [H_3O^+]}/[ HOAC]. Notes H_2O is omitted, since it is considered a constant. Ka = 1.75 × 10^-5 . Equating, {[OAC-] [H_3O^+]}/[HOAC] = 1.75 × 10^-5. Let x = concentration of H_3O^+. According to the reaction, [H_3O^+] = [OAC^-], thus, x = concentration of [OAC^-], also. If the initial concentration of HOAC is .1 and X (moles/liter) of [H_3O^+] are formed, then you have (.1-x) moles/liter of HOAC left. Substituting these variables into the equilibrium constant expression, you have [x^2/(.1 - x)] = 1.75 × 10^-5 . Solving, x = [H_3O^+] = 0.0013M

Question:

A man is exposed to a pathogenic bacterium. If inflammation occurs, has the body's first line of defense been penet-rated?

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Solution:

This answer will describe the body's first two lines of defense against invading pathogenic micro-organisms. The first line of defense is basically me-chanical, involving physical barriers such as the skin and mucous membranes. The tough layer of skin forms a pro-tective barrier as long as it is intact. The sticky mucous secretions of the respiratory and digestive tract, along with other tissues, collect and hold many pathogens until they can be disposed of. Cilia, small hairlike appendages lining the epithelial cells of the body cavities, help sweep bacteria way from susceptible areas. The mechanical processes of coughing, sneezing, tear shedding, and sali-vating provide a clearing mechanism that flushes away bacteria. Both enzymes, such as lysozyme in tears, and the acidity or alkalinity of body secretions prevent bacteria from entering further into the body. The second line of defense is cellular. If pathogens succeed in penetrating the first line of defense and enter deeper tissue, they may be engulfed and destroyed by cells in the body named phagocytes. This process is called phagocytosis. Phagocytes include macrophages and leukocytes (white blood cells). Inflammation results from damage to body tissue caused by the invading pathogen. There is an accumulation of serum and leukocytes in the area, and this produces the signs of inflammation: swelling, heat, pain and redness. Although the phagocytes destroy the bacteria by ameboid engulfment, some phagocytes are themselves also destroyed after the engulfment occurs. Dead phagocytes, bacteria and serum that result from an inflammation are together termed pus, which may be colored according to the pigment of the bacteria involved in the infection. We thus see that inflammation occurs when the body's first line of defense is broken and the second line of defense, phagocytosis, takes over.

Question:

Most bread is made from flour, water, fat, sugar and salt. Yeast is added as a leavening or "raising agent". Besides providing flavor, explain the function of salt in bread.

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Solution:

Question:

A hungry chimpanzee is able to solve the problem of getting a banana that is out of reach by stacking boxes on top of one another. Explain.

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Solution:

This type of behavior is called insight learning. Considered to be the highest form of learning, it is most prevalent in primates and man. Insight is the ability to respond correctly on the first and sub-sequent trials to problems not encountered before. The animal is able to apply its prior learning experience in other situations to a new problem and solve the problem without having to use a trial-and-error process. In the situation given, the chimpanzee used insight to solve a problem it had never encountered before. Insight is the ability to incorporate concepts and prin-ciples with past experience in order to solve problems. During his survey of the problem, the chimp might mentally or visually internalize a trial-and-error process. By mentally integrating several elements ori-ginally learned separately, the chimpanzee can figure out the solution without doing the actual trial-and-error process and learning by mistake. In humans, insight learning is often referred to as reasoning. Dogs and cats demonstrate lack of insight or reasoning by being unable to perform well on detour tests. Detour tests are the simplest techniques used to investigate problem- solution capacities in animals. In these tests, an animal is blocked from a direct approach to food which he can see and smell by a barrier. In order to get to the food he must first walk away from the food and detour around the barrier (Fig. 1). Monkeys and chimpanzees are usually successful initially without having to rely on a trial and error process.

Question:

What are plant hormones?Discuss the important effects of theauxins.

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Solution:

Plant hormones, like animal hormones, are organic substances which can produce striking effects on cell metabolism and growth even though present in extremely small amounts. They are produced primarily in actively growing tissues, especially in the apicalmeristemsof the plant. Like animal hormones, plant hormones usually exert their effects on parts of the plant body somewhat removed from the site of their production. Movement of plant hormones to target regions of the plant is made possible by the presence of phloem. Auxins may be regarded as the most important of the plant hormones, since they have the most marked effects in correlating growth and differentiation to result in the normal pattern of development. The differential distribution ofauxinin the stem of the plant as it moves down from the apex (where it is produced) causes the plant to elongate and bend toward light.Auxinfrom seeds induces the maturation of the fruit. Auxin from the tip of the stem passes down into the vascular cambium below and directs the tissue toward differentiating into secondary phloem and xylem.Auxinalso stimulates the differentiation of roots-by placing a cut stem in a dilute solution ofauxin, roots can be readily produced. The auxins , in addition, deter-mine the growth correlations of the several parts of the plant. They inhibit development of the lateral buds and promote growth of the terminal bud. Finally,auxinscontrol the shedding of leaves, flowers, fruits, and branches from the parent plant. By inhibiting the forma-tion of the abscission layer between leaf petioles and the stem or branch, theauxinsprevent the leaves from being shed.

Question:

An inductor of inductance 3 henrys and resistance 6 ohms is connected to the terminals of a battery of emf 12 volts and of negligible internal resistance, (a)Find the initial rate of increase of current in the circuit, (b) Find the rate of increase of current at the instant when the current is 1 ampere. (c) What is the instantaneous current 0.2 sec after the circuit is closed?

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Solution:

(a) Using Kirchoffs voltage law around the cir-cuit yields V = Ri(t) + L[{di (t)} / dt] or[{di (t)} / dt] = (V/L) - (R/L) i (t).(1) or[{di (t)} / dt] = (V/L) - (R/L) i (t).(1) Here, L (di/dt) is the EMF induced in the inductance due to the changing current. Equation (1) relates the rate of change in current at a time t, to the current at time t. At t = 0, the initial current is zero. Hence the initial rate of in-crease of current is (di/dt) = (V/L) = (12 V/3h) = 4 amp/sec. (b)When i = 1 amp, (di/dt) = (12 V/3h) - (6\Omega/3h) × 1 = 2 amp/sec. , (c) The current at any time can be found by solving equation (1) for i. The function i = (V/R) [1 - \epsilon^-(R/L)t ] can be shown by substitution into the differential equa-tion to satisfy both equation (1) and the initial condition i = 0 at t = 0. At t = 0.2 sec, we then have i = (V/R) [1 - \epsilon^-(Rt/L) ] = (12 V/6) [1 - \epsilon^-6×.2/3 ] = 2 [1 - \epsilon^-.4 ] amps = 2(1 - .672) amps = 0.65 amps.

The best way to attack this problem is by writing a series of flowcharts, each being more detailed than the previous one. The concept of refinement is used here, because there are many subtasks that must be accomplished throughout. The first flowchart to be written is only a skeleton of the entire algorithm and is shown in figure 1. We Include annotation nodes where the subsequent flowcharts should fit in. Let us call the six input items A, B, C, D, E, and F. Assume that the two dates are June 1, 1978 and May 15, 1979. Then, A, B, and C are 06, 01 and 78 respectively and D, E, and F are 05,15 and 79 respectively. To find intervening days in a single month, control passes to the box containing T \leftarrow T + (E - B). To find longer periods of time, you must write more complicated instructions. First, let us examine node 7. This section evaluates the number of days in the month given. February is set aside until we have to adjust for leap years. The flowchart for node 7 is given in figure 2. This section checks the data for all types of errors. The variable KMT is used in the error messages to indicate that the KNTth card is wrong. FLAG remains at zero until both pairs of dates, (A and B, D and E) are checked on each card. X and Y store the dates temporarily while error testing is done. The following relations are checked; if there is an error, a message is output. 1) No month code can be greater than 12. 2) If the month code is 4, 6, 9, or 11 (April, June, September, or November), a maximum of 30 days is allowed. 3) If the month is neither February nor one of the above, it must have a maximum of 31 days. The flag must be raised to 1 before control passes to the next card. When the next pair of dates is examined through the flowchart, KNT is increased by 1, and the main program continues. Node 3 rearranges the dates if the first group ABC follows DEF on the calendar. Year, month, and day are evaluated in that order, and appropriate action is taken. The flowchart for node 3 is given in figure 3. Node 5 calculates the number of days if the dates are not in the same year. Leap years are taken into account by adding 366 days to the elapsed time. Further refinements for the elapsed months and days will come in node 6. The flowchart for node 5 is shown in figure 4. To explain the next node, we must first introduce the following data table. It gives the month codes, the elapsed days in a year starting with the first of each month(s), the days remaining in a year starting with the last day of each month (W), and the number of days in each month (Z). For leap years, the entries in the table are adjusted in the program. The table is shown in figure 5 and the flowchart in figure 6. Month S W Z 1 0 334 31 2 31 306 28 3 59 275 31 4 90 245 30 5 120 214 31 6 151 184 30 7 181 153 31 8 212 122 31 9 243 92 30 10 273 61 31 11 304 31 30 12 334 0 31 fig. 5 Finally, if the dates are in the same year but not in the same month, we turn to node 4. The data table is used in this section to adjust the dates accordingly. The flowchart of node 4 is given in figure 7.

Selaginella is a member of the subphylum Lycophyta. Its dominant sporophyte generation generally consists of a branched, prostrate stem with short upright branches, normally only a few inches high. Both horizon-tal and upright stems are sheathed with small leaves in four longitudinal rows or ranks. At the ends of the upright branches, reproductive leaves, called sporophylls group to form cones or strobili. Sporangia grow in or near the axils of the sporophylls. Two types of sporangia are present in Selaginella, the microsporangia, which produce small microspores; and the megasporangia, which produce large megaspores. A single strobilus usually contains both types of sporangia and thus produces both types of spores. In a developing megasporangium, all but one of the spore mother cells degenerate. This remaining cell undergoes meiosis and forms four megaspores. Each megaspore is capable of giving rise to a female game-tophyte. In the microsporangium, only a few spore mother cells degenerate. Each of the 250 or so cells that remain gives rise to four microspores by meiosis. All microspores are capable of developing into the male microgametophyte. Repeated mitotic cell divisions of the megaspore result in the formation of the female gametophyte within the megaspore cell wall. The megaspore may be shed from the strobilus at any stage in the development of the megagametophyte, or may remain in the strobilus well after fertilization until the completion of early embry-onic development. The material stored in the megaspore is the major source of food for the female gametophyte and the developing embryo. The gametophyte increases in size, eventually rupturing the megaspore wall, and the small, colorless megagametophyte tissue protrudes, exposing the archegonia. The haploid microspore also undergoes mitosis and forms two cells. The smaller cell, the prothallial cell, remains vegetative and does not divide further. The other cell, by repeated divisions, develops into a gametophyte, or antheridium, composed of a jacket of sterile cells enclosing either 128 or 256 biflagellated sperm, the gametophyte remains in the microspore cell, which is released from the microsporangia and may drop near a megaspore either on the ground or on a strobilus. When wet by dew or rain, the microspore wall splits, and the sperm within are free to swim to the megagametophyte and fertilize the haploid egg, thus initiating the diploid sporophyte generation. Of the two cells formed by the first division of the zygote, only one develops into an embryo. The other grows into an elongated structure, the suspensor, which pushes the developing embryo into gametophytic tissue for food supply. The young Selaginella soon acquires its own food-making mechanism and becomes an independent sporophyte plant. In some species, as in the seed plants, the embryo is produced in the female gametophyte while it is still within the sporophyte.

A transistor operates as a switch. When the base voltage is high, > .7V base to emitter, the collector-emitter junction forms a closed switch and when the base voltage is low, < .7V base to emitter, the collector-emitter junction forms an open switch. Hence, when Vin 1, and Vin 2are low, both switches are open, preventing current from flowing through T_1 or T_2, and the output voltage is high (\approx 5V). When V_in 1 (or Vin 2) is high and other input is low, then T1 (T2) is closed. In this case the output voltage is low, because current flows through one of the switches, connecting V_out- to ground. When Vin 1and Vin 2are high, switches T1 and T2 close, connecting V_out to ground. This circuit behavior can be tabulated as follows: Vin 1 Vin 2 V_out L L H L H L H L L H H L Using positive logic, this table is seen to be the truth table for a NOR gate.

Question:

A metal disc is suspended from its center by a torsion bar, as shown in the figure. Find the period of oscilla-tion for small angular displacements from equilibrium. Torsion bars have the property of exerting a torque which is proportional to the angular displacement and oppositely directed.

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Solution:

The angular displacement \Phi is opposed by a torque \tau = -\cyrchar\cyrk \Phi where \cyrchar\cyrk is the torsion constant, and corresponds to the force constant of a tensile spring. The angular acceler-ation \alpha and the moment inertia of the disc are related to \tau by \tau = I \alpha = I[(d^2\Phi)/(dt^2)] henceI [(d^2\Phi)/(dt^2)] =\Elzbar\cyrchar\cyrk \Phi. The resulting differential equation for \Phi is [(d^2\Phi)/(dt^2)] + \Omega^2 \Phi = 0 where \Omega = \surd(\cyrchar\cyrk/I). This equation defines a simple harmonic motion, given by\Phi(t) = \Phi_0 sin \Omegat, \Phi_0 is themaximum angular displacement and we assumed that \Phi = 0 at t = 0. The period T is T = 2\pi/\Omega = 2\pi\surd(I/\cyrchar\cyrk).

The energy acquired in the fall is just the potential energy: PE =mgh = (1 g) × (980 cm/sec^2) × (1 cm) = 980 ergs The energy of each yellow photon is l =hv= (hc/\lambda) = [{(6.6 × 10^-27 erg) × (3 × 10^10 cm/sec)}/(6 × 10^-5 cm)] = 3 × 10^-12 ergs/photon whichrepresents one photon having a total energy of 3 × 10^-12 ergs. Since nphotons have a total energy of 980 ergs, we may set up the following proportion: (nphotons/1 photon) = [(980 erg)/(3 × 10^-12 ergs)] From which we haven = 3 × 10^14 photons.

Question:

The mean length of a Rowland ring Is 50 cm and Its cross section is 4 cm^2. Use the permeability curve given below to compute the mag-netomotive force needed to establish a flux of 4 × 10^-4 weber in the ring, (a) What current is required if the ring is wound with 200 turns of wire? (b) If an air gap one millimeter in length is cut in the ring, what current is required to maintain the same flux?

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Solution:

(a) As can be seen from Fig. 2, a Rowland ring is a torus of a given ferromagnetic material with two colls around. The first long coil is used to set up the circular magnetic field B Inside the ring. As the current I in the first coil changes, an induced e.m.f. will be set up in the second coil. Thus, by measuring the voltage at the terminals of the second coil, we can measure B. If the ferromagnetic material is not present, the magnetic field B_0 in the toroidal coil is given by (similar to a solenoid) B_0 = \mu_0 n I where n is the number of turns of the first coil per unit length. We can find B_0 by measuring I. As we insert the ferromagnetic core in-side the coil, B_0 will induce a magnetic field (called the magnetiza-tion M) in the direction of B_0, in the core. The resultant magnetic field B will be the sum of these fields, (for magnetic materials, M is often much larger than B^0). Defining the magnetic intensity H as B_0 = \mu_0 H , the expression for B is given by B = \mu_0 (H + M) = \muH where M is the magnetization and \mu the permeability of the core. From the magnetization curve, the H required to set up a magnetic field is 2 × 10^2 amp/m. At this point the permeability is \mu = (B/H) 5 × 10^-3 w/amp.m B = (\varphi/A) = [(4 × 10^-4 w)/(4 × 10^-4 m^2 )] = 1 w/m^2 . In analogy with electric circuits, one can represent a magnetic material subject to a magnetic flux as a magnetic circuit. The circuit for this problem is given in Fig. 3, where the loop variable is magnetic flux and the magnetomotive force that gives rise to \varphi_B in a coil is taken to be nIl. With these definitions, the reluctance of our ring with \varphi_B through it, is given by R = (\scorpio/\varphi_B ) = (nIl/BA) The magnetic field B inside the ring is B = \muH = \mu(B_0 /\mu_0 ) = \munl , Thus R = (1/\muA) = [(0.5 m)/{5 × 10^-3 (w/amp.m) × 4 × 10^-4 m^2}] = 2.5 × 10^5 amp/w, and since \scorpio = \varphiR, the required magnetomotive force is M = 4 × 10^-4 w × 2.5 × 10^5 (amp/w) = 100 amp-turns. Using the expression for the magnetomotive force \scorpio = NI where N is the total number of turns in the coil, the required cur-rent is obtained as I = (\scorpio/N) = [(100 amp turns)/(200turns)] = 0.5; amp. (b) The air gap will introduce a new reluctance R_a into the magnetic circuit, in series with the reluctance of the ring (see Fig. 4). Since the same flux \varphi_B goes through both of the reluctances, the loop equation for this magnetic circuit is \scorpio = (R + R_a ) \varphi_B where R_a = (l_a /\mu_0 A) = [(10^-3 m)/(12.57 × 10^-7 w/amp.m × 4 × 10^-4 m^2 )] = 20 × 10^5 amp/w . Hence, assuming that the change in length of the ring is negligible and \varphi_B in the ring is kept constant, the new magnetomotive force for this arrangement is \scorpio = (2.5 × 10^5 + 20 × 10^5 ) × 4 × 10^-4 = 900 amp-turns. Therefore the current in the first coil should be increased to I = (\scorpio/N) = [(900 amp-turns)/(200turns)] = 4.5 amp .

On examining the flow chart, we see that we shall need the following registers: A register for storing the variable N:say,Register2. A register for the NUMBER to be checked:say,Register3. A register for storing the constant '1':say,Register4. A register for storing the constant 'O':say,Register5. Registers for multiplication and division:say,Registers , 6 and 7. A utility register:say,Register8. With these registers alloted, we can now write the program as follows. MAINSTART0 PRINTNOGEN INITIAL L2, = F '2'SET REGISTER 2 TO '2',FOR AN INITIAL VALUE OF N = 2 L4, = F '1'SET REGISTER 4 TO '1' L5, = F'0'SET REGISTER 5 TO 'O' RWD3READ THE VALUE OF THE NUMBER \textasteriskcenteredTO BE CHECKED INTO REGISTER 3 LOOPL7, 4LOAD REG. 7 WITH CONTENTS OF REG. 4 MR6, 3MULTIPLY CONTENTS OF REGS. 6 & 7 BY REG. 3 (i.e., MULTIPLY 1 BY NO. TO BE CHECKED) & STORE THE RESULT IN REG. 6 AND REG.7, AS A COUBLE-WORD. DR6, 2DIVIDE CONTENTS OF REG. 6 \textasteriskcentered& 7, i.e., THE NUMBER, BY \textasteriskcenteredCONTENTS OF REG. 2, i.e., N. \textasteriskcentered(THE QUO-TIENT GETS \textasteriskcenteredSTORED IN REG. 7 & THE REMAINDER IN REG. 6) CR5, 6COMPARE REMAINDER WITH `0' BMCHKIF NEGATIVE, BRANCH TO LABEL CHK WWD4IF NOT NEGATIVE, \textasteriskcenteredREMAINDER IS ZERO, HENCE, \textasteriskcenteredNUMBER IS NOT A PRIME, HENCE, WRITE '1' AS AN INDICATION BOUTBRANCH TO END THE PROGRAM CHKAR2, 4ADD CONTENTS OF REGISTERS 2 & 4, i.e., INCREMENT N TO N + 1. LR8, 2LOAD INCREMENTED 'N' IN REG. 8 SR8 , 3SUBTRACT 'NUMBER' FROM 'N' BMLOOPIF NEGATIVE, IT MEANS \textasteriskcentered'NUMBER' IS GREATER THAN \textasteriskcentered'N'. HENCE, REPEAT THE LOOP. WWD3IF NOT NEGATIVE, IT MEANS 'NUM-BER' = 'N' HENCE, \textasteriskcentered'NUMBER' IS A PRIME \textasteriskcenteredNUMBER, AND HENCE, PRINT \textasteriskcenteredTHE 'NUMBER' ITSELF AS AN IN-DICATION. OUTEXIT ENDMAIN

Question:

What is the difference in the levels of a liquid in two connecting capillaries of diameters D_1 and D_2? The surface tension of the liquid is \sigma. The edge angles of the surface films are zero.

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Solution:

The surface forces, S_1 add S_2, acting on the water surfaces in capillaries 1 and 2 (as shown in the figure), respectively are S_1 = \piD_1\sigma,S_2 = \piD_2\sigma. If W_1^\ding{217} and W_2^\ding{217} are the weights of the water columns in tubes 1 and 2, the net force (neglecting the air pressure), at the bottom of each capillary is F_1^\ding{217} = W_1^\ding{217} + S_1^\ding{217}, F_2^\ding{217}= W_2^\ding{217} + S_2^\ding{217} Corresponding pressures are P_1 = F_1/A_1 = W_1/A_1 - S_1/A_1 = gh_1\sigma - (S_1/A_1) P_2 = F_2/A_2 = W_2/A_2 - A_2/A_2 = gh_2\sigma - (S_2/A_2) where g is the gravitational acceleration, \sigma is the density of the liquid, and A_1 and A_2 are the cross-sectional areas of the tubes. The hydrostatic pressures at the bottoms of the tubes should be equal, hence we have P_1 = P_2 giving gh_1\sigma - (S_1/A_1) = gh_2\sigma - (S_2/A_2), or\sigmag (h_2 - h_1) = (S_2/A_2) -(S_1/A_1), \sigmag (h_2 - h_1) = [(\piD_2\sigma) / {(1/4)\pi D^2_2 }] - [(\piD_1\sigma) / {(1/4)\pi D^2_1}] = 4\sigma[(1/D_2 ) - (1/D_1)]. The difference in the levels of the water columns is therefore, (h_2 - h_1) = [ {4\sigma(D_1 - D_2)} / (\sigmag D_1D_2) ].

Breathing requires the coordinated contrac-tion and relaxation of several muscles. This is achieved by the respiratory center, which is composed of special groups of cells in the medulla and pons of the brain. From this respiratory center, nerve impulses are rhythmically discharged to the intercostal (rib) muscles, resulting in their periodic contraction every 4 to 5 seconds. The breathing movements are automatic and occur without voluntary control under normal conditions. We can voluntarily hold our breath, but not indefinitely, since then the automatic center eventually takes over and forces us to exhale. By experimentation, neurons have been found in the medulla which propagate action potentials in perfect synchrony with inspiration. A smaller number of other neurons have been discovered which discharge synchronous-ly with expiration. These two types of neurons are called inspiratory and expiratory neurons, respectively. These may also be referred to as inspiratory and expiratory centers. Electrical stimulation of the inspiratory neurons can produce a maximal inspiration. Conversely, electrical stimulation of the expiratory neurons shuts off inspiration abruptly, and produces contraction of the expiratory muscles. The question arises as to what induces firing of the medullary inspiratory neurons. It appears that these neurons have inherent autorhythmicity - the capacity for periodic self-excitation. However, synaptic input from other neurons plays an essential role in regulating the rhythmicity of these inspiratory neurons. There are three vital inputs to the medullary inspiratory neurons, which play a role in modulating respiratory rhythm. They are (1) direct intracranial connections between the medulla and the pons (2) reciprocal connections with the medul-lary expiratory neurons, and (3) afferent input from stretch receptors in the lung (see Figure). The connections between the inspiratory and expiratory centers are inhibitory in nature. Thus at the beginning of inspiration, when the inspiratory neurons are firing, the expiratory center is prevented from firing so that expiration is inhibited. When inspiration ceases, ex-piratory inhibition also stops, and expiration is able to occur which then, in turn, inhibits inspiration. These reciprocal connections serve to synchronize inspiration and expiration. As was previously mentioned, the medullary inspiratory center is connected to the pons. This area of the pons is often called the pneumotaxic center, and destruc-tion of this center produces profound changes in respira-tion. The medullary neurons receive neural input from the pons, which exerts a tonic effect upon the inspiratory neurons. It is also likely that the pneumotaxic center serves as a central relay station for the respiratory in-hibition initiated by the lung stretch receptors. As the lungs expand during inspiration, these receptors are stimulated, and impulses travel up the afferent nerves to the brainstem, where they aid in terminating inspiration. To summarize, the medullary inspiratory neurons primarily control the cycle of ventilation. These inspiratory neurons innervate the inspiratory muscles. The spontaneous increase in the firing rate of these inspiratory neurons is a crucial factor for initiating ventilation. Due to the inhibitory impulses from medullary expiratory neurons and pulmonary stretch receptors which act through higher brain centers, these neurons will stop firing. The cessation of activity by the inspiratory neurons releases the inhibition on the expiratory neurons so that expiration can passively occur. Only in forced expiration are the expiratory muscles themselves used. This active expiratory movement is synchronized with the passive component of expiration. This is possible because of the reciprocal connections between the medullary inspiratory and expiratory centers.

Binary trees are very similar to a linked list. Consider the following structure struct binary _ tree { int number; struct binary _ tree \textasteriskcenteredright _ node; struct binary _ tree \textasteriskcenteredleft _ node; }; }; A single element in a binary tree is visualized as follows Binary trees are commonly used to sort information using the following methods them in ascending order is as follows: #define TRUE 1 #define FALSE 0 #define NULL '\textbackslash0' structbinary _ tree { intnumber; structbinary _ tree \textasteriskcenteredright _ node; structbinary _ tree \textasteriskcenteredleft _ node; } ; main ( ) { intnumber _ entered; /\textasteriskcenterednumber entered or \textasteriskcentered/ /\textasteriskcenteredread in\textasteriskcentered/ intnumber _ of _ values; /\textasteriskcenterednumber of values to be read in\textasteriskcentered/ inti; intlocation _ not _ found; /\textasteriskcenteredto track the location of the value \textasteriskcentered/ char \textasteriskcenteredcalloc ( ); struct binary _ tree \textasteriskcenteredroot, \textasteriskcenteredcurrent _ node, \textasteriskcenterednew _ node; root = (struct binary _ tree \textasteriskcentered) calloc (1, sizeof (structbinary _ tree)); number _ of _ values = 50; /\textasteriskcenterednow get the 1st value \textasteriskcentered/ current _ node = root; current _ node \rightarrow number = get _ integer ( ); current _ node \rightarrow right _ node = NULL; current _ node \rightarrow left _ node = NULL; /\textasteriskcenteredsince the 1st value has been read in \textasteriskcentered/ /\textasteriskcenteredwe are left with 49 more values\textasteriskcentered/ for (i = 1; i < 50; i++) { /\textasteriskcenteredtake the next value\textasteriskcentered/ number _ entered = get integer ( ); /\textasteriskcenteredstart at the top i.e. the root \textasteriskcentered/ current _ node = root; location _ not _ found = TRUE; new _ node = (structbinary _ tree \textasteriskcentered) calloc (1,sizeof (structbinary _ tree)); new _ node \rightarrow number = number _ entered; new _ node \rightarrow right _ node = NULL; new _ node \rightarrow left _ node = NULL; /\textasteriskcenterednow we have to find the location \textasteriskcentered/ /\textasteriskcenteredin the tree for new value\textasteriskcentered/ while (loation _ not _ found) { if (number _ entered < current _ node \rightarrow number) if (current _ node \rightarrowleft _ node == NULL) location _ not _ found = FALSE; else current _ node = current _ node \rightarrow left _ node; else if (current _ node \rightarrow right _ node == NULL) location _ not _ found = FALSE; else current _ node = current _ node \rightarrow right _ node; } /\textasteriskcenteredend while \textasteriskcentered/ /\textasteriskcenterednow to place the new node in the tree\textasteriskcentered/ /\textasteriskcenteredat the correct location\textasteriskcentered/ if (number _ entered < current _ node \rightarrow number) current _ node \rightarrow left _ node = new _ node; else current _ node \rightarrow right _ node = new _ node } print _ tree (root); /\textasteriskcenteredprint _ tree prints out the sorted values\textasteriskcentered/ } print _ tree (node) structbinary _ tree \textasteriskcenterednode; { if (node \rightarrow Left _ node ! = NULL) print _ tree (node \rightarrow left _ node); printf ("%d", node \rightarrow number); if (node \rightarrow right _ node ! = NULL) print _ tree (node \rightarrow right _ node); } Explanation of Routines Used: 1) get _ integer ( ) This is an assumed routine (i.e. not present in the li-brary) which gets next integer from the input. As an excercise write this routine. 2) print _ tree (node) The function print _ tree ( ) invokes itself recursively to print the values contained in the tree. This routine searches each left node until node \rightarrowleft equals NULL. It then prints the value contained at the node and returns up one level in the tree and prints the value at that node. The routine further checks whether the node has a right node and if so starts the same process at that node.

Question:

Design an 8 × 3 encoder.

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Solution:

An encoder is a device with 2^n input lines (in this case eight) and n output lines (in this case 3). Only one of the input lines can be excited at any one time. The encoder generates on its output lines a code corresponding to the input line that was excited. The truth table for the 8 × 3 encoder is shown in figure 1. Bits I_0 through I_7 are the input lines and bits A_0 through A_2 are the output lines. INPUTS I_0I_1I_2I_3I_4I_5I_6I_7 OUTPUTES A_0A_1A_2 10000000 000 01000000 001 00100000 010 00010000 011 00001000 100 00000100 101 00000010 110 00000001 111 Fig. 1 \rule{1em}{1pt} truth table output lines. A_0 is most significant output bit; A_2 is least significant output bit. From the truth table it is seen that A_0 = I_4 + I_5 + I_6 + I_7 A_1 = I_2 + I_3 + I_6 + I_7 A_2 = I_1 + I_3 + I_5 + I_7 The logic diagram of the encoder is obtained from these equations. Figure 2 shows the encoder.

The figure shows the situation. The steel is held in the water by a string with tension T. We wish to find the force exerted by the system on the scale. By Newton's Third Law, this is equal in magni-tude to the force exerted by the scale on the system, shown in figure (A). Since this system is in equi-librium, the net force acting on it is zero and T + N - (M + m)g = 0 N = (M + m)g - T(1) However, in order to solve for N, we must know T. We can obtain T by applying Newton's Second Law to the system shown in figure (C). Since this system is in equilibrium, we may write T + F_B - mg = 0 where F_B is the buoyant force on the block. By Archimedes' Principle, F_B is equal to the weight of water displaced by the steel. Hence, F_B = \rho_wgv F_B = \rho_wgv where V is the volume of the cube and \rho_w is the density of water. Then, T = mg - F_B = mg - \rho_wgV(2) Using (2) in (1), N = (M + m)g + \rho_wgV - mg orN = Mg + \rho_wgV This is the "weight" registered by the scale. If the tank were weighed without the steel, we'd find N' = Mg The difference in these 2 readings is N - N' = Mg + \rho_wgV - Mg = \rho_wgV or the buoyant force.

The Lorentz transformation, of the special theory of relativity, may be used to relate the rest lifetime of the neutral particle, t_0 to the relative lifetime, t, as measured in the laboratory frame of reference. The laboratorylifetime t is given by. t = (distance traveled between creation and annihilation , d) / (speed of the particle, v) d / v = (1 × 10^-1 m) / (2.24 × 10^8 m/s) = 4.5 × 10^-10 s Using the relativistic expression for time dilation, the neutral particle's lifetime in its rest frame is t_0 = t \surd{1 - (v^2c^2)} = 4.5 × 10-10\surd [1 - {(5 × 10^16)/(9 × 10^16)s}] = 4.5 × 10-10\surd(4/9)s = (2/3)(4.5 ×10^-10) = 3.0 × 10^-10 s

Question:

Gyroradius :- What is the radius of the cyclotron orbit in a 10-kiloguass field for an electron of velocity 10^8 cm/sec normal to B?

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Solution:

In the cyclotron , a particle is launched from pointS . A voltage v is applied across the \textquotedblleftgap\textquotedblright (G) . A magnetic field B exerts a centripetal force towards the centre . The particle is accelerated by the voltage v each time it crosses the gap , which increases the radius of the orbit . Thus the particle describes a spiral . The centripetal force which keeps the particle in orbit is given by F \ding{217} = (q/c) v = (q/c) v \ding{217} × B \ding{217} The magnetic field is perpendicular to the linearvelocity of the particle so that the force exerted on the particle is towards the center of the cyclotron. The magnitude of the force is (q/c) v B sin \texttheta, and since sin 90 o = 1 , we have F = q/c vB . Since within each orbit we can think of the particle as describing a circle, its acceleration is simply \omega2R or (v^2 /R) where v is the linear velocity. 2 R or (v^2 /R) where v is the linear velocity. Since F = ma (q/c) vB = m (v^2 /R) R = (mvc)/qB Substituting our values we have: R = [(0.911 × 10^-27 gm 10^8 cm/sec 3 × 10^10 cm/sec)/(4.8 × 10^-10 esu 10^4 guass)] \approx5 . 7×10-4cm 5 . 7 10 4 cm

Since we do not know the actual frictional force acting on the cylinder, we cannot use dynamical methods to solve for the final position of the cylinder. Our only other recourse is to use the prin-ciple of conservation of energy to relate the energy of the cylinder at point A to its energy at point C. By doing this, we will find an expression for the final position of the cylinder, and the problem will be solved. Since friction acts along path AB, but not along path BC, we ap-ply the principle of energy conservation in 2 steps. First, we relate the cylinder's energy at points A and B. Then, using the data ob-tained from the first step, we relate the cylinder's energy at points B and C. The cylinder begins from rest at point A, and therefore has only potential energy. Taking the reference level for potential energy at y = r (see figure), we obtain E_A = mgy_1(1) for the cylinder's energy at A. In traveling from A to B, friction is present. However, this force does no work because we are given the fact that the cylinder is not slipping. By definition, this means that the velocity of the con-tact point of cylinder and surface is zero. Under these conditions the velocity of the cylinder's center of mass, v, is related to the angular velocity by v = \omegar(2) where r is the cylinder radius. Hence, the energy of the cylinder at B is E_B = (1/2) mv^2 + (1/2) I\omega^2(3) where I is the cylinder's moment of inertia. The first term repre-sents the cylinder's translational motion, and the last term represents its rotational motion. Now, in going from B to C, no friction acts. As a result, the cylinder cannot roll, and the rotational motion it had at B is pre-served throughout the trip to C. At C, the cylinder's center of mass has no velocity, but it is still spinning, with angular velocity \omega , about its center of mass. The energy at C is then E_C = mgy_2 + (1/2)l\omega^2(4) By the principle of conservation of energy E_A = E_B and E_B = E_C. Therefore, using (1), (3) and (4) mgy_1 = (1/2) mv^2 + (1/2) I\omega^2(5) mgy_2 + (1/2) I\omega^2 = (1/2) mv^2 + (1/2) I\omega^2(6) form (2) \omega = v/r Substituting this in (5) mgy_1 = (1/2) mv^2 + (1/2) I(v^2/r^2)(7) Solving for v^2 mgy_1 = (v^2/2) [m + (I/r^2)] or v^2= (2mgy_1)/(m + I/r^2)(8) From (6) mgy_2 = 1/2 mv^2(9) We may eliminate v^2 from (9) by substituting (8) in (9), whence mgy_2 = 1/2 m[(2mgy_1)/(m + I/r^2)] then y_2 = (my_1)/(m + I/r^2) for a cylinder, I = (1/2) mr^2 and y_2 = (my_1)/[m + (1/2)mr^2 /r^2] = (my_1)/[(3/2) m] y_2 = (2/3)y_1. For a sphere, the above analysis still holds. Since I = 2/5 mr^2 y_2 = (my_1)/[m + (2/5) mr^2/r^2] = (my_1)/[(7/5) m] y_2 = (5/7) y_1. Hence, the sphere travels further.

Question:

Compute and print out a table of the function Y = 1/(4 - X^2) inthe range 0 \leq X < 2 using a variable step interval. Start out witha step width of 0.1 (i.e. X = 0,0.1, 0.2...1.9). Reduce the stepwidth by a factor of 10 whenever X would reach 2 duringnext iteration, or whenever two successive values differby more than 10. Stop the tabulation when Y reaches 100. Details of printout: The x and Y values are to be printedin two parallel columns, in the 2nd and 4th field. Headings are to be printed as shown below. An extra blank lineis to be inserted whenever the step interval is changed. FUNCTION TABLE OF F (X) = 1/(4 - X\textasteriskcentered\textasteriskcentered2) 2 blank lines \rightarrow { VALUE OF XVALUE OF F(X) 2 blank lines{ 0\textbullet00000E + 002\textbullet50000E - 01 \textbullet\textbullet \textbullet\textbullet \textbullet\textbullet \textbullet\textbullet

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The forces acting on the car at the highest point of the bridge are its weight W^\ding{217} = mg^\ding{217} downward and the normal force N^\ding{217} exerted by the bridge upward. These cannot be equal, since there must be a net downward force to provide the acceleration necessary to keep the car traveling in a circle. Thus mg - N = mv^2/r, by Newton's Second Law. N= m [g - (v^2/r)] = W/g [g - (v^2/r)] = W [1 - (v^2/rg)]. where W is the weight of the car. Here v = 30 mph = 44 ft/s. \thereforeN= 2 tons[1 - {(44^2 ft^2/s^2)/(88 × 32 ft/s^2 ) } ] = 2 [ 1 - (11/16)] ton = 5/8 ton. But action and reaction are equal and opposite. Thus, if the road exerts a force of 5/8 ton on the car, the car exerts the same force on the road. The car loses contact with the road when N = 0, that is, when v^2 = rg. Thus the speed required is v= \surdrg = \surd(88 ft × 32 ft/s^2) = 16 \surd11 ft/s = 53.1 ft/s = 36 mph.

Question:

Bubbles of air escaping from a cylinder of compressed air at the bottom of a pond 34 ft deep are observed to expand as they rise to the surface. Approximately how much do they increase in volume if it can be assumed that there is no change in temperature?

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Solution:

Recognizing this as a situation involving Boyle's Law, it follows that V_t/V_b=p_b/pt where the subscripts b and t refer to bottom and top respectively. Furthermore, p and V represent the pressure and volume, respectively. But p_t is atmos-pheric pressure (1 atmosphere = 14.7 lb/in.^2). The pressure at the bottom of the pond is P_b = P_t + (\rhog)h where \rho is the density of water, g = 32 ft/s^2, and h is the depth of point b relative to the surface of the pond (point t). Therefore, p_b = 1atm+ (62.4 lb/ft^3) (34 ft) p_b = 1atm+ 2121.6 lb/ft^2 Since 1 lb/ft^2 = 1/144 lb/in^2 2121.6 lb/ft^2 = (2121.6/144) lb/in^2 = 14.73 lb/in^2 Because 14.7 lb/in^2 = 1atm 2121.6 lb/ft^2 \approx 1 atm. Hencep_b= 2 atmospheres (29.4 lb/in.^2) It follows thatV_t/V_b= 2/1

Question:

A uniform eccentric drive wheel is circular and of radius 4 in. It has a circular hole cut in it, of radius 1/2 in., for the drive shaft. The center of the hole is 1/2 in. from the center of the wheel. What is the location of the center of gravity of the drive wheel?

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Solution:

By symmetry the center of gravity must lie on the diameter AB which passes through O and X, the centers of the circular wheel and circular hole. Set AB horizontal and let Y, which is a distance x from O, be the location of the center of gravity of the drive wheel. Weights acting vertically are now at right angles to AB. If the circular piece removed to form the hole were replaced, the resultant of the weight W\ding{217} - w\ding{217} of the drive wheel plus the weight w\ding{217} of the piece replaced would have to be the weight W\ding{217} of the whole circle which acts at O. But the moment of the re-sultant weight about any point in the plane of the wheel must be equal to the sum of the moments about the same point of the individual forces making up the resultant. For simplicity let this point chosen be O. The weight of the circle replaced must act at X (see figure). The moment of the resultant weight about O is zero. Hence (W - w) x - w × (1/2) in. = 0. \therefore x = (1/2) in. × [w/(W - w)](1) Butw = mg W - w = (M - m)g where m is the mass of the piece originally occupying the space of the hole, and M - m is the mass of the drive wheel minus the hole. Then x = (1/2) in. × [m/(M - m)](2) If \sigma is the surface density of the material of which the drive wheel is composed m = \sigma \pi (1/2 in.)^2 M - m = \sigma \pi [(4 in.)^2 - (1/2 in.)^2] Using this in (2) x = 1/2 in. × [{\sigma \pi (1/4 in^2.)}/{ \sigma \pi (16 - 1/4) in^2.}] x = 1/2 in. × [(1/4)/(63/4)] = 1/126 in. The center of gravity of the drive-wheel is thus (4 - 1/126)in.= 3.992 in. from point A in the figure.

Question:

Bone, like other connective tissues, consists of cells and fibers, but unlike the others its extracellular components are calcified, making it a hard, unyielding substance ideally suited for its supportive and protective function in the skeleton. Describe the macroscopic and microscopic structure of bone.

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Solution:

Upon inspection of a long bone with the naked eye, two forms of bone are distinguishable: cancellous (spongy) and compact. Spongy bone consists of a network of hardened bars having spaces between them filled with marrow. Compact bone appears as a solid, continuous mass, in which spaces can be seen only with the aid of a microscope. The two forms of bone grade into one another without a sharp boundary. (See Fig.1.) In typical long bones, such as the femur or humerus, the shaft (diaphysis) consists of compact bone surrounding a large central marrow cavity composed of spongy bone. In adults, the marrow in the long bones is primarily of the yellow, fatty variety, while the marrow in the flat bones of the ribs and at the ends of long bones is primarily of the red variety and is active in the produc-tion of red blood cells. Even this red marrow contains about 70 percent fat. The ends (epiphyses) of long bones consist mainly of spongy bone covered by a thin layer of compact bone. This region of the long bones contains a cartilaginous region known as an epiphyseal plate. The epiphyseal cartilage and the adjacent spongy bone constitute a growth zone, in which all growth in length of the bone occurs. The surfaces at the ends of long bones, where one bone articulates with another are covered by a layer of cartilage, called the articular cartilage. It is this cartilage which allows for easy movement of the bones over each other at a joint. Compact bone is composed of structural units called Haversian systems. Each system is irregularly cylindrical and is composed of concentrically arranged layers of hard, inorganic matrix surrounding a microscopic central Haversian canal. Blood vessels and nerves pass through this canal, supplying and controlling the metabolism of the bone cells. The bone matrix itself is laid down by bone cells called osteoblasts. Osteoblasts produce a substance, osteoid, which is hardened by calcium, causing calcification. Some osteoblasts are trapped in the hardening osteoid and are converted into osteocytes which continue to live within the bone. These osteocytes lie in small cavities called lacunae', located along the interfaces between adjoining concentric layers of 'the hard matrix. Exchange of materials between the bone cells and the blood vessels in the Haversian canals is by way of radiating canals. Other canals, known as Volkmann's canals, penetrate and cross the layers of hard matrix, connecting the different Haversian canals to one another. (See Fig. 2.) With few exceptions, bones are invested by the periosteum, a layer of specialized connective tissue. The periosteum has the ability to form bone, and contri-butes to the healing of fractures. Periosteum is lacking on those ends of long bones surrounded by articular cartilage. The marrow cavity of the diaphysis and the cavities of spongy bone are lined by the endosteum, a thin cellular layer which also has the ability to form bone (osteogenic potencies). Haversian type systems are present in most compact bone. However, certain compact flat bones of the skull (the frontal, parietal, occipital, and temporal bones, and part of the mandible) do not have Haversian systems. These bones, termed membrane bones, have a different architecture and are formed differently than bones with Haversian systems.

(1) Here, you have an acid, H_2SO_4 , reacting with the alcohol, ethanol. There exist two reactions that can occur. Which ever one predominates depends on the conditions. If the temperature is raised to 140\textdegreeC, you have the formation of ethyl ether, C_2H_5OC_2H_5. Let us investigate why. When the H_2SO_4 is added to ethanol, it protonates the OH, i.e., The protonated alcohol undergoes the following reaction: A water molecule is expelled. In the solution you have other ethanol molecules, thus, the following can occur; The C_2H_5+ needs electrons, which can be supplied by the oxygen atom in another alcohol molecule. Once this occurs, a proton is expelled to obtain the ether. The proton reacts with the existing HSO_4-, so that H_2SO_4 is not diminished. You have the following: If the temperature is raised above 160\textdegreeC, a different set of reactions occur. Again, the acid protonates the alcohol to form Water departs to yield C_2H_5+. Let us write this as This time, however, it does not react with another alcohol molecule but expels a proton immediately to form the alkene, ethylene or ethene. This is the final product, ethylene, at a temperature greater than 160\textdegreeC. (2) Here you have citric acid reacting with a base, KOH. This suggests the formation of a salt and water. To predict which salt, write out the structure of citric acid. The structure is, To understand how it reacts, you need to know the origin of its acidic nature. The acidic nature is determined by the H of the carboxylic group, COOH. Thus, you can anticipate the following reaction:

Question:

A ball of mass 100 g is attached to the end of a string and is swung in a circle of radius 100 cm with a constant linear velocity of 200 cm/sec. While the ball is in motion, the string is shortened to 50 cm. What is the change in the velocity and in the period of the motion?

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Solution:

We define the initial state of the ball to be the circular path of 100 cm radius and the final state to be the path of 50 cm radius. The initial angular momentum of the object is L = mvr = (100 g) × (200 cm/sec) × (100 cm) = 2 × 10^6 g-cm^2 /sec The initial period is T = 1/f where f is the frequency of rotation of the ball in its initial motion. But 2\pif = \omega where \omega is the initial angular velocity of the ball. Hence T = 1/f = 2\pi/\omega Also, \omega = v/r where v is the velocity of the ball and r is its distance from the axis of rotation. Therefore T = 2\pi/(v/r) = 2\pir/v = [{2\pi × (100 cm)}/{200 cm/sec}] = \pi sec An alternate derivation is as follows. The velocity v of the object about the initial circular path is constant. Hence, v = (distance)/(time) In one period, however, the object moves a distance of one circumf-erence length. Therefore v = (2\pir)/T orT = (2\pir)/v Shortening the string does not apply any torque to the ball because the applied force lies along the line connecting the ball with the center of rotation. Therefore, the torque is zero. However, \tau= dL\ding{217}/dt \ding{217} Then O = dL\ding{217}/dt \ding{217} orL\ding{217}= constant. \ding{217} Hence the final angular momentum, L', is equal to the initial angular momentum L: L' = mv'r' = L Thus, the final velocity is v' = L/mr' = [2 × 10^6 g-cm^2/sec]/[(100 g) × (50 cm)] = 400 cm/sec . The new period is T' = 2\pir'/v' = [2\pi × (50 cm)]/(400 cm/sec) = \pi/4 sec. Therefore, decreasing the radius by a factor of 2 has increased the linear velocity by the same factor but has decreased the period by a factor of 4.

Question:

Find all integers less than 50,000 that equal to the sum of the factorials of their digits. Write a computer program to solve this problem.

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Solution:

Let the reaction be represented by A + BC \rightarrow B + AC where BC means a molecule consisting of atoms B and C. In the reaction, atom C attaches itself to atom A to form AC. In the inertial frame of Figure A, let us write the equation of conservation of energy before and after the collision (here, the final total energy should also include the energy lost during collision) E_i = E_for (1/2)M_Av_A^2 + (1/2)(M_B+ M_C)v_BC^2 = (1/2)M_Bw_B^2 + 1/2(M_A + M_C)w_AC^2 + \Delta\epsilon Here \Delta\epsilon represents changes in the binding energy of the molecules taking part in the reaction. In a second inertial frame moving with velocity V with respect to the first the new velocities of the particles will be their velocities in the old frame minus the velocity of the old frame relative to the new one. With this re-placement of v_A \ding{217} by v_A \ding{217} - V \ding{217} , etc. the law of conservation of energy in the new frame is (Fig. B). (1/2) M_A(v_A - V)^2 + (1/2) (M_B + M_C) (v_BC - V)^2 = (1/2) M_B(w_B - V)^2 + (1/2)(M_A + M_C)(w_AC - V)^2 + \Delta\epsilon On writing out the squares of the quantities in parentheses and comparing this equation with the previous one, we see that the two equations are con-sistent if M_Av_A + (M_B + M_C)v_BC = M_Bw_B + (M_A + M_C)w_AC which is exactly a statement of the law of conservation of linear momentum.

We use an approximation method known as bracketing. Of course, calculus could give us a precise solution, but we will demonstrate the usefulness of this method. First we know that the surface area S of the cylinder is given by S = 2\piR2 + 2\piRH, where R is the radius and H is the height. We also know that volume V = 58 in.3 = \pi R2H. We can solve for H by dividing both sides by \piR2 : H = V/\piR2 = 58 in.3/\piR2 Substituting into the surface area formula, we have S = 2\piR2 +2\piR(58/\pitR2) After combining terms, S = (2\piR2) + (116/R) We seek to minimize S by trying various values for R. We begin with very small and very large R's, and we gradually approach the solution from both sides. A flow chart illustrates the plan of the program: K is a counter; D is the increment on radius R. We begin by initializing the radius to 1, and then we calculate the surface area. If the calculated S is less than OLDS, an arbitrary initial guess at the surface area, OLDS is replaced by the calculated S, and the radius is incremented again. If S is greater than or equal to OLDS, the radius is too big. In this case, the increment D must be reduced to one-tenth its previous size. The variable K serves as a counter of precision. When 8 digits of precision have been reached, the solution of R is printed. Below is the program segment: R = 0.0 D = 1.0 OLDS = 1000.0 K = 0 2\OR = R + D S = (2.0\textasteriskcentered3.14159\textasteriskcenteredR\textasteriskcentered\textasteriskcentered2) + (116.0/R) IF (S.GE.OLDS) GO TO 3\O OLDS = S GO TO 2\O 3\OR = R - D D = D/10.0 K = K + 1 IF (K.GE.8) GO TO 4\O GO TO 2\O 40WRITE (5,\textasteriskcentered) R [Note use of FORMAT-free WRITE statement.]

Question:

Fiddler crabs are a dark brown color during the day and change to a pale brown color at night. Is it possible to reverse this pattern and make the crabs turn dark at night and light during the day? How?

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Solution:

Fiddler crabs undergo a daily cycle of color change. However, this is not just a simple adjustment to conditions of light and dark. When the crabs are brought into a room, which is kept dark continuously, they still turn dark brown at dawn and light brown at sunset. These crabs have an internal sense of time, called a "biological clock." It is possible to reset this biological clock by artificially producing periods of light and darkness. For example, we can keep the crabs in a room in which a bright light is turned on at sunset and shut off at sunrise. After several days of this treatment, the crabs begin to turn dark brown at sunset and pale brown at sunrise. Even when the crabs are exposed to constant light or constant darkness, the 24 hour rhythm continues. But this cycle is 12 hours out of phase with the true day, showing that the crabs' internal clock has been reset. Rhythms of this sort, with approximately 24- hour periods, are called circadian rhythms. The biological clock of the fiddler crab can also be reset by exposing it to ice water. This type of ex-perimental manipulation seems to stop the clock. When it is taken out of the ice water, the clock restarts, but it has lost time. This manipulation seems to support the belief that biological clocks are controlled by metabolic rhythms, since metabolism slows down under very cold conditions.

Question:

A young's experiment is set up with the following characteristics: Monochromatic source (\lambda = 0.55\mu), slit separation d = 3.3mm, distance from slits to screen D = 3m (see figure). 1) Calculate the fringe spacing. 2) Place a sheet of glass with plane parallel faces and thickness e = 0.01 mm in front of slit F_1. a) Determine the direction of the displacement of the fringes and the formula giving the relationship for their displacement. b) Knowing that the fringes are displaced by 4.73 \pm .01 mm find the index of refraction of the glass and its error.

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Solution:

Here we need to concentrate on interference between two slits. From a standard optics text we find that the intensity of the interference pattern on N slits is, I = (I_O / N^2) (sin^2N\delta / sin^2\delta)(1) where I_O is the initial intensity; I is the intensity at the screen; N is the number of slits; and \delta is given by, \delta = (\pid / \lambda )sin \texttheta(2) where \lambda is the wavelength of coherent monochromatic light from the source; d is the slit separation, and \texttheta is the angle between the normal to the slit system and the point on the screen. For small angles, sin\texttheta = (x / D) where x is the distance from the center of the screen to a point at which we wish to measure the intensity and D is the distance from slits to screen. Substituting into equation (2) gives the result \delta = (\pid / \lambda )(x / D)(3) For two slits N = 2. Substituting this into equation (1) we realize we can expand the sin^2 2\delta term because, sin2\delta = 2 sin\delta cos\delta.(4) Squaring both sides results in , sin^2 2\delta = 4 sin^2\delta cos^2\delta .(5) Substituting this into equation (1) and cancelling terms, I = I_O cos^2\delta .(6) The fringe spacing i is the distance between two maxima or two minima . We therefore let i = x_2 - x_1 and \delta = \pi . Applying equation (3), \pi = (\pidi) / (\lambdaD).(7) Solving for i, i = (\lambdaD) / (D)(8) Substituting the given values for \lambda, D, and d into equation (8), i= [{(.55 × 10^-6)(3)} / (3.3 \bullet 10^-3)] = 5 × 10^-4m .(9) Now we insert a sheet of glass of thickness 0.01 mm and index of refraction n in front of one of the slits. The light of this slit will now travel a different optical distance to a position x on the screen. Initially before the glass plate was inserted in front of the slit, the path difference was, \delta_1 = (F_1M) - (F_2M) = x(\lambda / i).(10) The glass plate will add an additional path difference (n-1)e 30 the new path difference is, \delta = x (\lambda / i) + (n - 1)e(11) so, the new \delta for a maximum will be p\lambda (p being an integer) or, p\lambda = x (\lambda / i) + (n - 1)e .(12) Solving for x, x = (i / \lambda) [(p\lambda) - (n - 1)e](13) so the shift in fringes will be the change in x: ∆x =-(i / \lambda)(n - 1)e .(14) To measure the index of refraction we solve for n: n = 1-(\lambda / i)(∆x / e) .(15) So we do the measurement as follows. Observe the fringe system with no glass plate, record the positions x of the maxima, then insert the glass plate, record the new positions of the maxima and determine the shift ∆x of the maxima. In our problem ∆x = -4.73 mm, substituting values into equation (15), n = 1 - [(0.55 x 10^-3) / (0.5)] [(-4.73) / (10^-2)]= 1.5203.(16) To determine the error in n we want to look at the relationship d(∆x) / ∆x. Differentiating equation (15) we have dn = -(\lambda / ie)d(∆x).(17) Solving equation (14) for -(\lambda / ie) , -(\lambda / ie)= (n - 1) / (∆x).(18) Substituting for -(\lambda / ie)in equation (17), dn = (n-1)[d(∆x) / (∆x)] . Substituting values dn = 0.5 [(2 × 10^-2) / (4.73)] = 2 × 10^-3 , so n = 1.520\pm0.002.

The current i in the straight wire produces cir-cular magnetic lines of force. In the vicinity of the loop these lines are perpendicular to the plane of the loop. Be-cause the current in the long wire is time dependent, the magnetic field which it sets up at the site of the circu-lar coil will change with time. Furthermore, being that magnetic flux is defined as \varphi = \int B^\ding{217} \textbullet ds^\ding{217},(1) (where B^\ding{217} is the magnetic induction and the integral is a surface integral), \varphi will also be time dependent. Then, by Faraday's Law E.M.F. = [(d\varphi)/(dt)](2) an E.M.F. (electromotive force) will be induced in the coil of wire. If the resistance of this loop is R, then by Ohm's Law, the current in the loop is i_l= (E.M.F.)/R = -(1/R) (d\varphi/dt)(3) To evaluate this current, we must find \varphi , and, hence, B^\ding{217}. To calculate the latter, we use Ampere's Law, or \int B^\ding{217} \bullet dl^\ding{217} = (\mu_0 i enc(4) This integral must be evaluated over a closed path, i_enc is the net current passing through this path and \mu_0 = 4\pi × 10^-7 Weber/amp. m. The lines of magnetic induc-tion, B^\ding{217}, for a long wire are circles centered on the wire. Hence, to evaluate (4) , we use a circular path of radius r centered on the wire. In this way, B and di will be paral-lel, and, from (4) B2\pir = \mu_0 i_enc andB = (\mu_0 i_enc /2\pir) We may next find \varphi. Because the area elements ds^\ding{217} of the circular loop are all parallel to the lines of B^\ding{217} set up by the long wire, f reduces to \varphi = \int B^\ding{217} \textbullet ds^\ding{217} = \int Bds. Since the radius a of the loop is very small compared with its distance r from the wire, it is reasonable to assume that the magnetic field has this value at all points inside the loop. The flux through the loop is therefore \varphi = BA = [(\mu_0 i_enc )/(2\pir)] (\pia^2 ) But i_enc = 3.36 (1 + 2t) × 10^-2 amp/s and \varphi = [{(\mu_0 (3.36) (1 + 2t)10^-2 amp/s a^2 )}/(2r)] Using(3) i = - (1/R) (d/dt) [{\mu_0 (3.36) (1 + 2t)10^-2 amp/s) a^2 }/(2r)] = - (1/R) [\mu_0 (3.36) (2) {10^-2 (amp/s)} a^2 ]/(2r) = {[4\pi × 10^-7 (Weber/amp\bulletm)] (6.72) [10^-2 (amp/s)] (10^-6 m^2 )}/{(2) (1m) (8.99 10^-4 ohms)} = [(4.22 × 10^-14 Weber)/(8.99 × 10^-4 ohms \bullet s)] = 4.70 × 10^-11 (Weber/ohms \bullet s) . But 1 Weber = 1 (Newton \bullet m/amp) and 1 ohm = (1 volt/amp) = 1 joule/amp coul) . i_l = 4.70 × 10^-11 (Newton \textbullet m/amp s) \bullet (amp \bullet coul/joule) = 4.70 × 10^-11 [(Newton \textbullet m \textbullet coul)/(Newton \bullet m \bullet s)] = 4.69 × 10^-11 amp The magnetic field produced by the straight wire goes through the loop into the page in the figure and it is increasing with time. The current i_1 in the loop tries to counteract this by producing a magnetic field coming out of the page. The induced current must therefore flow counterclockwise round the loop.

Question:

What is the minimum length L of a wall mirror so that a person of height h can view herself from head to shoes?

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Solution:

This is not easily solved by a diagram. We suppose that the person stands a distance x from the wall and that her eyes (E) are a distance y from the top of her head (H). To look at her toes (F) she looks at point A which is the point of reflection of a light ray from her foot. A must be at a height halfway between her eyes and feet (so that the angle of incidence equals the angle of reflection). Similarly to look at the top of her head she looks at point B. If OP = h and BP = y/2, then the length of the mirror is AB = PO - BP - OA = h - y/2 - 1/2 (h - y) = h/2. Thus the minimum length of the mirror is h/2, and this does not depend on the distance x that the person is standing away from the mirror.

Question:

a) Calculate the ratio of the volume of a hepatocyte to that of an E. coli cell. Assume the hepatocyte to be a cube 20\mu on an edge, b) Calculate the ratio of their surface areas, c) Calculate the (surface / volume) ratios for each.

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Solution:

A hepatocyte is an eucaryote, a large complex cell, and an E. coli cell is a procaryote, a small simple cell. An E. coli cell is cylindrical with a length of 2.0\mu and a cross-sectional diameter of 1.0\mu. The volume of a cylinder is equal to the product of its length and the area of its cross- section. Volume of an E. coli cell = 2.0\mu × \pi × (0.5\mu)^2 = 1.57\mu^3 Volume of a hepatocyte = (20\mu)^3 = 8.0 × 10^3\mu^3 The volume ratio is (hepatocyte / E. coil) = 8.0 × 10^3\mu^3 = 5.1 × 10^3 : 1 b) The surface area of a solid cylinder is equal to 2\pirl + 2\pir^2, where r is the cross-sectional radius and l is the length. Surface area of an E. coli = 2\pi (0.5\mu) (2.0\mu) + 2\pi(0.5\mu)^2 = (6.28\mu)^2 + (1.57\mu)^2 = 7.85\mu^2 The surface area of a cube is 6s^2, where s is the length of the edge. Surface area of a hepatocyte = 6(20\mu)^2 = 2.40 × 10^3\mu^2 The ratio of their surface areas is (hepatocyte) / (E.coil) = (2.40 × 10^3 u^2) /(7.85 \mu^2) = 306 : 1. (c) Surface to volume ratiost E. coli; (suface) / (volume) = (7.85\mu^2) / (1.57\mu^3) = 5\mu^-1 Hepatocyte; (suface / volume) = (2.40 × 10^3\mu^2) / (8.0 × 10^3\mu^3) = 0.3\mu^-1.

Question:

Two devices, whose resistances are 2.8 and 3.5\Omega, respective-ly, are connected in series to a 12-V battery. Compute the current in either device and the potential applied to each.

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Solution:

Resistances in series add. The equivalent re-sistance of the two devices is given by R = 2.8\Omega + 3.5\Omega = 6.3\Omega . The current supplied by the battery can be computed using Ohm's law, I = (V/R) = [(12 V)/(6.3\Omega)] = 1.9 A . This current I flows through both devices since they are in series. Ohm's law may be applied to calculate the potential applied to each device: (see figure) V_1 = IR_1 = (1.9 A) (2.8\Omega) = 5.3V V_2 = IR_2 = (1.9 A) (3.5\Omega) = 6.7V. Note that the sum of these potentials is equal to the battery potential, 12V.

Question:

What is meant by embryonic development? Describe the various stages of embryonic development.

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Solution:

Embryonic development begins when an ovum is fertilized by a sperm and ends at parturition (birth). It is a process of change and growth which transforms a single cell zygote into a multicellular organism. The earliest stage of embryonic development is the one-celled, diploid zygote which results from the fer-tilization of an ovum by a sperm. Next is a period called cleavage, in which mitotic division of the zygote results in the formation of daughter cells called blastomeres. At each succeeding division, the blastomeres become smaller and smaller. When 16 or so blastomeres have formed, the solid ball of cells is called a morula. As the morula divides further, a fluid-filled cavity is formed in the center of the sphere, converting the morula into a hollow ball of cells called a biastula. The fluid filled cavity is called the blastocoel. When cells of the blastula differentiate into two, and later three, embryonic germ layers, the blastula is called a gastrula. The gastrula period generally extends until the early forms of all major structures (for example, the heart) are laid down. After this period, the de-veloping organism is called a fetus. During the fetal period (the duration of which varies with different spec-ies) , the various systems develop further. Though de-velopmental changes in the fetal period are not as dramatic as those occurring during the earlier embryonic periods, they are extremely important. Congenital defects may result from abnormal de-velopment during this period.

Memory units, such as fullwords, are quite valu-able, but if we are unable to alter or operate on them, they become quite useless. The accumulator facilitates this. Let's for the moment assume that each unit of memory can hold four decimal digits. The Accumulator may be regarded as a special unit of memory in the computer. For simplic-ity, we will also assume that it also holds four decimal digits. 1) The AC has the power to copy data from any word in memory without destroying the word's contents. This process is analogous to making a photostat copy. This copy can then be used in other operations. The CLA (clear and add) instruction will do this copying. It first clears the ac-cumulator and then adds the contents of the specified loca-tion to it, as shown in Example 1. 2) The AC can also reverse the above process; ;it can store (STO) its contents in a specified word, with the use of the STO instruction. 3) The AC also performs arithmetic instructions. Thus, an ADD (add) instruction adds the contents of any specified word to the contents of the AC and retains the accumulated sum in the AC. To enable the AC to perform some other task, the new sum is then stored in a convenient location. Storage may be done as shown in Example 2. Since subtraction is merely another form of addition, the AC can also be used in subtractions. Using the SUB (subtract) instruction, this calculation may be done as in Once again, the sum will be stored in a convenient location of memory. The above are the basic properties of the accumulator. But the great flexibility that results from the AC can be seen in much more depth by studying its use in conjunction with another arithmetic register referred to as an MQ. The Multiply Quotient (MQ) register is generally combined with the AC when multiplications and divisions are to be done. If we assume that each register has a length of 17 bits, then in a multiplication or division, they merge to form a simple 34-bit field. In this 34-bit field, the MQ occupies the low order 17 bits. After a multiplication or division, the result is stored starting with the MQ register. In other words, the results are right-justified. In the case of division, the remainder is stored in the AC field.

The satellite experiences a force in the radial direction, towards the center of the earth. This force, the gravitational force on the satellite, provides the centripetal force that will keep the satellite in its circular orbit. Using Newton's Second Law, F = ma, we may write ma = mg' where g' is the acceleration due to gravity at a height of 135 miles above the earth's surface. Because the satellite is in a circular orbit, a = v^2/d (see figure). Then mv^2/d= mg' andv= \surd(g'd) v= \surd[(30 f/s^2)(3960 mi + 135 mi)] But 1 mi = 5280 ft v= \surd[(30 f/s^2)(4095)(5280 f) v= \surd(6.49×10^8 f^2/s^2) = 2.55×10^4 f/s To find the period of the satellite, we note that its speed is constant. Hence v = 2\pid/T where 2\pid is the distance travelled by the satellite in 1 revolution and T is the time required for this traversal, T = 2\pid/v = [{(2)(\pi)(4095 mi)(5280 f/mi)}/{(2.55×10^4 f/s)}] = (1.36×10^8 f)/(2.55×10^4 f/s) = 5.33×10^3 s

Question:

A vibrating tuning fork of frequency 384 cycles \bullet s^-1 is held over the end of a vertical glass tube, the other end of which dips into water. Resonance occurs when the top of the tube is 21.9 cm and also 66.4 cm above the water surface. Calculate the speed of sound in air and the end correction of the tube. Assuming that the temperature is 13\textdegreeC, what is the velocity of sound at 0\textdegreeC (v_0)?

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Solution:

Figure (C) shows the tuning fork in vibra-tion. In order to send a compression wave down the tube, one tine of the fork is pulled back to position c, and then is let go. During one oscillation, the tine moves from c to b and then back to c. Now, the pipes with which we are dealing have water at one end. This end must then be a displacement node, for the air at this end of the column is always at rest. Alternatively, it is a pressure antinode. Similarly, the open end of the tube is a displacement maximum, or a pressure node. Now, suppose the fork starts at point c and sends a compression wave down the tube. When this wave reaches the bottom of the tube, it must be reflected as a compression if that end is to remain a pressure anti-node. The reflected compression travels back up the tube. If resonance is to occur, the compression must reach the tine of the tuning fork when the latter is ready to move from b to c. (In this situation, the tine is preparing to send a rarefaction down the tube). If this does not occur, the top of the tube will not remain a pressure node, for the reflected pressure wave and the new pressure wave will not cancel. Hence, in 1/2 oscillation of the tuning fork, the wave travels a distance 2L (twice the tube length). In general, resonance will occur if the wave travels a distance 2L in an odd number of half oscillations of the fork. If the wave velocity is v and the frequency of the fork is v_source, the resonance condition is 2L = v(n + 1/2)/ v_source(n = 0, 1, 2 ...)(1) Since the wave frequency equals the source frequency, we may write v_wave = v_source= v Butv_wave \lambda = v where \lambda is the wavelength of the wave. Hence, (1) becomes 2L = v(n + 1/2)/v Solving this for v v = (2Lv)/(n + 1/2)(n = 0, 1, 2 ...)(2) Therefore, we may measure v if we know the tube length, the source frequency, and the value of n. In practice, the tube is put in the water, and slowly drawn out as the fork vibrates. When the first reso-nance occurs, n = 0, and the length of tube extending out of the water is measured. Again, the tube is drawn out of the water. When the second resonance occurs, n = 1, and the tube length is again measured, etc. Each resonance will give us the same value for v, as equation (2) indicates. Using the relation v = v\lambda where \lambda is the wavelength of the wave, we may also write (2) as v = [2L/(n + 1/2)] (v/\lambda) orL = (n + 1/2) (v/\lambda)(n = 0, 1, 2, ...)(3) In practice, an antinode never occurs quite at the end of an open pipe. Its position is just beyond the end of the pipe, the maximum displacement slightly overshooting the end (see figs. (a) and (b)). Thus for the first reso-nance (n = 0 in (3)) the length of the tube will be almost a quarter of a wavelength, and \lambda/4 = L + E, where L is the length of the air column and E is the end correction. Similarly, for the second resonance, 3\lambda/4 = L' + E, where L' is the length of the air column when the second resonance occurs. Thus, using our data, \lambda/2 = L' - L = (66.4 - 21.9) cm = 44.5 cmor\lambda = 89 cm. Therefore the velocity of sound is v = f\lambda = 384 s^-1 × 89 cm = 34176 cm \bullet s^-1 = 341.8 m \bullet s^-1. Further, E = \lambda/4 - L = (22.25 - 21.9) cm = 0.35 cm. The velocity of sound in a gas is proportional to the square root of the absolute temperature. Hence v_0/v = \surd[T_0/T] = \surd(2730K / 286\textdegreeK) = 0.977. \thereforev_0 = 0.977 × 341.8 m \bullet s^-1 = 333.9 m \bullet s^-1.

To understand what a reflex arc is, we must know something about reflexes. A reflex is an innate, stereotyped, automatic response to a given stimulus. A popular example of a reflex is the knee jerk. No matter how many times we rap on the tendon of a person's knee cap, his leg will invariably straighten out. This ex-periment demonstrates one of the chief characteristics of a reflex: fidelity of repetition. Reflexes are important because responses to certain stimuli have to be made instantaneously. For example, when we step on something sharp or come into contact with something hot, we do not wait until the pain is experienced by the brain and then after deliberation decide what to do. Our responses are immediate and automatic. The part of the body involved is being withdrawn by reflex action before the sensation of pain is experienced. A reflex arc is the neural pathway that conducts the nerve impulses for a given reflex. It consists of a sensory neuron with a receptor to detect the stimulus, connected by a synapse to a motor neuron, which is attached to a muscle or some other tissue that brings about the appropriate response. Thus, the simplest type of reflex arc is termed monosynaptic because there is only one synapse between the sensory and motor neurons. Most reflex arcs include one or more interneurons between the sensory and motor neurons (see Figure 1). An example of a monosynaptic reflex arc is the knee jerk. When the tendon of the knee cap is tapped, and thereby stretched, receptors in the tendon are stimulated. An impulse travels along the sensory neuron to the spinal cord where it synapses directly with a motor neuron. This latter neuron transmits an impulse to the effector muscle in the leg, causing it to contract, resulting in a sudden straightening of the leg (see Figure 2).

Question:

What is the energy of a photon of green light (frequency = 6 × 10^14vps)?

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Solution:

Planck's hypothesis states that E =hu, where u is the frequency of the radiation, and h is Planck's constant. Therefore, E = (6.63 x 10^-34 joule-sec) (6 x 10^14vps) E = 3.98 x 10^-19 joules.

Question:

Compare the teeth of mammals with those of lower vertebrates.

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Solution:

The chief differences between mammalian teeth and those of lower vertebrates is their respective embryonic origins. Of the three main types of embryonic tissues (ectoderm, mesoderm and endoderm), mammalian teeth are made up of two (ectoderm and mesoderm) while teeth of the lower vertebrates are derived solely from the ectoderm. Mammalian teeth result from outgrowths of the embryonic bone- producing tissue - the mesoderm. They are covered with a hard enamel derived from the embryonic ectoderm. Mammalian teeth are composed of living cells, supplied with blood vessels and nerves (see Figure 1). They are also highly evolved, consisting of different shapes adapted for different functions. The foremost teeth of most mammals are chisel-shaped incisors (see Figure 2). These are used for biting and cutting food. The incisors of rodents and horses grow continually and are worn down during eating. Meat eaters have sharp canine teeth that are used for tearing. All of a dog's teeth are sharp and pointed, and highly specialized for chewing raw meat. Herbivorous mammals have flat grinding teeth in the rear of their mouth that help crush and scrape the tough cellulose cell walls of plant cells. Figure 2. Structure and arrangement of teeth in different animals. (A) Snake: thin, sharp, backward-curved teeth that have no chewing function (the snake skull is here shown disproportionately large in relation to the other three). (3) Beaver (gnawing herbivore): few but very large incisors, no canines, premolars and molars with flat grinding surfaces. (C) Dog (carnivore): large canines, premolars and molars adapted for cutting and shearing. (D) Deer (grazing and browsing herbivore): six lower incisors (three on each side), but no upper incisors-these are functionally replaced by a horny gum; premolars and molars with very large grinding surfaces. Notice the large gap between the incisors and premolars. All the teeth of the lower vertebrates are sharp and pointed. The teeth of fish, reptiles, and amphibians have the same general shape, whether they are meat or plant eaters. The triangular, conical shape of these teeth in-dicates their evolutionary origin. The teeth of these vertebrates have evolved from scales. They are formed from the ectoderm (exterior tissue) of the embryo and are re-placed when lost. These teeth are not made up of living cells and do not contain blood vessels or nerve endings. Thus mammalian teeth are highly specialized, living organs while the teeth of the lower vertebrates are non-living, less evolved extensions of their skin tissue.

Question:

You are given 3-bromohexane. Draw the stereoisomers of this compound; specify the R and the S enantiomers.

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Solution:

Non-superimposable mirror-image stereo- isomers are called enantiomers. It is not enough that a molecule has a mirror image, for everything has a mirror image; the mirror images must not be able to be superimposed on each other. The arrangement of atoms that characterize a particular stereoisomer is called its configuration. You want to specify the configuration in terms of R and S. Let us see what this means. In assigning configuration, you do the following after determining the chiral carbon: (1) Assign.a sequence of priority to the four atoms or groups of atoms attached to the chiral center. (A chiral center consists of a carbon atom to which 4 different groups are attached.) (2) Visualize the molecule so that the group of lowest priority is directed away from you. Now, observe the arrangement of the remaining groups. If, in going from the group of highest priority to the group of second priority your eye travels in a clockwise direction, the configuration is R; if counterclockwise, the configuration is S. Now, how do you determine priority? If the four atoms bonded to chiral center are all different, priority depends on atomic number. The atom of higher atomic number receives the higher priority. If they are isotopes, the higher mass number has priority. Now, if you can't determine priority from this, because you have 2 or more atoms that are the same attached to the chiral center, make a similar comparison of the next atoms (i.e., their atomic numbers) as you move out from the chiral center. With this in mind, you can proceed as follows: (a) 3-bromohexane. Recall, enantiomers are mirror image isomers so that you can say the enantiomers I and II, where the asterisk designates the chiral carbon. To specify R and S, the lowest priority group must be drawn away from you. H has the lowest priority since it has the smallest atomic number. It can be drawn away from you by placing it in a vertical position. By con-vention, groups in vertical positions are considered away from you. Thus, you can rewrite these enantiomers as shown in figure B. The hydrogens attached to the chiral center are now vertical, and, as such, away from us. The other groups surrounding the chiral center are Br, CH_2CH_3, and CH_2CH_2CH_3. Of the three, Br has highest priority because its atomic number is higher than carbons. The other two have carbons attached to the chiral center, so that no priority is possible here. You must move out from the chiral center. The next atoms reached are, again, carbon. No priority determination is possible. If you move out to the next atom, you find a carbon atom in CH_2CH_2CH3and a hydrogen atom in CH_2CH_3 (it doesn't have another carbon). Thus, because carbon has a larger atomic number than hydrogen, CH_2CH_2CH_3 has priority over CH_2CH_3. In summary, for the groups or atoms attached to the chiral center, the order of increasing priority can be shown by Recalling step (2), in going from highest to next lowest, then to next lowest, in the enantiomer on the left, you move in a clockwise direction. For enantiomer I, you move in a counterclockwise direction. Thus it is S and enantiomer II is R.

The block diagram of Figure 1 shows the various elements of the PDP-8 computer. As shown in Figure 1, the PDP-8 memory has 4096 words or locations. Each word of the memory has 12 bits. Memory Address :Each word of the memory is numbered sequentially from 0 to 4095. This number associated with a word is called the "address" of the word. Thus, the "address" of a word shows the physical location of the word in memory. Actually, the PDP-8 uses the octal number system rather than the decimal number system. Hence, the address of a memory word (in octal) can be anything from 0 to 7777_8. To see this note that 4095 = 2^12 - 1 = 111111111111_2 = 111 111 111 111_2 = 7 7 7 7 _8 Thus, if the address of a word is 7777_8, the word is located at the last physical location in the memory. And, if the address of a word is 0, it is located at the first physical location in the memory. PC :The PDP-8 has a Program Counter (PC). The PC stores the address of the instruction to be executed (i.e., carried out) by the computer. At any given time the PC contains the address of the next instruction to be car- ried out, after the completion of the instruction which is presently being executed. The PC has 12 bits. MAR :The PDP-8 has a register called Memory Address Register (MAR). The MAR is used whenever any information is to be read from or written into the memory. The pro-cedures are as follows: If there is a memory location from which information is to be read from or written into, then the address of this location must be entered into the MAR. The MAR then communicates with the memory and the MBR to get the information transferred as de-sired. The MAR holds 12 bits. The number of bits in MAR depends on the size of the program counter. In this case the program counter has 12 bits. Therefore the number of bits in MAR is 12. MBR :The PDP-8 has another register called Memory Buffer Register (MBR) . Any information which is read from the memory is read into the MBR first. Then, from the MBR the information can be sent to wherever it is needed. Similarly, any information which is to be written into the memory must first be placed into the MBR. Then, from the MBR the information is written into the memory. In many instances, the main function of the MBR is to momentarily store data that is being transferred be-tween the general purpose registers and main memory. Registers manipulate data much faster than memory de-vices, and as such, memory buffer registers (MBR) were created to bridge the timing gap between registers and main memory. The outstanding feature of the MBR is that it can regulate its speed. When information is to be passed to memory, the MBR accepts this information from the high-speed registers and passes it at a slower speed to the memory device. The reverse is done when informa-tion is to be transferred from memory to a register. The MBR accepts this information from the low speed memory device and passes it at a higher speed to the register. Thus, MBR acts as a buffer between the memory and other registers in the computer. The bit size of the MBR depends on the bit size of one computer word. In the case of the PDP-8 one word is 12 bits long so MBR is also 12 bits long. IR :The PDP-8 has a register called Instruction Register (lR). At any given time the IR contains the Operation Code (OP) of the instruction which is presently being executed by the computer. The Operation Code present in the IR is coded information telling the computer what is to be done with the various data present in the different registers of the computer. The IR has 3 bits. AC :The PDP-8 has a general purpose register called the Accumulator (AC). The computer carries out addition, subtraction, multiplication, division, etc. by storing the intermediate and final operands and sums in the ac-cumulator. The AC has 12 bits. L :The PDP-8 has a 1-bit register called a Link Register (L). The Link register works in association with AC. Sometimes, the Link register and the AC are referred to by the combined mnemonic LAC, forming a combined 13-bit register. The Link register (normally referred to as the Link bit) can be set to 1 or cleared to 0 independently of the AC. But, while doing register shift-left and shift-right operations, the L bit works in association with the 12- bit AC, as LAC, to store the bit shifted into or out of the AC register.

Question:

An electron in an oscilloscope tube is situated midway between two parallel metal plates 0.50 cm apart. One of the plates is maintained at a potential of 60 volts above the other. What is the potential gradient between the plates? What is the force on the electron?

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Solution:

The difference of potential between points a and b, is, by definition V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}b\int_a E^\ding{217} \bullet dl^\ding{217}(1) where the integral is evaluated over an arbitrary path between a and b and E^\ding{217} is the electric field intensity. In this example, E^\ding{217} is constant between the plates (see figure), and V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}E^\ding{217} \bullet^b\int_adl^\ding{217}(2) Also, we evaluate (2) over the path shown in the figure, since E^\ding{217} and dl^\ding{217} are in the same direction for this path. Then V_(x)2 \rule{1em}{1pt} V_(x)1 = \rule{1em}{1pt}E^\ding{217} \bullet ^(x)2\int_(x)1dl^\ding{217}= \rule{1em}{1pt}E(x_2 \rule{1em}{1pt} x_1) andE =(v_(x)2 v_(x)1) / (x_2 \rule{1em}{1pt} x_1) = \rule{1em}{1pt}(∆v / ∆x)(3) By definition, ∆v / ∆x is the potential gradient. Hence, for this case, by(3) \rule{1em}{1pt}E = ∆V / ∆x = \rule{1em}{1pt}(60 Volts / .5 × 10^\rule{1em}{1pt}2m) \rule{1em}{1pt}E = ∆V / ∆x = \rule{1em}{1pt}(60 Volts / 5 × 10^\rule{1em}{1pt}3m) = \rule{1em}{1pt}12 × 10^3 (V / m) Therefore, ∆V / ∆x = \rule{1em}{1pt}12 × 10^3 (V / m) E = 12 ×10^3 (V/m) By definition, the force on the electron is the product of E and the electron charge, or F = Eq = (1.2 × 10^4 nt / coul) × (\rule{1em}{1pt}1.60 × 10^\rule{1em}{1pt}19 coul) = -1.92 × 10^\rule{1em}{1pt}15 nt This force is toward the plate of higher potential.

Question:

If one increases the angle away from the Z axis, how does the wave function for a 2P_Z electron change?

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Solution:

Question:

Show that the moment of inertia of a body about any axis is equal to the moment of inertia about a parallel axis through the center of mass plus the product of the mass of the body and the square of the distance between the axes. This is called the parallel-axes theorem. Prove also that the moment of inertia of a thin plate about an axis at right angles to its plane is equal to the sum of the moments of inertia about two mutually perpendicular axes concurrent with the first and lying in the plane of the thin plate. This is called the perpendicular-axes theorem.

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Solution:

Let I be the moment of inertia of the body about an arbitrary axis and 1_G the moment of inertia about the parallel axis through the center of mass G, the two axes being distance h apart. (See fig. (A)). By definition of the center of mass of a body relative to an arbitrary axis through a point P, we obtain I = \sum_i m_i r^2_i = \sum_i m_i(x_i^2 + y_i^2) = \sum_i m_i x^2_i + \sum_i m_i y_i^2 where the sum is carried out over all mass particles m_1 of the body, and r_i^2 is the distance form P to m_i. Now, x_i = xʹ_i + a y_i = yʹ_i + b as shown in figure (A). Here, (xʹ_i, yʹ_i) locates m_i relative to G, the center of mass. Then I = \sum_i m_i (xʹ_i + a)^2 + \sum_i m_i (yʹ_i + b)^2 I = \sum_i m_i (xʹ_i + a)^2 + \sum_i m_i (yʹ_i + b)^2 I = \sum_i m_i (xʹ_i + yʹ_i^2) + \sum_i m_i (a^2 + b^2) + 2a\sum_i m_ixʹ_i + 2b\sum_im_iyʹ_i But xʹ_i^2 + yʹ_i^2 = rʹ_i^2 and a^2 + b^2 = h^2, whence I = \sum_i m_i rʹ_i^2 + \sum_i m_i h^2 + 2a\sum_i m_ixʹ_i + 2b\sum_im_iyʹ_i By definition of the center of mass, however, \sum_i m_i xʹ_i = \sum_i m_iyʹ_i = 0,and I = \sum_i m_i rʹ_i^2 + \sum_i m_i h^2 = I_G + Mh^2 where M( = \sum_im_i) is the net mass of the body. This is the parallel-axes theorem. Although we derived this theorem in 2 dimensions, it is equally applicable in three dimensions. Take, in the case of the thin plate, the axes in the plane of the plate as the x- and y-axes, and the axis at right angles to the plane as the z-axis (see fig. (B) ). Then the moment of inertia of the plate about an axis perpendicular to the plate (the z-axis) is I_z = \sum_im_i r_i^2 where r_i locates m_i relative to 0. But r_i^2 = x_i^2 + y_i^2 where x_i and y_i are the x and y coordinates of m_i. Then I_z = \sum_im_i x_i^2 \sum_im_iy_i^2 But \sum_im_i x_i^2 = I_y and \sum_im_iy_i^2 = I_x, whence I_z = I_x+ I_y This is the perpendicular axes theorem.

Question:

Electrons are located at definite energy levels in orbitals around the nucleus of an atom. Sometimes electrons move from one orbital to another. What happens to cause such a change?

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Solution:

On a long straight road a car accelerates uniformly from rest, reaching a speed of 45 mph in 11 s. It has to maintain that speed for 1(1/2) mi behind a truck until a suit-able opportunity for passing the truck arises. The car then accelerates uniformly to 75 mph in a further 11 s. After maintaining that speed for 3 min, the car is brought to a halt by a uniform deceleration of 11 ft/s^2. Illustrate the motion on a suitable diagram, and calculate (a) the total distance traveled, (b) the total time taken, (c) the average speed, and (d) the average acceleration in the first 142 s.

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Solution:

A velocity-time diagram should be drawn. During the first 11 s the car accelerates uniformly to a speed of 45 mph = 66 ft/s. This part of the diagram is therefore a straight line OA inclined to the t-axis at an angle whose tangent is 66/11. The distance traveled, s_1, is the area under this portion of the graph. Thus s_1 = (1/2) × 11 s × 66 ft/s = 363 ft. In the second portion of the motion, the car travels for 1 (1/2) mi at a constant speed of 45 mph. This part of the graph, AB, is a straight line parallel to the t-axis, its length being t_2 = [(1(1/2) mi]/[45 mi/hr] × 60 min/hr × 60 s/min = 120 s. In the third portion of the motion, the car increases its speed by 30 mph = 44 ft \textbullet s^-1 at uniform acceleration in 11 s. This part of the graph is thus a straight line BC of slope 44/11. The distance traveled in this 11 s, s_3, is the area under this part of the graph, i.e., the shaded portion. s_3 = (1/2) × 11 s × 44 ft/s + 11 s ×66 ft/s = 968 ft. The next portion of the graph is again a straight line parallel to the t-axis. The time t, is 3 min = 180 s, and thus s_4 = 75 mi/hr × (3/60) hr = 3.75 mi. In the final part of the motion, the car is brought to rest from a speed of 110 ft \textbullet s^-1 by a uniform de-celeration of 11 ft/s^2. This portion of the graph, DE, is thus a straight line with a negative slope of 110/11. The time taken to come to rest, t_5, and the dist-ance traversed, s_5, are t_5= [(110 ft/s)/(11 ft/s)] = 10 s and s_5 = (1/2) × 10s × 110 ft/s = 550 ft. = (29,601/332) ft/s = 89.16 ft/s × (60 mph)/(88 ft/s) = 60.8 mph.

Question:

Upon the addition of heat to an unknown amount of KCLO_3, .96 g of oxygen was liberated. How much KCLO_3 was present?

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Solution:

The key to answering this question is to write a balanced equation that illustrates this chemical re-action. From this, you can employ the mole concept to determine the weights of the substances involved. Given that oxygen is liberated and that oxygen gas exists as O^2 and not O, you can write the balanced equation 2KCLO_3 \rightarrow 2KCL + 3O_2. All atoms are accounted for. The coefficients indicate the relative number of moles that react. For example, every two moles of KCLO_3 yield 3 moles of oxygen. A mole is defined as weight in grams of a substance divided by its atomic or molecular weight. Therefore, you have [(.96 g) / (3 g/mole)] = .030 moles of oxygen. To calculate the weight of KCLO_3, you look at the balanced equation. You find that the number of moles of KCLO_3 is 2/3 the number of moles of oxygen. As such, the number of moles [(KCLO_3) / (3 moles O_2)] = .20 moles of KC1O_3. Recalling the definition of a mole, the number of grams of KC1O_3 (MW of KCLO_3 = 122.55) is .020 × (122.55 g/mole) = 2.5 g.

Question:

In the Bohr model of the hydrogen atom, the electron is considered to move in a circular orbit around the nuclear proton. The radius of the orbit is 0.53 × 10^-8 cm. What is the velocity of the electron in this orbit?

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Solution:

If we assume that the only force acting is the electro-static force, F_E = -[(q_1q_2)/r^2] = (e^2/r^3) = [(4.80 × 10^-10 statC)^2 / (0.53 × 10^-8 cm)^2 = 8.2 × 10^-3 dyne In order to maintain a circular orbit, the electron must be subject to a centripetal acceleration: a_c = (v^2/r). The mass of the electron times this acceleration must equal the force on the electron: m_ea_c = [(m_ev^2) / r] = F_E or, solving for v, v = \surd[(rF_E) / (m_e)] = \surd[(0.53 ×10-8cm × 8.2 × 10^-3 dyne)] / (9.11 × 10^-28 g) = 2.18 × 10^-8 cm/sec Hence, the electron velocity is about 1 percent of the velocity of light.

(a) Isolate the cylinder, and indicate forces acting on it. There is no need to tabulate x- and y-components since they are all up or down forces. Set\sumF = ma Where \sumF is the net force acting on the cylinder, a is the acceleration of the cylinder's center of mass, and m is its mass. (Note that in problems such as this one, it is convenient to take the direction of motion as positive.) mg - T = ma\therefore a = g - (T.m)(1) Now, consider rotation. Consider torques about the center of mass 0, and set \sumL = l\alpha, where \sumL is the net torque about 0, I is the moment of inertia at the cylinder about 0, and \alpha is its angular acceleration. Hence \sumL = rT = l\alpha \thereforeT = I(\alpha/r) (Clockwise rotation corresponds to downward motion, already assumed to have the positive direction.) But I for a cylinder = 1/2 mr^2 , where r is the cylinder radius. In this problema = r\alpha T = (1/2)[(mr^2)/(r)] [a/r] = [ma/2] whereupon, using (1), a = [(-ma)/(2m)] + g = [(-a)/2] + g \therefore3a = 2g a = (2/3)g = 21.3 ft/sec^2(downward) (b) Now since a = constant (2/3 g), the linear motion is uniformly accelerated, such that the ve-locity of the center of mass is v^2 = v^2_0 + 2as,where s becomes the drop, h, that the cylinder experiences, and v_0 is its initial velocity. v^2 = 0 + (2) (21.3 ft/sec^2) (6 ft) v^2 = (12) (21.3 ft^2/sec^2) \thereforev = \surd256 = 16 ft/sec (c) The angular velocity \omega = v/r = (16 ft/sec)/[(1/2) ft] = 32 rad/sec (d) From (1) T = m(g - a) = m[g - (1/2) g] T = (1/3) mg = (1/3)(3 lb) = 1 lb

The human sex act is generally divided into four phases. These are the excitement, plateau, orgasm and resolution phases. The excitement phase in the female is characterized by a rise in heart rate and blood pressure. Muscle tension also increases as the breasts swell and the genitalia become filled with blood. A lubricating fluid is secreted during this phase by the vaginal wall. In the male, the excitement phase is also characterized by increased blood pressure, heart rate and muscle tension. The penis becomes rigid as the erectile tissue becomes engorged with blood. The plateau phase is characterized by increased breathing rate, pulse rate and muscle tension in male and female. The orgasm phase is characterized by muscular contrac-tions of the vagina in the female and by the muscles in the genital area of the male. The orgasm phase is initiated by frictional force against the vaginal wall and clitoris in the female. In the male, orgasm is caused by pressure and friction against the nerve endings in the glans penis and in the skin of the shaft of the penis. The resolution phase is the final stage in the sex act. During this period body functions return to normal levels. Heart beat, pulse rate and breathing levels return to the precoital state.

Question:

In a certain engine, fuel is burned and the resulting heat is used to produce steam which is then directed against the vanes of a turbine, causing it to rotate. What is the efficiency of the heat engine if the temperature of the steam striking the vanes is 400\textdegree K and the temperature of the steam as it leaves the engine is 373\textdegree K?

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Solution:

The efficiency E of a heat engine is the ratio of the net work W done by the engine in one cycle to the heat Q, absorbed from the high temperature source in one cycle. E = W/Q1 For the Carnot cycle, which describes the operation of a reversible heat engine, we know the efficiency to be, E = W/Q_1 = (Q_1 - Q_2)/Q_1 = (T_1 - T_2)/T1 where T_2 is the low temperature reservoir. Carnot stated that the efficiency of all Carnot engines operating be-tween the same two temperatures is the same and that no irreversible engine working between these two temperatures can have a greater efficiency. This means that the maxi-mum efficiency of this heat engine is given by, \varepsilon = (T_1 - T_2)/T_1 = (400\textdegreeK - 373\textdegreeK)/400\textdegreeK = 0.068 = 6.8%.

The human brain is the enlarged, anterior end of the spinal cord. This enlargement is so great that the resemblance to the spinal cord is obscured. The adult human brain has six major regions: the medulla, pons and cerebellum which constitute the hindbrain; the thalamus and cerebrum; both of which are in the forebrain; and the midbrain. The most inferior part of the brain, connected immediately to the spinal cord, is the medulla. Here the central canal of the spinal cord (spinal lumen) enlarges to form a fluid-filled cavity called the fourth ventricle. The medulla has numerous nerve tracts (bundles of nerves) which bring impulses to and from the brain. The medulla also contains a number of clusters of nerve cell bodies, known as nerve centers. These reflex centers control respiration, heart rate, the dilation and constriction of blood vessels, swallowing and vomiting. Above the medulla is the cerebellum, which is made up of a central part and two hemispheres extending side-ways. The size of the cerebellum in different animals is roughly correlated with the amount of their muscular activity. It regulates and coordinates muscle contraction and is relatively large in active animals such as birds. Removal or injury of the cerebellum is accompanied not by paralysis of the muscles but by impairment of muscle co-ordination. A bird, with its cerebellum surgically removed, is unable to fly and its wings are seen to thrash about without coordination. The pons is an area of the hindbrain containing a large number of nerve fibers which pass through it and make connections between the two hemispheres of the cerebellum, thus coordinating muscle movements on the two sides of the body. The pons also contains the nerve centers that aid in the regulation of breathing. In front of the cerebellum and pons lies the thick- walled midbrain. The midbrain is an important integrating region and contains the centers for certain visual and auditory reflexes. A cluster of nerve cells regulating muscle tone and posture is also present in the midbrain. A small canal runs through the midbrain and connects the fourth ventricle behind it to the third ventricle in front of it. The midbrain, pons and medulla are collectively called the brainstem. Ten of the twelve cranial nerves originate in the brainstem. The thalamus of the forebrain serves as a relay center for sensory impulses. Fibers from the spinal cord and parts of the brain synapse here with other neurons going to the various sensory areas of the cerebrum. The thalamus seems to regulate and coordinate the external signs of emotions. By stimulating the thalamus with an electrode, a sham rage can be elicited in a cat - the hair stands on end, the claws protrude, and the back becomes humped. However, as soon as the stimulation ceases, the rage responses disappear. The hypothalamus, located under the thalamus, is a collection of nuclei (a cluster of nerve cell bodies located within the central nervous system), concerned with many im-portant homeostatic regulations. Electrical stimulation of certain cells in the hypothalamus produces sensations of hunger, thirst, pain, pleasure, or sexual drive. The hypothalamus is also important for its influence on the pitui-tary gland, which is functionally under its control. Cells of the hypothalamus synthesize chemical factors that modu-late the release of hormones produced and stored in the an-terior pituitary. The hypothalamus has neural connections with the posterior pituitary. The hypothalamus and thalamus are collectively called the diencehpalon. The cerebrum, consisting of two hemispheres, is the largest and most anterior part of the human brain. In human beings, the cerebral hemispheres grow back over the rest of the brain, hiding it from view. Each hemisphere contains one cavity (one contains the first and the other the second ventricle, collectively known as the lateral ven-tricles), which is connected to the third ventricle. The outer portion of the cerebrum, the cortex, is made up of gray matter which comprises the nerve cell bodies. The grey matter folds greatly, producing many convolutions of the cerebral surface. These convolutions increase the surface area of the gray matter. The inner part of the brain is the white matter which is composed of masses of nerve fibers. Each cerebral hemisphere is divided into four lobes. The occipital or posterior lobe receives visual information. The temporal lobe, located on the side of the cerebrum just above the ears, receives auditory information. The frontal and parietal lobes are demarcated by the fissure of Rolando. Just in front of this fissure in the frontal lobe is the primary motor area, while the primary sensory area lies be- hind it in the parietal lobe. When all the areas of known functions are plotted, they cover only a small part of the total area of the human cortex. The rest, known as association, areas, are regions responsible for the higher intellectual faculties of memory, reasoning, learning and imagination, all of which help to make up one's personality. In some unknown way, the association regions integrate into a meaningful unit, all the diverse impulses constantly reaching the brain, so that proper response is made.

Question:

Sulfur exists as S_2 vapor at temperatures between 700\textdegreeC and 1500\textdegreeC. At 1473 k it combines with hydrogen according to the equation H_2 (g) + (1/2)S_2 (g) \rightarrow H_2 S (g) . At 750\textdegreeC the equilibrium constant is 1.07 × 10^2 and at 1200\textdegreeC it is 4.39. Determine the heat of reaction in the temperature range 750\textdegreeC to 1200\textdegreeC, and the change in free energy at each of these temperatures.

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Solution:

The change in free energy ∆G is determined from the equation ∆G\textdegree = - 2.303 RT log k, where R is the gas constant, T the absolute temperature, and k the equilibrium constant. From this equation and the expression ∆G\textdegree = ∆H\textdegree - T∆S\textdegree, where ∆H\textdegree is the change in the heat of reaction and ∆S\textdegree the change in entropy, a relation between ∆H and T will be found. At T = 750\textdegreeC = 1023\textdegreeK,k = 1.07 × 10^2and ∆G\textdegree = - 2.303 RT log k = - 2.303 × 1.987 [(cal)/(mole-deg)] × 1023\textdegreeK × log (1.07 × 10^2) = - 9500 cal/mole = - 9.5 Kcal/mole. At T = 1200\textdegreeC = 1473\textdegreeK,k = 4.39and ∆G\textdegree = - 2.303 RT log k = - 2.303 × 1.987 [(cal)/(mole-deg)] × 1473\textdegreeK × log (4.93) = - 4330 cal/mole = - 4.33 Kcal/mole. To find an expression relating ∆H to T, consider the following two equations: ∆G\textdegree = - 2.303 RT log k,and∆G\textdegree = ∆H\textdegree - T∆S\textdegree. Setting these equal gives - 2.303 RT log k = ∆H\textdegree - T∆S\textdegree, or,log k = [(- ∆H\textdegree)/(2.303 RT)] + [(∆S\textdegree)/(2.303 R)]. For two different temperatures, T_1 and T_2 with equi-librium constants k_1 and k_2, respectively, log k_1 = [(- ∆H\textdegree)/(2.303 RT_1)] + [(∆S\textdegree)/(2.303 R)]and log k_2 = [(- ∆H\textdegree)/(2.303 RT_2)] + [(∆S\textdegree)/(2.303 R)]. Subtracting the first equation from the second, log k_2 - log k_1 = [{(- ∆H\textdegree)/(2.303 RT_2)} + {(∆S\textdegree)/(2.303 R)}] - [{(- ∆H\textdegree)/(2.303 RT_1)} + {(∆S\textdegree)/(2.303 R)}], or,log (k_2 /k_1) = [(- ∆H\textdegree)/(2.303 R)] [(1/T_2) - (1/T_1)]. Let T_1 = 1023\textdegreeK, k_1 = 1.07 × 10^2, T_2 = 1473\textdegreeK,andk_2 = 4.39. Then log (k_2 /k_1) = [(- ∆H\textdegree)/(2.303 R)] [(1/T_2) - (1/T_1)] log [(4.39)/(1.07 × 10^2)] = [(- ∆H\textdegree)/(2.303 × 1.987 cal/mole-deg)] [(1/1473\textdegreeK) - (1/1023\textdegreeK)] log [(4.39)/(1.07 × 10^2)] = [(- ∆H\textdegree)/(2.303 × 1.987 cal/mole-deg)] (0.000679\textdegreek^-1 - 0.000978\textdegreek^-1) or, ∆H\textdegree = [{log (4.39/107)}/(2.303 × 1.987 cal/mole-deg)] (0.000978\textdegreek^-1 -0.000679\textdegreek^-1) = - 21, 300 cal/mole = - 21.3 Kcal/mole. Thus, the heat of reaction in the temperature range 1023\textdegreeK to 1473\textdegreeK (750\textdegreeC to 1200\textdegreeC) is - 21.3 Kcal/mole.

Question:

Contrast the development of an amphibian, bird, and starfish egg up to the point of gastrulation.

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Solution:

In the isolecithal starfish egg, cleavage is radial (Figure 1). The first cleavage division passes through the long axis of the egg, splitting the egg into two equal daughter cells. The second cleavage plane is also vertical but at right angles to the first, separating the two cells into four. The third division is horizontal, at right angles to the other two, and splits the four cells into eight. Further divisions result in embryos containing 16, 32, 64, 128 cells, and so on, until a hollow ball of cells called the blastula, is formed. The wall of the blastula is a single layer of cells, the blastoderm, surrounding the blastocoel, the cavity in the center. It is this single-layered blastula which is converted later into a double-layered sphere, the gastrula. Amphibians also display radial cleavage but because of the unequal distribution of yolk in the telolecithal egg, the resulting blastomeres are not of equal size (Figure 2). The first cleavage begins as a shallow furrow at the animal pole and progresses gradually through to the vegetal pole. This first cleavage usually results in two equal blastomeres. The second cleavage is at right angles to the first. It begins at the animal pole even before the first cleavage has been completed at the vegetal pole. The result is four blastomeres of nearly equal size. The third cleavage is typically at right angles to the first two cleavages, and hence cuts horizontally across the egg's vertical axis. It passes well above the equator, so that the eight-celled stage commonly consists of four smaller animal blastomeres (micromeres), and four larger, yolk-laden vegetal blas-tomeres (macromeres). A cavity is present at the center of the group of blastomeres beginning at the eight-celled or 16-celled stage, which increases in size as cleavage progresses. The egg thud becomes a hollow ball of cells, namely, a blastula. Its blastocoel is roughly hemispherical. It has a dome of smaller animal cells and a floor of larger yolk-laden vege-tal cells. Telolecithal bird eggs also display radial cleav-age, but the cleavage is meroblastic. In bird eggs and other eggs which contain a large amount of yolk, cleavage occurs only in a small disk of cytoplasm in the animal pole (Figure 3). At first, all cleavage planes are vertical and all the blastomeres lie in a single plane. The cleavage furrows separate the blastomeres from each other but not from the yolk. The central blastomeres are continuous with the yolk at their lower ends, and the blastomeres at the circumference of the disk are contin-uous both with the yolk beneath them and with the uncleaved cytoplasm at their outer edge. As cleavage continues, more cells become cut off to join the ones in the center, but the new blastomeres are also continuous with the uncleaved underlying yolk. The central blastomeres eventually become separated from the underlying yolk either by cell divisions with horizontal cleavage planes or by the appearance of slits beneath the nucleated portions of the cells. Horizontal cleavages separate upper blastomeres and lower blastomeres. The upper blastomere is a cell with a complete plasma membrane, which is separated from its neighbors and from the yolk. The lower blastomere is a cell which remains connected with the yolk. The blastomeres at the margin of the disk, and the lower cells in contact with yolk, eventually lose the furrows that partially separated them and fuse into a continuous syncytium. This syncytium con-tains many nuclei and is termed the periblast. The periblast is believed to break down the yolk, thereby making its nutrients available for the growing embryo. The free blastomeres (with complete plasma membranes) become incorporated into two layers, an upper epiblast, and a lower hypoblast. Between these two layers is a cavity, the blastocoel. Below the hypoblast and above the yolk is the shallow subgerminal cavity, which appears only under the central portion of the blastoderm. The area of the blastoderm over the subgerminal space is relatively transparent (due to lack of yolk) and is called the area pellucida, whereas the more opaque part of the blastoderm (which contains some yolk) that rests directly on the yolk is called the area opaca.

In a circular orbit of radius r an earth satellite of mass m has a velocity v. The distance, d, the satellite moves in one revolution equals 2\pir. Its period T is the time it takes for the satellite to make one revolution. Therefore T = d/v = 2\pir/v(1) The centripetal force necessary to keep the satellite moving in a circle is supplied by the gravi-tational force exerted by the earth. By Newton's Second Law, G(M_Em/r^2) = mv^2/r where G is the gravitational constant and M_E is the mass of the earth. From equation (1), V^2/r^2 = (2\pi/T)^2 = 4\pi^2/T^2 andG(M_Em/r^2) = 4\pi2mr/T^2(2) The same arguments apply to the moon of mass M_M which moves in a circle of radius R with period T_M. G(M_EM_M/R^2) = 4\pi^2M_MR/T^2_M ;(3) Solving for GM_E in equations (2) and (3), we have respectively GM_E = 4\pi^2r^3/T^2 GM_E = 4\pi^2R^3/T^2_M Equating the above two expressions, 4\pi^2r^3/T^2 = 4\pi^2R^3T_M^2 orT^2/T^2_M = r^3R^3 Therefore, the period of the satellite can be found by substitution of the numerical values given. Using r = 22,300 mi + R_E = 22,300 mi + 4000 mi = 26,300 mi the period is T = T_M\surd(r^3/R^3) = 27.3 days\surd[{(2.630 × 10^4)^3 mi^3}/{2.39 × 10^5)^3 mi^3}] = 1.00 day. Such a satellite therefore rotates about the center of the earth with the same period as the earth rotates about its axis. In other words, if it is rotating in the equatorial plane, it is always vertically above the same point on the earth's surface.

Question:

To measure magnetic fields, rotating coils are often used. A cir-cular coil of radius 1 cm and with 100 turns is rotated at 60 cps in a magnetic field. The emf induced in the coil has a maximum value of 12.3 volts. Calculate the Intensity of the field.

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Solution:

The magnetic flux (\varphi) through one coil is defined as follows: \varphi_m = \int_A B^\ding{217} \bullet dA^\ding{217} = \int_A B cos \texttheta dA = B cos \texttheta \int_\texttheta dA = BA cos \texttheta B is a constant in this problem, and \texttheta is a function of time, so they were taken out of the integral. \texttheta is the angle between the plane of the coil and the direction of B^\ding{217}, and A is the area of the loop, (= \piR^2 ). Since the coil is rotating with a constant angular velocity \omega, \texttheta is related to time by the following relation; \texttheta = \omegat Therefore, the flux through N coils is, N times the flux through one coil, \varphi_m = NBA cos \omegat According to Faraday's law of induction, a change in flux through a coil with N turns will induce an EMF in the coil. The induced EMF is \epsilon = (d\varphi_m /dt) The negative sign signifies the fact that the EMF is induced in such a direction as to oppose the change in flux that created it. The EMF Induced in the coil is then \epsilon = \omega NBA sin \omegat The sine function has values ranging from +1 to -1. \epsilon will then attain its maximum value when sin \omegat = \pm 1 Then \vert\epsilon_max \vert = \vert\omega NBA \vert or, since \omega = 2\pin where n is the number of revolutions per second, \verte_max \vert = 2\pin NBA = 2\pin NB(\piR^2 ) orB = [(\epsilon_max )/{(2\pinN)/(\piR^2 )}] = [(12.3 volts)/(2\pi × 60 sec^-1 × 100 × \pi × (.01m)^2 ] = 1.04 [(volt-sec)/(m^2 )] = 1.04 (Weber/m^2 ) The unit of length of the radius, R, was changed to meters to be con-sistent with the MKS system being used.

Because of the rotation of the earth the body at the equator moves in a circle whose radius is equal to the radius of the earth (see figure). r = radius of earth = 6.37 × 10^6 meters We are going to use the MKS System of units. One revolution, which is 2\pi radians, takes 1 day or [(24 hr)/(1 day)] × [(60 min)/(1 hr)] × [(60 sec)/(1 min)] [(24 hr)/(1 day)] × [(60 min)/(1 hr)] × [(60 sec)/(1 min)] = 24 × 60 ×60 secs/day = 24 × 60 ×60 secs/day (a) Since the frequency of revolution is f = 1/T, where T is the period (the time for one revolution), then 2\pif = \omega = 2\pi/T . This equals the number of radians traveled per unit time, or the ang-ular velocity \omega. \omega = 2\pi/(24 × 60 × 60) = 7.27 × 10^-5 radians per second (b) The linear velocity is, by definition v= \omegar = (7.27 × 10^-5) × (6.37 × 10^6)(red/s)\bulletm = 4.64 × 10^2 in/sec . Since 1 mph = 0.447 m/sec v= [4.64 × 10^2(m/sec)] [{1/(.447)}(sec/m)] = 1040 mph (c) The acceleration toward the center of the earth is, since the motion is circular, a= v^2/r = (4.64 × 10^2 m/sec)^2/(6.38 × 10^6 m) = 3.37 × 10^-2 m/sec^2

Question:

A man hangs from the midpoint of a rope 1 m long, the ends of which are tied to two light rings which are free to move on a horizontal rod (see the figure). What is the maximum possible separation d of the rings when the man is hanging in equilibrium, if the relevant co-efficient of static friction is 0.35?

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Solution:

Since the man hangs from the midpoint of the rope, by symmetry the tensions in the two portions of the rope must be equal and have magnitude T, and each portion will be inclined at the same angle \texttheta to the vertical. Thus the system of forces acting on each ring will be the same. Now consider one of the rings. Three forces are acting on it: the tensional pull on the ring due to the rope, the normal force exerted upward by the rod, and the frictional force attempting to prevent motion of the ring toward its fellow. Since the ring is light, its weight may be ignored. If the ring is too far out, slipping will occur. At the maximum distance apart, each ring is just on the point of slipping. Hence F = \mu_sN. When we resolve T\ding{217} into its horizontal and vertical components, the equations for equilibrium become \sum perpendicular force= N - T cos \texttheta = 0 \sum parallel forces= T sin 0 - F = 0 where we take the positive perpendicular direction as pointing upward and the positive parallel direction as pointing to the right. Then: N= T cos \textthetaandF = \mu_sN = T sin \texttheta. \mu_s= (\mu_sN)/(N) = (T sin\texttheta)/(T cos \texttheta) = tan \texttheta = 0.35 or\texttheta = 19.6\textdegree Finally, we solve for d: sin \texttheta= sin 19.6\textdegree =(d/2)/([(1/2)m] = d m^-1 0.33= d m^-1 d = 0.33 m which is the maximum separation permissible. Note that \texttheta and d do not depend on T and therefore the ring separation is not dependent on what it is that is hanging from the midpoint of the rope.

If the boat were pointed directly at the opposite bank, then during the crossing it would drift downstream and it would not reach the other bank at a point directly opposite the starting point. It must therefore be pointed in a direction tilted in the up-stream direction as shown in the figure. As illustrated in the vector diagram PQR, the result of adding the velocity of the boat relative to the water to the velocity of the water must be a resultant velocity v\ding{217} pointing directly toward the opposite bank. We can-not draw this triangle of vectors immediately be-cause we do not know the angle \texttheta between the direction of motion and the direction straight across the stream. However, inspecting the triangle PQR and remembering that in trigonometry the since of the angle \texttheta is defined as sin \texttheta = (QR/PQ) = 5/10 = 0.5 we refer to table of since and find that the angle whose sine is 0.5 is 30\textdegree. The boat must therefore be pointed upstream at an angle of 30\textdegree from the direction perpendicular to the bank. Applying Pythagoras' theorem to the triangle PQR PQ^2 = QR^2 + PR^2 orPR^2 = PQ^2 - QR^2 that is;v^2 = 10^2 - 5^2 = 75 v = \surd75 = 8.66 m/sec The boat therefore crosses the river at a speed of 8.66 m/sec. Since the distance across the river is 400 m, the time taken is, since v = constant, t = d/v = (400 m)/(8.66 m/sec) t = 46.2 sec.

Question:

The hard shell of crustaceans (lobsters, etc,) and insects (roaches, etc.) is a polysaccharide called chitin. On enzymatic hydrolysis of chitin, N-acetylglucosamine is obtained. This molecule resembles glucose except that at C\rule{1em}{1pt}2 a is attached instead of -OH. (a) Write an open chain formula for N-acetylglucos amine. (b) The structure of chitin is analogous to that of cellulose. Draw a formula containing two joined N-acetylglucosamine units, (c) If the molecular weight of chitin is 150,000, how many units are in the polymer?

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Solution:

A polysaccharide is a chain made up of many simple sugars. Chitin is a long unbranched chain of many molecules of the sugar, N-acetylglucosamine. (a)One is given that the only difference between glucose and N-acetylglucosamine is that agroup replaces the \rule{1em}{1pt}OH on C\rule{1em}{1pt}2. The open-chain formula of N- acetylglucosamine can therefore be written as shown in Figure A (b) From the open chain formula, one can see that the ring structure can be written as shown in Figure B for N-acetylglucosamine. For the ring structure to form, the O in the \rule{1em}{1pt} OH on C\rule{1em}{1pt}5 joins C\rule{1em}{1pt}l. From the figure of cellulose, Figure C, one can see that the structure of two N-acetylglucosamine joined to-gether can be written as shown in Figure D. Because the O in the -OH on C-l joins to a C-4 atom on the adjacent molecule and the ring structures are called glycosides, this is termed a 1,4- glycosidic linkage. (c) To determine the number of units in one strand of chitin, one can divide the total weight of the molecule by the weight of one glycoside. From the ring structure of a glycoside joined to another glycoside, one sees that there are 8c's, 5 0's, 13 H's and 1 N. The weight of one mole of this glycoside is (8 × 12) + (5 × 16) + (13 × 1) + (14) = 203 (g / mole). No. of units= (150,000 g) / (203 g/mole) (unit) = 739 units.

Question:

A gallon of gasoline will deliver about 110,000 Btu when burned. To how many foot-pounds is this equivalent?

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Solution:

1 Btu = 778 foot-pounds 110,000 Btu = 778[(ft-Ib)/Btu] × 110,000 Btu = 85,580,000 foot-pounds.

Question:

An electron, charged - 1.6 × 10^-19 coul, moves at some instant of time in the +x-direction with velocity v = 0.8 c. A magnetic field B = 10 W/m^2 is present in the +y-direction. What is the direction and magnitude of the magnetic force?

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Solution:

To find the magnetic force acting on a charged particle moving through a magnetic field, one uses the formula F_m^\ding{217} = qv^\ding{217} × B^\ding{217} where F_m^\ding{217} is the resulting magnetic force, v^\ding{217} is the velocity of the particle and B^\ding{217} is the magnetic field vector. In this problem v^\ding{217} is 0.8 c = 0.8 × 3 × 10^8 m/sec in the +x-direction, B^\ding{217} is 10 W/m^2 in the +y-direction, and q is e ־ = - 1.6 × 10^-19 coul. So: F_m^\ding{217} = (- 1.6 × 10^-19 coul.) (0.8) (3 × 10^8 m/sec) \^{\i} × (10 W/m^2 ) \^{\j} = - 3.84 × 10^-10[(coul. - m - W)/(m^2 - sec)] (\^{\i} × \^{\j}) = - 3.84 × 10^-10 N K^^ where we used the fact that \^{\i} × \^{\j} = K^^,\^{\i},\^{\j}, and K^^ being the unit vectors in the +x, +y, and +z-directions, re-spectively.

(a)R = (V/I) = [(6.0 volts)/(1.5 amp)] = 4.0 ohms (b)I= (V/R) = [(8.0 volts)/(4.0 ohms) = 2.0 amp . In part a of this solution we have used merely the de-finition of resistance. But in part b we have used Ohm's law, that is, the fact that R is constant.

Question:

Explain the locomotory pattern of the earthworm.

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Solution:

In the earthworm, two layers of muscle in the body wall and the four pairs of short bristles or chaetae of each body segment are the structures involved in locomotion. Circular muscle occupies the outer layer of the body wall, and just beneath it is the longitudinal muscle layer. Contractions of these two sets of muscles account for the changes in shape of the worm. Each segment of the body is filled with coelomic fluid that is in compressible. Thus the contraction of either layer of muscle results in a change of shape. Contractions of the longitudinal muscles make the worm shorter and thicker, while contractions of the circular muscles make it thinner and correspondingly longer (see Figure). Circular muscle contraction is most important in crawling, and always generates a pressure pulse in the coelomic fluid. Longi-tudinal muscle contraction is more important in burrowing. When the worm is crawling, longitudinal contraction may not be strong enough to generate a coelomic pressure pulse, but it does so in burrowing. Chaetae are extended during longitudinal muscle contraction, functioning to anchor the worm in its burrow and to determine the direction of locomotion. They are retracted during circular muscle contraction. Because of the combined effects of the chaetae and longitudinal muscle contraction, each segment of the body moves in steps of 2 to 3 cm, at the rate of seven to ten steps per minute. The direction of contraction can be reversed, thus enabling the worm to reverse directions. It should be remembered that without the chaetae, no forward move-ment can take place, but only a change in shape.

Question:

Give examples of the following control constructs as they would be rendered in FORTRAN: a) IF-THEN-ELSE b) DO-WHILE c) REPEAT-UNTIL d) DO-FOR

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Question:

A 5.0 - lb ball is fastened to the end of a flat spring. A force of 2.0 lb is sufficient to pull the ball 6.0 in. to one side. Find the force constant and the period of vibration.

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Solution:

The restoring force on the ball as it is displaced from the equilibrium point 0 (see the figure) is, by Hooke's law, F\ding{217}= -kx\ding{217} \ding{217} \ding{217} where x is the displacement and k is the spring constant. Hence, k = \vertF\ding{217}\vert / \vertx\ding{217}\vert \ding{217} \ding{217} = 2.0 lb/0.50 ft = 4.0 lb/ft The mass of the ball is m = W/g where W is the ball's weight. Hence, m = (5.0 lb)/(32 ft/sec^2) = 0.16 slug The period of oscillation is given by T = 2\pi\surd(m/k) = 2\pi\surd[(0.16 slug) / (4.0 1b/ft) = 1.2 sec.

In order to use the conditions of equilibrium to compute an unknown force, we must consider some body which is in equilibrium and on which the desired force acts. The hanging block is one such body and the tension in the vertical cord supporting the block is equal to the weight of the block. The inclined cords do not exert forces on the block, but they do act on the knot at 0. Hence, we consider the knot as a small body in equilibrium, whose own weight is negligible. The free body diagrams for the block and the knot are shown in figure B, where T_1\ding{217}, T_2\ding{217}, and T_3\ding{217} represent the forces exerted on the knot by the three cords and T'_1\ding{217}, T'_2\ding{217} , and T'_3\ding{217} are the reactions to these forces. Consider first the hanging block. Since it is in equilibrium, T'_1 = w = 50 lb Since T_1\ding{217} and T'_1\ding{217} form an action-reaction pair, T'_1 = T_1 HenceT_1= 50 lb. To find the forces T_2\ding{217} and T_3\ding{217}, we resolve these forces (see fig. C) into rectangular components. Then, from Newton's second law, \sumF_x= T_2 cos \texttheta_2 - T_3 cos \texttheta_3 = 0, \sumF_y= T_2 sin \texttheta_2 + T_3 sin \texttheta_3 - T_1 = 0 We have T_2 cos 30\textdegree - T_3 cos 60\textdegree = 0 T_2 sin 30\textdegree + T_3 sin 60\textdegree = 50 or0.866 T_2 - 0.500 T_3 = 0 0.500 T_2 + 0.866 T_3 = 0 Solving these equations simultaneously, we find the tensions to be T_2 = 25 lb,T_3 = 43.3 lb. Finally, we know from Newton's third law that the inclined cords exert on the ceiling the forces T'_2 and T'_3\ding{217}, equal and opposite to T_2\ding{217} and T_3\ding{217}, respectively.

The recruit's path is shown in the figure, where each division on the graph represents one mile. We find R\ding{217} by first adding the components of his individ-ual displacements which we regard as vectors. We will let E\ding{217} and N\ding{217}, representing east and north, be our unit vectors, regarding western and southern displacements as being negative eastern and negative northern dis-placements, respectively. Assume north and east are given equal weights. Then N\ding{217}E\ding{217} is as shown in the diagram. Thus, the sum of the components is: E\ding{217} N\ding{217} - 5 mi 4 cos 45\textdegree mi 0 mi 0 mi 4 sin 45\textdegree mi 3 mi (4 cos 45\textdegree - 5)mi (4 sin 45\textdegree + 3)mi R\ding{217} = [ 4 (1/\surd2) - 5 ] mi E\ding{217} + [ 4 (1/\surd2) + 3 ] N\ding{217} = (2.8 - 5)mi E\ding{217} + (2.8 + 3)mi N\ding{217} = - 2.2 mi E\ding{217} + 5.8 mi N\ding{217} The recruit's final distance from the starting point will be the magnitude of R\ding{217}: R = \surd{(- 2.2 mi)^2 + (5.8 mi)^2} = 6.20 mi

a) The block diagram of a clocked, master-slave SC type flip-flop is shown in figure 1. The master-slave flip-flop is a combination of two flip-flops, the master getting the direct clock pulse and the slave get-ting the inverted clock pulse. The input condition is that the Boolean product S\bulletC = 0. b) The timing diagram is shown in figure 2. Timings for four clock pulses t_1, t_2, t_3 and t_4 are shown. We assume that initially, before t_1 arrives, x and y are each = 1. Note that x and x' change state, whenever required to do so, at the leading edge of the clock pulse, but y and y' change state at the trailing edge of the clock pulse due to the inverter. c) The Karnaugh Map of the master-slave SC flip-flop is shown in figure 3. The Karnaugh Map in figure 3 is for y\rightarrow, where y\rightarrowdenotes the next state of \rightarrow \rightarrow y. Note also that as the input with both S=1 and C=1 is not allowed, and hence does not ever' occur, we insert a 'd' in the squares for S, C = 1,1. The 'd' repre-sents a "don't care" condition. d) The Karnaugh Map leads to the formulation of the exci-tation equation, i.e., the equation for y\rightarrowin terms of S, C and y. \rightarrow \therefore y\rightarrow= Sy' + C'y . \rightarrow e) The Karnaugh helps us to draw the state table, Figure 4, SCy y\rightarrow \rightarrow REMARKS 000 001 0 1 NO CHANGE 010 011 0 0 CLEARS 100 101 1 1 SETS 110 111 d d NOT ALLOWED Fig. 4 The state diagram is drawn in Figure 5: Note in the above that the input condition 11 does not occur at all.

The equation of heat conduction is dQ = - KA (dT/dx)(1) wheredQ/dtis the rate at which heat is transferred across a cross-section A of a material with coefficient of thermal conductivity K.dT/dxis the temperature gradient in the material. In the steady state, the temperature at each point of the material remains constant in time. Hence, the rate of heat transfer across a cross-section is the same at all cross-sections. As a result of (1),dT/dx must be the same at all cross-sections. If T_1 is the temperature at a cross-section at x_1,and T_2 is the temperature at x_2, we obtain dT /dx= ∆T/∆x = (T_2 - T_1)(x_2 - x_1)(2) (Note that this is a direct consequence of the fact thatdT/dxis constant). Using (2) in (1) dQ/dt= -K A [(T_2 - T_1)(x_2 - x_1)] = K A [(T_1 - T_2)(x_2 - x_1)] But x_2 - x_1 is equal to L, the length of the material across which the heat conduction is taking place. dQ/dt= [K A (T_1 - T_2)]/L For the pan of glass, T_1 = 70\textdegreeF@x_1 = 0 mm T_2 = 0\textdegreeF@x_2 = 2 mm Furthermore, K = .0015 cal/cm\bullets\bullet\textdegreeC for glass, whence dQ /dt= [(.0015 cal/cm\bullets\bullet\textdegreeC)(2500 cm^2)(70\textdegreeF - 0\textdegreeF)]/(2mm - 0 mm) Since 70\textdegreeF = 5/9 \textbullet 70\textdegreeC = 350\textdegreeC/9, we obtain, dQ /dt= [(.0015 cal/cm\bullets\bullet\textdegreeC)(2500 cm^2)(350\textdegreeC/9)]/(.2 cm) dQ /dt= 729 cal/s Note that, by convention, temperature decreases as x increases. Hence, T_2 < T_1 and x_2 > x_1. As a result,dQ/dt> 0 in the direction in which dT /dx< 0.

Question:

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Solution:

To predict the products, you need to identify the type of reaction that will take place and its mechanism. The first two reactions, (a) and (b), involve addition across a double bond, and the last reaction, (c) concerns the dehalogenation of an alkyl halide using the base, KOH. In other words, it is an elimination reaction. The general mechanism of addition to a double bond is as follows: Note: From the mechanism, observe that the reaction goes through a species called a carbonium ion. The stability of a carbonium ion follows the order: 3\textdegree>2\textdegree>1\textdegree. This is shown schematically as A reaction will proceed so that the product formed in greatest amounts has gone through the mechanism that gives the most stable carbonium ion possible. Thus, you have You do not have to consider carbonium ion formation here, since the final product would be the same in either case, however, for (b), the carbonium ion must be con-sidered. There are two possible reactions: Two different products are obtained. Notice that in (1), the reaction proceeds through a 3\textdegree carbonium, whereas in (2) the reaction proceeds through a 1\textdegree carbonium ion. Because 3\textdegree is more stable than 1\textdegree, (c) In considering this part, remember that potassium hydroxide (KOH) eliminates the halogen from an alkyl halide to give alkenes, compounds containing double bonds. With this in mind, two possible reactions and products are poss-ible: The first product, greatly predominates over the second. The reason stems from the fact that alkenes are more stable when the double bond is surrounded by the maximum number of alkyl groups.

Question:

An aircraft wing requires a lift of 25.4 lb/ft^2. If the speed of flow of the air along the bottom surface of the wing is to be 500 ft/s, what must be the speed of flow over the top surface to give the required lift? The density of air is 2.54 × 10^-3 slug/ft^3.

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Solution:

Below an aircraft wing, there is little disturbance of the air flow. Because of the shape of the wing, the streamlines crowd together above it, effectively decreasing the cross-sectional area A for the air flow. The equation of continuity states that for an incompressible fluid (such as air), Av = constant. Therefore, the velocity v of the air above the wing must increase. Consider two points in the air flow at the same height, the first at a point to the left of the wing (see figure),where the flow has yet been disturbed and the second at a point above the wing. Bernoulli's equation states that P_1 + (1/2) \rhov^2_1 = P_2 + (1/2) \rhov^2_2 where the \textquotedbllefty\textquotedblright terms in the equation cancel since the two points are at the same elevation. From this equation, we see that since v_2 is greater than v_1, p_2 must be less than p_1. Since p_1 is also the pressure of the air under the wing, we see that the pressure is less above the wing than below it. This pressure differential gives rise to the lift on the wing. In Bernoulli's equation, let the subscript 1 refer to the lower surface and the subscript 2 to the upper surface of the wing. Thus the dynamic lift per unit area is P_1 = P_2 = 25.4 lb/ft^2 Solving for the speed of flow v_2 over the top of the wing in the first equation, we get v^2_2 = [2(p_1 - p_2)]/\rho+ v^2_1 = [(2)(25.4 lb/ft^2)]/[(2.54) (10^-3) slug/ft^3] + (25) (10^4)ft^2/sec^2 = [(2) (10^4) + (25) (10^4)] ft^2/sec^2 = (27) (10^4) ft^2/sec^2 v_2= 519.6 ft/sec.

(a) As the armature rotates between the poles F and F' of the magnet (see figure 1), an emf is induced in the loops C. According to Lenz's Law, this emf will always be in opposition to the emf applied to run the motor. This induced emf is called the "back emf" since its polarity is opposite to that of the line voltage (the voltage which runs the motor. See Figure 2). Kirchoff's Voltage Law states that the sum of the voltage drops around a circuit loop must be zero. Using this law along with Ohm's Law, \epsilon + IR - V_line = 0 Hence\epsilon = V_line - IR = 120 V - (2\Omega) (4A) = 112 V and opposesV_line , as shown in the diagram. (b)By definition, the power delivered to the motor is equal to the product of the line voltage and the current or P_s = IV_l = (4 amp) (120 volt) = 480 watts. (c)The power dissipated in the motor is P_d = l^2 R = (16 amp^2 ) (2 ohms) = 32 watts. (d)The mechanical power developed is the power supplied to the motor (P_s) less the power dissipated (P_d) P_m = P_s - P_d = 448 watts. The mechanical power developed may also be found from Mechanical Power = back emf × current = (112 v) × (4 amp) = 448 watts.

One must know two important things to answer this question: (1) The wave function expression for a one-electron atom for 3s, and (2) The fact that at a node, \psi or the wave function is zero. If \psi^2 is the probability of finding an electron in a given area and a node is an area of zero probability, then \psi must also be zero. The wave function expression relates \psi to radius r. The wave function for a 3s electron is \psi_300 = [(1)/{81\surd(\pi)] [Z/a_0]^3/2 [27 - {(18 Zr) / a_0} + {(2Z^2 r^2 )/(a^2_0)}] e^-Zr/3(a)0 For \psi300 to be equal to zero the factor [27 - {(18 Zr) /a_0} + {(2Z^2 r^2 )/(a2_0)}] must be equal to zero because the other factors in the equation can never be equal to zero. Using O = 27 - [(18 Zr)/(a_0)] + [(2Z^2 r^2)/(a2_0)] solve for r. O = 27 - [{(18 (11) (r)} / (.0529)] + [{(2(11)^2 (r)^2} / (.0529)^2] = 27 - 3.74 × 10^3 r + 8.67 × 10^4 r^2 Using the quadratic equation one finds that r is equal to 0.0092 or 0.034 nm.

(a) When the anemometer is moving at maximum speed, the cups are moving at 120 mph = 53.64 m \bullet s^-1 . The angular velocity of rotation of the cups is thus \omega = (v/r) [(53.64 m \bullet s^-1 )/(0.25 m)] = 214.6 rad \bullet s^-1 . The plane of the coil is vertical, and thus the vertical component of the earth's magnetic induction produces no effect during rotation, as shown in Fig. 1. Only the horizontal component is involved. The maximum emf induced during rotation of a coil in a magnetic field of induction B is \epsilon_max= n\omegaAB, where n is the number of turns of the coil and A is its area. In this case must not exceed 15 mV. Hence the number of turns on the coil is n = [(\epsilon_max )/(\omegaAB)] = (15 × 10^-3 V)/(214.6 s^-1 × 10^-2 m^2 × 1.5 × 10^-5 Wb \bullet m^-2) = 470 turns. (b) The rod which holds the cups is rotating at right angles to the vertical component of the earth's magnetic induction, and is therefore acting as a crude form of Faraday disk dynamo. (The horizontal component is in the plane of rotation and thus has no effect.) Let us consider an element, dy of the rod at a dis-tance y from the center of the rod as shown in Fig. 2. This element is moving with velocity perpendicular to the vertical component of the earth's magnetic field, he electric field Ey^\ding{217} set up in this element is, Ey^\ding{217} = V_y^\ding{217} × B_v^\ding{217} acting in a radial direction. Since V_y = \omegay, its magnitude is given by Ey = V_y B_v = y \omega B_v . The e.m.f. established between the center and either end of the rod has the magnitude \epsilon = ^a\int_ody Ey = \omegaB_v ^a\int_o dy y = (1/2) \omegaB_v a 2 = (1/2) × 214.6 s^-1 × 5.5 × 10^-5 wb \bullet m^-2 × (0.25)^2 m^2 = 3.69 × 10^-4 v.

Question:

Write a PL/I program that reads in pairs of points on a Car-tesian coordinate system and classifies them according to the quadrant in which they lie. Make sure you take care of the possibility that one of the pairs lies on the axes of the coordinate system.

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Solution:

To solve this problem, we will use three if state-ments, each taking care of a specific task. The first, la-beled IF1, decides whether the calculation should terminate by going to the statement FINISHED, or if the coordinate point lies on the Y-axis. IF2, the second if statement, looks for points that lie on the X-axis. Finally, IF3 classifies all remaining points by quadrant location. As a refresher, the following coordinate system will point out the X- and Y-axes and the quadrant names: Assume that the program ends when the origin point is encountered, which is when X = 0. and Y = 0. Output is in tabular form. POINT_CLASSIFY: PROC OPTIONS(MAIN); PUT PAGE LIST('b'CLASSIFYING POINTS'); PUT SKIP(2) LIST ('X-COORDINATE','Y-COORDINATE', 'CLASSIFICATION'); SEARCH: GET DATA(X,Y) IF1:IF X = 0 THEN IF Y = 0 THEN GO TO FINISHED; ELSE DO; PUT LIST(X,Y,'Y-AXIS')SKIP; GO TO SEARCH; END; IF2: IF Y = 0 THEN DO; PUT LIST(X,Y,'X-AXIS')SKIP; GO TO SEARCH; END; IF3: IF X>0 THEN IF Y>0 THEN PUT LIST(X,Y,'I')SKIP; ELSE PUT LIST(X,Y,'IV')SKIP; ELSE IF Y>0 THEN PUT LIST(X,Y, 'II')SKIP; ELSE PUT LIST(X,Y, 'III')SKIP; GO TO SEARCH; FINISHED:PUT LIST('ORIGIN ENCOUNTERED: DONE')SKIP; END POINT_CLASSIFY;

Question:

Removal of the pituitary in young animals arrests growth owing to termination of supply of growth hormone. What are the effects of growth hormone in the body? What is acromegaly?

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Solution:

The pituitary, under the influence of the hypothalamus, produces a growth-promoting hormone. One of the major effects of the growth hormone is to promote protein synthesis. It does this by increasing membrane transport of amino acids into cells, and also by stimu-lating RNA synthesis. These two events are essential for protein synthesis. Growth hormone also causes large increases in mitotic activity and cell division. Growth hormone has its most profound effect on bone. It promotes the lengthening of bones by stimulating protein synthesis in the growth centers. The cartilaginous center and bony edge of the epiphyseal plates constitute growth centers in bone. Growth hormone also lengthens bones by increasing the rate of osteoblast (young bone cells) mitosis. Should excess growth hormone be secreted by young animals, perhaps due to a tumor in the pituitary, their growth would be excessive and would result in the production of a giant. Undersecretion of growth hormone in young animals results in stunted growth. Should a tumor arise in an adult animal after the actively growing cartilaginous areas of the long bones have disappeared, further growth in length is impossible. Instead, excessive secretion of growth hormone produces bone thickening in the face, fingers, and toes, and can cause an overgrowth of other organs. Such a condition is known as acromegaly.

Question:

Explain the following statements. (1) An impulse will be initiated only when the stimulus is of a certain intensity. (2) Although a neuron exhibits all-or-none response, an animal is able to detect different degrees of intensity of a stimulus.

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Solution:

The two statements above come from experiments done on an isolated nerve. Two electrodes are placed at points several centimeters apart on the surface of a nerve. The electrodes are connected to recording devices, which are able to detect any electrical changes that may occur at the points on the nerve the electrodes are in contact with. An extremely mild electrical stimulus is applied to the nerve initially. The recording device shows no change. Next, a slightly more intense stimulus is applied, and still there is no change. The intensity of the stimulus is increased further and this time the device records sin electrical change occurring at the point of the membrane in contact with the first electrode and a fraction f a second later, a similar electric change at the point in contact with the second electrode. The nerve is now stimulated, as evidenced by the wave of electrical changes that moves down the fiber, passing first one electrode and then the other. This shows that an impulse, will be generated in a nerve fiber only if the stimulus reaches a certain intensity. If a more intense stimulus is applied next, an impulse is again triggered and a wave of electrical changes moves down the nerve. However, the intensity and speed of this impulse are the same as those of the one recorded from the previous stimulation. Based on these experiments, we are ready to account for the two statements. (1) A potential stimulus must be above a critical intensity if it is to generate an action potential. This critical intensity is known as the threshold value, and it varies in different neurons. Physiologically, what happens is that below the threshold value, the membrane permeability to sodium ions, though increased, is still less than that of potassium ions. This means that the inside of the membrane is still too negative, and no action potential is generated. Only when the intensity of the stimulus reaches the thresh-old value does the membrane permeability of Na^+ ions in-crease so much that its inflow exceeds the potassium out-flow. This inflow of sodium ions is now sufficient to depo-larize the membrane potential to a critical level, known as the threshold potential. Consequently, an action potential is triggered. Once threshold depolarization is reached, the action potential is no longer dependent upon stimulus strength. Thus, action potentials generated by threshold stimuli are identical to those generated by above-threshold (or suprathreshold) stimuli. (2) Increasing the intensity of the stimulus above the threshold value does not alter the intensity or speed of the nervous impulse produced, that is, the nerve fires maximally or not at all, a type of reaction commonly called an all-or-none response. An animal can detect the different degrees of intensity of a stimulus in two ways. The more intense the stimulus, the more frequent are the impulses moving along the fiber. This is called the frequency code. Second, a more intense stimulus ordinarily stimulates a greater number of nerve fibers than does a weak stimulus. This is called re-cruitment or the population code. Hence, the frequency of firing of impulses and the number of axons stimulated can be interpreted by the brain, and the strength of the stimulus can be detected.

Question:

A parallel plate capacitance is charged (linearly) to 10^\rule{1em}{1pt}3 coul in a time of 10^\rule{1em}{1pt}3 sec. Calculate the displacement current between the plates.

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Solution:

A plot of charge on one plate versus time is shown in the figure. The charge density on a surface of the parallel plate capa-citance, as a function of time, is given by, \sigma(t) = q(t) / A where q is the charge on a plate of A. The field as a function of time is E(t) = \sigma(t) / \epsilon_0 = q(t) / \epsilon_0A The electric flux is then \Phi_E = EA = q(t) / \epsilon0 A change on charge on the plate ∆q will then produce a change of electric flux given by ∆\Phi_E = ∆q / \epsilon0 Dividing both sides by ∆t, we have ∆\PhiE/ ∆t= (1 /\epsilon0)( ∆q / ∆t) This expression relates a change in flux with time to a change in charge with time, ∆q / ∆t can be viewed as a current through the capacitance. It is called a dis-- placement current, I_D to distinguish it from an actual movement of charge, From the graph we see that ∆q = 10\rule{1em}{1pt}3coul \rule{1em}{1pt} 0 coul for ∆t = 10^\rule{1em}{1pt}3 sec\rule{1em}{1pt} 0 sec Therefore I_D = ∆q / ∆t =10^\rule{1em}{1pt}3 / 10\rule{1em}{1pt}3=1 coul / sec

Assume the following program is in the active workspace: \nablaVARIANCE [1]'ENTER SAMPLE ELEMENTS' [2]S \leftarrow X1 X2 .... XN [3]AVG \leftarrow (+/S) \div \Rhos [4]RANGE \leftarrow (\Gamma/S) - L/S [5]NS \leftarrow AVG × (+/S) [6]S [7]S2 \leftarrow S\textasteriskcentered2 [8]SSQ \leftarrow (+/S2) [9]VAR \leftarrow SS2 - NS [10]'THE VARIANCE IS:';VAR [11]\nabla This program finds the variance of a set of numbers. The procedure is as follows: 1) Let X1, X2,......, XN be a series of observations: X =^N\sum_i= 1X_i / N VAR = [\sum(X_i - x)^2]/ [N] = \sumX_i^2 - [\sumX_i]^2 / N. The second expression is more easily evaluated and is also more amenable to treatment by APL than the first. i ) Suppose it is desired to change NS to MSQ (Mean Squares) a more descriptive acronym. Key in \nabla VARIANCE [5 [] 10]. Here [5] is the line number and 10 is the print position, where editing is to begin. Actually NS may be in any print posi-tion (<10) . Backspace until the type ball is below N and enter two slashes followed by a 3. The two slashes erase NS and the 3 allows 3 characters for MSQ. Now, depress the RETURN key. At this point, the fifth line looks as follows: [5]\leftarrow AVG × (+/S) Key in MSQ and then ask for a display of [5]: [5 [] ], which should be [5]MSQ \leftarrow AVG × (+/S) A similar procedure is followed for [9]. ii) Next, suppose it is required to find the standard deviation of the set of numbers. The header can be changed from \nabla VARIANCE to \nablaSDEV as follows: \nabla VARIANCE [0 [] 10] [0]VARIANCE Backspace until the ball is under V and print 8 slashes, followed by 4. Depress the RETURN key to display a blank [0]-line with 4 vacant spaces for a new name. [0] Type SDEV and display the line [0] for a check. It should look like the following [0]SDEV Type DEL to terminate the procedure. To add a line that finds the standard deviation, key in \nablaSDEV[9.5] SDEV \leftarrow VAR\textasteriskcentered.5\nabla iii) Line [10] should be amended to 'THE STANDARD DEVIATION IS:';SDEV Key in \nablaSDEV[10 [] ] to get a display of the tenth line: [10]'THE VARIANCE IS:';VAR Key in [10]'THE SDEV IS:';SDEV [11]\nabla iv) Suppose SDEV is no longer required. The command ) ERASE SDEV deletes it from the active workspace.

Question:

Discuss the absorption of glucose and amino acids. How does it differ from the mechanism of absorption of fatty acids?

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Solution:

Almost all the nutrient products obtained from digestion are absorbed through the walls of the small intestine. The total volume occupied by the small intes-tine is relatively small. However, the internal surface area of the small intestine is immense, allowing for very efficient absorption of digested nutrients. This large surface area is a result of the folds and ridges of the small intestine, compounded with the villi and microvilli that cover its internal surface. The villi are finger-like processes that extend into the lumen of the intestine. The microvilli are hair-like projections of the membrane of cells lining the villi. It is here that specific ab-sorption takes place. When glucose or free amino acids approach the cell membrane, they are bound by special groups of enzymes that are within the membrane. These enzymes quickly transport the molecules through the membrane into the cytoplasm. The nutrients are then transported across the intestinal epithelium into the blood. This occurs in part by simple diffusion, in part by facilitated diffusion, and in part by active transport. The presence of sodium in the lumen is required for monosaccharide transport. The various hexoses are absorbed by active transport. Amino acids are also absorbed by means of active transport. These nutrients are then circulated to the cells of the body, or transported to the liver for storage. Glycerol and fatty acids are not selectively absorbed. After fats are digested in the intestinal lumen, the majority of their products, primarily fatty acids and monoglycerides, simply diffuse across the cell membrane because of their high solubility in membrane lipids. These are then brought into the endoplasmic reticulum (E.R.) of the epithelial cells, where they are resynthesized into triglycerides. These lipids, in turn, combine with speci-fic proteins, forming lipoproteins. The lipoproteins aggregate into droplets called chylomicra within the E.R. The E.R. then fuses with the cell surface and empties the chylomicra into the extracellular space. Here, they are absorbed by the lacteals of the lymphatic system and even-tually transferred into the bloodstream. After a high- fat meal, one's blood may become "milky-looking", due to the high concentration of chylomicra. Note that no energy is required for fat absorption. Another important aspect of absorption is the uptake of partially digested material. When disaccharides and dipeptides are within the vicinity of the cell membrane, the membrane extends itself, forming a pocket around the partially digested molecules. This pocket invaginates into the cell in a process known as pinocytosis. The pinocytotic vesicle fuses with another vesicle called a lysosome. The lysosome contains the cells own digestive enzymes. As these enzymes digest the vesicle's contents, the degraded nutrients diffuse through the membrane into the cytoplasm and are used by the cell.

A sinusoidal wave that maintains its overall shape between two termination points is called a standing wave. Its amplitude does change with time. For such a wave, the end points have zero amplitude at all times. Zero amplitude points are called nodes and occur every half wavelength. The wave of lowest frequency has the longest wave-length. For a given distance L, the standing wave of longest wavelength has \lambda = 2L with the only nodes occurring at the termination points. Therefore, \lambda = 2L = 2 × 25 ft = 50 ft The frequency of the wave is v = \upsilon/\lambda = (1100 ft/sec) / (50 ft) = 22 Hz which is close to the lowest frequency that can be heard by a human ear. Therefore, a room somewhat larger than 25 ft is necessary in order to set up standing waves of the lowest audible frequency (for example, organ notes of 16 Hz).

Question:

A mixture consisting only of KOH and Ca(OH)_2 is neutralized by acid. If it takes exactly 0.100 equivalents to neutral-ize 4.221 g of the mixture, what must have been its initial composition by weight?

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Solution:

One equivalent of an acid is the mass of acid required to furnish one mole of H_3O^+ ; one equivalent of a base is the mass of base required to furnish one mole of OH^- or to accept one mole of H_3O^+. Here one must find the mixture of KOH and Ca(OH)_2 that contains 0.100 equivalent of base. There is one equivalent for each OH^- in a molecule of base. Thus, there is 1 equivalent of base for each KOH. There are two for each Ca(0H)_2. Since one is given that there are 0.100 equivalent in 4.221 g of the mixture, one should solve for the equivalent per gram for KOH and Ca(OH)_2. for KOH: MW = 56.1 equiv = 1 (equiv/g) = (1 / 56.1) = 1.78 × 10^-2 equiv/g For Ca(OH)_2 MW = 74.06 equiv = 2 (equiv/g) = (2 / 74.06) = 2.70 × 10^-2 equiv/g. There must be 0.100 equiv present. Let x = number of grams of KOH 4.221 - x = number of grams of Ca(OH)_2 0.100equiv. = [1.78 × 10^-2 (equiv/g)] (x) + [2.70 × 10^-2 (equiv/g)] × (4.221 - x) 0.100equiv. = 0.0178x (equiv/g) + .114 equiv - .0270x (equiv/g) 0.100equiv = - .0092x (equiv/g) + .114 equiv -.014equiv. = - .0092x (equiv/g) [(- .014 equiv)/(- .0092 equiv/g)] = x 1.52 g = x 2.701 g = 4.221 - x The original mixture contains 1.52 g-KOH and 2,701 g Ca(OH)_2 .

Question:

There are genes on theE.colichromosome which determine amino acid synthesis. For one particular strain of E .coli, how can one determine the order of genes for the synthesis of threonine (thr^+), methionine (met^+), histidine (his^+) and arginine (arg^+)? Use your knowledge of con- jugation.

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Solution:

To become an Hfr bacterial cell, the F factor must become integrated into the chromosome. When an Hfr cell and an F^- cell begin conjugation, the F factor portion of the circular Hfr chromosome initiates synthesis of a linear chromosome. This linear chromosome carries a frag- ment of F factor on both its ends, with one end acting as the origin for the transfer of the chromosome. Since the other end of the chromosome contains the remaining portion of the F factor, and since the whole chromosome rarely gets transferred, the F-cell usually does not receive a complete F factor to become F^+. In a particular Hfr strain, the F factor inserts at the same place on the chromosome, so that the origin of transfer is at the same site in all bacteria of this strain. During conjugation, the origin is the first part to travel through the conjugation tube and enter the F^- cell. Usually the conjugating cells separate before the complete transfer of the chromosome. The part that enters the F^- cell may or may not be incorporated into its chromosome. If it does become incorporated, recombination occurs and the F^- cell acquires new traits that are determined by the newly transferred Hfr genes. To determine the order of the genes on the chromosome of a particular Hfr strain, one can use the interrupted mating technique. We must first acquire an F^- strain that is auxotrophic (demonstrating a nutritional requirement) for the four amino acids but is resistant to streptomycin, an antibiotic which killsE.coli. The F^- genotype in con- sideration would be as follows: thr^-, met^-, his^-, arg^-, str^+. The Hfr strain must be prototrophic (can synthesize the amino acids in question) and be sensitive to strepto-mycin. It is also necessary that in the Hfr strain selected, the gene for streptomycin sensitivity be located far from the origin of transfer of the linear, chromosome to avoid its transfer. The Hfr genotype would be thr^+ , met^+ , his^+, arg^+, str^-. The reason for selecting streptomycin sensi-tive and resistant strains will be made evident later. The two cultures of F^- and Hfr cells are then mixed and incubated. At specific time intervals (e.g. every ten minutes), samples are removed from the conjugating mixture and agitated in a blender. This separates the conjugating bacteria to prevent genetic transfer. Since the linear Hfr chromosome always enters the F^- cell in a regular sequence, we can order the genes according to the length of time it took for the Hfr cells to transfer the genes to the F^- cells. For example, in the first ten minutes, only the thr^+ gene might be transferred. In the first fifty minutes, both the thr^+ and his^+ genes get transferred, the arg^+ gene after 80 minutes. The genes can then be located se-quentially on the chromosome by noting their time of trans-fer. For this particular Hfr strain, the chromosome map is as follows: In order to find out which gene is transferred after a certain period of time, the bacterial mixture is plated on four special media: The sample produced after ten minutes will only grow on plate I since the only gene transferred is the thr^+ gene. It cannot grow on the others since it is still auxotrophic for the other three amino acids. The sample after the fifty minutes time interval would grow on both plates I and II. We thus know that the his^+ gene follows the thr^+ gene. The reason why all the plates contain streptomycin is that any Hfr bacteria remaining in the sample placed on the special media plates will be killed. Although F^- cells are strepto-mycin resistant, the only F^- cells that will grow on the special plates are those that have conjugated with Hfr cells and have received and incorporated the genes for determin-ing amino acid synthesis. The streptomycin thus acts as a control to allow only newly-made recombinants to live. An-other relevant point is that in an actual experiment, more media plates with different combinations of amino acid supplements would be required. For example, Plate I, which selected for thr^+, should also check for his^+, or for arg^+, or for met^+ , since we would not know which gene is really first.

Question:

List the antagonistic activities of the sympathetic and parasympathetic systems.

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Solution:

Below is tabulated the antagonistic activities of the sympathetic and parasympathetic systems. Organs innervated Sympathetic action Parasympathetic action Heart Strengthens and speeds up heart beat Weakens and slows down heart beat Arteries constricts lumen and raises blood pressure No innervations Digestive track slows peristalsis speeds peristalsis Digestive glands decreases secretion increases secretion Urinary bladder relaxes bladder constricts bladder Bronchial muscles dilates passages, facilitating breathing constricts passages Muscles of iris dilates pupil constricts pupil Muscles attached to hair causes erection of hair No innervations Sweet glands increases secretion No innervations In general, the sympathetic system produces the effects which prepare an animal for emergency situations, such as quickening the heart and breathing rates and dilating the pupil. These alert responses are together termed the fight-or-flight reactions. The parasympathetic systems reverse the fight-or-flight responses to restore an animal to a calm state.

Question:

Describe the parts of a typical flower. What are their functions?

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Solution:

The flower of an angiosperm is a modified stem on which concentric rings of modified leaves are attached. A typical flower consists of four such rings of structures borne on the receptacle, the expanded end of the flower stalk. The stalk of the flower is also known as the pedicel. The outermost ring, usually green and most like ordinary leaves, is composed of sepals. These enclose and protect the flower bud until it is ready to open. Collectively, the sepals are known as the calyx. Internal to the calyx are the petals, which are often brilliantly colored in order to attract those insects or birds which promote pollination. Flowers which are pollinated by the wind need not be attractive to birds and insects, and so their petals tend to be less showy, and may even be absent. Inside the corolla lie the stamens, the male reproductive organs of the flower. Each stamen consists of a slender stalk called the filament, which supports an expanded anther at the tip. The immature anther is composed primarily of pollen sacs. After reaching maturity, the anther contains numerous pollen grains, which produce haploid male gametes upon germination. In the center of the flower is a pistil (carpel) or pistils, the female reproductive organ. Each pistil is composed of a swollen portion, the ovary, at its base; a long, slender stalk which rises from the ovary, the style; and on top of this, an enlarged flattened crown called the stigma. The ovary contains one or more ovules, which are the future seeds of the angiosperm. The stigma's function is to secrete a moist, sticky substance to which the pollen grains will adhere. The style provides a lubricated pathway through which the pollen tube of the germinated pollen grain can grow on its way towards the egg cell.

Consider the two weights as a system. This implies that the cable does not stretch and no internal forces have to be considered, since they consist of action-reaction forces and therefore cancel. The external forces acting on the system are the frictional force on the 32 lb block and the gravitational force on both weights. Since the frictional forceF_f is proportional to the normal force, we first find N. The 32 lb block has no movement perpendicular to the plane. Setting the sum of the forces in this direction equal to zero, we get from the diagram, N - mg cos 37o = 0 Therefore N = 32 cos 37o and F_f = \muN = (0.2)(32 cos 37o) with direction down the plane, since it opposes the motion. All the cable does is change the direction of forces which are applied to each weight and which are in line with the cable. In the same direction as the 64 lb. force on m_1, (the cable makes it so), the 32 lb, weight experiences the frictional force and the component of the gravitational force parallel to the cable or 32 sin 37\textdegree. Applying Newton's second law to the system, we have \sumF = (m_1 + m_2)a since both weights experience the same acceleration. Then 64 - 32 sin 37o - (0.2)(32 cos 37o) = {(64/32) + (32/32)}a 64 - 19.2 - 5.1 = 3a 39.7 = 3a a = 13.2 ft/sec^2(a) The tension is obtained by isolating the 64 lb weight and noting that the only two forces acting on it are T and the force of gravity. Calling the downward direction positive, we obtain from Newton's second law, 64 - T = m_1a = (64/32)(13.2) = 26.4 T= 64 - 26.4 + 37.6 lbs.(b)

Let r be the radius of the horizontal circle traversed by the bob of mass m, l be the length of the string, and T^\ding{217} be the tension which the string exerts on the mass. The forces acting on the bob are the weight mg^\ding{217} downward and the tension T^\ding{217} at an angle \texttheta to the vertical. Resolve T^\ding{217} into horizontal and vertical com-ponents. Applying Newton's Second Law to the horizontal direction of motion F_net = ma where a is the horizontal acceleration of m, and F_net is the net horizontal force on m. Since m is in uniform circular motion, T sin \texttheta provides the centripetal force necessary to keep the bob in the circle. Thus T sin \texttheta = mv^2/r, where v is the velocity of the bob. But v is the distance traveled in 1 s. That is, v = n × 2\pir = 2\pirn, where n is the angular speed in rev\bullets^-1. Also, from the figure, sin \texttheta = r/l. \thereforeT= ( T= ( 4\pi^2 rmn^2)/(r/l) = 4\pi^2mln^2 = 4\pi^2 × 1/2 slug × 10 ft × (1/2 s^-1)^2 = 49 lb. The bob stays in the same horizontal plane, so that the vertical forces must balance. Thus, from Newton's Second Law, T cos \texttheta = mg. \therefore cos \texttheta= mg/(4\pi^2mln^2) \therefore cos \texttheta= mg/(4\pi^2mln^2) = 32/[4\pi^2 × 10 ft × (1/2 s^-1)^2] = 32/[4\pi^2 × 10 ft × (1/2 s^-1)^2] = 0327; = 0327; \therefore\texttheta= 71\textdegree.

Question:

The molecular weight of nicotine, a colorless oil, is 162.1 and it contains 74.0 % carbon, 8.7 % hydrogen, and 17.3 % nitrogen. Using three significant figures for the atomic weights, calculate the molecular formula of nicotine.

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Solution:

A molecular formula gives the actual composition in number of atoms per molecule. It also summarizes the weight composition of the substance. Hence, to calculate such a formula, the molecular weight, the weight composition of the substance, and the atomic weights of the constituent elements must be known. The weight composition of the substances in nicotine, assuming there are 100 g of it, are C = 74.0 g, H = 8.7 g, N = (110 - 74.0 - 8.7) = 17.3 g. The atomic weights of these substances are C = 12.0 g / mole, H = 1.01 g / mole, N = 14.0 g / mole. The number of moles of each constituent is, therefore, moles C= [(74.0 g) / (12.0 g / mole)] = 6.17 moles moles H= [(8.7 g) / (1.01 g / mole)]= 8.61 moles moles N= [(17.3 g) / (14.0 g / mole)] = 1.24 moles Thus, the mole ratios of C : H : N are 6.17 : 8.61 : 1.24. The empirical formula is C_6.17 H_8.61 N1.24or C_[(6.17)/(1.24)] H_[(8.61)/(1.24)] N_[(1.24)/(1.24)] = C_5H_7N_1. To determine the molecular formula, which is related to the formula weight of the empirical formula by a whole number, divide the molecular weight (MW Nicotine = 162.1) by the formula weight and multiply this whole number times the ratio of atoms in the empirical formula. Thus, the formula weight of C_5H_7N_1= 81.05 the molecular weight of nicotine= 162.1 the whole number =(1.62 / 81.05)= 2. The molecular formula is (C_5H_7N_1)_2 = C_10H_14N_2.

Question:

What is meant by an antagonistic muscle? Give examples.

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Solution:

Muscle can exert a pull but not a push. For this reason, muscles are typically arranged in antagonistic pairs: one pulls a bone in one direction and the other pulls it in the opposite direction. The biceps, for example, bends or flexes the arm and is termed a flexor. Its antagonist, the triceps, straightens or extends the arm and is termed an extensor (see accompanying figure.) Such pairs of opposing extensors and flexors are found at the wrist, ankle and knee, as well as at other joints. When either the flexor or the extensor contracts, its antago-nistic muscle must relax to permit the bone to move. The proper coordination of nerve impulses is necessary for antagonistic pairs to function properly. Other antagonistic pairs of muscles are adductors and abductors, which move parts of the body toward or away from the central axis of the body, respectively; levators and depressors raise or lower parts of the body; and while pronators rotate parts of the body downward and backward, supinators rotate them upward and forward.

The opening and closing of a stoma is regulated by changes in the turgor pressure within the two guard cells that surround the stomatal opening. Each bean-shaped guard cell has a thicker wall on the side toward the stoma called the inner wall, and a thinner wall on the side away from the stomatal opening called the outer wall. Increased turgor pressure causes the cells to bulge. Since the outer walls are thinner than the inner walls, the former stretch more than the latter, causing the cells to bow in and the stoma to open. When the turgor pressure in the guard cells decreases, the inner walls regain their original shape and the cells unbend, closing the stoma as a result. Thus the opening and closing of a stoma depends on a mechanism which varies the turgor pressure within the guard cells. This variation of turgor involves in part the production of glucose and other osmotically active substances in the guard cells themselves. It is believed that light initiates a sequence of enzymatic reactions that lead to the conversion of starch stored in the guard cells into osmotically active glucose molecules, thus increasing the turgor pressure of the guard cells. As light increases the rate of photosynthesis, carbon dioxide is utilized and the decrease in carbon dioxide concent-ration, increases the pH of the guard cells (remember that guard cells contain chloroplasts and so they are able to undergo photosynthesis). This increased basicity stimulates the enzyme phosphorylase to convert starch to glucose-l-phosphate. The glucose-l- phosphate is sub-sequently converted to glucose, increasing the concentration of glucose in the guard cells and causing the influx of water into the cells by osmosis. (Another osmotically active solute may be K+ -potassium ions- which are pumped into the guard cell from the epidermal cell as CO_2 levels decrease.) This phenomenon raises the turgor pressure in the guard cells and causes the stoma to open (see Fig.) In the dark the process is reversed. Thus under normal conditions, the stomata in many species open regularly in the morning and close in the evening.

The forces acting on the trapdoor of length a are its weight Mg acting downward at the center, since it is uniform, the hunter's weight mg acting downward at the end, and the normal force N exerted by the pivot upward (see the figure ). When we take moments about the pivot, counterclockwise moments being taken as positive, the moments causing rotational acceleration are -Mg(a/2 - x) - mg(a - x), and thus using the rigid body analog of Newton's second law, if T is the resultant torque acting on the rigid bar, I_T, the moment of inertia of the man-bar system about the pivot, and \alpha its angu-lar acceleration, then T = I_T\alpha a or -Mg(a/2 - x) - mg(a - x) = I_T\alpha The moment of inertia of the trapdoor about a horizontal line parallel to the pivot and passing through the center of gravity is (1/12)Ma^2 . By the parallel-axis theorem, if I_cmis the moment of inertia of the bar (or any rigid body) about its center of gravity and I_b is its moment of iner-tia about any axis, where h is the distance separating the two axes, then I_b = I_cm + Mh^2. Thus, I-_b about the pivot is (1/12) Ma^2 + M[(a/2) - x]^2. The moment of inertia of the hunter about the pivot is I_m = m(a-x)^2 . Hence T = (I_b + I_m)\alpha for I_T = I_b + I_m -Mg(a/2 - x) -mg(a - x) = \alpha[(1/12)Ma^2 + M(a/2 - x)^2 + m(a - x)^2].(1) If the hunter and the end of the trapdoor are to have a linear acceleration g downward, then -g = (a - x)\alpha. Where (a - x) is the radius of the circular arc path the man fol-lows about the pivot (see the figure). Therefore, upon division of both sides of equation (1) by \alpha and using the above relation between g and \alpha, M(a/2 - x)(a - x) + m(a - x)^2 = (1/12)Ma^2 + M(a/2 - x)^2 + m(a - x)^2. M(a/2 - x)(a - x) = (1/12)Ma^2 + M(a/2 - x)^2 a^2/2 - ax - ax/2 + x^2 = (1/12)a^2 + a^2/4 - ax + x^2 (a^2/2 - a^2/4) - (3/2)ax + ax = (1/12)a^2 a^2/4 - ax/2 = (1/12)a^2 \therefore a/2(a/2 - x) = (1/12)a^2 \therefore 3a^2 - 6ax = a^2 \therefore x = (1/3)a The pivot must be located one-third of the length of the trapdoor from the end.

Question:

What is the Lorentz contraction of an automobile traveling at 60 mph ? (60 mph is equivalent to 2682 cm/sec.)

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Suppose we are given two frames of reference moving relative to one another with a velocity v (see figure). If we are dealing with classical physics and want to relate the coordinates of an event occurring in the S - frame (x, y, z , t) to the coordinates of an event occurring in the S'-frame (x', y', z', t'), we use the Galilean trans-formation , or x' = x - vt y' = y (If v is in the x-direction only.) z' = z t' = t. In relativistic physics, this transformation is invalid, and must be replaced by the Lorentz transformation , or x' = (x - vt) / \surd{1 - (v^2 / c^2)} y' = y z' = z t' = {t - (vx / c^2)} / \surd{1 - (v2/ c^2)} . Now, we may relate distances measured in S' to distances as measured in S. Let us imagine the measurement of a dis-tance parallel to the x'-axis in the S' frame. In order to measure the length of a rod in S, we must locate both ends of the rod (x_1, x_2) at the same time (t_1 = t_2 ) in S. Hence, the length in S' is x'_2 -x'1= [(x_2 - x_1) - {v (t_2 - t_1)}] / [\surd{1 - (v^2 / c^2)}] . But t1= t2, therefore x'_2 -x'1= (x_2 - x_1) / \surd{1 - (v^2 / c^2)} . Hence (x_2 - x_1) = x'2-x'1\surd{1 - (v^2 / c^2) }.(1) Since\surd{1 - (v^2 / c^2)} < 1 we then have x_2 - x1< x'2-x'_1 . The observerin S measures a smaller rod length (contracted) than the observer in the rod's rest frame, S'. Now, we calculate the length of the car in S, ( x_2 - x_1)If V_r is 2682 cm/sec. Vr/ c = 2682 / (3 × 10^10) = 8.94 × 10^-8 {V_r/ c}^2 = 8.0 × 10^-15 . When x is very much less than 1, \surd(1 - x) = [1 - {(1/2)x} approximately . Therefore , \surd{1 - (v_r / c)^2 } \approx [1 - (4.0 × 10^-15)]. Substituting in (1) x_2 - x_1\approx (x'2-x'_1) (1 - 4.0 × 10^-15) . This means that the change in length of a meter rule is only 4.0 × 10-15 meters, or 4.0 × 10^-13 cm. Since the diameter of an atom is about 10^-8 cm, the diameter of a nucleus is about 10^-12 cm and the size of the electron is about 10^-13 cm, this contraction is clearly negligible. Again we see that the difference between relativistic and classical physics is not important for the velocities we are normally concerned with.

Use FUNCTION subprograms with t as the argument. The parameters can be read in, Case 1:FUNCTION CONST(T) READ, H CONST = H RETURN END Case 2:FUNCTION PULSE (T) READ, H, T1, T2 PULSE = 0 IF(T.GE.T1. AND. T.LT,T2) PULSE = H RETURN END Case 3:FUNCTION SINE (T) READ A, OMEGA, PHI SINE = A \textasteriskcentered SIN(OMEGA\textasteriskcenteredT + PHI) RETURN END Case 4:The formula for E_4(t) is a recursion formula that defines E_4(t) for t < t_1 + PW and equates E_4(t) with E_4(t-PW) for t \geq t_1 + PW, thus requiring repeated subtraction of PW until t < t_1 + PW. Then t will be mapped into t\textasteriskcentered wheret_1 \leq t\textasteriskcentered < T_1 + PW. This mapping is a translation of t along the time axis to the point t\textasteriskcentered; and the distance of the translation is t - t\textasteriskcentered = nPW(1) where n is the number of subtractions required. (See Fig. 2). But repeated subtraction is the operation of division. Therefore, the same translation mapping can be achieved by dividing the distance t - t_1 by PW: (t - t_1) / PW = n + (r +PW)(2) where r is the remainder. Multiplying both sides of equation (2) by PW: t - t_1 = nPW + r . Substituting, using equation (1) t - t_1 = t - t\textasteriskcentered + r - t_1 = - t\textasteriskcentered + r r=t\textasteriskcentered - t_1.(3) Equation (3) indicates that the remainder gives t\textasteriskcentered's distance from t_1. Therefore, adding r to t_1 gives the absolute location of t\textasteriskcentered. FUNCTION TRAIN(T) READ H, T1, W, P TIME = T IF (T.GE. (T1 + P\textasteriskcenteredW)) GO TO 5 GO TO 6 TIME = AM0D ((T-T1)/(P\textasteriskcenteredW))\textasteriskcenteredP\textasteriskcenteredW + T1 IF( (TIME.LT.T1) .0R. (TIME.GE. (T1-W))) TRAIN = 0 . IF((TIME. GE.T1). AND. (TIME.LT. (T1+W))) TRAIN = H. RETURN END

Both free protons (H^+) and carbon dioxide (CO_2) are known to promote the release of oxygen from hemo-globin. This is known as the Bohr effect. Myoglobin shows no change in oxygen binding over a range of pHs or carbon dioxide concentrations. But in hemoglobin, acidity enhances the release of oxygen. Lowering the pH shifts the oxygen dissociation curve to the right (see Figure). A shift to the right enhances oxyhemoglobin dissociation. For example, at any partial pressure of oxygen, the curve at the lower pH is lower, signifying that less oxygen is bound to the hemoglobin molecule (less % saturation) as a result of a decreased O_2 affinity. An increasein the CO_2 concentration also lowers oxygen affinity. This could be due to, in part, a pH effect. CO_2 can react with the water in the erythrocyte to form carbonic acid: CO_2 + H_2O \rightarrow H_2CO_3 Carbonic acid then dissociates to form H^+ ions which lower the pH: H_2CO_3 \rightarrow H^+ +HCO^-_3 However, increasing the CO_2 concentration, while keeping pH constant, also lowers the oxygen affinity. Both H^+ and CO_2 bind to the hemoglobin molecule at special regulatory sites spatially distinct from the heme site. The binding alters the structure of the molecule by changing the arrangement of its polypeptide chains. This altered conformation lowers hemoglobin's affinity for oxygen. Since H^+ and CO_2 affect hemoglobin's binding of oxygen by allosteric interaction between the hemoglobin subunits, they have no similar effect on the myoglobin molecule, since it has only one polypeptide chain. Now that we understand the Bohr effect, what is its physiological significance? Rapid metabolism occurs in highly active tissue such as contracting muscle. Here much CO_2 is produced from cellular respiration. Acid production is also high, resulting from the increased CO_2 concentration and the reduction of pyruvate to lactic acid when the amount of oxygen is limiting. Thus, the presence of high levels of CO_2 and H^+ in actively metabolizing tissue enhances the release of oxygen from oxyhemoglobin. More oxygen is thus provided to the tissues where it is needed most. The opposite process occurs in the alveoli of the lungs. The increased concentration of oxygen drives off H^+ and CO_2, so that more oxygen may bind.

The manufacture of most explosives involves the use of concentrated sulfuric acid. The action of concentrated nitric acid on such compounds as glycerin seems to be markedly hastened by concentrated sulfuric acid because water is re-moved from the reaction. Most of the compounds formed by the action of nitric acid on organic substances are explosive. Thus, for part (a), An explosion occurs as a result of a rapid chemical reaction attended by the formation of a large volume of gas. The equation for the detonation of nitroglycerine is 4C_3H_5 (NO_3)_3 \ding{217} 12CO_2 + 10H_2O + 6N_2 + O_2 For the detonation of C_2H_4 (NO_3)_2, the products CO_2, N_2, and H_2O are known, thus only balancing need be done. after balancing: C_2H_4 (NO_3)_2------------------\rightarrow 2CO_2 + N_2 + 2H_2O.

In this circuit point B is grounded. When dealing with circuits, the grounded point is considered as zero potential and we express potentials relative to this refer-ence level. The direction of current flow is counterclock-wise so we will traverse the circuit in this direction. Kirchoff's loop theorem states that the sum of the changes in potential around any closed loop must be equal to zero. Note that this is just a statement of conservation of energy. Applying this theorem gives the following equation: \epsilon - ir - i (3 ohms) - i (1 ohms) = 0 Observe that although there is 10 volts across the battery by itself, when it is connected to the circuit its internal resistance, r, causes a potential drop equal to -ir. Substituting 1\Omega for r and solving for i, yields 2 am-peres as the value for the current. Since the potential at point b is zero, we start there and proceed to points a or c to find V_a or V_c respectively. We may go in either direction so long as we take into account the proper sign for the current (negative if counterclockwise, positive if clockwise). Proceeding counterclockwise from b to c yields: Vc = -i (1\Omega) = -2 volts Proceeding clockwise from b to a and remembering that in this direction the current is -2 amp, we see that V_a = - (-2)(3) = +6 volts. That is, point a is 6 volts above ground and point c is 2 volts below ground. The potential difference V_ac can now be found by subtraction ground. The potential difference V_ac can now be found by subtraction V_ac = V_a - V_c = 6 - (-2) = +8 volts

Question:

The density of concentrated sulfuric acid is 1.85 g/ml. What volume of the acid would weigh 74.0 g?

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Solution:

Density is defined as weight per unit volume. density = (weight / volume) = g/ml. Therefore: volume =(weight / density) . Solving for the volume: volume = (74.0 g / 1.85 g/ml) = 40.0 ml.

Question:

Describe the events of meiosis and account for its importance .

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Solution:

Meiosis is the process of gamete formation in which two cell divisions occur in succession to give rise to four haploid daughter cells with each bearing only one copy of an individual chromosome, unlike the diploid parent cell which carried two copies. The sequence of events in meiosis begins as in mitosis. Chromosomes that have already duplicated in interphase begin to shorten in the first meiotic prophase to form con-densed, easily visible structures; each consists of two identical chromatids joined together at the centromere. It is during this condensation of the chromosomes that the first unique event of meiosis occurs: the synapsis (pairing) of homologous chromosomes, each made up of two chromatids. Each joined pair of homologous chromosomes is known as a tetrad. As the chromatids become more and more condensed, it is possible to see points at which they connect to one another, forming cross-bridges called chiasmata. These may be considered to be the visible manifestations of crossing over between a pair of chroma-tids. Only two of the four chromatids (non-sister chromatids) present in the tetrad can be involved in the formation of a chiasma. The end of the first meiotic prophase is marked by the disappearance of the nuclear membrane and the nucleolus. At the first meiotic metaphase, the homologous pairs (tetrads), having reached maximum condensation, line up along the equator of the meiotic spindle. This is followed by the first meiotic anaphase, during which the chromosomes separate from their homologs without separation of the sister chromatids, which remain attached at the centromere. Gradually, the duplicated chromosomes, now known as dyads, move to the two poles of the meiotic spindle. At telophase, when the chromosomes have reached the two poles, cytoplasmic division occurs. At this point, the chromosomes may become less condensed and enter into a brief, second meiotic interphase, during which the nuclear membrane forms again. The chromosomes grow diffuse but do not replicate. A second prophase follows rapidly, and the chromosomes again condense. There is, however, no synapsis, because the homologous chromosomes had separated at the first meiotic anaphase. The nuclear membrane and the nucleolus now disappear as during the first meiotic division. Those species which do not exhibit a second interphase go directly from the first telophase into a brief second prophase, during which spindle formation occurs. During metaphase, the two members of each dyad line up at the equator. At the second meiotic anaphase, the sister chromatids separate at the centromere, one going to each pole. When they have reached the poles, the nuclear membrane and nucleolus reappear. Cytoplasmic cleavage follows, giving rise to distinct daughter cells. After the second meiotic division, the number of chromosomes in each cell is half the original number, and reduction is complete. The process of meiosis began with a single cell (2n) which doubled its chromosomes, resulting in four copies of each type of chromosome (4n). Two meiotic cell divisions then produced four cells (1n) from this one cell. Because there was no chromosome doubling between the two divisions, there is necessarily a halving of the original chromosome number. In any sexual organism, there are two genetic forms, the haploid (1n) and the diploid (2n), which alternate from generation to generation. The diploid generation gives rise to the haploid generation by meiosis, and the haploid generation gives rise to the diploid generation by fertilization, which is the fusion of two haploid cells. Meiosis is thus important in ensuring that, after fertilization, the original diploid state is retained. Another extremely important factor in meiosis is the possibility of getting new combinations of genes as a result not only of crossing over between synapsing chromosomes in the first meiotic division, but also of the random segregation of the homologous chromosomes during the first anaphase. These factors contribute to the genetic variation which forms the basis for natural selection and thus evolution. If the new combination of genes confer some advantage to the organism possessing it, that organism will have a better chance of survival.

As shown in the figure, the \lambda_max appearing in Wien's displacement law is the value of \lambda at which \psi (\lambda) is a maximum. In order to show that Wien's Law follows from Planck's radiation law, we calculate the values of \lambda for which \psi (\lambda) is a maximum. This will yield the value of \lambda _max as a function T. Planck's radiation law connects the monochromatic energy density in an isothermal blackbody enclosure with the wavelength and absolute temperature T \psi = [(8\pich\lambda^-5)/{exp(ch/\lambdakT) - 1}] . where h is Planck's constant, and k is Boltzmann's constant. Thus (d\psi/d\lambda) = - [(40\pich\lambda^-6)/{exp(ch/\lambdakT) - 1}] - [8\pich\lambda^-5 × { - (ch/kT\lambda^2) exp (ch/\lambdakT)}/{exp (ch/\lambdakT) - 1}^2] = [(8\pich\lambda^-6)/{ exp (ch/\lambdakT) - 1}] [ - 5 + {(ch/\lambdakT) exp (ch/\lambdakT)} / { exp (ch/\lambdakT) - 1}] . For maxima or minima of \psi, d\psi/d\lambda must be zero. The term outside the bracket becomes zero for either \lambda = 0 or \lambda = \infty. Both of these are minima, as can be verified by differentiating again. When the expression inside the bracket becomes zero, we have a maximum, and for this [{(ch/\lambda_max kT) exp (ch/_max kT)} / {exp (ch/\lambda_max kT) - 1}] = 5 (ch/\lambda_max kT) exp (ch/\lambda_max kT) = 5 [exp (ch/\lambda_max kT) -1] exp (ch/\lambda_max kT) (5 - ch/\lambda_max kT) = 5 orexp (- ch/\lambda_max kT) = 1 - ch/5\lambda_max kT(1) This is a transcendental equation, which must be solved graphically. Letting x = -ch/\lambda_max kT equation (1) becomes exp (x) = 1 + (x/5)(2) In order to find x, we draw a graph of y = exp (x), and another graph y = 1 + x/5. The intersection points of these 2 graphs are the values of x which satisfy (2). From figure (2), we obtain x = - 4.965 as the solution set of (2). Thus for a maximum of the radiation curve, exp(- ch/\lambda_max kT) = 1 - ch/5\lambda_max kT and this implies that ch/\lambda_max kT = 4.965. \lambda_max T = (ch/4.965 k) = const, which is Wien's displacement law. Experimentally, \lambda_max T = 2.891 × 10^-3 m \textbulletK deg. Solving for h h = [{(4.965 k) (\lambda_max T)} / c] h = [(4.965 k × 2.891 × 10^-3 m \textbullet k deg) / c] = [(4.965 × 1.380 × 10^-23 J \textbullet k deg^-1 × 2.891 × 10^-3 m\textbulletk deg)/(2.998 × 10^8 m \textbullet s^-1)] = 6.607 × 10^-34 J \textbullet s.

In Thompson's apparatus [figures (a) and (b)] a gas of the element

Pascal's Law states that pressure applied to an enclosed fluid is transmitted throughout the fluid in all directions without loss. In the hydraulic press shown, this means that the pressure applied to the smaller piston is transmitted unchanged to the larger piston. Since it has a larger area, it experiences a greater force since F = PA. Hence, we have P_2 = P_1F_2/A_2 = F_1/A_1 F_2 = (A_2/A_1) F_1 = [\pi(4.0 in.)^2]/[ \pi(0.50 in.)^2] 120 lb = 7.7 × 10^3 lb

Question:

The ionization constant for acetic acid is 1.8 × 10^-5 . a) Calculate the concentration of H^+ ions in a 0.10 molar solution of acetic acid. b) Calculate the concentration of H^+ ions in a 0.10 molar solution of acetic acid in which the concentration of acetate ions has been in-creased to 1.0 molar by addition of sodium acetate.

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Question:

Using a clocked JK flip-flop construct a; a) D flip-flop.b) T flip-flop. c) Clocked T flip-flop.d) SR flip-flop.

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Solution:

The construction shown in fig. 1 gives a D flip-flop realization. Note that both S and R inputs are zero, and the circuit obeys the truth table given below; b) By applying 0 to both inputs S and R and 1 to inputs J and K, a T flip-flop may be constructed as shown in fig. 2. Note that the C (clock) input becomes the T input and the circuit constructed obeys the truth table of a T flip-flop. c) Shown in figure 3 is a realization of the clocked T flip-flop. The control signal T_c allows the clock pulses to be selectively applied to the input terminal. Each clock pulse that arrives at T causes the flip-flop to a state change. d) Applying 0 to J, K, and Clock inputs gives an SR flip-flop realization, shown in fig. 4.

Consider a photon emitted by nucleus A located at the lower gravitational potential. Its energy upon emission is E_1 = hf_0 where h is Plank's constant and f_0 is the frequency associated with the photon. As the photon rises, it gains potential energy. The law of conservation of energy demands that the total energy of the photon remain con-stant in time. The frequency of the photon must then decrease to offset the increase in potential energy to maintain the total energy constant. After travelling a height H, its energy is E_2 = hf + mgH m is the mass equivalent of its energy which is given by Einstein's mass- energy relationship m = (hf/c^2) To satisfy the conservation of energy, we must have E_1 = E_2 hf_0 = hf + (hfgH)/c^2 or(f_0/f) = 1 + (gH/c^2) = 1 + [(980 cm/sec^2 × 2.2 × 10^3 cm)/{(3 × 10^10 cm/sec)^2}] = 1 + 2.3 × 10^-15 The frequencies differ by 2.3 parts in 10^15. This minute change in frequency is termed the gravitational red shift (for the frequency decreases and shifts towards the red part of the spectrum as the light leaves the Earth).

Question:

Calculate the most probable speed ѵ_p the arithmetic mean speedѵ, and the root mean square speed ѵ_rms for hydrogen molecules at 0\textdegreeC.

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Solution:

Before beginning this problem, we must first understand the basic differences in the three types of speeds. The arithmetic mean speedvis obtained by summing all the speeds and dividing by the total number of molecules N. v= 1/N^N\sumi = 1v_i The symbol \sum v_i indicates the sum of the speeds v_1, v_2, v_3, ... v_N of all the N individual molecules. The equation for mean speed is v= [(8kT)/(\pi m)]^1/2 =[(8RT)/(\pi M)]^1/2 where k is Boltzmann's constant, R is the gas constant, T is the absolute temperature, m is the weight of 1 molecule, and M is the molecular weight. The most probable speed is obtained from the plot of f(v) versus v, where f(v) is the fraction of molecules with speed v. The equation of this curve f(v) = [m /(2\pikT)]^3/2 e-(mv)2 / 2kT4\piv^2 is differentiated with respect to v and set equal to zero [{df(v)}/{dv} = 0]. The shape of the curve of f (v) versus v is indicated in the accompanying figure. {df(v)} / {dv} = [m/2\pikT]^3/2 e-(mv)2 / 2kT[8\piv + 4\piv^2 {(-mv)/(kT)}] This derivative vanishes at v = v_p , called the most probable speed, where v_p = [(2kT)/(m)]^1/2 = [(2RT)/(M)]^1/2 The root-mean-square speed, v_rms, is defined by v_rms = [1/N ^N\sumi =1v2_i]^1/2 The symbol \sum v2_i indicates the sum of the squares ofthe the velocities,v_1, v_2, v_3, ..., v_n of all the N individual molecules. Since the velocity distribution is continuous, the root-mean-square velocity is obtained by multiplying each velocity squared by the probability of that velocity, integrating over all velocities, and taking the square root: v_rms = [^\infty\int_o v^2f(v) dv]^1/2 Substituting for f (v) and simplifying one obtains v_rms = [(3kT)/m]^1/2 = [(3RT)/M]^1/2 At any given temperature, these velocities are in-versely proportional to the square root of the molecular weight. Lighter molecules move more rapidly so that their average kinetic energies are exactly equal to those for the heavier molecules. To solve this problem for hydrogen at 0\textdegreeC one must 1) find the molecular weight of hydrogen in kg, 2) convert 0\textdegreeC into the absolute temperature scale, 3) substitute these values into the equations for v_p,v, and v_rms. The molecular weight of hydrogen is 2.016 g/mole, which is 2.016 × 10^-3 kg/mole. The temperature in degrees Kelvin is 273\textdegreeK. Other pertinent information R = 8.314 J/\textdegreeK -mole, \pi = 3.142. Thus, v_p = [(2RT)/M]^1/2 = [(2) (8.314 J/0K-mol) (273\textdegreeK) / (2.016 × 10^-3 Kg/mol)]^1/2 = 1.50 × 10^3 m/s v= [(8RT)/(\piM)]^1/2 = [(8)(8.314 J/0K-mol) (273\textdegreeK) / (3.142) (2.016 × l0^-3 Kg/mol)]^1/2 = 1.69 × 10^3 m/s. v_rms = [(3RT)/M]^1/2 = [(3) (8.314 J/0K-mol) (273\textdegreeK) / (2.016 × 10^-3 Kg/mol)]^1/2 = 1.84 x 10^3 m/s. The relative ratios of v_p :v: v_rms are 1 ; 1.13 : 1.23.

Question:

Write a program that into an array. Maximum length of the string is 60 characters.

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Solution:

Before defining the program, the following points are to be noted: a) #include :It is a standard file used for reference during character I/0 b) #define statement:It is a statement used for symbol and constant declarations. c) Print a string usingTo print a string usingprintf( ), we just have to printf( ):provide the name of the array (toprintf( )) containing the string. #include /\textasteriskcenteredconstants for character I/O\textasteriskcentered/ #define MAXLINE 61 /\textasteriskcenteredmaximum size of the string\textasteriskcentered/ #define EOL '\textbackslashn' /\textasteriskcenteredassume carriage return is the EOL\textasteriskcentered/ #define NULL '\textbackslash0' /\textasteriskcenteredappend NULL to the end of the string\textasteriskcentered/ main ( ) { int index; /\textasteriskcenteredindex to count characters \textasteriskcentered/ int letter; /\textasteriskcenteredletter entered by the user\textasteriskcentered/ char string [MAXLINE]; /\textasteriskcenteredstring to be read in and stored in an array\textasteriskcentered/ /\textasteriskcenteredFollowing operations are to be performed\textasteriskcentered/ /\textasteriskcentered1. Read a character into string [ ] until\textasteriskcentered/ /\textasteriskcentered an EOL is encountered.\textasteriskcentered/ /\textasteriskcentered2. Append a NULL character to the string.\textasteriskcentered/ /\textasteriskcentered3. Useprintf( ) to write the string.\textasteriskcentered/ for (index = 0; (letter = get char ( )) ! = EOL; ++index) string [index] = letter; /\textasteriskcenteredput letter in the string\textasteriskcentered/ string [index] = NULL /\textasteriskcenteredappend a NULL character to the string\textasteriskcentered/ printf ("The string entered was %s\textbackslashn", string); /\textasteriskcenteredwrite the string\textasteriskcentered/ printf ("The number of characters is %d\textbackslashn", index); /\textasteriskcenteredwrite the no. of characters in the string\textasteriskcentered/ }

Question:

A cylinder contains an ideal gas at a pressure of 2 atm, the volume being 5 liters at a temperature of 250\textdegreeK. The gas is heated at constant volume to a pressure of 4 atm, and then at constant pressure to a temperature of 650\textdegreeK. Calculate the total heat input during these processes. For the gas, C_v is 21.0 J \bullet mole^-1\bullet K deg^-1. The gas is then cooled at constant volume to its original pressure and then at constant pressure to its original volume. Find the total heat output during these processes and the total work done by the gas in the whole cyclic process.

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Solution:

The number of moles, n, originally present at point (1) in the P-V diagram can be calculated from the gas equation n= pV/RT = (2 atm × 5 liters)/(0.0821 liter\textbulletatm\bulletmole^-1\bulletK deg^-1 × 250\textdegreeK) = 0.487 mole. The specific heat C_p at constant pressure is C_p = C_v + R = (21.0 + 8.317)J \bullet mole^-1 \bullet K deg^-1 = 29.317 J \bullet mole^-1 \bullet K deg^-1. In going from (1) to (2), V is constant. There-fore in the first change, P/T remains constant: (from the universal gas law) P_1/T_1 = P_2/T_2 orT_2 = T_1(P_2/P_1) SinceP_2/P_1 = 4 atm/2 atm = 2, we have T_2 = 2T_1 = 2 × 250\textdegreeK = 500\textdegreeK The heat input along the change of state from (1) to (2) is H_12 = nC_v(T_2 - T_1) = 0.487 mole × 21.0 J\bulletmole^-1\bullet K deg^-1 × (500 - 250) \textdegreeK = 2558 J In the second change, from (2) to (3), P and there-fore V/T are constant V_2/T_2 = V_3/T_3 orV_3 = (T_3/T_2)V_2 = (T_3/T_2)V_1 = (650\textdegreeK/500\textdegreeK) 5 lit = 6.5 lit. Heat input during this change is H_23 = nC_p(T_3 - T_2) = 0.487 mole × 29.317 J\bulletmole^-1\bulletK deg^-1 × (650 - 500)\textdegreeK = 2143 J. The total heat input during these two processes is thus H = H_12 + H_23 = 4701 J. During the change from (3) to (4), the gas cooled at constant volume. Hence P_3/T_3 = P_4/T_4 orT_4 = (P_4/P_3)T_3 SinceP_4/P_3 = 1/2 we get T_4 = (1/2)T_3 = (1/2) × 650\textdegreeK = 325\textdegreeK. The heat rejected by the gas during this process is H_34 = nC_v(T_3 - T_4) = 0.487 mole × 21.0 J\bulletmole^-1\bulletK deg^-1 × (650 - 325)\textdegreeK = 3325 J In the second cooling process; from (4) to (1), P is kept constant; V_4/T_4 = V_1/T_1 orT_1 = (V_1/V_4)T_4 SinceV_1/V_4 = V_1/V_3 = 5 lit/6.5 lit, we get T_1 = (5/6.5) × 325\textdegreeK = 250\textdegreeK as expected. The heat output during this change is H_41 = nC_p(T_4 - T_1) = 0.487 mole × 29.317 J\bulletmole^-1 K deg^-1 × (325 - 250)\textdegreeK = 1072 J. The total heat output during the cooling processes is thus H' = H_34 + H_41 = 4397 J. The difference between heat input and heat output is 304 J. This must appear as work done by the gas, since the internal energy of the gas must be the same at the beginning and at the end of a cyclic process. The mechanical work done during the cycle is given by W = ^2\int_1 P dV + ^3\int_2 P dV + ^4\int_3 P dV + ^1\int_4 P dV, which is the area enclosed by the rectangular figure in the P-V diagram. This is a rectangle of height 2 atm and length 1.5 liters. The area under the curve is thus W = 2 × 1.013 × 10^6 dynes\bulletcm^-2 × 1.5 × 10^3 cm^3 = 3.04 × 10^9 ergs = 304 J, which agrees with the net heat input.

Question:

(a) Set up the truth table for each of the following Boolean expressions: (i)\textasciitilde A (ii)A + B(inclusive or) (iii)A \textbullet B(and) (b) Show the equivalent switching circuits for (i) - (iii). (c) Show the gate representations of (i) - (iii).

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Solution:

(a) A truth table is set up by specifying all the possible combinations of values of the Boolean variables, then evaluating each term according to the rule of relation of the operation. There are 2 possible values for each variable: 0 or 1. Then, if there are n variables, there will be 2^n possible combinations of the variables, and therefore, the truth table will have 2^n rows of entries. (i) Since there is only one variable, the truth table has only 2^1 = 2 rows of entries. A \backsimA 0 1 1 0 The \backsim operation (called complement or invertion ) changes 0 to 1 or 1 to 0. (ii) Since there are 2 variables, the truth table has 2^2 = 4 rows of entries: A B A + B 0 0 1 1 0 1 0 1 0 1 1 1 The + operation (called disjunction) yields a value of 0 if and only if both variables have a value of 0. Otherwise, it yields a value of 1, (iii) Since there are 2 variables, the truth table has 2^2 = 4 rows of entries: A B A \bullet B 0 0 1 1 0 1 0 1 0 0 0 1 The \textbullet operation (called conjunction) yields a value of 1 if and only if both variables have a value of 1. Otherwise, it yields a value of 0. (b) There is a one-to-one correspondence between Boolean expressions and switching circuits. An open switch or nonconductor can represent the 0-condition and a closed switch or conductor can represent the 1- condition. Boolean algebra can be used to determine what circuits are necessary to perform a specific logic function. The action of a two-position switch with an open position and a closed position can be represented by a Boolean variable. (i) If A represents a switch in the closed position, then \textasciitilde A would represent the same switch in the open position, as shown in figure 1. (ii) The + operation yields a value of 0 if and only if both vari-ables have a value of 0. Therefore, the switching circuit needed to perform the + operation must be open (nonconducting) if and only if both switches are open. The parallel connection is the switching circuit realization of the + operation; as shown in figure 2. (iii) The \bullet operation yields a value of 1 if and only if both variables have a value of 1. The switching circuit needed to perform the \bullet operation, therefore, must be closed (conducting) if and only if both switches are closed. The series connection is the switching circuit realization of the \bullet operation, as shown in figure 3. (c) Gates are switches that are sensitive to high and low voltages. If 0 represents a low voltage and 1 represents a high voltage, the output of a gate is either 0 or 1, depending on the type of gate. (i) The \textasciitilde operation changes 0 to 1 and 1 to 0. The gate needed to perform the \textasciitilde operation, therefore, must output a 1 for a 0 input and output a 0 for a 1 input. This is the operation of a NOT gate shown in figure 4. (ii) An OR-gate outputs a 0 if and only if all inputs are 0. Other-wise, it outputs a 1. This corresponds to the + operation and is shown in figure 5 (iii) An AND-gate outputs a 1 if and only if all inputs are 1. Otherwise, it outputs a 0. This corresponds to the \textbullet operation; shown in figure 6.

The average velocity of an object in motion is the distance d it travels divided by the time t it takes in transit. v_av = d/t Looking at the figure: from A to Bv_av= d_AB/t_AB = (3 m)/(2 s) = 1.5 m/sec from A to Cv_av= d_AC/t_AC = (5 m)/(6 s) = 0.83 m/sec.

Question:

The variable capacitor in the figure is connected to a battery of e.m.f. V and internal resistance r. The current in the circuit is kept constant by changing the capacitance. (Assume that we are able to increase the capacitance in-definitely in order to accomplish this). a)Calculate the power supplied by the battery. b)Compare the rate at which energy is supplied by the battery with the rate of change of the energy stored in the capacitor.

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Solution:

a)The current in the circuit is related to the voltage across the capacitor as C (dV_c /dt) = I As the capacitor charges up, V_C rises toward its limiting value and its rate of change goes to zero. Therefore, in order to keep I constant, capacitance, C, should compensate for this drop in (dV_c /dt) (observe that, as time t \rightarrow \infty, (dV_c /dt) \rightarrow 0 and we should have C_\infty \rightarrow \infty ). Ignoring the practical difficulties of maintaining a constant I, asymp-totically we have a continuous charge flow in the circuit [(dq)/(dt)] = I, andV_C = V. The battery supplies a constant power since its e.m.f. is constant, P_B = VI . b)The capacitor is being continuously charged. Once the voltage across the capacitor stabilizes at t \rightarrow \infty, we have V = Ir + V_C, \infty orV_C, \infty= V - Ir. The energy stored in the capacitor becomes W = (1/2) q_\infty V_C, \infty = (1/2) q_\infty (V - Ir). The rate of change of W with time is P_C = [(dE)/(dt)] = (1/2) [(dq_\infty )/(dt)] VC , \infty= (1/2) I (V - Ir) The rate of dissipation of energy in the resistor r is P_r = I^2 r, Therefore the power supplied directly to the capacitor is P = P_B - P_r = IV - I^2 r = I(V - Ir) , which is twice as large as that used in charging the capacitor: P = 2P_C . The battery supplies twice as much energy to the capacitor as is stored in the capacitor. The difference is the work done by the capacitor against the external agent that is changing the capacitance.

Droplets on a leaf commonly seen early in the morning are usually assumed to be dew, which is water that has condensed from the air. Actually, these droplets may have resulted fromguttation, and come from the leaf itself through special openings on the leaf margins (and not through the stomata). Like trans-piration,guttationis the loss of water through plant leaves, yetguttationoccurs only in those plants with short stems. The phenomenon ofguttationcan be explained by the effect of root pressure. Root pressure is a term given to the forces opera-ting at the junction of root and stem in pushing a column of water up the xylem. The existence of root pressure has been experimentally demonstrated by removing the upper part of the stem of a well-watered plant, and attaching an air-tight piece of glass tubing to the remaining stump. Water actually is seen to rise in the tubing from the stump up to a height of one meter or more. If roots are killed or deprived of oxygen, root pressure falls to zero, indicating that the pushing force must be generated by the root. Root pressure has been found to equal 6 to 10 or more atmospheres of pressure even in plants that raise water to a height of less than 1 meter. Guttation usually occurs in plants when little or no transpiration is taking place. At night, for instance, when the stomata are closed, transpiration is prevented or kept to a minimum. When the air is very moist or when the air currents are still, transpiration occurs only slowly. In these cases,guttationreplaces trans-piration, and is the chief force pushing water up the stem to the leaves. A small part of the water is actually utilized by themesophyllin photosynthesis, and a large part of it passes through special pores of the leaf and appears as droplets on its surface.

Question:

In a car which is accelerating, a plumb line hanging from the roof maintains a constant angle of 30 with the vertical. What is the acceleration value?

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Solution:

Since the plumb line maintains a constant angle, the acceleration of the car must be constant. There are only two forces acting on the bob of the plumb line, the weight W^\ding{217} = mg acting downward, and the tension T^\ding{217} in the string. Splitting T^\ding{217} into its vertical and horizontal components (see figure) one obtains for the vertical direction T cos 30o = mg,(1) This results because the bob does not move in the vertical direction, hence the vertical forces must balance. By Newton's second law, T sin 30o = ma,(2) since the horizontal force must produce acceleration a to match the motion of the car. Dividing equation (2) by equation (1) T sin 30o/T cos 30o = tan 30o = a/g Thusa = g tan 30o = g(1/\surd3) = (32 ft/sec^2)/(1.732) = 18.47 ft/sec^2 .

(a) The restoring force on the pendulum for a small angular displacement \texttheta is F = - mg sin \texttheta \approx -mg \texttheta where m and g are the mass of the pendulum and the gravitational acceleration, respectively. The frictional force is f = - bv where b is a proportionality constant, and v is the velocity. The minus sign in the expressions for F and f means that they oppose an increase in \texttheta and v, respectively. The displacement along the arc is x = l\texttheta The equation of motion for displacement x is m[(d^2x)/(dt^2)]= F + f = - mg \texttheta - bv = - mg (x/l) - b (dx/dt) which gives the following differential equation for x: [(d^2x)/(dt^2)] + (b/m)(dx/dt) + (g/l)x = 0(1) In the weak damping limit, b is small and the amplitude of the oscillation is damped as shown in Fig. b. Therefore, we try the solution x(t) = A e^-t/\tau sin \omegat in (1), where e^-t/\tau is the equation of the amplitude envel-ope. We assumed that at t = 0, x = 0. We have [(d^2x)/(dt^2)] = A e^-t/\tau [{(1/\tau^2) - \omega^2} sin \omegat - (2\omega/\tau) cos \omegat] (b/m)(dx/dt) = A e^-t/\tau [(b\omega/m) cos \omegat -(b/m\tau) sin \omegat] (g/l) × = A e^-t/\tau (g/l) sin \omegat. Substituting in (1), we obtain A e^-t/\tau [{(1/\tau^2) - \omega^2 - (b/m\tau) + g/l} sin \omegat + {(b\omega/m) - (2\omega/\tau)} cos \omegat] = 0 The coefficients of sin \omegat and cos \omegat must equal zero separately; (b\omega/m) - (2\omega/\tau) = 0 (1/\tau^2) - \omega^2 - (b/m\tau) + (g/l) = 0 or(1/\tau) = (b/2m)(2) \omega^2 = (g/l) + (1/\tau^2) - (b/m\tau) = (g/l) + (b^2/4m^2) - (b^2/2m^2) = (g/l) - (b^2/4m^2). Hence the angular velocity of the motion is \omega = \surd{\omega^2 _0 - (b^2/4m^2)} = \omega_0 \surd{1 - (1/2m\omega_0)}^2,(3) where \omega^2 _0 = g/l. For weak damping, (b/2m\omega_0) = (1/\omega_0\tau) < < 1, and (3) becomes \omega \approx \omega_0 {1 - (b/m\omega_0)} = \omega_0 - (b/m).(4) (b) The kinetic energy \cyrchar\cyrk is \cyrchar\cyrk= 1/2 mv^2 = 1/2 m(dx/dt)^2 = 1/2 m[A e^-t/\tau {\omega cos \omegat - (1/\tau) sin \omegat}] = 1/2 m A^2 e^-t/\tau [\omega^2 cos^2 \omegat - (1/\tau^2) sin^2 \omegat - (2\omega/\tau) sin \omegat cos \omegat](5) Now, let us consider the time average of \cyrchar\cyrk. For this we substitute the following trigonometric identities in (5). 2 sin \omegat cos \omegat= sin 2\omegat, cos^2 \omegat= (1 + cos 2 \omegat)/2 sin^2 \omegat= (1 - cos 2 \omegat)/2 \cyrchar\cyrk = 1/2 m A^2 e^-2t/\tau[1/2 \omega^2 + (1/2\tau^2) + 1/2 \omega^2 cos 2\omegat - (1/2\tau^2) sin 2\omegat - (\omega/\tau) sin 2\omegat] The time average of a pure sine wave is zero, i.e. < cos 2\omegat > = < sin 2\omegat > = 0 Therefore, only constant terms in the expression for \cyrchar\cyrk survive when we take its time average < \cyrchar\cyrk > = 1/4 mA^2 e^-2t/\tau[\omega^2 + (1/\tau^2)]. We see that the average kinetic energy decays exponentially. The potential energy U is U= mgh = mgl (1 - cos \texttheta) \approx 1/2 mgl \texttheta2 \approx 1/2 mgl (x/l)^2 = 1/2 (mg/l) x^2 \approx 1/2 (mg/l) A^2 e^-2t/\tau sin^2 \omegat \approx 1/4 (mg/l) A^2 e^-2t/\tau (1 - cos 2\omegat) The time average of U is \approx 1/4 (mg/l) A^2 e^-2t/\tau Therefore, the time average of total energy is < E >= + < \cyrchar\cyrk > = 1/4 A^2 e^-2t/\tau[(mg/l) + m\omega^2 + (m/\tau^2)] = 1/4 A^2 e^-2t/\tau[(mg/l) + (mg/l) - (b^2/4m) + (b^2/4m)] = 1/2 \omega^2 _0 A^2 e^-2t/\tau. The rate of change of energy with time has a negative sign since the energy decreases. Therefore, the rate of dissipation, P, is positive since it is the rate of decrease (negative rate of increase) of the energy = -(d/dt) < E > = {(\omega2_0 A^2)/\tau} e^-2t/\tau = (2 < E >/\tau).

DNA, deoxyribonucleic acid, may be described as the genetic blueprint of cells and is responsible for heredity. DNA strands consist of a sugar (deoxyribose) - phosphate "backbone\textquotedblright and four important bases.The bases are divided into two main classes: purines and pyrimidines. There exists two bases under each class. The purines, are adenine and guanine? the pyrimidines are cytosine and thymine. In a particular DNA strand, these bases alter-nate with each other in a specific way - the sequence. DNA structures consist of 2 strands, each the complement of the other. Each complement refers to the fact that the bases on one strand pair up with the bases on the other strand in a particular fashion through hydrogen bonding. This bonding holds the two strands together. Adenine (A) will pair with thymine (T) . Guanine (G) will pair with cytosine (C) . Only these pairings produce effective hydrogen bonding. Thus, to determine the complementary DNA strand sequence of ATGGACGTATTC, determine the bases that corre-spond to the bases present for effective hydrogen bond-ing to occur. In the complementary strand thymine will occupy the same spaces as adenine, adenine will occupy same positions as thymine, quanine the same positions as cytosine and cytosine the same as guanine. Thus the complementary strand is TACCTGCATAAG. This is seen in the accompanying figure.

Different arrangements of atoms that can be converted into one another by rotation about single bonds are called conformations. In the anti conformation of n-butane the methyl groups (CH_3) are as far apart as possible. The gauche conformations of n-butane are with the methyl groups only 60\textdegree apart. This can be seen more clearly by using Newman projections for the conformations of n-butane. See figure A. Notice that in the gauche conformations, the methyl groups are closer together than in the anticonformation. When these groups are close together, as in the gauche conformations, steric hindrance occurs between them. This means that there is not really enough room for these two groups to be so close together. This is due to the fact that around any nuclei there is a certain volume which is taken up by the atoms electrons. This volume is determined by the van der Waal's radii of a particular atom. When atoms are too close to permit this volume, a repulsive force called van der Waals' repulsion occurs. This force acts to create a greater distance between the nucleii. This, then, raises the energy (level and lowers the stability) of the gauche conformation. In the anti conformation, the methyls are much further apart, so that you don't have van der Waal's repulsion, thus the molecule is much more stable.

The cross wind causes the plane to travel 50 mi/hr to the east in addition to its speed of 200 mi/hr to the north. To find its speed with respect to ground, use vector addition. Vectors are quantities that have both magnitude and direction; and velocity fits this specification. Using the Pythagorean theorem, we can find the magnitude of the resultant velocity v. This magnitude is the plane's speed. Speed does not have direction (note that speed is not a vector). v = \surd(v^2_airplane + v^2_wind) = \surd[(200)^2 + (50)^2] = \surd42,500 = 206 mi/hr.

The hemichordates are marine animals, often called acorn worms. They live in U-shaped burrows in sand or mud. The body is bilaterally symmetrical and consists of a conical proboscis, a short collar, and a long trunk. The hemichordates' embryonic development is similar to that of the chordates. The echinoderms, hemichordates, and chordates are deuterostomes. The characteristics shared by these phyla are the formation of the anus from the embryonic blastopore. They also undergo radial and in-determinate cleavage, with the origin of mesoderm being from pouches of endoderm, and the formation of the coelom as the cavities in the mesodermal pouches. The other in-vertebrate phyla (mollusks, annelids and arthropods) are protosomes. In protosomes, the mouth forms from the blasto-pore, cleavage is determinate and spiral, mesoderm arises from ingrowth of a single cell near the blastopore, and the coelom developes from a split in an initially solid mass of mesoderm. The Deuterostomia and the Protosomia represent two important diverging lines of development from a primi-tive bilaterally symmetrical ancestor. Adult hemichordates resemble chordates in two other ways. The hemichordates have pharyngeal gill slits. Water drawn into the mouth is forced out over these slits in the wall of the pharynx. Oxygen is removed from the water and carbon dioxide is released into it from the blood. Capillary beds are located in the septa between the gill slits. Only chordates and hemichordates have pharyngeal clefts. Hemichordates also have a thick, hollow, dorsal nerve cord in the collar. The anterior region contains a ventral nerve cord as well as a dorsal nerve cord. All chordates have a dorsal hollow nerve cord. The echinoderms are related to the hemichordates. Hemichordates have a hydrostatic skeleton similar to the water vascular system of echinoderms. In addition, there are similarities between their larval stages. The dipleurula larvae of echinoderms and hemichordates are quite alike, and differ from the trochophore larvae of molluscs and annelids. Chordates have no ciliated larval stages similar to both of these. However, this ciliated larvae link between echino-derms and hemichordates is additional indirect evidence of the link between echinoderms and chordates.

DeMorgan's theorem states that a logical expression can be complemented by complementing all its variables and exchanging AND (\bullet) operations with OR (+) operations. For example the complement of isF = AB F=A+B. Another expression for F is found by complementingF. Thus a) Complementing variables V, A, and L and changing + to \textbullet , we get Complement the entire expression to find F, Thus,V + A + L=VAL b) Again, complementing A,B,C, and D and changing + to \textbullet we get F= ABCD c) This time exchanging \textbullet for +, we have d) This time split the function into two parts F = X + D where X =ABC. Apply DeMorgan's theorem to X. thus,X =A+B+C F =A+B+C+ D

Question:

(a)If the inputs to half-adder are x and y, show that the sum output could be given by the Boolean expression: (x + y) \textbullet (\simx + \simy)(1) and the carry output could be given by: \sim (\simx + \simy)(2) (b)Set up a new circuit for a half-adder using the alternate expressions for sum and carry, given by (1) and (2) and OR, AND and NOT gates. (c)Show that (x + y) \textbullet \sim(x \bullet y)(3) is equivalent to (1) and set up a new half-adder circuit using x \bullet y as the expression for the carry output and (3) as the expression for the sum output.

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Solution:

(a) To show that the sum output is given by expression (1) and the carry output is given by expres-sion (2), we could set up their truth tables and note that the results agree with the addition table. But since this approach is used to show that the sum output could be given by; (x \textbullet \simy) + (x\sim \textbullet y)(4) and that the carry output could be given by; x \bullet y(5) we will use the laws of Boolean algebra to show that (1) is equivalent to (4) and (2) is equivalent to (5) : (x + y) \bullet (\simx + \simy) = [x \bullet (\simx + \simy)] + [y \bullet (\simx + \simy)] Distributive law of \textbullet over + = [ (x \textbullet \simx) + (x \textbullet \simy)] + [(y \textbullet \simx) + (y \textbullet \simy)] Distributive law of \textbullet over + = [0 + (x \textbullet \simy)] + [(y \textbullet \simx) + 0] Law of the excluded middle = (x \textbullet \simy) + (y \textbullet \simx) Identity of + = (x \textbullet \simy) + (\simx \textbullet y) Commutativity of \textbullet To show that (2) is equivalent to (5) : \sim(\simx + \simy) = \sim(\simx) \textbullet \sim(\simy) de Morgan's Law = x \bullet yInvolution Law (b)Setting up the terms in parentheses of equation 1 first, we create the circuit of Figure 1. Since both x + y and \simx + \simy are operands of the\textbackslash operation they will be inputs to an AND gate whose output is the sum digit. Using \simx + \simy as input to a NOT gate gives an output that is the carry digit. The complete implementation of equation (4) is shown in Figure 2. (c)De Morgan's law allows us to write: (x + y) \textbullet \sim(x \bullet y) = (x + y) \textbullet (\simx + \simy) The left and right terms in parentheses of (3) are simply outputs of OR and AND gates, respectively, where x and y are the inputs. But x \bullet y is also the expression for the carry digit. So the complete half-adder circuit using (3) is the simplest of those we've constructed.(See Figure 3).

Question:

A train whose speed is 60 mi/hr rounds a curve whose radius of curvature is 484 ft. What is its acceleration?

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Solution:

For uniform circular motion, we have an acceleration directed towards the center of curvature of magnitude a = v^2/R where v is the speed of the object, and R is the radius of the circle. To keep the units consistent, we have to con-vert the speed from mi/hr to ft/sec, since R is in feet. v = [60(mi/hr)][(5280 ft/mi)/(3600 sec/hr)] = 88 ft/sec. Substituting the appropriate values, we find the accelera-tion to be a = v^2/R = (88 ft/sec)^2 / (484 ft) = 16 ft/sec^2 .

Question:

A 10-g lead bullet is traveling with a velocity of 10^4 cm/sec and strikes a heavy wood block. If, in coming to rest in the block, half of the initial kinetic energy of the bullet is transformed into thermal energy in the block and half into thermal energy in the bul-let, calculate the rise of temperature of the bullet. (The block remains stationary during the collision.)

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Solution:

This problem involves the transfer of energy of one form (kinetic energy) into energy of another font (heat). The total energy of the bullet- block system before the bullet strikes the block is just the kinetic energy of the bullet. Or E_T = (1/2)mv^2 = (1/2) × (10 g) × (10^4 cm/sec)^2 = 5 × 10^8 ergs After the bullet strikes the block, it loses its kinetic energy. The law of conservation of energy demands that this energy appear as ther-mal energy in the block and bullet. Or E_T = KE = Q_block + Q_bullet = 2Q_bullet for we are given that Q_block = Q_bullet . Hence, Q_bullet = E_T/2 . But Q_bullet = cm∆T where c is the specific heat of the lead bul-let (the amount of heat required to raise the temperature of one gram of the substance one degree centigrade), m is the mass of the bullet, and ∆T is the rise in temperature of the bullet due to the thermal energy. The specific heat for lead is 0.0310 cal/g.c. We then have ∆T = (Q_bullet)/mc = {(1/2)ET}/mc = [{(1/2) × 5 × 10^8 ergs} / {(4.186 x 10^7 ergs/cal)(0.0310 cal/g- 0c) × 10 g}] = 19.30C

The green algae share a number of charac-teristics with the higher plants. The other algal phyla have unique characteristics which are not shared with higher plants. There are two types of chlorophyll present in most algae and plants. All photosynthetic organisms, except the photosynthetic bacteria, have chlorophyll a. The green algae and the higher plants have chlorophyll b .in addition to chlorophyll a. Neither the brown algae nor the red algae have chlorophyll b. Instead, the brown algae have chlorophyll c and the red algae possess chlorophyll d. The principal reserve material of the Chlorophyta (green algae) and of higher plants is usually starch. The dinoflagellates utilize starch and oils. The diatoms and the golden algae store food as oils and as the poly saccharide leucosin, which is a white substance. Brown algae utilize a variety of unusual carbohydrates as food reserves, one of which is laminarin. Red algae do not utilize starch, but a polysaccharide similar to it called the Floridean starch. The Euglenoids utilize the carbo-hydrate polymer paramylan as their reserve material. The cell walls of Chlorophyta and the higher plants are composed chiefly of cellulose, with very little other materials. The Euglenoids lack a cellulose cell wall, and have instead a proteinaceous pellicle, which is internal to the plasma membrane. The cell walls of the Chrysophyta (diatoms, yellow-green, and golden algae) usually consist of pectic materials impregnated with silica. These siliceous cell walls consist of two halves, which fit together like the two parts of a pill box. Some unicellular dinoflagellates may lack cell walls, but most have thick cellulose cell walls that fit together as interlocking plates to form a shell. The cell walls of red algae have large quantities of mucilaginous material. The higher plants are biochemically similar to the green algae and biochemically different from other algae. This leads biologists to propose that the ancestor of higher plants was similar to the present green algae. Also, among the Chlorophyta, there are many multicellular non-motile forms. Higher plants are always multicellular and non-motile. The Phaeophyta and Rhodophyta (brown and red algae) are also mostly multicellular and non-motile, but this is not true of the Euglenophyta, Chrysophyta, and dinoflagellates.

Question:

Hydrogen chloride (HCl) is used industrially in the manufacture of fabric dyes. The acidic properties of HCl arise because the H-Cl bond is polar. The observed dipole moment, u_obs, of HCl is 1.03 D and the H-Cl bond distance is 1.28 \AA. Estimate the percent ionic character of the HCl bond. Electron charge, e = 4.8 × 10^-18 esu.

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Solution:

A dipole moment is an indication of bond polarity. The magnitude of the dipole moment is the product of the electronic charge and the distance between the charge centers. To determine the ionic character of HCL, the magnitude of the dipole moment is needed. Percent ionic character = (U_obs) / (U_ionic) × 100, where Uionic= the magnitude of dipole moment for total ionic character. If HCl were ionic, one electron would be trans-ferred from H to Cl to give the ionic structure H^+Cl^-. Thus, the electron charge would be the charge on one electron, 4.8 × 10^-18 esu. The distance between the two atoms is given as 1.28 \AA or 1.28 × 10^-8 cm. (There are 10^-8 cm per ^---\AA.) Recalling the definition of dipole moment, u_ionic = 4.80 ×10^-10 esu × 1.28 ×10^-8 cm = 6.14 ×10^-18 esu-cm. Dipole moment is usually expressed, however, in Debyes. There are 10^18 Debyes per esu-cm. Thus, 6.14 ×10^-18 esu-cm = 6.14 Debyes = u_ionic. Therefore, percent ionic character = (1.03) / (6.14) × 100 % = 16.8 %.

Question:

Write a program which handles a credit card operation. For simplicity, assume that there are only six customers. The set ofsix data cards is 511'MARIAN GREEN'0.00 128'WILLIAM BROWN'-5.89 094'GARY WHITE'48.50 407'JUDY GREY'234.00 351'ALLEN BLACK'0.00 013'JOHN GREEN'127.14 wherethe first (three digit) integer is an account number, the seconditem is the customer's name, and the third num-ber is theunpaid balance of previous purchases. The minus sign meansBROWN overpaid and has positive balance. The second set of data cards is 013127.1483.25Yourprogram must Your program must 407100.000.0be able to handle 5110.0141.50up to 6 sets of data 1280.047.10here where the first integer is again an account number, the se-condnumber represents the payment since the last bill, andthe third number represents new purchases made with thecredit card. Write a PL/I program which reads in this data and printsout a five column table showing, for each customer, theaccount number, name, previous unpaid balance, the newunpaid balance, and the word DELINQUENT in a case wherethe payment made is less than one half the previous balance. Print out the above table in numerical order by accountnumber with the lowest number first:

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Solution:

This is a program which requires the use of ar-rays. The program illustrateshow to use DECLARE, GET LIST, manipulate data and PUT LIST statements.The problem also involves sorting. There are various waysof sorting, list; one technique is dealt with in the program. The programis as follows. Let the arrays beA- ACCOUNT NUMBER B - NAME OP CUSTOMER C - PREVIOUS UNPAID BALANCE G - NEW UNPAID BALANCE H - FOR REMARK Arrays A, C, G consist of numeric values and arrays B, H have character stringsas its element. D - SPECIFIES ACCOUNT NUMBER E - PAYMENT MADE FOR PREVIOUS BILL F - IS NEW PURCHASES MADE. These three D, E and F are not arrays, but simple variables as designated. SI, S2, S3, S4, S5 are simple variables used for sorting the creditcard operation in ascending order. The new unpaid balanceG[I] = Previous unpaid balance C[l] + Purchases F - Payment E. To check for condition of 'DELINQUENT' payment made E < (1/2)Previous unpaid balance{ -[C (I)/2]} Previous unpaid balance{ -[C (I)/2]} In sorting, the procedure applied here is to compare two adjacent elements, A[J] and A[J + 1]. IfA[J]>A[J+1] i.e. the number in A[J+1] is less than number in A[J], henceswitch these two numbers. For exampleA[1] = 5, A[2] = 3. SinceA[1] > A[2] therefore3 must be placed in A[1] and 5 in A[2] to have ascending sequenceorder. CREDIT_CARD:PROCEDUREOPTIONS(MAIN); DCL(((A,C,G)(6)) FIXED(6,2)),(B,H)(6)CHAR(15) VAR; DCL(D,E,F,S1,S3,S4) FIXED DEC(6,2),(S2,S5) CHAR(15) VAR; /\textasteriskcentered FORMS THE HEADING \textasteriskcentered/ PUT LIST ('ACCOUNTNUMBER' ,'NAME', 'PRE UNPAID BAL ' , 'NEW UNPAID BAL','REMARK'); PUT LIST ('- -----', '- ---' , '- --------' , '- --------' , '- -----'); GET LIST( (A (M) ,B(M) ,C (M) DO M = 1 TO 6)); /\textasteriskcentered TAKES IN THE FIRST SET OP CARDS \textasteriskcentered/ /\textasteriskcentered INITIALIZE ELEMENTS OF ARRAY H TO DELINQUENT EXCEPT FORC[I] \leq0 \textasteriskcentered/ DO K = 1 TO 6; IF ([K] > 0) THEN H [K] = 'DELINQUENT'; END; /\textasteriskcentered CALCULATES NEW UNPAID BAL, CHECKS FOR DELINQUENT \textasteriskcentered/ DO WHILE [1 = 1] ; ONENDFILE(SYSIN) GOTO LABEL; GETLIST[D,E,F]; DO I = 1 TO 6; IF D =A[I]THEN DO; G[I] = C(I) + F - E; IF EA(J + 1)THEN DO; S1 =A(J); A(J) = A[J+1] ; A[J + 1] = S1; S2 =B(J); B(J) = B[J + 1]; B[J + 1] = S2; S3 =C[J]; C[J] = C[J + 1]; C[J + 1] = S3; S4 =G(J); G[J] = G(J + 1); G(J + 1) = S4; S5 =H(J); H(J) = H(J + 1); H(J + 1) = S5; J = 0; END; END; DO N = 1 TO 6; PUT LIST(A(N),B(N),C(N),G(N),H(N)) SKIP (2); END; END CREDIT_CARD;

A tissue may be defined as a group or layer of similarly specialized cellswhich together perform a certain specific function or functions. Each kindof tissue is composed of cells which have a characteristic size, shape andarrangement, and which are bound together by an intracellular substanceallowing communication between adjacent cells. Some examplesof tissues are epithelial tissue, which separates an organism fromits external environments; muscle tissue, which contracts and relaxes; and nervous tissue, which is specialized to conduct information. An organ is composed of various combinations of tis-sues, grouped togetherinto a structural and functional unit. An organ can do a specific job: the heart pumps bloodthrough outthe circulatory system; or an organ canhave several different functions: the human liver produces bile salts, breaksdown red blood cells and stores glucose as glycogen. A group of organs interacting and cooperating as a functional complexin the life of an organism is termed an organ system. In the humanand other vertebrates, the organ systems are as follows: (1)thecirculatory system, which is the internal trans-port system of animals; (2)therespiratory system, which provides a means for gas exchange betweenthe blood and the environment; (3)thedigestive system, which functions inprocurringand processing nutrients; (4)theexcretory system, which eliminates the waste products of metabolism; (5)theintegumentarysystem, which covers and protects the entire body; (6)theskeletal system, which supports the body and pro-vides for body shapeand locomotion; (7)themuscular system, which functions in movement andlocomotions; (8)thenervous system, which is a control system essen-tial in coordinating andintegrating the activities of the other systems with themselves and withthe external environment; (9)theendocrine system, which serves as an additional coordinator of the bodyfunctions; and (10)thereproductive system, which functions in the production of new individuals for the continuation of the species.

Question:

An instrument that has been used to measure atomic masses with great accuracy is the 180\textdegree mass spectrometer shown in the figure. Ionized atoms from a source pass through a potential difference and enter a region in which there is a uniform magnetic field B. The magnetic force acts at right angles to the direction of the velocity vector and causes the ions to move along a circular path. After completing one-half revolution (180\textdegree), the ions strike a detector, usually a photographic film. (a) Show that the charge/mass ratio, q/m, of these ions is given by q / m = v / BR where v is the speed of the ions, B the magnetic field strength, and R the orbit radius. (b) In an experiment the orbit radius is observed to be 0.2 m for singly ionized atoms of speed 2.1 x 10^5 m/s in a magnetic field of 0.13 T. What is the charge/mass ratio of these ions? (c) What is the mass of these atoms?

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Solution:

(a) For circular motion the centripetal acceleration is a = v^2 / R. Newton's second law of motion requires that the force exerted on the ion be F = ma = m(v^2 / R^2). The magnitudes of the electric potential and magnetic fields determine the value of q/m for protons. The charge of the proton, q = 1.6 x 10^\rule{1em}{1pt}19 C, is known; therefore, the mass may be computed as follows: An ion of charge q is accelerated through a potential difference V. It attains a velocity given by (1 / 2)mv^2 = Vq since the potential energy of the ion at the source is Vq with respect to the inlet of the magnetic deflection area. The deflecting force on the charge due to the magne-tic field is F = qvB = m(v^2 / R^2) (by Newton's second law) giving V = qBR / m. Therefore the ratio q / m for the spectrometer is , q / m = v / BR. The value of q/m for protons from this type of experiment is approximately 9.57 × 10^7 c / kg. We find that for the proton q / m = 9.57 × 107C/kg but q = 1.67 × 10^\rule{1em}{1pt}19c m =(1.6× 10^\rule{1em}{1pt}19c) / (9.57 × 107c/ kg) = 1.67 ×10^\rule{1em}{1pt}27kg. (b) The given quantities are v = 2.1 x 10^5 m/s, B = 0.13 T, and R= 0.2 m. The charge/mass ratio for these ions is found to be q / m = (2.1 × 10^5m/ s) / [(1.3 × 10^\rule{1em}{1pt}1 T)(2 × 10^\rule{1em}{1pt}1m)] = 8.1 × 10^6 c / kg. (c) The charge q of the ions is 1.6× 10^\rule{1em}{1pt}19 c, therefore, the mass is given by m = q / (q / m) = (1.6 × 10\rule{1em}{1pt}19c)/ (8.1 × 106c/ kg) = 2.0 × 10\rule{1em}{1pt}26kg.

Question:

A chemist possesses 1 cm^3 of O_2 gas and 1 cm^3 of N_2 gas both at Standard Temperature and Pressure (STP). Compare these gases with respect to (a) number of molecules, and (b) average speed of molecules.

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Solution:

This problem requires the use of the kinetic theory of gases and Avogadro's principle. Avogadro proposed the principle that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Both gases have equal volumes, 1 cm^3, and are at the same temperature and pressure (STP). According to Avogadro's Principle, both have the same number of molecules. The average speed of these molecules can be found from kinetic theory of gases. The rate of diffusion of a gas is directly proportional to the average speed of the molecules. According to Graham's law, there is an inverse proportionality between diffusion rate and the square root of the molecular weight (mass). In other words, rate of diffusion of N_2 / rate of diffusion of O_2 = V_(N)2 / V_(O)2 = \surd[M_(O)2 / M_(N)2] , where V = velocity and M =masM_(N)2 = 28, M(O)2= 32. In this case, mass = molecular weight. Thus, V_(N)2 / V_(O)2 = \surd[32/28] = 1.07. Thus, if the average speed of O_2 is one and the average speed of N_2 is 1.07.

Question:

Draw the structure of 4-ethyl-3, 4-dimethyl-2-hexene.

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Solution:

To draw the structure of more complex compounds, such as this one, certain steps must be followed. (1) Identify the parent compound that associated with the longest carbon chain that contains the functional group. In 4-ethyl-3, 4-dimethyl-2 hexene, the parent compound is hexene. (2) Draw the parent carbon skeleton, in this case, a six carbon chain. Do not put any hydrogen atoms in yet. (3) Number the carbon atoms starting at either end. This is important; otherwise it may get confusing when one adds the functionality. (4) Add the suffix functionality, in this case -2-ene. "Ene" tells one that a double bond is present, while "2" indicates that it is at the second carbon. (5) Add the prefix functionality, starting at the beginning of the name and continuing until the parent name is reached. Here, the prefixes are 4-ethyl-3, 4-dimethyl-. Now, the hydrogen atoms can be added to give a complete structure.

The blue-green algae, (Cyanophyta or cyanobac-teria) are prokaryotic unicellular or filamentous organ-isms. The filamentous forms are strings of individual cells held together by fused walls. (See Figure 1.) In most species, there are no protoplasmic connections, such as plasmodesmata, between the adjacent cells. The uni-cellular forms exist as either single rods or spheres, and each individual cell is capable of carrying out all necessary life processes. The Cyanophyta are the most primitive chlorophyll- containing autotrophic organisms now living. The Cyanophyta are quite different from other algae, being structurally similar to bacteria. These similarities have formed the basis for grouping them with the bacteria in the kingdom Monera. All blue-green algae possess photosynthetic pigments, which are located in folds or convolutions of the cell membrane extending into the interior of the cell. These photosynthetic pigments are chlorophyll a, carotenoids, phycocyanin (blue pigment) and sometimes phycoerythrin (red pigment). Like the bacteria, the blue-green algae lack mitochondria, Golgi apparatus, a nuclear membrane, endoplasmic reticulum and the large cell vacuole charac-teristic of higher plants. Ribosomes are present along with many proteinaceous granules and granules of a stored carbohydrate material known as cyanophycean starch, which is very similar to glycogen. The nuclear region consists of a single circular chromosome composed of double--stranded DNA. The cell walls of blue-green algae contain some muramic acid and cellulose. Outside this cell wall is a sticky gelatinous sheath composed of pectic materials. It covers the entire cell. The cytoplasm of blue-green algae is unusually viscous, being composed of a very dense colloidal material. Cell division in blue-green algae is accomplished by binary fission. Reproduction also occurs frequently by fragmentation of filaments. Sexual reproduction has never been observed in the Cyanophyta but is thought to occur infrequently. No blue-green algae possess flagella, yet many species are capable of movement. The filamentous forms exhibit a peculiar slow gliding or oscillatory motion. The blue-green algae are found in many varied habitats. Most are found in fresh water pools and ponds. A few species are found in hot springs, at temperatures up to 85\textdegreeC. Some species are marine while others are common in soils, the banks of trees and the sides of damp rocks. Some species of blue-green algae exhibit a symbiosis with fungi, in which both act together to form a lichen.

First, we compute the field due to a cylindrical shell of charge. We note that the field must be radial since no other direction is preferred. We erect as a Gaussian surface, a cylinder, of radius r (r > R) and height l, concentric to the shell. (See figure (a)). By Gauss' law (since E is constant along any cylinder concentric to the axis): \int E^\ding{217} \textbullet dA^\ding{217} = 4\piq, 2\pirlE = 4\pi(s\piRl\sigma) where \sigma = q/area is the surface change density of the shell. We note that since the field is radial, the flux through the top and bottom of the Gaussian cylinder is zero. Thus: E = 4\piR\sigma/r = 2 [{2\piR (q/2\piRl)}/r] = [2(q/l)]/r = 2\lambda/r where \lambda = q/l is the linear charge density. This formula holds for a cylindrically symmetrical charge distribution since we can consider such a configuration as the superposition of many cylindrical shells of charge density \lambda_i , i = 1, 2, 3,... E = (2\lambda_1/r) + (2\lambda_2/r) + (2\lambda_3/r) +... = 2/r (\lambda_1 + \lambda_2 + \lambda_3 +...) where \lambda' = \lambda_1 + \lambda_2 + \lambda_3 + ... = q_1/l + q_2/l + q_3/l + ... = (q total)/l. Now, we compute the field due to an infinite line charge. Figure (b) depicts an infinite line of charge of linear charge density \lambda, with a cylindrical Gaussian surface of radius r and height l, surrounding it. Again, we see as with the cylindrical shell, that the field must be radial since no other direction is preferred. Thus, the flux through the top and bottom of the Gaussian surface is zero. Ap-plying Gauss's law: \int E^\ding{217} \textbullet dA^\ding{217} = 2\pirl E = 4\pil\lambda = 4\piq E = 2\lambda/r. The same result is thus obtained.

H\ding{217}= r\ding{217}× p\ding{217}where p\ding{217}is the linear momentum of the electron, and \ding{217} \ding{217} \ding{217} \ding{217} r\ding{217}is the vector from the proton to the electron. Because the orbit of the \ding{217} electron is circular, r\ding{217}is perpendicular to p\ding{217}. Hence \ding{217} \ding{217} \vertH\ding{217}\vert = H = rp = mrv \ding{217} But v =\cyrchar\cyromegar, where \cyrchar\cyromega is the angular velocity of the electron. The frequency of revolution is f = 1/T, where T is the period , or the time it takes for the election to make one revolution in its orbit. In T units of time, then, the electron goes through an angle of 2\pi radians. Hence \cyrchar\cyromega = 2\pi/T Therefore, H = mr^2\cyrchar\cyromega = (m)(2\pi/T)(r^2) = H = (9.11 × 10^-31 kg){(2\pi rad)/(1.51 × 10^-6 sec)} (0.53 × 10^-10m)^2 = 1.06 x 10^-34 kg-m^2 /sec.

(a)The magnetic field B inside the solenoid, due to current I flowing in it, is given by B = \mu_0 n I. Then, the magnetic flux \varphi_B inside the coil becomes \varphi_B = BA = \mu_0 nA I and the rate of change of this flux is [(d\varphi_B )/(dt)] = A [(db)/(dt)] = \mu_0 nA [(dI)/(dt)] = (4\pi × 10^-4 (w/a-m) [1000 (turns/m)] (4 × 10^-2 m^2) [100 (amp/sec)] = 16\pi × 10^-7 w/sec (b)A long solenoid can be thought of as a long cylindrical current with the magnetic field outside becoming negligible in comparison to the field inside (at least far away from the edges). Therefore, the problem reduces to that of finding the electric field Induced by a time varying uniform magnetic field B while it is Increasing (see Fig. 2). For this, we shall employ Faraday's law around the circle with radius d: \oint E^\ding{217}.dl^\ding{217} = - (d\varphi_B /dt ) The direction and shape of the electric field can be obtained by remembering that, it will tend to set up a new flux to oppose the increase in magnetic flux through the coll. The electric field is circular and it is constant around the path we defined for the above integral. Therefore it can be taken outside the integral sign. E^\ding{217} \oint dl^\ding{217} = El = E 2\pid = - (d\varphi_B)/(dt) orE = - [1/(2\pid)] [(d\varphi_B)/dt)] = 8 × 10^-5 (w/m \bullet sec) = 8 × 10^-5 (v/m).

Each register-reference instruction is speci-fied when the expression q_7I ̅c_2t_3 [MBR]_i is true (logical 1) where the i_th bit of the MBR is the high bit specified by each instruction. All the instructions can be exe-cuted during one timing signal t_3 . CIR:The circulate right instruction is explained with the use of Figure 1. The content of each bit is shifted to the next higher bit, E is shifted to bit 0 of the AC, and bit 15 of the AC is lost. q_7I ̅c_2 [MBR] _8 t_3 :n\leftarrow15 AC_n \leftarrow [AC]_n-1;for 1 \leq n \leq 15 A_C0\leftarrow [E] E\leftarrow0 F\leftarrow0 CIL:This is just the opposite of the CIR instruction, with E getting the bit shifted off the end. q_7I ̅c_2t_3 [MBR] _9 :n\leftarrow0 AC_n\leftarrow [AC]_n+1;for 0 \leq n \leq 14 AC_15\leftarrow E E\leftarrowAC_0 F\leftarrow0 SPA:The PC is incremented only if the AC is positive. The AC is positive when bit 0 is 0 and the AC is negative when bit 0 is 1 (2's complement notation). Note that when the AC is zero it is also positive (zero is treated as a positive number). q_7I ̅c_2t_3 [MBR]_11 [AC]_0 :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_11 [AC]_0 :F\leftarrow0 The SNA and SZA micro-instructions are written in a similar manner. SNA : q_7I ̅c_2t_3 [MBR]_12 [AC]_0 :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_12 [AC]_0 :F\leftarrow0 SZA:The AC is zero when A_0A_1 . . . A_15 = 0. q_7I ̅c_2t_3 [MBR]_13 [A_0+A_1+ . . . +A_15] :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_13 [A_0+A_1+ . . . +A_15] :F\leftarrow0 SZE:The PC is incremented if E = 0. q_7I ̅c_2t_3 [MBR]_14 E ̅ :PC\leftarrow[PC]+1 F\leftarrow0 orq_7I ̅c_2t_3 [MBR]_14 E :F\leftarrow0

Question:

A man and a woman are heterozygous for tongue rolling, and have three sons, The three sons marry women who are not tongue rollers. Assuming that each of the three sons has a different genotype, show by diagram what proportion of their children might have the ability to roll their ton-gues.

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Solution:

Let us again use the letter R to represent the dominant tongue rolling allele, and r to represent the recessive allele. Only three possible genotypes could result from the cross of the heterozygous parents: These alleles are:RR, Rr and rr. Let one of the sons be RR and therefore a tongue roller, the second son Rr and also a tongue roller, and the third son rr (homozygous recessive) and so unable to roll his tongue. All three sons have married woman who are not tongue rollers and who are therefore rr. Let us now determine the proportion of the offspring who will be able to roll their tongues. F_1Rr100% Rr: tongue rollers 50% Rr: tongue rollers F_1Rr;rr50%rr: non-tongue rollers F_1rr100% rr: non-tongue rollers As a result of the three crosses, we can see that of the total offspring from the three marriages, half will be tongue rollers and half will be non-tongue rollers. The tongue rollers consist of all the children of the first son plus half the children of the second son. The non-tongue rolling children consist of all the offspring of the third son's marriage plus half the offspring from the second son's marriage.

Question:

Using a specific case, show that the effect of a 10\textdegreeK rise in temperature will have a greater effect on the rate constant, k, at low temperatures than it does at high temperatures.

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Solution:

The best way to demonstrate this fact is to use the natural logarithm form of Arrhenius' equation, which gives an indication of the effect of temperature on reaction rate. The expression may be written lnk= lnA - E_a /2.303 RT, where k is the rate constant, A = the Arrhenius constant, E_a = activation energy, R = universalgas constant (8.314 J mol^-1 deg^-1) and T = temperature in Kelvin (Celsius plus 273). For a low temperature, consider room temperature (T = 300\textdegreeK). For a first-order reaction, A might be 1.0 × 10^14sec^-1 and E might be 80 k J/mole. Thus, at T = 300\textdegreeK, you have log k = log (1 × 10^14) - 80,000 / (2.303) (8.314) (300) = .07. Thus, k = 1.2 sec^-1. If you increase the temperature 10\textdegreeK to 310\textdegreeK, you find by exactly the same method of calculation, that k = 3.3 sec^-1. Thus, at a temperature of 10\textdegreeK higher (in a low temperature range), k was increased nearly three times. Now consider a higher temperature range, say T = 900\textdegreeK. For this, log k = (1 × 10^14) - 80,000/(2.303) (8.314)(900) = 9.358. Solving log k = 9.358 k = 2.28 × 10^9. Again, let us increase the temperature by 10\textdegreeK to 910\textdegreeK. Using the same type of calculations, you find that k becomes 2.56 × 10^9. The percent change = {(2.56 × 10^9 - 2.28 × 10^9)/ (2.28 × 10^9)} × 100= 12.28% . Thus, at the higher temperatures, the rate constant was increased only 12% as compared to 300% at the low tem-peratures .

Question:

A variable resistor in series with a 2-V cell and an ammeter is adjusted to give a full-scale deflection on the meter, which occurs for a current of 1 mA. What resistance placed in series in the circuit will re-duce the meter readings by 1/f? The meter is calibrated to measure resistance on this basis, but the emf of the cell drops by 5% and the variable resistor is readjusted so that the full-scale deflection again corresponds to the zero of the resistance scale. What percentage error is now given on a resistor which has a true resistance of 3800 \Omega?

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Solution:

The total resistance in the circuit when the meter is giving full-scale deflection is R = (\epsilon/I) = [(2 V)/(10^-3 A)] = 2000 \Omega, for we are given that full scale deflection on the ammeter occurs for a current of 1 mA = 1 × 10^-3 A. If an unknown resistance X is added to the circuit and produces a meter reading of (1/f) mA, then R + X = [(2 V)/{(1/f) × (10^-3 A)}] = 2000 f \Omega. \thereforeX = (2000f - 2000)\Omega = 2000(f - 1)\Omega.(1) The emf of the cell now drops to 95/100 (= 95%) of 2 V = 1.9 V. Alternatively since we are given that the emf of the cell drops by 5%, then the emf loss is 5/100 of 2 V = .1 V. The new emf is now 2V - .1 V = 1.9 V. For full-scale deflection the resistance in the circuit is now R' = [(1.9 V)/(10^-3 A)] = 1900 \Omega, and if a further resistance of 3800 \Omega is inserted in the circuit, the current is 1.9 V/(1900 + 3800)\Omega =1/3 mA. But from the meter calibration, when the current drops to one-third of its value (i.e. the meter read-ing is reduced by 1/f = 1/3, or f = 3) the inserted resistance should have a value given by (1) X = [2000 (3 - 1)]\Omega = 4000 \Omega. The error in the reading is thus (4000 - 3800)\Omega = 200 \Omega, and the percentage error is (200/3800) × 100% = 5.3% .

The game of life can be played either manually on a checkerboard square or on a CRT terminal. The game may be interpreted as a dynamic example of the laws of genetics. The rules of the game are as follows: 1) The game is played on a chessboard (imagined infinite) with counters. An initial distribution of counters is as-sumed. The pattern of the initial distribution changes ac-cording to the number of survivals, deaths and births. 2) Survivals: Every counter with two or more neighboring counters survives for the next generation. Neighbors are defined by the diagram 7 6 5 8 C 4 1 2 3 We see that counter C has eight potential cells on which neighbors may sprout. 3) Deaths: Each counter with four or more neighbors dies (is removed) because of over-population. Every counter with one or no neighbors dies from isolation Fig. 2 In Fig. 2 the circled counters die from isolation and are removed. The squared counter dies from lack of resources. 4) Births: Each empty cell adjacent to three and only three neighbors is a birth cell. A counter is placed on it on the next move. Fig. 3 In Fig. 3 the circled cell will have a counter placed on it in the next generation (move). Finally, note that births, deaths and survivals consti-tute a move. The game continues until a stable population is reached (no more births or deaths) or until the popula-tion dies out. An oscillating population is also possible. As an example, consider the following part game: InitialXX InitialXX Distribution :XXX X X Generation 1:XX XXXX XX X Generation 2:XX XXX XX XX Generation 3:XX XXX X XX Generation 4:XXX XXX X The program follows: 10PRINT "LIFE" PRINT "LIFE" 20PRINT "ENTER YOUR PATTERN. TYPE `DONE' WHEN FINISHED. " 30X1 = 1: Y1 = 1: X2 = 24: Y2 = 70 40DIM A (24, 70), B$ (24) 50C = 1 60INPUT B$ (C) 70IF B$ (C) = "DONE" THEN B$ (C) = " " : GO TO 120 80IF LEFT $ (B$ (C), 1) = " . " THEN B$ (C) = " " + RIGHT $ (B$ (C), LEN (B$ (C) ) - 1) 100C = C + 1 110GOTO 60 120C = C - 1: L = 0 130FOR X = 1 TO C - 1 140IF LEN (B$ (X) ) > L THEN L = LEN (B$ (X) ) 145REM SEE END OF PROGRAM 150NEXT X 160X1 = 11 - C/2 170Y1 = 33 - L/2 180FOR X = 1 TO C 190FOR Y = 1 TO LEN (B$ (X) ) 200IF MID$ (B$ (X), Y 1) <> " " THEN A (X1 + X, Y1 + Y) = 1: P = P + 1 220NEXT Y 230NEXT X 240PRINT: PRINT: PRINT 250PRINT "GENERATION: " ; G, "POPULATION: " ; P ; 260IF I9 THEN PRINT "INVALID"; 270X3 = 24: Y3 = 70: X4 = 1: Y4 = 1: P = 0 280G = G + 1 290FOR X = 1 TO X1 - 1: PRINT: NEXT X 300FOR X = X1 TO X2 310FOR Y = Y1 TO Y2 320IF A (X,Y) = 2 THEN A (X,Y) = 0: GOTO 400 330IF A (X,Y) = 3 THEN A (X,Y) = 1: GOTO 350 340IF A (X,Y) <> 1 THEN 400 350PRINT TAB (Y); " \textasteriskcentered " ; 360IF X < X3 THEN X3 = X 370IF X > X4 THEN X4 = X 380IF Y < Y3 THEN Y3 = Y 390IF Y > Y4 THEN Y4 = Y 400NEXT Y 410NEXT X 420FOR X = X2 + 1 TO 24: PRINT: NEXT X 430X1 = X3: X2 = X4: Y1 = Y3: Y2 = Y4 440IF X1 < 3 THEN X1 = 3: 19 = - 1 450IFX2 > 22 THEN X2 = 22: I9 = - 1 460IF Y1 < 3 THEN Y1 = 3: I9 = - 1 470IF Y2 < 68 THEN Y2 = 68: I9 = - 1 480P = 0 490FOR X = X1 - 1 TO X2 + 1 500FOR Y = Y1 - 1 TO Y2 + 1 510C = 0 520FOR I = X - 1 TO X + 1 530FOR J = Y - 1 TO Y + 1 540IF A (I,J) = 1 OR A (I,J) = 2 THEN C = C + 1 550NEXT J 560NEXT I 570IF A (X,Y) = 0 THEN 610 580IF C < 3 OR C > 4 THEN A (X,Y) = 2: GOTO 600 590P = P + 1 600GOTO 620 610IF C = 3 THEN A (X,Y) = 3: P = P + 1 620NEXT Y 630NEXT X 640X1 = X1 - 1: Y1 = Y1 - 1: X2 = X2 + 1: Y2 = Y2 + 1 650GOTO 250 660END If your computer does not accept conditional execution, change it into few simple statements with the same meaning. For example: 140 IF LEN (B$ (X)) > L THEN L = LEN (B$ (X)) can be changed into: 140IF LEN (B$ (X)) > L THEN 142 141GOTO 150 142L = LEN (B$ (X) )

The molecular formula of a substance indicates the relative number of atoms in a molecule of the substance. Therefore, to solve this problem, you must first calculate the ratios of the gram-atoms to each other, the empirical formula, and then extrapolate to the molecular formula via the molecular weight. The number of gram-atoms of phosphorus (P) is (wt. in grams of P) / (atom weight) = (10) / (30.97) = 0.323 gram - atoms P. For oxygen we have (7.77g) / (16.00) = 0.484 gram-atoms The weight in grams of oxygen is 7.77 because the final product weighs 17.77 g and the phosphorus weighs. 10.00 g. Since, the only other element is oxygen, its weight must be the difference. The ratio of the gram-atoms of P and O is respect-ively 1 : 1.5 or 2 : 3. Therefore, the empirical formula of the oxide is P_2O_3. To calculate the molecular formula, we must use the stated molecular weight of 220. We must look for a formula that totals to this molecular weightANDmaintains the 2 : 3 ratio of P : O as expressed in the empirical formula. With some arithmetic, we find that the only formula that meets these two requirements is P_4O_6 . 4 : 6 is the same as 2 : 3. The atomic weight of P and O is respectively 30.97 and 16. We have four P atoms for a total of 123.88 and we have 6 O atoms for a total of 96. Now add: 123.88 + 96 = 219.88, which is approximately 220. Another method for determining the molecular formula is to divide the molecular weight of the molecule, 220, by the weight of 1 P_2O_3, 110. no. of P_2O_3 = (220g) / (110g/mole of P_2O_3) = 2 mole of P_2O_3. The formula is, therefore 2 × P_2O_3 or P_4O_6.

Question:

As shown in the figure, a block of mass .5 slugs moves on a level frictionless surface, connected by a light flexible cord passing over a small frictionless pulley to a second hanging block of mass .25 slugs. What is the acceleration of the system, and what is the tension in the cord connec-ting the two blocks?

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Solution:

In order to find the system's acceleration, we must relate the net force on the system to the acceleration via Newton's Second Law. First we isolate the rope, and calculate its acceleration. By the second law, T_1 - T_2 = m_ropea where T_1 and T-_2 are in opposite directions. In this problem, we assume m_rope = 0, and T_1 = T_2 Hence, the rope acts only to transmit the force of tension to the block. Applying Newton's Second Law to the horizontal (x) and vertical (y) directions of motion of the block on the table, we obtain T = m_1a_(x)1 N - w_1 = a(y)1 where m_1 is the mass of the block on the table, and a_(x)1 is its horizontal acceleration. Noting that a_(y)1 = 0, since the block doesn't accelerate vertically, we find T = m_1a_(x)1(1) N = w1(2) We next apply the third law to the hanging block of mass m_2, and m_2g - T = m_2a_(y)2(3) where a_(y)2is the vertical acceleration of block 2. Now, since the 2-block system moves as a unit, a_(x)1 = a_(y)2 = a, and, using (1), (2) and (3) T = m_1a(4) N = w1(5) m_2g - T = m_2a(6) Substituting (4) in (6), and solving for a m_2g - m_1a= m_2a a= m_2-g/(m_1 + m_2) From (4),T= m_1m_2g/(m_1 +m_2) Substituting the given data in these equations a = [{(.25 sl)(32 f/s^2)} / {.75 sl}] = 10.7 f/s^2 T = (m_1m_2g) / (m_1 + m-_2) = m_1a = (.50 sl)(10.7 f/s^2) T = 5.4 lb

An assembler is a program that accepts a symbolic language and produces its binary machine language equiva-lent. The input symbolic program is called the source pro-gram and the resulting binary program is called the object program. The assembler is a program that operates on char-acter strings and produces an equivalent binary interpre-tation. An assembler must know the arrangment of the char-acters in memory so it can distinguish between symbols and lines. The symbolic program constitutes the input data for the assembler. Now we shall see how a two pass assembler translates the symbolic program of the previous problem. First Pass: A two pass assembler scans the entire sym-bolic program twice. During the first pass, it generates a table that correlates all user defined address symbols with their binary equivalent value. The binary translation of the program is done during the second pass. In order to keep track of the location of instructions, the assembler uses a memory word called location counter (LC). The ORG pseudo instruction initializes the LC to the value of the first location; in case ORG is missing LC is set to 0 initially. The tasks performed by the assembler during the first pass are described in the flow chart of Fig. 1. LC is initially set to 0. A line of symbolic code is scanned to see if it has a label (by presence of comma).If the line of code has no label, the assembler checks in the instruction field. If it contains ORG, it sets LC to the number followed by ORG. If it is END pseudo instruc-tion, the assembler terminates the first pass and goes to the second pass: If the line of code has a label, it is stored in the address symbol table together with its binary equivalent number specified by LC. Label LC MIN 100 SUB 107 DIFF 108 Fig. 2 On the first pass, the assembler encounters ORG in the example program (previous problem) and sets the LC to 100. It then scans the next line. If it has neither label nor END, LC is incremented by 1 to 101 & proceeds until HLT, when LC has value of 105. Now LC is incremented by 1 to 106. Upon scanning, it finds label MIN, then SUB and DIFF. The address symbol table is shown in Fig. 2. On encounter-ing the last instruction, END, the assembler goes to sec-ond pass. Second Pass: - Machine instructions are translated during the second pass by means of table-lookup procedures. A table lookup procedure is a search of table entries to determine whether a specific item matches one of the items stored in the table. The assembler uses four tables. 1) Pseudo-instruction table. 2) MRI table. 3) Non-MRI table. 4) Address-symbol table. The way the assembler proceeds in the second pass is described in the flow chart of Fig. 3. LC is initially set to 0. Labels are neglected during the second pass, so the assembler goes immediately to the instruction field. It first checks the pseudo-instruction table. For a match with ORG, the assembler sets LC to num-bers followed by ORG. A match with END terminates the translation process. An operand pseudo instruction causes a conversion of the operand into binary. This operand(e.g. DEC 50 from the previous problem) converted into its binary value is placed in the memory location specified by con-tents of LC i.e. at location 106. The LC is incremented by 1 and the assembler continues to analyze the next line of code. If the symbol encountered is not a pseudo-instruction, the assembler refers to the MRI table. If the symbol is not found in this table, the assembler refers to the non- MRI table e.g. in a scan of the second line, the symbol is CMA with LC = 101. The assembler stores the bit code of CMA into a memory location specified by LC i.e. at place 101. Again LC is incremented by 1 and a new line is analyzed. When a symbol is found in the MRI table, the assembler ex-tracts its equivalent code and inserts it in, say, bits 2-4 of the memory location specified by LC. The correspond-ing second symbol is an address. This is converted to bi-nary by searching the address symbol table. The first bit of instruction is set to 0 or 1, depending on whether the letter I is absent or present. For example the first line from the previous problem is LDA SUB. LDA is an MRI and its equivalent 3 bit code, say, 010 is set in bit position 2-4, SUB is searched in the address symbol table and is found to be at location 107. I is not present, therefore, LDA SUB has the HEX code 2107; its equivalent binary is stored in memory location specified by LC = 100. Therefore memory location 100 contains 2107. One important task of an assembler is to check for possible errors in the symbolic program. This was just the basics of how an assembler performs translation. A practical assembler is much more complicated.

We need the principal axes of the target to solve this problem. With the simplifying assumption that the bullet does not change the principal axes, note that the x, y, z axes shown are the principal axes. I = I_z = (1 / 2)ma^2. Find I' = I_x = I_y by use of the parallel-axis theorem. The moment of inertia of the disk with respect to an axis through it, parallel to the x or y axis in the figure, is (1 / 4)mr^2 . Since the mass of each disk is [(1/2)m] and the distance from the parallel axis to the axis in the figure is a, the moment of inertia of one disk is I' = I_x = I_y = (1/4)[(1/2)m]a^2 + [(1/2)m]a^2 . Due to symmetry, it is possible to multiply by two to find the Moment of Inertia of the total system. I' = I_x = I_y = 2[(1/4)[(1/2)m]a^2 + [(1/2)m]a^2] = (5/4)ma^2 The total angular momentum of the target-bullet system is conserved during collision and remains constant during the ensuing motion. Taking moments about the coordinate origin G yields H^\ding{217}_G = -a\^{\j} × m_0v_0kˆ + l\cyrchar\cyromega_0kˆ H^\ding{217}_G = -m_0v_0\^{\i} + l\cyrchar\cyromega_0kˆ(1) Now determine the angular velocity after impact by invoking the conservation of H_G. Using H^\ding{217} = K_x\cyrchar\cyromega_x\^{\i} + l_y\cyrchar\cyromega_y\^{\j}+ l_z\cyrchar\cyromega_zkˆ, the result is -m_0v_0a = l'\cyrchar\cyromega_x = (5/4)ma2\cyrchar\cyromega_x \cyrchar\cyromega_x = (4/5)[(m_0v_0) / (ma)]; 0 = I'\cyrchar\cyromega_y , \cyrchar\cyromega_y = 0; I\cyrchar\cyromega_0 = I\cyrchar\cyromegaz \cyrchar\cyromega_z = \cyrchar\cyromega_0. \cyrchar\cyromega^\ding{217} = -(4 / 5) [(m_0v_0) / (ma)]\^{\i} + \cyrchar\cyromega_0kˆ.(2) While this equation is correct it is not very helpful since the body axes x, y, z are moving in space in an unknown way. For the target considered, \cyrchar\cyromega_0 = 40 rpm = 4.19 (rad / sec), )(m_0 / m) = (1 / 100), a = 0.400 m, v_0 = 100 (m / sec); we find \cyrchar\cyromega_x = -2 (rad /sec), \cyrchar\cyromega_y = 0, \cyrchar\cyromega_z = 4.19 (rad / sec). This gives \cyrchar\cyromega = \surd[(\cyrchar\cyromega^2_x) + (\cyrchar\cyromega^2_z)] = 4.64 (rad / sec) = 44.3 rpm with angle \Upsilon = arctan [(-\cyrchar\cyromega_x) / (\cyrchar\cyromega_z)] = 25.5\textdegree to the z-axis. (Figure 2). The angle \texttheta shown in fig. 2 is between the precession axis and the z-axis. (H_G is fixed in space; the target wobbles around it.) \texttheta = arctan[(m_0v_0a) / (I\cyrchar\cyromega_0)] = arctan [(2 m_0v_0)( ma\cyrchar\cyromega_0)] \texttheta = 50.0\textdegree . Figure 3 shows \cyrchar\cyromega^\ding{217}, the space and body cones and \varphi ׂ and \Psi ׂ which, added vectorially, must equal \cyrchar\cyromega^\ding{217}. Using the law of sines, \mid[(\cyrchar\cyromega) / {sin(\pi - \texttheta)}]\mid = \mid[(\varphi ׂ ) / (sin \Upsilon )]\mid = \mid[(\Psi ׂ ) / {sin(\texttheta -\Upsilon)}]\mid; (\cyrchar\cyromega / sin \texttheta) = (\varphi ׂ / sin \Upsilon)= (\Psi ׂ / sin \texttheta -\Upsilon) . This gives \varphi ׂ = 24.9 rpm , \Psi ׂ = 24.0 rpm.

To solve this problem properly you need to know the definition of a unit cell and lattice point. You need, also, to know the actual structures of the unit cells given. You proceed as follows: A unit cell is that small fraction of a space lattice, which sets the pattern for the whole lattice. In other words, it is the smallest portion of the space lattice (which is just the pattern of points which describes the arrangement of atoms or molecules in the crystal), which, when moved repeatedly a distance equal to its own dimensions along the various directions, generates the whole space lattice equivalent to the original lattice. The lattice is, of course, just points that denote atoms or molecules; thus, the term lattice point. The accompanying illustrations give you the structures in question. From this, you can obtain justification for the fractions and find the net number of lattice points. For example, take the simple cubic structure. In a space lattice, each atom is shared with three other unit structures as seen in figure d. Note; This point is shared by 8 unit cells, if you place another 4 cubes on the side. Thus, each point is only 1/8 in the cell. This same sort of procedure and reasoning can be used to justify all fractions in all unit cells given. Calculations now become easy. Simple cubic: Have 8 lattice points. Each is only 1/8 inside cell. Thus, you have net of (1/8) (8) = 1. Face-centered cubic: 8 lattice points at corners, each only 1/8 inside cell. Have a point on each face. Each point is only 1/2 inside cell. Have six points. Thus, net = (8) (1/8) + (6)(1/2) = 4. Body-centered: Again have (8) (1/8) from lattice points at corners. 1 point in center, which is entirely in cell. Thus net becomes (8) (1/8) + (1)(1) =2. Tetragonal: Only have a net of (8) (1/8) from lattice points at corners.

Question:

In an electric shaver, the blade moves back and forth over a distance of 2 mm, with a frequency of 60 cps. This constitutes a simple harmonic motion with amplitude R = 1 mm. (a) Find the maximum accelera-tion. (b) Find the maximum velocity, (c) Calculate the velocity and acceleration at point C, a distance x = 0.5 mm from 0, the center point (see diagram).

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Solution:

For simple harmonic motion, the force acting on the blade must be proportional to its dis-placement x, but opposite in direction. The limits of the oscillation are equidistant from the equi-librium position. We can then write F = -kx = ma = m[(d^2x)/(dt^2)](1) where k is a constant, m is the mass of the blades, and the negative sign indicates that the force acts in the opposite direction to the displacement. Equation (1) can be written [(d^2x)/(dt^2)] = -(k/m)x(2) [(d^2x)/(dt^2)] = -(k/m)x(2) which requires a solution where the second derivative of x(t) is proportional to -x(t). The sinusoidal functions have such a property. Assuming that at t = 0 sec, the displacement is at its maximum ampli-tude R, we can write x = R cos \omegat(3) where \omega is a constant. Differentiating twice, dx/dt = - \omega R sin \omegat(4) d^2x/dt^2 = - \omega^2 R cos \omegat(5) Substituting equations (3) and (5) into (2), we find - \omega^2 R cos \omegat= -(k/m) R cos \omegat \omega2= k/m Therefore, if the value for \omega^2 is chosen to be k/m, then x = R cos \omegat is a solution of the equation obtained for the simple harmonic oscillation of the blade. From this solution, we see that \omega must be the angular frequency of the motion since the function cos \omegat repeats itself after a time 2\pi/\omega. The velocity and acceleration are dx/dt and d^2x/dt^2, respectively as found in equations (4) and (5). (a) Looking at the acceleration function a = -\omega^2 R cos \omegat(6) we note that the maximum value is attained when cos \omegat is one or \omegat = n\pi where n is an integer. This occurs when the displacement is a maximum. The angular frequency for the blade is \omega = 2\pif = (2\pi)(60 sec^-1) = 377 sec^-1 Therefore, the maximum acceleration is a_max= \omega^2R = (377 sec^-1)^2 × 1 mm = 1.42 × 10^5 mm/sec^2 = 1.42 × 10^2 m/sec2 (b) To find the maximum velocity, see that v = - \omegaR sin \omegat(7) has a maximum whenever sin \omegat = 1. This happens when-ever the blade is at its equilibrium position at which point \omegat = nr/2,where n is an integer. The maximum velocity is calculated to be v_max = \omegaR = 377 sec^-1 × 10^-3 m = 0.377 m/sec (c) To express the velocity function in terms of the displacement x, we square equation (7). Then, v^2 = \omega^2R^2 sin^2 \omegat = \omega^2R^2 (1 - cos^2 \omegat) = \omega^2(R^2 - R^2 cos^2 \omegat) = \omega^2(R^2 - x^2) Expressing this in terms of v_max = \omegaR, we write v^2 = \omega^2R^2[1 - (x^2/R^2)] = v^2_max [1 - (x^2/R^2)] At x = 0.5 mm v = v_max \surd[1 - (x^2/R^2)] = 0.377 m/sec \surd[1 - (0.5 mm/1.0 mm)^2] v = 0.377 \surd[{1 - (1/2)^2} m/sec] = 0.377 × 0.865 m/sec = 0.326 m/sec The acceleration function of equation (6) can be expressed as a = -\omega^2x and at x = 0.5 mm = (5)(10^-4m), the acceleration is a = \omega^2x = (377 sec^-1)^2 × (5 × 10^-4m) = 71 m/sec^2

Question:

There are N_0 = 10^25 uranium-238 nuclei in a sample. (a) What is the number of uranium nuclei remaining in the sample after 10^8 years if the decay rate per nucleus, \lambda, is 5 × 10\Elzbar18sec\Elzbar1? (b) What is the half-life of uranium?

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Solution:

(a) It is known that each nuclear disintegration takes place independently of any of the others. The radio-active decay rate therefore involves only the instantane-ous number N(t) of the decaying nuclei present ; [dN(t)] /dt] =\Elzbar \lambda N (t)(1) where the constant of proportionality \lambda, is known as the decay rate per nucleus. The solution of (1) is known to be N(t) = N_0- e\Elzbar\lambdat(2) where N_0 is the number of nuclei at t = 0. There are approximately 3.1 × 10^7 sec in one year, therefore 10^8 years equals (10^8 years) × (3.1 × 10^7 sec/year) = 3.1 × 10^15 sec. The number of uranium atoms remaining at the end of this period is N = 10^25 × e^\Elzbar (5 × 10\Elzbar18sec\Elzbar1) × (3.1 × 10^15 sec) = 10^25 × e^\Elzbar0.0155 = 10^25 × 0.984 = 9.84 × 10^24 which is practically the original number . (b) The half-life, T, is defined as the time after which the number of radioactive nuclei has decreased to half its original value. From equation(2) (1/2) N_0 = N_0 e\Elzbar\lambdaT, or (1/2) = e\Elzbar\lambdaT givingIn (1/2) =\Elzbar In 2 =\Elzbar\lambdaT, T = [(In 2) / \lambda] = [(0.7) / (5 × 10\Elzbar18sec\Elzbar1) ] = 1.4 × 10^17 sec. = [(1.4 × 10^17 sec ) / (3.1 × 10^7 sec/year )] = 4.5 × 10^9 years .

Question:

What is the magnetic field 10 cm from a long straight wire carrying a current of 10 amp?

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Solution:

The magnitude of the current in the CGS system of units is 10 amp = (10 A)3 × 10^9 statA/A = 3 × 10 statA The magnetic field, B^\ding{217} , due to the current I in the wire can be found by applying Ampere's law to the circular path shown in the figure, \oint B^\ding{217} \bullet dl^\ding{217} = (4\pi/c) I. From the symmetry of the configuration, the lines of B^\ding{217} form con-centric circles around the wire, then B^\ding{217} \bullet dl = B dl.B at a dis-tance r from the wire is B \int dl = (4\pi/c) I orB = (4\piI)/(c 2\pir) B = (2I/cr) = [2 × (3 × 10^10 statA)]/ [(3 × 10^10 cm/sec) × 10cm] = 0.2 gauss For comparison, this is approximately equal to the magnitude of the Earth's magnetic field at the surface of the Earth and at middle latitudes.

If we could somehow separate the last character from each word, it would be possible then to check that character and, if it were "K", output the corresponding word. Unfortunately, there is no straightforward way to perform the separation. However, SNOBOL provides a very convenient way of separating the first letter from the word. In order to apply this method, it is necessary to reverse the order of letters in the word. The format - free flow-chart of the solution is shown In the figure. There is no primitive (built-in) function in SNOBOL IV compiler which reverses a string; therefore, it has to be defined. SNOBOL IV provides the programmer with this capability. A function is defined by executing the primitive function DEFINE to specify the function name, formal arguments, local variables, and the entry point of the function. The entry point is the label of the first of a set of statements constituting the procedure for the function. Thus: DEFINE ( 'REVERSE (W)', 'A') defines a function REVERSE having one formal argument, W, and entry point A. As was explained in the previous problem LEN(I) is a primitive function which returns a pattern that matches any string of the length specified by the integer I. Therefore, function LEN (1) will return the first matched string of length one, which is the first letter of the input word. Thus the statements ASTRING = LEN (1) .LETTER BW STRING =: F (RETURN) REVERSE = LETTER REVERSE : (B) form a procedure that reverses the string W. The first statement of this procedure (labelled) A, which indicates the entry point of the function REVERSE) indicates that both STRING and LETTER will be assigned the first one-character string found (matched). The second statement assigns the first letter of the input word under variable name W to STRING and consequently, to LETTER, and deletes that letter. The third statement concatenates values of LETTER and REVERSE (which is now just a variable with initial value 0) assigns the result to REVERSE and sends the execution back to the second statement. The next letter is taken now and the procedure repeats. When all the characters of the input word are reversed, the second statement fails and returns the execution to the main program, after hitting the RETURN label in the transfer field of the second statement. The full solution for this problem looks as follows: STARTWORD = INPUT: F(END) DEFINE( 'REVERSE(W)', 'A') WORD 1 = REVERSE(WORD) WORD 1 LEN (1) . K K'K':F(START) OUTPUT = REVERSE (WORD1)(START) ASTRING = LEN (1) . LETTER BW STRING =: F(RETURN) REVERSE = LETTER REVERSE: (B) END The statement W0RD1 = REVERSE (WORD) calls the function REVERSE, performs it for the part of the input stored in variable WORD, and assigns the final (reversed) value to the variable WORD1. The next statement, W0RD1 LEN(1) . K takes the first letter of the reversed word (which is the last letter of the original word) and assigns it to variable K. The statement K 'K' : F(START), then, checks the variable K for occurrence of letter K, and transfers the execution to the first statement if it fails. The first state-ment reads the next word, and the procedure repeats. However, if the check succeeds, the computer executes the next statement, OUTPUT = REVERSE(W0RD1):(START), which reverses the word for the second time to give it its original form, outputs the word, and sends the control to the first statement. The program terminates when the last word of the input is processed. Note that a function has to be defined before it is called, while the procedure statements may appear later in the program. Also note, that the entry point label in the DEFINE statement could be omitted. In that case, the entry point to the procedure is taken to be the same as the function name. For example: DEFINE ( 'REVERSE (W) ') could have the procedure: REVERSESTRING = LEN(l). LETTER BWSTRING =: F (RETURN) REVERSE = LETTER REVERSE : (B)

A carbon atom that has four different atoms or groups attached to it is termed an asymmetric carbon atom or a chiral center. These four groups can give two spatial arrangements, whose relationship to each other is that of an object and its mirror image. For example, see the accompanying figure. The two arrangements are not identi-cal, because they are not superimposable. In such cases, these molecules,i.e. those which differ only in the positioning of groups around asymmetric carbon atoms, are called optical isomers. Such isomers can rotate plane polarized light. Plane polarized light vibrates in only one plane. When substances can rotate plane polarized light, they are optically active. Each of the two optical isomers rotates the light in a different direction. The dextrorotary isomer rotates it to the right, while the levorotatory rotates it to the left. If both isomers are present in equal amounts, the light will not be rotated at all. For one rotates it to the left and the other rotates it to the right, which means there is a cancellation of the rotations. When plane-polarized light is not rotated, the solution is optically inactive. With this information, one can now answer the question. The student formed CH_3CHOHCO_2H, which he believed was lactic acid. The solution was optically inactive, and yet, it is known that lactic acid is optically active. The only possible explanation for this is that the student formed equal amounts of the two possible optical isomers. As mentioned, this results in optical inactivity. There is every reason to believe that formed in equal amounts. The star (\textasteriskcentered) indicates the asymmetric carbon. Thus, the student did form lactic acid. It is just that a mixture of the two optical isomers in equal proportions is present.

To understand PKU, or phenylketonuria, we have to trace metabolic pathways. When we digest protein foods, they are broken down into their component amino acids, one of which is phenylalanine. Eventually, these amino acids are taken up by the cells. Within the cells, some enzymes cause the phenylalanine to be combined with amino acids to form structural proteins that become part of the cell protoplasm. If phenylalanine is not used this way by the body cells, it remains in the blood and as it passes through the liver, it may be acted on by an enzyme which converts it into tyrosine, another amino acid. Still another enzyme will convert some of the, phenylalanine into phenylpyruvicacid. Each arrow in the diagram indicates a metabolic pathway mediated by an enzyme. Each enzyme is produced by a gene and a mutation in any one of these genes can break the chains of conversions. In PKU, the gene which codes forphenylalanine H-monooxygenase. the enzyme responsible for conversion of phenylalanine to tyrosine, is altered so that it no longer codes for that protein. Individuals homozygous for this recessive gene cannot produce the mono-oxygenase. Since the conversion of phenylalanine to tyrosine is blocked, there is an increase in the levels of the other by products of phenylalanine metabolism, the important one here being phenylpyruvate. Phenylpyruvate accumulates in blood, at levels from 50 to 100 times the normal values, and is excreted in large amounts in the urine. In childhood, excess circulating phenylpyruvate impairs normal brain development, causing severe mental retardation. Since the disease results from a genetic defect, it can be inherited and passed from generation to generation.

When the tube is submerged in mercury with both ends open, mercury rises in the tube until the levels of mercury in the tube and in the container are the same (Fig. a). We expect no change in the level of the mercury inside the tube if we close the top of the tube since the air trapped inside was previously in equilibrium with the atmospheric pressure P_a. Closing the top does not change the state of the air inside so that the air pressure on the mercury inside the tube is still P_a (Fig. b). When the tube is in air with its top still closed, the mercury flows out until the pressure at the bottom of the tube equals air pressure (Fig. c). If P is the pressure of the air in the tube and h is the length of the remaining mercury, we can write P_A = P + [(weight of the mercury)/(cross-sectional area)] = P + (W_Hg/A) = P + [(\rho_Hg A\bullet h)/A] = P + h\rho_Hg(1) where \rho_Hg is is the specific weight of mercury. We can find P by using Boyle's law for the air trapped in the tube. Since the temperature remains constant, we have, (P_a) × (Volume in Fig. b) = (P) × (Volume in Fig. c) orP_a (1/2)LA = P (L - h)A givingP = [LP_a] / [2(L - H)].(2) Substituting (2) in (1), we get P_a = P_a{L/[2(L - H)]}+ h\rho_Hg.(3) We are told that P_a is equal to the weight of H cm long mercury column with unit cross-sectional area, P_a = H\rho_Hg Hence, (3) becomes H\rho_Hg = H\rho_Hg{L/[2(L - H)]} + h\rho_Hg orH = H{L/[2(L - H)]} + h.(4) If we solve (4) for h, we get 2H(L - h) = HL + 2h(L - h) 2HL - 2Hh = HL - 2hL + 2h^2 = 0 2h^2 - 2h(H + L) + LH = 0. The roots of this equation are, by the quadratic formula, h = {H + L\pm\surd[(H + L)^2 - 2LH]} / 2 = [H + L \pm \surd(H^2 + L^2)]/2 Since P < P_a, we see that h\rho_Hg < H\rho_Hg, or h < H. The root [H + L + \surd(H^2 + L^2)]/2 is greater than H, therefore it can not be the physical choice. Hence h = [H + L - \surd(H^2 + L^2)]/2.

Question:

Write a program to read in a man's earning for the day, cal-culate his salary for the week and print it.

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Solution:

APWKDAYS,= P'0'INITIALIZE COUNTER READREADCARD CARDIN,CRDEOFREADS IN EARNINGS FOR DAY GOES TO LABEL 'CRDEOF' WHEN NO MORE DATA IS FOUND. PACKWKERN, CRDERNCONVERT EBCDIC DATA TO PACK DECIMAL FOR CALCULATIONS APWKSUM, WKERNADDS DAY'S EARNINGS TO SALARY APWKDAYS , = P '1'ADDS 1 TO NUMBER OF DAYS WORKED CPWKDAYS, = P '6'MAN WORKS 6 DAYS PER WEEK BESALFND BREAD\textasteriskcenteredif the program reaches this statement it means the pre-vious test has failed, it tells the computer that days worked does not yet equal 6, therefore another card must be read. \textasteriskcenteredAT THIS POINT SALARY FOR WEEK IS FOUND SALFNDZAPWKDAYS, = P'0'RESET WKDAYS TO ZERO MVC PRTSUM, WKSUMPREPARE TO PRINT SALARY FOR THE WEEK PRINTLINPRTSAL, 132PRINT SALARY \textbullet \textbullet \textbullet CRDEOF . . .END OF DATA; DO END OF FILE PROCEDURES \textasteriskcenteredTHE DATA DEFINITION FOR THE INPUT AREA CARDINDS\varphiCL80 \varphi \varphi CRDERNDSCL4 DSCL76 \textasteriskcenteredTHE DATA DEFINITION FOR THE PRINT AREA. PRTSALDS\varphiCL132 \varphi \varphi PRTSUMDSCL6 6 DS126C" DS126C" \textasteriskcenteredDATA DEFINITION FOR WORK AREAS WKERNDSPL3 WKSUMDSPL4 WKDAYSDSPL2 READCARD CARDIN, CRDEOF: The first operand of this instruc-tion specifies the name of the data definition for this area. The data definition indicates how the data should be read in. On a card, the data must be punched exactly in the format specified by the data definition. You should also observe that there is a data definition for the print area. PACK WKERN, CRDERN: The data is read in EBCDIC form and cannot be operated on in this form. Hence, we convert it to packed decimal which is an operable form. ZAP WKDAYS, = P'O': The man has worked for six days, so after the sixth day we reset WKDAYS by zeroing it. MUC PRTSUM, WKSUM: Here the salary WKSUM is moved into the print area prior to printing. It is generally a good idea to edit (convert back to EBCDIC) data in packed decimal be-fore printing it. PRTLIN PRTSAL, 132:PRTSAL is the name of the data defini-tion for the print area, and 132 is the length' of the print area. You should note that the general format of a data definition is NAME DS\varphiCLX. The\varphiindicates \varphi \varphi that the symbols that follow are sub-area of the field. X is the length of the field, 80 for card, 132(or 133) for a print line.

Question:

Distinguish between a soap and a detergent. How does a soap clean? Explain the function of phosphate in a detergent.

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Solution:

Soaps are usually potassium or sodium salts of fatty acids (long carbon chain carboxylic acids of formula R-COOH, where R is a long alkyl group). They are produced commercially by heating a fat with an aqueous sodium or potassium hydroxide solution. The structure of soap explains its cleansing action. Soap is a long hydrocarbon (non-polar) group with an ionic group at one end. (See figure A.) The ionic group is water-soluble and the hydrocarbon group is oil soluble. Dirt particles are oil-containing. The hydrocarbon end of the soap is attached to and dissolves in dirt particles, leaving the ionic end exposed to water. The dirt particle is dispersed in the water as a stable emulsion and removed from the article being cleaned. It is known that the cleansing action of soap is reduced by hard water, which contains calcium or magnesium ions. To avoid this problems, soaps have been replaced by detergents; they are more resistant to precipitation by hard water. The active ingredients, or surfactants, of detergent possess the same principal features as soaps. Detergents have a long hydrocarbon chain and an ionic group, usually a sulfonyl group. One example is the sodium salt of alkylbenzene sulfonate, (shown in Fig. B). In addition to surfactants, detergents possess builders. The most widely used builder is Na_5P_3O_8. The phosphate in the detergent functions as a water softener by tying up hard-water ions. It aids the surfactant by dispersing the suspended dirt. It can also act as a buffer, thereby maintaining the pH of the solution at the optimum point for effective washing action.

Each cell is surrounded by a selective membrane, a complex elastic covering that separates the cell protoplasm from the external environment. The structure of this covering, called the plasma membrane, has been under major investigation for many years. Studies of membrane permeability, electron microscopy, and biochemical analysis have enabled biologists to better understand the structure and composition of the plasma membrane. The plasma membrane contains about 40 percent lipid and 60 percent protein by weight, with considerable variation from cell type to cell type. The different types and amounts of lipids and proteins present determine to a great extent the characteristics of different membranes. As seen in electron micrographs, all membranes appear to have a similar fundamental structure. The plasma membrane is revealed by electron microscopy to resemble a railroad track in cross-section - two dark lines bordering a central lighter line. The membranes of cellular organelles also display this charac-teristic. The two dark lines were suggested to corres-pond to two layers of protein and the light middle layer to lipid. It was soon revealed that the lipid actually exists in two layers. The lipid molecules of the plasma membrane are polar, with the two ends of each molecule having different electric properties. One end is hydrophobic ("fear of water"), which means it tends to be insoluble in water. The other end is hydrophilic ("love of water"), which means it has an affinity for water (see Figure 1). The lipid molecules arrange themselves in two layers in the plasma membrane so that the hydrophobic ends are near each other, and the hydrophilic ends face outside toward the water and are stabilized by water molecules (see Figure 2). In this bilayer, individual lipid molecules can move lateral-ly, so that the bilayer is actually fluid and flexible. Protein molecules of the plasma membrane may be arranged in various sites but embedded to different degrees in relationship to the bilayer. Some of them may be par-tially embedded in the lipid bilayer, some may be present only on the outer surfaces, and still others may span the enitire lipid bilayer from one surface to the other (see Fig. 3). The different arrangements of proteins are determined by the different structural, conformational, and electrical characteristics of various membrane proteins. Like the lipid bilayer, the protein molecules tend to orient them-selves in the most stable way possible. The proteins are usually naturally folded into a globular form which enables them to move laterally within the plane of the membrane at different rates. Certain proteins can actually move across the membrane. Thus membrane proteins are not static but dynamic. The functions of the plasma membrane are highly specific and directly related to its structure, which is in turn, dependent on the specific types and amounts of proteins and lipids present. The discriminating permea-bility of the membrane is its primary function. It allows certain substances to enter or leave the cell, and prevents other substances from crossing it. Whether or not a molecule can cross a membrane depends on its size after hydration, electric charge, shape, chemical proper-ties, and its relative solubility in lipid as compared to that in water. This selective permeability of the plasma membrane gives the cell the ability to keep its interior environment both chemically and physically dif-ferent from the exterior environment. The plasma membrane is also found to be particularly important in cell adherence. Because of the specificity of protein molecules on the membrane surface, cells can recognize each other and bind together through some inter-action of their surface proteins. Surface proteins are believed to provide communication and linkage between cells in division so that cells divide in an organized plane, rather than in random directions giving rise to an amorphous mass of cells as in cancer. Surface proteins of the plasma membrane are also proposed to recognize foreign substance due to their remarkable specificity; they can bind with the foreign substance and inactivate it. Membrane proteins are further suggested to interact with hormones, or convey hormonal messages to the nucleus so that a physiological change can be effected. The plasma membrane also is involved in the conduction of impulse in nerve cells. The axon of nerve cells transmits impulse by a temporary redistribution of ions inside and outside the cell, with a subsequent change in the distribution of charges on the two surfaces of the membrane.

Question:

(A) Explain the purpose of a coliform bacteria count in natural water. (B) What is meant by hard or refractory organic compounds?

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Solution:

(a) One of the most important problems associated with polluted water is the possible presence of disease-causing bacteria. A test for the presence of bacteria is made by measuring the amount of common coli form bacteria in water. (Coliform bacteria are harmless and reside in the large intentines of humans.) Coliform bacteria are present in plentiful quantities in feces, which become part of sewage wastes. Since coliform bacteria do not thrive inahaquatic environment, they give an in-dication of how recently and to what extent sewage pol-lution has occurred by their count (which is proportional to the amount of disease-causing bacteria present). If the count is high, human sewage contamination has taken place, possibly accompanied by the presence of dangerous bacteria. (b) Hard or refractory organic compounds are nondegradable substances that decay very slowly in water. Degradation refers to the breaking down of substances into simpler forms. Examples of nondegradable substances are certain pesticides and detergents.

Question:

In rabbits, spotted coat (S) is dominant to solid color (s), and black (B) is dominant to brown (b). In a large population, brown spotted rabbits are mated to solid black ones and all the offspring are black spotted. What are the genotypes of the parents? What would be the appearance of the F_2 if two of these F_1 black spotted rabbits were mated? Illustrate your answer with a diagram.

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Solution:

The first thing to note about this problem is that the cross is dihybrid, and so deals with two separate traits. We can determine the genotypes of the parents by comparing their phenotypes with those of the F_1 offspring. The brown spotted parents have two possible genotypes, bbSS and bbSs. The solid black parents have two possible genotypes, BBss and Bbss. There are four possible matings then: 1)bbSS×BBss, 2)bbSS×Bbss, 3)bbSs×BBss,and 4)bbSs×Bbss. We can see that crosses 2, 3, and 4 could produce off-spring with either or both of the homozygous recessive traits, brown and solid. But we are told that we are dealing with a large sample size and that all the off-spring were black spotted. Therefore the genotypes in 1) must be the parental genotypes. Looking at the cross: BS Bs bS Bs BS BBSS BBSs BbSS BbSs BS BBSs BBss BbSs Bbss bS BbSS BbSs bbSS bbSs bs BbSs Bbss bbSs bbss Phenotypically: BBSS BBSs BbSS BbSs BBSs9 black, spotted BbSs BbSS BbSs BbSs BBss Bbss3 black, solid Bbss bbSS bbSs3 brown, spotted bbSs bbss1 brown; solid That is, a9:3:3:1ratio blackblackbrownbrown spottedsolidspottedsolid Note that whenever we come across a 9 : 3 : 3 : 1 ratio in a dihybrid cross, it indicates that the cross involved dominant-recessive characteristics, and was between two heterozygotes.

To add two 4 bit numbers, A = A_4 A_3 A_2 A_1 and B = B_4 B_3 B_2 B_1, take four full adders F_4, F_3, F_2 and F_1 and connect at the input of the four full adders A_4 B_4, A_3 B_3, A_2 B_2, and A_1 B_1 to F_4, F_3, F_2 and F_1 respectively. Also, the output carry from the previous full adder is connected to the input carry of the next full adder. The complete circuit is shown in Figure 1. For implementing subtraction using Full Adders, we use 2's complement technique, i.e. take a complement of the subtrahend, add it to the minuend, and add 1 in the least significant position of the result, i.e. Difference.For example 8100010001's complement method -5-01011010 3001110010 +1 0011 or810001000 -810000111 01111 +1 0000 The subtraction circuit is shown in Figure 2. In Figure 3 a circuit realizing adding and subtracting functions is shown depending on the control line state. If 1 is placed on the control line, the two numbers are added, and if a 0 is placed on the control line, the two numbers are subtracted.

A graph is a collection of nodes and branches in which each branch links one node to another. The particular graph above is called an undirected graph because there is no particular direction associated with any of the branches. As you can see, the nodes are numbered. Two nodes that are connected by branches may be represented by pairs; for this graph, we notice the pairs: (1, 2), (1, 3), (1, 4), (2, 3), and (2, 5). One way of representing the graph is to use a list of the branches as defined by the pairs of nodes. We will look at the representation of the graph through an adjacency matrix. We define an adjacency matrix as one whose (i, j)th element is nonzero if nodes i and j are joined by a branch. Other-wise, the (i, j)th element is zero. For this graph, we have the adjacency matrix below: Notice that the nonzero elements are found at both the (i, j)th elements and the (j, i) th elements. This is because the graph is undirected: nodes may be accessed along branches from either side. For this simple graph, we store nodes in a 5 by 5 array. Other data structures may be used to represent graphs; a chained list could be used if the nodes are to be changed often. The pseudocode is presented below. After the array is filled with zeros, the (i, j)th elements and the (j, i)th elements are changed to ones. C TO CONVERT FROM PAIRS TO THE ADJACENCY MATRIX, C ASSUME THE BRANCHES ARE GIVEN AS PAIRS A(K) , B(K) integer A, B, I, J, K, m (5, 5) INPUT (A (1), B (1)), (A (2), B (2)) , ... , (A (5),B (5)) do for I \leftarrow 1 to 5 do for J \leftarrow 1 to 5 m (I, J) \leftarrow 0 end do for end do for C NOW CHANGE ELEMENTS TO NONZERO FOR PAIRS A (K), B (K) C FOR EACH PAIR K indicates the branch number and A and B the nodes of that branch. do for K \leftarrow 1 to 5 m (A (K), B (K)) \leftarrow 1 m (B (K), A (K)) \leftarrow 1 end do for output m (K, K) end program

Question:

Calculated the resultant force, on the charge q_3 in the figure.

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Solution:

The force exerted by q_1 on q_3 is F_1 = q_1q_3 / R^21 = {(50) (\rule{1em}{1pt}5)} / (10)^2 = \rule{1em}{1pt}2.5 dyne The negative sign denotes an attractive force. The force exerted on q_3 by q_2 is F_2 = q_2q_3 / R^22 = {(20) (\rule{1em}{1pt}5)} / (30)^2 = \rule{1em}{1pt}100 / 900 = \rule{1em}{1pt}0.111 dyne Since q_2 is positive and q_3 is negative this force is attractive and is directed to the right toward q_2. The resultant force on q_3 is F_R = F_1 \rule{1em}{1pt} F_2 = \rule{1em}{1pt}2.5 \rule{1em}{1pt} (\rule{1em}{1pt} 0.111) = \rule{1em}{1pt}2.389 dyne and is directed to the left.

Question:

What is modular programming?Name the six functional Name the six functional modulesof a structured program.

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Solution:

Modular programming is a term which describes a disciplined meansof writing computer programs. By writing program segments which performspecific tasks, you will find that it is easier to debug your program. This concept is a step to take some of the "art" out of computer programmingand to replace It with more logical, intelligent, and efficient techniques. There are six types of modules needed in almost every type of program. The table below gives an overview of these modules. FUNCTIONAL MODULE DESCRIPTION 1) INPUT Performs the input of data necessary to accomplish the task 2) INPUT DATA VALIDATION Checks the validity of the incoming data and diagnoses what type of error if any, has occurred 3) PROCESSING Completes the basic task requested of the algorithm 4) OUTPUT Performs output of information needed by the user 5) ERROR SPECIFICATION Analyzes problematical points in the execution of the program and takes appropriate action 6) CLOSING Finalizes program execution The above modules should stand alone in their meaning. That is, If definedthoroughly enough, the module should be readily adaptable to anotherprogram with only minor changes in variable names. In addition, eachmodule should have only one point of entry and one point of exit. Multiple returns and branches, if used excessively, tend to bewilder the readerof your program. The term robustness is used in programming to mean the degree of completenessa program offers. A robust algorithm takes care of almost all possibilitieswith swift and precise action. The idea of modular programming, if followed carefully, yields robustness.

Question:

A small pebble lies at the bottom of a tank of water 24 feet deep. Determine the size of a piece of cardboard which, when floating on the surface of the water, directly above the pebble, totally obscures the latter from view.

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Solution:

We can consider the pebble as a point source of light. If the cardboard is big enough for its purpose, then all rays of light from the pebble which would be refracted into the air at the surface must be blocked off by the cardboard, and all rays striking the surface of the water outside the cardboard must be totally internally reflected (see figure). The cardboard must obviously be circular, and, if its center is directly above the pebble, a ray of light striking the edge of the cardboard must do so at an angle \textphi \geq the critical angle. By Snell's Law, n_w sin \textphi = n_a sin \texttheta where \textphi is the angle of incidence, \texttheta is the angle of refraction, n_a is the index of refraction of air (n_a = 1) and n_w is that of water. Then, n_w sin \textphi = 1 because \texttheta = 90\textdegree to satisfy the requirement that \textphi be a critical angle. Therefore, using the figure n_w [r/\surd(r^2 + d^2)] = 1 r^2 + d^2 = n^2_wr^2orr^2 = [d^2/(n^2_w - 1)] r = [24 ft/\surd{(16/9) - 1}] = (72 ft/\surd7) = 27.2 ft.

The maximization problem may be written as follows: Max {X.(P-C)} X where P = 35 - O.1X C = 15 + 0.001X^2 , X \geq 0 When X, the quantity produced is zero, there is zero pro-fit. Similarly, when the price of X equals the cost of X, profits are zero. This occurs when P = C or, 35 - 0.1X = 15 + 0.001X^2 Solving for X gives X = 100. Thus profits are zero when X = 0 and X = 100. Assuming profit is a continuous function of output Rolle's theorem may be applied to conclude that the maximum profit point lies in the interval (0,100). Numerical approximation of this maximum is performed on the computer using the GOLDEN SECTION method. To illustrate the method, suppose the graph of profit versus production looks as follows: In the interval [0,100] there is only one maximum. Suppose we pick two points, X_1 and X_2 in that interval and find their ordinates on the curve. If f(X2) > f(X1), then the interval of search may be reduced from [0,100] to [X_1,100] and the interval [0,X_1] can be eliminated. The original interval [X_L X_u] is equal to 1 (Fig. 2). Let \tau equal the ratio of the long subinterval to the total interval, i.e. (X_2 - X_L) /( X_u - X_L) = \tau Since and X1 and X2 have been located symmetrically about the center of the interval, the distance between X_1 and X_u is also \tau. From Fig. 2, the interval [X_L X1] will be elimi-nated after the first iteration since f(X_2) > f(X_1). In order for X2 to be properly located within the interval [X_1,X_u] for the next iteration, the following must be true: (X_u - X_2) / (X_u - X_1) = \tau But, from Fig. 2, X_u - X_2 = 1 - \tau and X_u - X_1 = \tau. Thus, [(1 - \tau) / \tau] = \tau,or \tau^2 + \tau - 1 = 0.(1) Solving (1) for \tau we obtain \tau = [(\surd5 - 1) / 2] = 0.618. Thus, according to the GOLDEN SECTION rule, X_2 should be 0.618 from X_L and X_1, should be 0.618 from X_u . For the given problem, the selection of points in the search interval would proceed as follows: X_1 = 100 - 0.618 × 100 = 100 - 61.8 = 38.2; X_2 = 0.618 × 100 = 61.8. From Fig. 1, f(X_2) > f(X_1) so the interval [0,38.2] is eliminated. The next point is located at 0.618 × (length of interval) = 0.618 × (61.8) = 38.2 from the left end of the interval, or X_3 = 38.2 + 38.2 = 76.4. Since f(X_2) > f(X_3), [76.4,100] is eliminated. X_4 is located at 0.618 × (length of interval) = 0.618 × (38.2) = 23.6 from the right end or 76.4 - 23.6 = 52.8. After three iterations the interval within which the optimum is located has been reduced from [0,100] to [52.8]. These values are now assigned to X_L and X_u. The process is continued until the length of the interval is less than some preassigned \epsilon > 0. The average of f(X) at final X_L and X u is then the required approximation to the optimum. The most efficient method of locating the points is the GOLDEN SECTION technique.

According to the de Broglie wave theory of matter every particle has wave-like characteristics. Electrons then should display dif-fraction effects when passing through a narrow slit, as do light waves. The wavelength associated with the moving electron is the de Broglie wavelength. \lambda = (h/p)(1) where p is the momentum of the electron and h is Planck's constant. The p appearing in (1) is the relativistic momentum of the electron. This quantity is related to the energy E of the electron by E^2 = p^2c^2 + m^2_0 c^4 where m_0 is the rest mass of the electron. The energy E is equal to the kinetic energy of the electron, T, plus the rest energy, m_0c^2 . E = T + m_0c^2 Hence, (T + m_0c^2)^2 = p^2c^2 + m^2_0c^2 T^2 + 2m_0c^2T = p^2c^2 Therefore p^2 = [(T^2 + 2m_0c^2T)/c^2] p = [\surd{T(T + 2m_0c^2)}/c] Now,m_0c^2 = (9.31 × 10^-31 kg)(9 × 10^16 m^2/s^2) = 83.79 × 10^-15 Joules because 1Joule = 6.24 × 10^18 eV m_0c^2 = .5 × 10^6 eV Hence, p = [\surd{(5 × 10^6 eV)(5 × 10^6 eV + 1 × 10^6 eV)} / (3 × 10^8 m/s)] p = [\surd{(30.0 × 10^12 eV^2)} / (3 × 10^8 m/s)] p = [(5.48 × 10^6 eV) / (3 × 10^8 m/s)] p = 1.83 × 10^-2 (eV\textbullets/m) whence,\lambda = (h/p) = [(6.63 × 10^-34 J\textbullets)/{1.83 × 10^-2 (eV\textbullets/m)}] since 1Joule = 6.24 × 10^18 eV \lambda = [{(6.63 × 10^-34)(6.24 × 10^18)m} / (1.83 × 10^-2)] = 22.6 × 10^-14 m = 2.26 × 10^-13 m The diffraction pattern due to the electron waves may be treated as if it were caused by the diffraction of light waves. We may then use the formula for the maxima produced by diffracting light rays to analyze the maxima produced by electron waves. With respect to the diffraction pattern (see the figure), we have d = 10^-6 m and L = 3 m. x_1 = (\lambdaL/d) = [(2.26 × 10^-13 m × 3m)/(10^-6 m)] = 6.8 × 10^-7 m The width of the central maximum will be 2x_1 = 1.4 × 10^-6 m, which would be difficult to observe. If the electron beam were of lower energy, 50,000 eV for example, the central maximum would have a width of 3.2 × 10^-5 m (0.032 mm), which is more easily measured.

Question:

What is meant by the statement "ontogeny recapitulates phylogeny"? With the advent of knowledge in genetics, how has this theory of recapitulation been modified and what significance does its present form have?

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The idea of recapitulation was first proposed by Ernst Haeckel in 1866, in the principle that "ontogeny recapitulates phylogeny.\textquotedblright Haeckle claimed that embryos, in the course of development, recapitulate, or repeat, the evolutionary history of their ancestors in some abbreviated form. Thus, if mammals evolved from reptiles, which evolved from amphibians, which in turn evolved from fish, then a mammalian embryo should first resemble a fish, then an amphibian, then a reptile, and finally a mammal. Through research and experimental findings, later workers modified Haeckle's idea and developed the modern theory of recapitulation. This theory states that the embryos of the higher animals resemble the embryos of lower forms, not the adults, as Haeckle had believed. An embryo passes through developmental stages similar to some of those which lower animal forms pass through. Ontogeny does not repeat the adult stages of phylogeny, but it does repeat, in an altered form, some of the ontogenic characteristics of ancestral forms. Inherent in the theory of recapitulation is the existence of an evolutionary relationship between animal groups. There is also the implication that a species arises from some previous species and may give rise to new ones. Because embryonic development must primarily be controlled by genes, ontogeny can be rightfully thought of as an expression of the inherited genetic composition of an organism. Differences in ontogeny can be related, then, to differences in genetic composition. The modern theory of recapitulation thus also holds that a given species expresses a slightly altered version of the ontogenic program that it inherited genetically from the previous species, which in turn acquired its developmental program genetically from its ancestor. According to this, a species will closely resemble, both genetically and ontogenically, its recent ancestors and less closely its more distant ones. In other words, the more remote the ancestry, the fewer the similarities. It is also observed that in remote relationships, the similarities are restric-ted to the very early stages of embryonic development. Beyond those early stages evolutionary changes have grossly altered the direction of development. Thus, the ontogenies of fish and mammals resemble each other only in their earliest stages, while the ontogenies of reptiles and mammals show a close resemblance through later stages, the reptiles being a more recent ancestor of the mammals than the fish (see Figure). The theory of recapitulation, as we can see, is significant for three reasons. First, it confirms that evolution does exist; second, it shows that evolution is a genetic occurrence; and third, taken with due caution, it may indicate the evolutionary hierarchy of some animals.

A cistron is defined as the genetic unit of biochemical function. One cistron corresponds to one functional unit (one gene coding for a polypeptide chain, protein, or enzyme) on a chromosome. A cistron can be defined by means of complementation tests. Complementation refers to the production of a normal phenotype through the combined activities of two chromosomes, each of which alone is incapable, due a recessive mutation, of producing a normal phenotype. If two homologous chromosomes contain mutations in different cistrons, the normal gene of one can compensate for the mutated gene of the other, and vice-versa, when the two chromosomes are in combination with each other in the organism (see figure 1) . If each chromosome of the pair contains a mutation within the same cistron, neither can compensate for the other, and a mutant phenotype will result. A cistron is thus the unit within which two chro-mosomes can have mutations and not complement each other when they occur in combination. It follows then that a cistron is the smallest unit capable of producing a gene product, and is equivalent to a gene. A muton is defined as the smallest unit which, if altered, gives rise to a mutant phenotype. A muton may thus be as small as a single nucleotide, since a change in a single nucleotide can cause a mutant phenotype (think, for example, of sickle-cell anemia). Of course, some single -point mutations may not result in a mutant phenotype. If a given codon were changed via a single base substitution, for example, from CCA to CCG, that codon would still specify proline, due to the degeneracy of the genetic code. A recon is defined as the smallest unit that is interchangeable, but not divisible, by recombination. In other words, recons are the smallest units between which crossing over can occur. This is known to be equivalent to a single nucleotide. This is because the breakage point in crossing over is between any two given nucleotides. Given two crossovers, it is thus possible for a single nucleotide to be exchanged (see figure 2). FIGURE 2.Double crossover between two strands of DNA, the base sequences of which are shown. Since a cistron consists of a sequence of mutons or recons, it is the largest of the three. A recon may be as small as a single nucleotide, as may be a muton.

The maximum voltage we need to read is 200 volts. We therefore want maximum deflection to occur when the voltage across the voltmeter is 200 volts. This occurs when the current through the galvanometer is 2ma or 0.002 amp. To accomplish this, we can put a resistor R in series with the galvanometer. The total resistance in the circuit, R + r (see figure), must then be, by Ohm's Law, R + r = (V/I) = [(200 volts)/(0.002 amp)] = 100,000 ohms andR = 100,000 - 5 = 99,995 ohms Since the voltage across the terminals is proportional to the current flowing through the galvanometer, each division of the meter corresponds to [(200 volts)/(100 divisions)] = 2 volts/division.

Question:

The executioner in charge of the lethal gas chamber at the state penitentiary adds excess dilute H_2 SO_4 to 196 g (about (1/2) lb) ofNaCN. What volume of HCN gas is formed at STP?

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Solution:

The equation for this reaction is H_2 SO_4 + 2NaCN \rightarrow 2HCN + Na_2SO_4 This means that for eachNaCNreacted, one HCN is formed. After one knows the number of moles of HCN formed, one can find the volume that the gas occupies by multiplying the number of moles by 22.4 liters (volume of one mole of gas at STP). STP (Standard Temperature and Pressure) is defined as 273\textdegreeK and 1 atm. The number of moles ofNaCNreacted can be found by dividing 196 g by the molecular weight ofNaCN. (MW ofNaCN= 49.) number of moles = [(196 g)/(49 g \bullet mole)] = 4.0 moles From thestoichiometryof the reaction, we see there will be 4.0 moles of HCN formed. The volume of the gas can now be found. volume = 4.0 moles × 22.4 liters/mole = 89.6 liters.

Question:

Describe the structure of a centriole . What hypothetical function does it serve ?

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Solution:

Centrioles are small bodies located just outside the nucleus of most animal cells and some plant cells in a specialized region of cytoplasm that has been known to play a role in cell division. Centrioles usually occur as a pair in each cell oriented at right angles to each other. Each centriole of the pair is composed of nine groups of tubules arranged longitudinally in a ring to form a hollow cylinder. Each group is a triplet composed of three closely associated tubular elements called micro-tubules. (See Figures.) The space immediately surrounding a pair of centrioles is called the centrosome. The centrosome appears to be clear or empty in the light microscope. Centrioles do not occur in most higher plant cells, although they are found in some algae and fungi and in a few reproductive cells of higher plants. The centrioles seem to play some part in directing the orderly distri-bution of genetic material during cell division. At the beginning of cell division the centrioles replicate, and the two pairs of centrioles that result move to opposite poles of the dividing cell. Under the electron microscope each pair is seen to send out spindle fibers, structures involved in separating and moving chromosomes to the oppo-site ends of the cell. This observation leads to the hypothesis that centrioles are needed in the formation of the spindle fibers. However, the existence of cells with-out centrioles yet capable of spindle formation seems to refute this hypothesis.

To understand the complementing system in computers and why this system is used, consider the following example: 12 + 7 = 19 - 12 + 7 = - 5 Each of these is defined as an addition between two operands (numbers), but because one of the numbers is actually negative in the latter case, it is really a subtraction. This means that, even though we instructed the computer to do an addition, it would first have to examine the numbers and, if one of them is negative, do a subtraction instead. This makes the job of the computer harder, because it has to examine the numbers, make a decision on what to do, and then do it. For this reason the complementing system was developed, so that addition could work for both positive and negative numbers. For decimal numbers there are two complementing systems - 10's and 9's. 10's complement system: Consider the example 5-3 = 2 \equiv5 + 10 - 3 - 10 = 2 \equiv5 + (10 - 3) - 10 = 2 \equiv5 +7 - 10 = 2 \equiv12 - 10 = 2 e., for - 3, take its 10's complement, 10 - 3 = 7. Now add 7 to 5 = 12 and drop the last 1. This gives12 = 2, the necessary answer. 9's complement method: Here we subtract the negative number from 9. Perform addition and the last digit is again added back to the result. 5 - 3 = 2 1st step9 - 3 = 6 2nd step5 + 6 = 11 3rd step11= 1 + 1 = 2 One of the methods for constructing a 1's complement for a binary number is the interchange of zeros and ones, e.g., decimalsubtraction 5101 - 3- 011 2010 = 2 Now the 1's complement of 011 = 100. Thus, This is equivalent to the 9's complement in decimal. Similarly, there is a 2's complement method equivalent to the 10's complement method. Here we take the complement of a binary number and add one to it. 011 = 1001st step complement + 1 1012nd step add 001 to it Now 5101 -3\equiv+100 21010drop out last bit as in 10's complement method. Now consider the case where the result of the subtraction is negative. For example, 6 - 7 - 1(decimal subtraction) or, 6 = 0110 in binary, and, 7 = 0111 in binary. 1's complement of 7 in binary = 1000. Hence, As no carry is generated in the last stage, it means the result is negative. Hence, take the complement 1110 \rightarrow 0001 Hence, the answer is: - 0001, i. e, - 1 (binary). Also, 2's complement of 7 in binary = 1000 +1 1001 Hence, the answer = - 0001, i. e., - 1. Hence, the answer = - 0001, i. e., - 1.

Question:

The property of skeletal muscle contraction in which the mechanical response to one or more successive stimuli is added to the first is known as summation. What is the underlying explanation for this phenomenon?

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A possible explanation of this phenomenon, based on the role of calcium in excitation-contraction coupling, is that the amount of calcium released from the sarcoplasmic reticulum during a single action poten-tial is sufficient to inhibit only some of the troponin-tropomyosin in the muscle. Multiple stimulation would then release more calcium so that more troponin-tropomyosin would be inhibited, allowing for further contraction. However, the truth is that more than enough calcium is released by the first action potential to inhibit all the troponin-tropomyosin, so this proposal must be discarded. The explanation of summation involves the passive elastic properties of the muscle. Tension is transmitted from the cross bridges through the thick and thin fila-ments, across the Z lines, and eventually through the extracellular connective tissue and the tendons to the bone. All these structures have a certain amount of elasticity, analogous to a spring that is placed between the contractile components of the muscle and the external object. In the muscle, the contractile elements in their fully active state begin to stretch the passive elastic structures immediately following calcium release. Only when the elastic structures are all taut, can increasing contraction by the muscle occur. Summation occurs because a second stimulus is given, very close in time to the first, while the elastic structures are still a bit taut and not yet slack. Under this condition, the active state of the contractile proteins is maintained and the result is contractions that are stronger than any single simple twitch. Should sustained stimulation occur, the elastic elements would never have time to relax at all, and it is at this point that maximal force by the muscle fibers is attained; the individual contractions are indistinquishably fused into a single sustained contraction known as tetanus, (see accompanying figure) If stimula-tion of the muscle continues at this frequency, the ulti-mate result will be fatigue and possibly complete cessation of activity due to exhaustion of nutrients. It is not surprising to note that cardiac muscle has an extremely long refractory period, allowing the elastic components to relax and thus avoiding tetanus, which would result in death due to loss of pumping action of the heart.

Observe that f(x) is a continuous function and that f(-1) = 1 - e^-1 = 0.632 > 0 and f(0) = -1 < 0 so that f(x) has a root between -1 and 0. By graphing f(x) we can get a good pre-liminary estimate of where the root lies. Values plotted are given in the Figure. Note that there is a root near x = -0.7. We can use this root as an input estimate to the Newton-Raphson method. This is an iterative method in which we solve the equation x_n+1 = x_n -f(x_n) / f'(x_n), wheref(x_n) = x^2_n -e^(x)n and f'(x_n) = 2x_n -e^(x)n . We make an initial guess x_0 (in this case, x_0 = -0.7), and evaluate the function. Next, we compute the error as an absolute value. This value is compared to the convergence condition, a previously establish-ed number that we have chosen. If our error is less than the conver-gence condition, then we write the solution x_n+1 . If not, we return to the start of the loop and continue. A sample program solution is given below: PROGRAM EXPLANATION 0010 INPUT X,L Input initial guess x_0 and limiting no. of iterations 0020 F = (X\uparrow2 - EXP (X)) Evaluate f(x) 0030 D = (2\textasteriskcenteredX - EXP(x)) Evaluate D = f'(x) 0040 A = F/D Evaluate f(x)/f'(x) 0050 E = ABS (A) E = \vertA\vert = \vert x_n+1 - x_n \vert = error 0060 IF E < 0.000005 THEN 110 Convergence condition 0070 X = X - A x_n+1 = x_n - f(x_n)/f'(x_n) 0080 N = N + 1 Increment loop counter 0090 IF N = L THEN 110 Check for limit on itera-tions 0100 GO TO 20 loop back 0110 PRINT X,N,A output results 0120 END end of program

Question:

Assuming that the density of water is .9971 (g / cm^3)at 25\textdegreeC and that of ice at 0\textdegree is .917 (g / cm^3), what percent of a water jug at 25\textdegreeC should be left empty so that, if the water freezes, it will just fill the jug?

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Density = [(mass) / (volume)]. When water freezes, the mass is constant, but its density decreases.This means, therefore, that an in-crease in volume occurs. To solve this problem, you must determine the change in volume of water from 25\textdegree to 0\textdegree. Because you are given the densities of the substance at each temperature, you can say for 1 mole of H_2O (MW = 18) (mass / volume) = 18 [(g / mole) / (y) = (0.997g / cm^3) for water at 25\textdegreeC and (mass / volume) = 18 [(g / mole) / x] = (0.997g / cm^3) for ice at 0\textdegreec Rewriting, you have .997 y = 18 and .917 x = 18. Both expressions equal 18, which means .997 y = .917 x, or (x / y) = [(.997) / (.917) = 1.087. This means, then, that at 0\textdegreeC the volume is 1.087 times greater than the original at 25\textdegreeC. If you let B equal the fraction of water to be put in the jug, then you can say 1.087 B = 1, or B = .920. In other words, only 92% of the jug should be filled at 25\textdegree to obtain a completely filled jug at 0\textdegree. This means that 8 % is empty space.

Question:

If 5 joules of work are done in moving 0.025 coulomb of positive charge from point A to point A', what is the difference in potential of the points A and A'?

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To solve this problem we use our formula for the definition of the volt, E = W/Q . The work is 5 joules and the charge Q is 0.025 coulomb. Then the potential difference E = (5 joules)/(0.025 coulomb) = 200 volts

The IBM/360-370 processors have many ways of in-terpreting groups of bits stored in memory. When storing fixed point integers, these computers use two forms; a short form and a long form. In the short form, two bytes (16 bits) are used to store the integer; for the long form, four bytes (32 bits) are used. In either case, if the processor determines that it has encountered an integer, the sign of this integer is indicated by using the leftmost bit of the 2 or 4 byte field. If the integer is positive this bit is zero, if it is negative the bit is one. The computers store packed decimal numbers by placing two decimal digits in each byte (8 bits) field. This is done for all the bytes of the field with the exception of the rightmost. For this byte, the rightmost four bits in-dicate the sign of the number. If the number is positive, the sign is an 'F' or a 'C', if negative, it is a 'D'. The length of a packed decimal field can be between 1 and 16 bytes. Assuming this, the field can be a minimum of 1 digit and a sign or a maximum of 31 digits and a sign.

Question:

List the arithmetic operators used in BASIC and state their priorities of execution in a statement. Also give a list and briefly explain the functions of the most popular statements.

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Arithmetic operations in BASIC are represented by the following symbols: +-ADDITION --SUBTRACTION AND NEGATION \textasteriskcentered-MULTIPLICATION /-DIVISION \textasteriskcentered\textasteriskcentered or \uparrow-EXPONENTIATION Exponentiation has the highest priority of execution; multiplication and division are executed next; addition and subtraction are done last. If 2 or more equally "powerful" operators are present in the statement, their execution goes from left to right. Operations in parenthesis are always done first. The most popular BASIC statements are: LET variable = expression (The arithmetic assignment statement) INPUT variable list (Statement for data entry from terminal into a program during the execution phase) READ variable, variable, etc. (Statement for data entry from the DATA statement) DATA constant, constant, etc. (Information supplier statement for READ statement) PRINT variable, constant, "junk", etc. (Primary output statement) REM comments (Comments, insertion statement) GO TO n ("Sends" the computer to the statementlabelledn) More complicated statements will be explained in later problems.

Question:

When carbon dioxide combines with water, carbonic acid is produced and rapidly dissociates to the bicarbonate ion and free hydrogen ions: CO_2 + H_2O\rightarrowH_2CO3 H_2CO_3\rightarrowH^+ + HCO_3^\rule{1em}{1pt} When carbon dioxide is carried by the blood, why doesn't the blood become acidic owing to the presence of free H^+?

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Not only does the blood transport oxygen from the lungs to the tissues (via hemoglobin), it also transports carbon dioxide in the reverse direction, from the tissues to the lungs. Some of the carbon dioxide (\sim 7%) is carried as gas dissolved in the plasma. A slightly larger proportion (\sim 23%) is carried by hemoglobin; the carbon dioxide-hemoglobin complex is called carbaminohemoglobin. The loosely bound CO_2 is combined with the terminal amino groups of all four polypeptide chains; each hemoglobin molecule having four polypeptide chains, can bind four CO_2 molecules. Since CO_2 does not combine with hemoglobin at the same site as oxygen, hemoglobin can transport both oxygen and carbon dioxide at the same time. However, most of the CO_2 (70%) is transported as the bicarbonate ion, HCO_3^-. Carbon dioxide enters red blood cells where it reacts with water to form carbonic acid: This reaction can only occur because of an enzyme called carbonic anhydrase. The carbonic acid quickly dissociates into hydrogen and bicarbonate ions: H_2CO_3\rightarrowH^+ + HCO_3^- The presence of free H^+ ions should drop the pH of the blood significantly. Since cells can only live within a narrow range of pHs, this change would be physiologically harmful. But the blood exhibits only very slight variations (\pm.04 pH units) around the normal pH of 7.4. What accounts for this phenomenon? The pH would not drop if the H^+ ions could be combined with some substance, so as to remove them as free ions from solution. The H^+ ions combine with an ionized form of hemoglobin (Hb^-) to form acid hemoglobin (HHb) while the HCO_3^- leaves the red blood cell and enters the plasma. HCO_3^- + H^+ + Hb^-\rightarrowHCO_3^- + HHb Acid hemoglobin dissociates to a much lesser degree than does carbonic acid, so that most H^+ ions remain bound, and therefore effectively removed from solution. Acid hemoglobin acts as a buffer in the blood. A buffer is a substance which prevents large changes in pH. When there is an excess of H^+ ions(when the pH drops), hemoglobin binds H^+ ions to remove them from solution, thus restoring the pH. Hb^-+H^+\rightarrowHHb When the blood becomes alkaline because of an excess of hydroxide ions (OH^-) or a decrease of H^+, the acid hemo-globin dissociates, releasing H^+ and restoring the pH: OH^- + HHb \rightarrow OH^- + H^+ +Hb^- \rightarrow H_2O + Hb^- Hemoglobin and some plasma proteins act to prevent the large deviations from the normal pH which would occur in carbon dioxide transport. A summary of CO_2 transport is provided in the following illustration: The reverse process occurs in the lungs. Carbaminohemoglobin dissociates to release CO_2. Acid hemoglobin dissociates to produce H^+ ions which combine with bicar-bonate ions to form carbonic acid. This acid, then dissoci-ates to form water and CO_2, which is expired from the lungs.

a) The only acceptable characters in a variable name in any FORTRAN statement are the twenty-six letters of the alphabet and the ten digits 0 to 9 (some systems also accept the $ included in the variable name). Algebraic signs cannot be included, and therefore M + 1 is an invalid variable. b)For FORTRAN statements involving arithmetic operations, we use the following symbols: -for subtraction +for addition {_\ast}{_\ast}for exponentiation {_\ast}for multiplication /for division No arithmetic operations are allowed on the left side of the "=" sign. c) In the WRITE statement the first number (3) indicates the device that is to be used to print the values obtained at the end of the program. (Typewriter, printer). The second number (13) is the number of the corresponding FORMAT statement (will be discussed in later chapters). The listing of the variables should not be followed by a period. This is an error. d) FORTRAN has no symbol \textquotedblright×" (presumably we mean multiplication, in which case the correct FORTRAN statement is K = 2 {_\ast} I . (Z^Y)^W or (Z)^(Y)W. These are not always equal. For example, if Z = 2, Y = 3, W = 4: (2^3)^4 \not = (2)^(3)4.

Question:

Design an experiment which demonstrates that both positive and nega-tive ions move in electrolytic conduction.

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The accompanying figure illustrates the set up of an ex-periment which will demonstrate the migration of both positive and nega-tive ions in electrolytic conductivity. A U- tube is initially half filled with a deep purple solution of Cu(MnO_4)_2 in water. The color of the blue hydrated Cu^2+ ions is masked by the purple of the MnO^-_4 ions. A colorless aqueous solution of HNO_3 floats on top of the Cu(MnO_4)_2 solution in each arm of the U- tube. If an electric field is established for a period of time across the solution by 2 electrodes, the blue color, characteristic of hydrated Cu^2+ ions, moves into the region marked A, suggesting a migration of positive ions toward the negative electrode. At the same time, the purple color, characteristic of MnO^-_4 , moves into the B region, indicating a migration of nega-tive ions toward the positive electrode. Thus, there is empirical evi-dence of the migration of both positive and negative ions.

If all capacitors are joined in parallel, the resultant capacitance is 5 \muF. (For the resultant capacitance of a set of capacitors connected in parallel equals the sum of the individual capacitances). If the capacitors are connected in series, the resultant capacitance is 1/5 \muF (for the resultant capacitance of a set of capacitors connected in series equals the reciprocal of the sum of the reciprocals of the in-dividual capacitances). The connection is thus more complicated. Suppose that n capacitors are connected in parallel and 5 - n in series. The resultant capacit-ances are thus such that C1 = (1 + 1 + . . .) \muF = n \muF n times and(1/C_2 ) = [(1/1) + (1/1) + . . .] = (5 - n) \muF^-1 (5 - n) times orC_2 = (1/5 - n) \muF. If C_1 and C_2 are connected in parallel, then C_r = C_1 + C_2 = [n + (1/5 - n)] \muF = (3/7) \muF. 5n - n^2 + 1 = [(15 - 3n)/(7)] 35n - 7n^2 + 1 = 15 - 3n or7n^2 - 38n + 8 = 0 By the quadratic formula, n = [{- (- 38) \pm \surd{(- 38)^2 - 4 (7) (8)}} / {2 (7)}] There is thus no integral solution for n. But if C_1 and C_2 are connected in series (see figure), then (1/C_r ) = (1/C_1 ) + (1/C_2 ) = [(1/n) + 5 - n] \muF^-1 = (7/3) \muF^-1 . \therefore3 + 15n - 3n^2 = 7nor3n^2 - 8n - 3 = 0. \therefore(3n + 1) (n - 3) = 0. This has an integral solution for n, n = 3. Thus the required capacitance is given if 3 capacitors are connected in parallel, and the combination is connected in series with the other two.

Question:

Find K_b andpK_bfor the acetate ion CH_3COO^-, The ionization constant of CH_3COOH is K_a = 1.75 × 10^-5;K_w= 1.00 × 10^-14.

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Question:

A cable drum of inner and outer radii r and R is lying on rough ground, the cable being wound round the inner cylinder and being pulled off from the bottom at an angle \texttheta to the horizontal. An inquiring student stroll-ing by notes that when the cable is pulled by a workman, with \texttheta a small angle, the drum rolls without slipping toward the workman. Whereas if \texttheta is large, the drum rolls without slipping in the opposite direction. He works out a value for the critical angle \texttheta_0 which separates the two types of motion. What is the value of \texttheta_0?

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We define the drum's acceleration to be positive when the drum moves toward the workman. Hence the mathematical condition that the drum roll toward or away from the worker is that a (the acceleration) be positive or negative, respectively. The critical con-dition distinguishing the 2 types of motion is that a = 0. Therefore, to find the critical angle \texttheta_0, we find a as a function of \texttheta, set it equal to zero, and solve for \texttheta. We can use Newton's Second taw to relate the net force on the drum to its acceleration. (We do this for the vertical and horizontal directions separately.) Figure (1) shows the drum with the forces acting on it. The force applied by the workman acts tangentially to the inner cylinder at such a position that the angle between this tangent, and the horizontal is \texttheta. It follows that the angle between the corresponding radius and the vertical is \texttheta also, and that this radius is at right angles to the tangential force F^\ding{217}. The other forces acting are the weight Mg^\ding{217} of the cable drum, the normal forces exerted by the ground on the drum at the two points of contact, which combine into a resultant N^\ding{217} passing through the center of gra-vity, and the frictional forces at the same points of contact which combine to form a single resultant force f^\ding{217} . There is no movement in the vertical direction. Hence N = Mg - F sin \texttheta.(1) The forces in the horizontal direction produce an acceleration a. Thus F cos \texttheta - f = Ma.(2) Further, the moments of the forces about the center of mass, O produce a rotational acceleration about that point. The only forces whose lines of action do not pass through the center of gravity are F and f. Hence Fr - fR = l\alpha(3) where I is the moment of inertia of the drum about its center of mass. At the points at which the drum touches the ground no slipping occurs. Therefore instantaneously these points are at rest. But all points of the drum have an acceleration a forward and in addition the points of contact, due to the rotation about the center of mass, have a further linear acceleration R\alpha forward. Thus a + \alphaR= 0 or\alpha= - a/R(4) We wish to eliminate f from (2). Solving (3) for f (Ff - I\alpha)/R= f(5) Substituting (4) in (5) f = (Ff - Ia/R)/R f = [(Fr)/R] + [(Ia)/R2](6) Inserting (6) in (2) F cos \texttheta - [(Fr)/R] - [(Ia)/R^2 ] = Ma Solving for a [F cos \texttheta - {(Fr)/R}]/[M + (I/R^2)] = a ora = [F(cos \texttheta - r/R)]/[M + (I/R^2 )](7) Since we do not know F, we solve (1) for F and insert this in (7) F = (Mg - N)/(sin \texttheta) anda = [{(Mg - N)/(sin \texttheta)} {cos \texttheta - (r/R)}] / [M + (I/R^2)](8) The critical value of \texttheta, (\texttheta_0), is found by setting (8) equal to 0, whence a = 0 cos \texttheta_0 - (r/R) = 0 cos \texttheta_0 = (r/R)(9) If cos \texttheta > r/R, the drum rolls towards the workman, and vice versa. This result could be obtained more easily by con-sidering rotation about A, the line of the drum in-stantaneously at rest. The only force that does not pass through A is F^\ding{217}, the applied force. If the line of action of F^\ding{217} cuts the ground to the left of A, the moment of F^\ding{217} about A causes the drum to roll to the right. If the line of action of F^\ding{217} cuts the ground to the right of A, the moment of F^\ding{217} about A causes the drum to move to the left. If the line of action of F^\ding{217} passes through A, the drum is stationary and \texttheta has the critical value \texttheta_0. Figure (2) shows this situation. Since the line of action of F^\ding{217} is tangential to the inner cylinder, OB and AB are at right angles and \angle AOB is \texttheta_0. \therefore cos \texttheta_0 = r/R .

From the diagram y^2 + (R - x)^2 = R^2 (R - x)^2 = R^2 - y^2 = (115^2 - 75^2)ft^2 = 87^2ft^2 R - x = 87 ft x = (115 - 87)ft x = 28 ft The distance from the vertex of the dome to the ground sur-face is x + 30 = 58 ft, the same as the focal length. Thus the focal point of the mirror lies on the ground surface at the center of the arena. As a result of this there will be a tendency for spec-tator noise to be focused at the center of the ground sur-face, and the noise there is liable to be deafening. There exists a hockey arena that has been designed this way (accidentally), and in the center ice region the noise is of such intensity that the players can not hear the whistles of the officials.

Question:

Obtain polar solutions for the following polar equations, in which the ordered pair (R, T) is interpreted as R = length of the vector going from the origin to the point R, and T = angle between polar axis and that vector: (a) R = 1 - 2 sin T (b) R = 2 + 2 sin T (c) R = 1 + 2cosT - sin^2T

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Question:

What is an anhydro bond? Give examples of these bonds in common biological molecules.

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In biology, the most important covalent bonds joining molecules together are anhydro bonds. These are formed by removing a molecule of water to join two other molecules together. In other words, an -OH group is removed from one molecule and an -H group from the other to leave the two molecules bonded to each other usually by an oxygen or nitrogen atom. Examples of anhydro bonds in common biological molecules include: glycoside bonds, peptide bonds, and ester bonds (see figure). Glycoside bonds are the anhydro bonds of carbohydrates and are formed by removing an H from an alcohol group (-OH) on one sugar and an OH from the other sugar. Peptide bonds are the anhydro bonds of proteins and are formed by removing an -OH from the carboxyl group (-COOH) of one amino acid and an H from the amino group (-NH_2) of another amino acid. Ester bonds are an-hydro bonds of fats, formed by removing an OH from the carboxyl group of a fatty acid and an H from an alcohol group of glycerol. Other ester bonds that have great biological significance are phosphate esters [formed by removing an OH from phosphoric acid (HO-PO_3H_2) and an H from a sugar] and thioesters (formed by removing an OH from the carboxyl group of an acid and an H from a -SH group).

The figure shows the situation. The plane has a velocity relative to the air, V\ding{217}_PA, of 120 mi/hr in a direction of \texttheta degrees west of north. The air has a velocity relative to the earth of V\ding{217}_AE (50 mi/hr east). We require the plane to travel with speed v_pa due north. From the figure tan \texttheta = (V\ding{217}_AE/V\ding{217}_PE) = [V_AE/(V_PA^2 - V_AE^2)^1/2] Here, we have used the Pythagorean theorem. tan \texttheta = [(50 mi/hr)/({(120 mi/hr)^2 - (50 mi/hr)^2})^1/2] = [(50 mi/hr)/(109.09 mi/hr)] = tan^-1 (.4584) \approx 24\textdegree38'.

These sessile marine organisms are known as the tunicates, or sea squirts, which belong to the subphylum Urochordata. There is nothing in the external appearance of the adult that suggests it is a member of the chordates. The adult body is covered by a tough, leathery tunic, composed principally of a kind of cellulose. Within the tunic, the greater portion of the body is comprised of a pharynx perforated by numerous gill slits. The tunic has two openings: the incurrent and excurrent siphens. Water and nutrients enter the body via the incurrent siphon. After entering the pharynx and having the food particles digested, the waste products are carried by the water to the excurrent siphon where they are released from the body. During reproduction, the gametes leave the animal via the excurrent siphon. Tunicates are filter- feeders, removing food particles from the stream of water passing through the pharynx. The food particles are trapped in mucus secreted by part of the pharynx called the endostyle, and are carried down by ciliated cells into the esophogus and digestive gland. The flow of water through the tunicate is maintained by the contraction of the body wall. Whereas the adult tunicate is a sessile organism, the tunicate larva is motile, and shows a greater re-semblance to the other chordates. They possess a well- developed dorsal hollow nerve cord, a notochord beneath it in the tail region, and pharyngeal gill slits. When the larvae settle down and undergo metamorphosis into the adult form, the notochord is completely lost. Most of the nerve cord is also lost, the rest remaining only as a ganglion or "brain" above the pharynx.

The image I has to be real since it appears on the wall. Therefore, we must use a concave mirror and the object 0 must be placed beyond the focal point F for a real image (see the figure). We rule out a convex mirror since one cannot obtain images that can be shown on a screen with a convex mirror. If p and q are the object and image distances from the mirror respect-ively, the first mirror equation is (1/p) + (1/q) = (1/f) where f = (R/2) is the focal distance of the mirror. Therefore, we have p = 10 cm,q = 300 cm. (1/10 cm) + (1/300 cm) = (2/R), and R = 19.4 cm. The radius R has a positive sign, as it is required for a concave mirror. The optical magnification m is m = - (q/p) where the negative sign is required because we want m to come out negative to signify an inverted image. The image height is obtained as follows. h_1 = h_0m = - h_0 (q/p) = -0.5mm × (300 cm/10 cm) = - 15 cm. The image is therefore 30 times bigger than the object and is inverted.

A flowchart giving the program logic is given below. Consider the operation of this program. Statements 2\O to 8\O provide for the entering of the names in arbitrary order. Next a pair of nested loops is set up. The outer one runs from statement 9\O to statement 18\O, and the inner one runs from 11\O to 15\O. In statement 1\O\O, we set S$ = "ZZZZZZZZZZZZZZZ" Since we assume all names are 15 letters or less, S$ repre-sents the largest possible numerical representation of a name. Now consider the inner loop. The variable with the smallest numerical value corresponds to the first name of the alphabetical list. Thus after the inner loop is completely cycled, J1 will be the number of the name that is alphabetized to be first. That is, A$(J1) will be the first name in the alphabetical list. After control leaves the inner loop, A$(J1) is printed. Next, we set A$ (J$) = "ZZZZZZZZZZZZZZZ" and cycle the loop again. The first name in the alphabetical list has now been moved to the end. Now, when the inner loop is cycled again, the second name in the alphabetical list will be printed. Hence, when the outer loop goes through a complete cycle, the complete list Will be printed in alpha-betical order. 1\OREM ALPHABETIZING PROGRAM 2\ODIM A$ (1\O\O) 3\OPRINT "ENTER TOTAL NUMBER" 4\OINPUT N 5\OPRINT "ENTER NAMES" 6\OFOR I = 1 TO N 7\OINPUT A$ (I) 8\ONEXT I 9\OFOR I = 1 TO N 1\O\OLET S$ = "ZZZZZZZZZZZZZZZ" 11\OFOR J = 1 TO N 12\OIF S$ < = A$ (J) THEN 15\O 13\OLET S$ = A$ (J) 14\OLET J1 = J 15\ONEXT J 16\OPRINT A$ (J1) 17\OLET A$ (J1) = "ZZZZZZZZZZZZZZZ" 18\ONEXT I 999END

This question deals with proteolytic enzymes, which may be defined as protein degradation catalysts. Two factors that affect enzymic activity are pH, and temperature. Whether these enzymes damage the stomach or not will depend, to a great extent, on the conditions in the stomach. The accompanying figure shows how pH can alter enzyme activity of pepsin and trypsin. From the diagram, you see that these enzymes do not function well at extremely low pH. Due to the presence of HCl in the stomach as a hydrolysis agent, the acidity of the stomach is high, which means an environment of low pH exists. It becomes doubtful that these proteolytic enzymes would have much activity at the low pH of the stomach. This means that little, if any, damage would be done to the stomach. Heating alters enzymic activity, also. It is likely that most of the enzyme would be inactivated by the cooking of the meat.

Cleavage in a mammalian egg takes place as the egg moves slowly down the oviduct, driven along by cilia and muscular contraction of the wall of the duct. The journey takes about four days in the case of cells, the morula, which divides and rearranges into a hollow ball of cells, the blastocyst. The blastocyst subdi-vides into an inner cell mass from which the embryo devel-ops, and an enveloping layer of cells, the trophoblast. The cavity of the blastocyst may be compared to the blastocoel, but the embryo is not a blastula, for its cells are diffe-rentiated into two types. The cells ot the inner cell mass differentiate further into a thin layer of flat cells, the hypoblast, which is located on the interior surface of the mass adjacent to the blastocoel, and which represents the endoderm. The remaining cells of the inner cell mass become the epiblast. The cells of the hypoblast, spreading along the inner surface of the trophoblast, eventually surround the cavity of the blastocyst, forming a "yolk sac," which is filled with fluid, not yolk. As the hypoblast spreads out, the inner cell mass also spreads and becomes a disc shaped plate of cells similar to the blastodisc of bird and reptilian eggs. The blastodisc becomes delimitated from the rest of the embryo. Gastrulation begins with the formation of a primitive streak and Hensen's node in which cells migrate downward, laterally and anteriorly between the epiblast and hypoblast. Those cells which remain between these two layers comprise the mesoderm. Those cells which join the hypoblast become endoderm. A crevice appears between the cells of the inner cell mass, which then enlarges to become the amniotic cavity. The cavity of the blastocyst becomes filled with mesodermal cells, and is comparable to the subgerminal cavity in the bird. The embryonic disc comes to lie as a plate between the two cavities, connected to the trophoblast at the posterior end by a group of extraembryonic mesoderm cells, the body stalk or allantoic stalk. The non- functional endodermal part of the allantois, which develops as a tube from the yolk sac, is rudimentary and never reaches the trophoblast. Thus, after two weeks of development the human embryo is a flat, two-layered disc of cells about 250 microns across, connected by a stalk to the trophoblast.

Question:

Write a FORTRAN program to solve linear congruences of the form ax = b (mod m).

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Solution:

Write a FORTRAN program that accepts two positive integers and out-puts a message indicating whether the integers between the two limits are perfect, abundant, or deficient.

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Solution:

The proper divisors of a positive integer include all divisors except the integer itself. For instance, the proper divisors of 12 are 1,2,3,4, and 6. We use three terms to classify integers according to the sums of their divisors. An integer is perfect if the sum of its proper divisors is equal to that integer. There exist only 6 perfect numbers less than 10 billion, all of which are even. One of them is integer 6. Its proper divisors are 1,2 and 3, the sum of which is equal to 1 + 2 + 3= 6. The rest of the integers may be either abundant or deficient. An integer is abundant if the sum of its proper divisors is greater than that integer. Conversely, an integer is deficient if the sum of its proper divisors is less than an integer itself. Perfect numbers are related toMersenneprimes, which take the form 2^K -1. We can say that if 2^K - 1 is prime, then n = 2^K - 1(2K - 1) is perfect. Note that we refer only to even perfect numbers. No odd perfect numbers have yet been demonstrated. The program uses a subprogram called NSIGMA, named for the \sigma-function. If q is a prime, then\sigma(q) = q + 1. In general, \sigma \sigma \sigma \sigma(q)^K = [(q^K+1 -1) / (q -1)] for q = 1, 2, 3,... \sigma \sigma An array, NPRIM, is used to store all primes between 2 and 181. Although this example uses a DATA statement for the primes, you could also use a subprogram for the Sieve of Eratosthenes to generate primes. The main program reads in upper and lower bounds (HI and LO, respectively), an increment term (I), and a case number (K), which indicates the following according to the values assigned to it: K = -2 output deficient numbers only K = -1 output deficient and perfect numbers only K = 0 output all numbers K = 1 output abundant and perfect numbers only K = 2 output abundant numbers only Since the highest prime number stored in NPRIM is 181, the upper bound cannot exceed (181)2 = 32761, otherwise some of the divisors may end up missing. The program looks as follows: INTEGER HI READ (5,100) LO, HI, I, K 100FORMAT (4(I5)) DO 60 N = LO, HI, I CUSE NSIGM& TO CHECK FACTORS IF (NSICMA (N) - 2\textasteriskcenteredN) 15, 20, 25 15IF (K) 30, 30, 60 20IF (IABS(K) - 1) 40, 40, 60 25IF (K) 60, 50, 50 30WRITE (6,101) N 101FORMAT (1H, I5, 13H IS DEFICIENT) GO TO 60 40WRITE (6, 102) N 102FORMAT (1H ,I5, 11H IS PERFECT) GO TO 60 50WRITE (6, 103) N 103FORMAT (1H , I5, 12H IS ABUNDANT) 60CONTINUE STOP END CFUNCTION SUBPROGRAM NSIGMA FUNCTION NSIGMA DIMENSION NPRIM (42) CFILL ARRAY WITH PRIMES FROM 2 TO 181 DATA NPRIM/2,3,5,7,11,13,17,19,23,29,31,37,41, 143,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113, 2127,131,137,139,149,151,157,163,167,173,179,181/ INT = N RN = INT NROOT = SQRT(RN) NSIGMA = 1 MSIGMA = 1 DO 40 I = 1,42 NF = 0 NP = NPR(I) IF (NP - NRT) 32, 32, 50 32KINT = INT/NP IF (INT-KINT\textasteriskcenteredNP) 36, 34, 36 34INT = KINT NF = 1 WSIGMA = NSIGMA\textasteriskcenteredNP + MSIGMA GO TO 32 36IF (NF) 40, 40 , 38 38MSIGMA = NSIGMA RN = INT NROOT = SQRT (RN) 40CONTINUE 50IF (INT-1) 60, 60, 56 56NSIGMA = NS IGMA\textasteriskcenteredINT + NSIGMA 60RETURN END

Question:

Suppose that three devices are connected in parallel to a 12-V battery. Let the resistances of the devices be R_1 = 2\Omega, R_2 = 3\Omega, and R_3 = 4\Omega, What current is supplied by the battery and what is the current in each device?

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Solution:

Since the resistors are in parallel, we have for the equivalent resistance R, (1/R) = (1/R_1 ) + (1/R_2 ) + (1/R_3 ) = (1/2\Omega) + (1/3\Omega) + (1/4\Omega) = (13/12)\Omega^-1 The equivalent resistance, therefore, is R = (12/13)\Omega = 0.92\Omega. The voltage across each device is 12 volts since they are in parallel to the battery. Therefore, using Ohm's law and the figure, I_1 = (V/R_1) = [(12 V)/(2\Omega)] = 6 amp I_2 = (V/R_2) = [(12 V)/(3\Omega)] = 4 amp I_3 = (V/R_3) = [(12 V)/(4\Omega)] = 3 amp The current supplied by the battery is found by applying Kirchoff's node equation at point A. Hence I = I_1 + I_2 + I_3 = (6 + 4 + 3) amp I = 13 amp.

The nucleic acid in a virus particle is surrounded by a protein coat, called a capsid, made up of protein subunits called capsomeres. A complete virus particle is called a virion and it may be covered by an envelope. Those that do not have envelopes are termed naked virions, while those virions that do have envelopes are termed enveloped. A typical naked and an enveloped virion are shown in Fig.1. Some of the protein coats are complex, containing numerous layers and various different proteins. Some also include lipid and carbohydrate molecules. Others, however, contain only one or a few types of protein molecules in their coats. The use of a large number of identical proteins in the protective coat of viruses is often mandatory. The extremely small viruses have a limited nucleic acid content which restricts the total number of amino acidsthat can be used to synthesize its protein. For example, there are approximately 6000 nucleotides in a TMV RNA chain. The TMV RNA can therefore code for a protein 2000 amino acids in length. (Three nucleotides are needed to code for one amino acid.) Since each amino acid weighs about 125 daltons (a dalton is a unit of weight equal to the weight of a single hydrogen atom), this protein molecule of 2000 amino acids should weigh 2.5 × 10^5 daltons. However, since we know the protein coat of TMV weighs 3.5 × 10^7 dal- tons, about 140 (3.5 × 10^7 \div 2.5 × 10^5) of these protein molecules would be needed. The virus does not use all its nucleic acids to code for a single protein. Since only a very small part of the RNA chain codes for the coat protein, the proteins consist of fewer amino acids. Therefore, many smaller identical protein molecules are used to construct the protein coat. Since the number of identical protein molecules used to form the TMV protein coat is about 2150, each protein molecule consists of about 128 amino acids {[(3.5 × 10^7 daltons) / (2150 proteins)] × [(1 amino acid) / (125 daltons)] = (128 amino acids) / (proteins )} The protein subunits either possess helical symmetry or cubical (or quasi-cubical) symmetry. The TMV shell con-sists of identical protein molecules helically arranged around a central RNA molecule (see Fig. 2) . The Adenovirus is an icosahedral virion, exhibiting quasi-cubical sym-metry (see Fig. 3). An icosahedron is a twenty-sided geometric figure.

Since equation (1) is of the form d^2y/dx^2 = g_1(y), we can use the EULER1 subroutine developed in a previous problem. First, rewrite equation (1) as a system of simultaneous first-order equations: ẏ = dy/dt = z(2) \.{z} = dz/dt = -Ay(3) with initial conditions y(0) = y_0 and z(0) = y'(0) = z_0. Now use the modified Euler method to find solutions to both equations (2) and (3): REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A COMMON A READ,N,TFIN,ACCUR,Y0,Z0, A CALL EULER1(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G1(W) REAL G1,W,A COMMON A G1 = -A\textasteriskcenteredW RETURN END

From the given table we observe that the gross return from the investment is $175,000 - $75,000 = $100,000. However, this is assuming that the money invested has no alternative uses, e.g. it could not be invested in a bank at 10% interest. Taking the time value of money into account, to obtain $20,000 after 1 year we would have to invest $20,000 = X + .IX now. Thus, X = (20,000 / 1.1) = $18182 Similarly, $30,000 two years from now is worth $30,000 = (X+.1X)(1+.1) = X(1+.1) (1+.1) = X(1.1)2 Or, X = [$30,000 / (1.1)^2] = $24,794 Applying the same analysis to the remaining returns: Year Return Present Value 1 2 3 4 5 20,000 30.000 35,000 40,000 50,000 20,000/1.1= 18,182 30,000/(1.1)2= 24,794 35,000/(1.1)3= 26,296 40,000/(1.1)4= 27,321 50,000/(1.1)^5= 31, 047 Adding the present values of returns over the five years gives the present value of returns from the investment. PV = $127,640. The net present value of the investment when the current interest rate is 10% is therefore NPV = PV - I = $127,640 - $75,000 = $52,640. This means that the $100,000 return over five years is actually worth $52,640 to the firm now. Note that as the interest rate changes continuously from 0% to 100%, the NPV changes continuously from $100,000 (the gross return becomes the net present value when the interest rate is 0%) to $-59,063. Thus, there must be an interest rate at which the NPV equals zero. This interest rate value is called the DISCOUNTED CASH FLOW interest rate. A graph of the NVP versus interest rate might look as shown in Fig. 1. To actually compute the DCF we must find a root of the equation f (r) = 0 where NVP = f (r) . We cannot use the Newton-Raphson method because the functional relationship is not in polynomial form. The bisection method, however, may be applied. The procedure is as follows: We know that f(.10) = 52,640 and f(1.00) = -$59,063. Compute the value of f at the mid-point of the interval [10, 1.00], f(.55). If f(.55) \geq 0 the root must lie in the interval [.55, 1.00] (since f(.10) is positive and f(1.00) is negative). If f(.55)<0, then the root lies in the interval [.10, .55]. In either case the interval of search has been reduced by 1/2. Repeat the above procedure with the new interval i.e. take the mid-point and evaluate f at this point. A flowchart for computing the DCF is as shown in Fig. 2

If we examine closely the structure of the chlorophyll molecule (see Figure) we will see that its atoms constitute a conjugated system, with double and single bonds alternating around the ring. Such a conjugated system provides for many possible resonance structures and is essentially a resonating system in which the pi electrons are spread, or delocalized, about the carbon atoms of the ring. In other words, a pi electron is no longer associated with a single atom or bond but with the conjugated system as a whole. This property, called resonance, gives the chlorophyll molecule considerable stability. As a conse-quence of resonance stabilization, only a small amount of energy is required to raise the pi electrons of the carbon atoms to a higher energy level. The chlorophyll molecule absorbs energy by having its pi electrons excited and "pumped" to a new, more energetic orbital (energy level). The energy of excitation comes from the sun in the form of visible light energy, and is converted to chemical energy in the light reactions of the photosynthetic process. The absorbed energy is always in a definite, specific quantity, known as a photon. In order to raise an electron to a given new energy level, the energy of the photon must just equal the difference in energy content of the electron's old and new orbitals. Unless a photon can raise the energy of the electron by just the right amount for a defined energy level, that photon will not be absorbed. The adaptive importance of the chlorophyll molecule lies therefore in its unique capacity to function in capturing the energy of visible light with a high degree of efficiency.

E.coliare usually able to utilize the mono-saccharide galactose as a coarbon and energy source. However, mutations arise which affect an enzyme necessary for galac-tose utilization. TheseE.colimutants are unable to utilize this sugar, and are called gal^-mutants. To transform some of these gal^-mutants into the wild type, gal^+, one can use a lysogenic virus such as the \lambda (lambda) bacteriophage. One can infect a culture of prototrophicE.coli(gal^+) with the lysogenic phages. The viruses will inject their DNA into the bacterial cells. In the host, the viral DNA molecule changes from a linear structure to a circular one. At a specific attachment site on the bacterial chromosome, the viral DNA pairs with the bacterial chromosome and integrates into it after recombination. The viral DNA, now called a prophage, remains incorporated within theE.colichromosome, until conditions are favorable for the excision of the prophage and its induction to a vegetative virus which replicates and lyses the cell. The gene necessary for galactose utilization is very near the incorporated prophage. In the case of the \lambda phage, its proximity to the bacterial gal^+ gene allows rare errors to occur in which excision of the prophage includes the gal^+ gene. The \lambda prophage may coil in such a way that the recombination event leading to excision of the circular viral DNA includes the gal gene. To remain approxi-mately the same size, the circular viral DNA leaves behind some of its own genes, which were replaced by the substituted bacterial region. Since it now lacks some necessary genes, the virus, now called a transducing particle or \lambda gal, is considered to be defective. These transducing particles can lyse the bacterial cell, but they cannot establish a lysogenic relatonship with, or cause lysis of, subsequent bacterial cells. The excision of the prophage which has the incorporated gal^+ gene can be illustrated as follows: One can then harvest these transducing particles and add them to a culture of gal^-E.colimutants. These phages attach to the mutant bacteria and inject their DNA (containing the gal^+ gene). The gal^+ region of the viral DNA may then recombine with the gal^- region of the bacterial chromosome. The gal^+ gene may be incorporated into theE.colithrough the recombination process, transforming them to the wild type, which are able to utilize galactose. The whole viral DNA rarely becomes incorporated, since the transducing phage lacks a complete genome. The viral chromosome, containing the gal^- gene, cannot be inserted into the bacterial chromosome and so is lost. This process by which a virus transfers genes from one bacterial cell to another is called transduction. In specialized transduction, as opposed to generalized transduction, only one gene is transferred. Transduction thus serves as a mechanism for recombination in bacteria.

Question:

A light horizontal bar is 4.0 ft long. A 3.0-lb force acts vertically upward on it 1.0 ft from the right- hand end. Find the torque about each end.

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Solution:

Since the force is perpendicular to the bar, the moment arms are measured along the bar. About the right-hand end L_r = 3.0 lb × 1.0 ft =3.0 lb-ftclockwise About the right-hand end L_1 = 3.0 lb × 3.0 ft = 9.0 lb-ftcounterclockwise The torques produced by this single force about the two axes differ in both magnitude and direction. This causes the bar to twist through an angle \texttheta which is proportional to the torque.

Question:

How do the algae differ from the fungi in their method of obtaining food?

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Solution:

Algae areautotrophs, while the fungi areheterotrophs. Certain algae contain the green pigment chlorophyll, others have the accessory pigments. These pigments enable them to utilize solar energy to synthesize organic compounds from CO_2 and H_2O. Fungi, lacking photo synthetic pigments, must obtain organic compounds direct-ly. The eucaryotic algae (all algae except the blue-green) contain chlorophyll in chloroplasts,subcellularorganelles where photosynthesis occurs. Solar energy is trapped by the chlorophyll and converted to ATP, the ATP is used by the chloroplast to convert carbon dioxide and water to organic compounds, and oxygen is liberated. Algae obtain water and CO_2 from their environment, as well as the other minerals needed for their organic con-stituents. The CO2and minerals are dissolved in the aquatic environment, and are absorbed through the cell wall and cell membrane of the algae. Fungi live where organic compounds are present. Some fungi are animal or plant parasites but most are saprophytes. Mostsarophyticfungi excrete digestive enzymes into their environment, and break down food materialextracellularly. The products of digestion are then absorbed through the cell wall and cell membrane by structures called thehautoria. Saprophytic fungi de-compose vastquatitiesof dead organic material. Some cause spoilage of foodstuffs (bread, fruit, vegetables) and deterioration of leather goods, paper, fabrics, and lumber. Parasitic fungi may carry out extracellular digestion or they may directly absorb organic materials produced by the body of their host. However, many fungi are benefical to man.

Question:

a notochord, a dorsal hollow nerve cord, a backbone, hinged jaws, lungs and large, fleshy paired pelvic and pectoral fins. A marine biologist concluded that the animal very closely represented the direct ancestor of the land vertebrates. What kind of animal was this?

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Solution:

By assigning the observed characteristics to a class of organisms in a progressively specialized manner, we may deduce the identification of this marine animal. The presence of a notochord and a dorsal hollow nerve cord indicates that the animal belongs to the phylum Chordata. The presence of a backbone and hinged jaws leads us to reason that the animal is a vertebrate; more specifically, it is a bony fish (Osteichthyes). The existence of lungs and large, fleshy fins further classifies the animal as a lobe-finned fish. Since the animal was thought to represent the ancestors of the land vertebrates, our conclusion is confirmed. The animal is a lobe-finned fish, otherwise known as a coelacanth.

As the electron is passing through the slit, we lose sight of it. Therefore its position at the slit has an x-axis uncertainty \Deltax = d. As a result of it, the momentum of the electron after it passes through the slit can have an x-component \Deltap_x = p sin \alpha . In order to observe the wave characteristics of the electron, we should be able to display its diffraction pattern. Therefore, the slit should have a size consistent with the single slit condition for observing the first fringes of the diffraction pattern sin \alpha \geq (\lambda/d) where \lambda. is the wave length (de Broglie wave length) of the electron. \lambda = (h/p). Now we calculate \Deltax \Deltap_x , to obtain the uncertainty relation for the electron; \Deltax \Deltap_x \approx dp sin \alpha \geq dp (\lambda/d) = dp (h/pd) \geq h.

a) This flowchart is somewhat fictitious, and it does not satisfy the one-entry one-exit rule of modules in structured program-ming. We will assume that this flowchart represents a program segment, since we are not shown where the four nodes (written as circles enclosing numbers) lead. The problem lies in the module for tasks A-F. It has two points of entry, an unstructured means of transferring program control. b) Recalling the three logical structures of programming -- sequencing, decision-making, and looping \textemdash you should recognize that the flow-chart can be rendered into proper pseudocode easily. We need the IF-THEN-ELSE construct, along with some kind of nesting of decisions. The pseudocode is rather straightforward: if (P) then if (Q) then tasks G-N else tasks A-F else if (R) then tasks O-Z else tasks A-F end if then else c) Notice that tasks A through F are referenced twice in the program. If these tasks take up many lines of instructions, it would be tedious to write out all of these instructions twice. As an alternative, you may want to call a subroutine to perform these tasks. By doing this, you need write out tasks A through F only once.

The partial pressure of each of the components in a mixture of gaseous substances will be equal to the product of the mole fraction of that component in the gas and the total pressure of the system. Given the total vapor pressure of the gases, one can calculate the mole fraction of the water vapor. Let P_A = partial pressure of the water vapor, N_T = total number of moles of gases produced, N_A = number of moles of water vapor, and P_T = total pressure. This law can now be expressed in these terms: P_A = [N_A / N_T] P_T . From thestoichiometryof the equation, the total number of moles of products is 3. Out of these 2 moles are water vapor. Thus, substituting, P_A = 2/3 (6.25torr) = 4.16torrof water vapor.

Question:

A block of mass m slides down a plane inclined at an angle \texttheta, the surface of which has coefficient of sliding friction \mu. The mass collides with a spring, having force constant k, after it has slid a distance d from rest. The spring will then reduce the speed of the block, which, will come to a momentary halt, when the spring has been compressed a distance X. The block will then be pushed back up the incline and begin a frictionally damped harmonic oscillation. Calculate X.

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Solution:

We will use the law of energy conservation. The kinetic energy of the block at the moment it collides with the spring is equal to the sum of its loss of gravitational potential energy and the work done on it by friction. Since the frictional force, f, acts in a direction opposite to the motion, the work due to friction is negative. Then (1/2) mv^2 = mgh - \muNd where f = \muN, and h = d sin \texttheta is the vertical height through which the block falls, (see figure). N = mg cos \texttheta is the force the plane exerts, normal to its surface, on the block to cancel the component of the block's weight perpendicular to the plane. Thus: (1/2) mv^2 = mg d sin \texttheta - \mu mg d cos \texttheta = mg d(sin \texttheta - \mu cos \texttheta) Once the spring has brought the block to rest, the spring's potential energy is equal to the sum of the block's kinetic energy, the further gravitational po-tential energy the block loses as it falls through height h' = X sin \texttheta, and the additional (negative) work done by friction on the block (see figure). (1/2) kX^2 = (1/2) mv^2 + mgh'-\mu NX (1/2) kX^2 = mgd(sin \texttheta - \mu cos \texttheta) + (mg sin \texttheta)X - (\mu mg cos \texttheta)X (1/2) kX^2 - mg (sin \texttheta - \mu cos \texttheta)X - mgd (sin \texttheta - \mu cos \texttheta) = 0 From the quadratic formula: x = [- b \pm \surd(b^2 - 4ac)] / 2awith a = (1/2) k, b = - mg (sin \texttheta - \mu cos \texttheta), and c = -mg d (sin \texttheta - \mu cos \texttheta) X = [mg(sin \texttheta - \mu cos \texttheta) \pm \surd{m^2g^2 (sin \texttheta - \mu cos \texttheta)^2 - 4(1/2 k)}] / [2{(1/2)k}] [-mgd (sin \texttheta - \mu cos \texttheta)] / [denominator] X = [mg(sin \texttheta - \mu cos \texttheta) \pm \surd{m^2g^2 (sin \texttheta - \mu cos \texttheta)^2 + 2mgdK(sin \texttheta - \mu cos \texttheta)}] / (k) This yields two answers for X. If we had been given numerical values for m, d, \mu, \texttheta, and k, we would find that one of the solutions for X would be negative and unacceptable. The other solution would be correct for the problem at hand.

There are four forces acting on the sled, as shown in the diagram; the weight W\ding{217}, the normal force N\ding{217} exerted by the slope, the frictional force F_f\ding{217} exerted by the snow down the slope opposing the motion, and the upward pull P\ding{217} exerted by the boy at an angle \texttheta to the slope. The sled is moving with uniform velocity up the hill and thus the forces are in equilibrium, and F_f = \mu_kN, where \mu_k is the coefficient of kinetic friction. Let us now resolve all the forces into components parallel and perpendicular to the surface of the slope. We will take forces going up the slope as positive in the parallel direction, and forces pointing upward out of the slope as positive in the perpendicular direction. Imposing the conditions of equilibrium, we have: \sumF_\parallel = P cos \texttheta - W sin \alpha - \mu_kN = 0(1) \sumF_\perp = N + P sin \texttheta - W cos \alpha = 0(2) where \alpha is the angle of the slope's incline. We can thus solve for P. From equation (2) N = W cos \alpha - P sin \texttheta From equation (1) P cos \texttheta = W sin \alpha + \mu_kN = W sin \alpha + \mu_k(W cos \alpha - P sin \texttheta) P(cos \texttheta + \mu_k sin \texttheta) = W sin \alpha + \mu_kW cos \alpha P = (W sin \alpha+ \mu_kW cos \alpha)/(cos \texttheta + \mu_k sin \texttheta) We see now that P is a function of \texttheta alone since W, \alpha, and \mu_k are constants. Thus, to find the angle \texttheta at which the boy must exert the minimum force, we use the calculus to find the value of \texttheta at which a minimum occurs for P. These minima occur when the derivative of P with respect to \texttheta is zero. Since the numerator is constant with respect to \texttheta: (dP/d\texttheta) = - [(W sin \alpha + \mu_kW cos \alpha)/{(cos \texttheta + \mu_k sin \texttheta)^2}](- sin \texttheta + \mu_k cos \texttheta) = 0 orsin \texttheta = \mu_k cos \texttheta tan \texttheta = \mu_k = 0.10 \texttheta = 5.7\textdegree Thus the boy should drag his sled up in such a way that the rope makes an angle of 5.7\textdegree with the slope. It might have seemed at first sight that a force parallel to the slope would be most efficient. It is now clear that this is not so. Any component of the pull P\ding{217} at right angles to the slope decreases the normal force N\ding{217} and thus the frictional force F\ding{217}. The best compromise between maximum forward force and least frictional force is achieved at the angle 5.7\textdegree.

The cartilaginous fish of the class Chondrichthyes are distinguished by, as their name implies, their cartilaginous skeletons. No bone is present in this group. The presence of a cartilaginous skeleton in this class must be regarded. as a specializa-tion because there is evidence that the ancestors of the group, the placoderms, had bony skeletons. Included in the class Chondrichthyes are the sharks, skates, and rays. The sharks are streamlined, fast-swimming, voracious feeders. One kind, the whale shark, is the largest known fish. It may attain a length of 50 feet or more. Rays are sluggish, flattened bottom feeders. Chondrichthyes have neither swim bladders nor lungs. Osmoregulation in some groups is unusual, in-volving retention of high concentrations of urea in the body fluids. Fertilization is internal, and the eggs have tough, leathery shells. Most species are predatory but a few are plankton feeders. Like the cartilaginous fish, the bony fish, belonging to the class Osteichthyes, arose from the placoderms. In most of them, the adult skeleton is composed largely of bone. Unlike cartilaginous fish, fertilization is external in most bony fish. The gills of bony fish are covered by a hinged bony plate, the operculum, on each side of the phcirynx. In most bony and cartilaginous fish the body is covered with scales. Unlike the cartilaginous fish, most bony fish have a swim bladder, which arose as a modification of the lungs of the earliest Osteichthyes. There are still some bony fish today that possess lungs instead of swim bladders. By modifying the gaseous content of the swim bladder, the fish can modify its buoyancy. Cartilaginous fish have no swim bladder and are denser than the sea water around them. They would sink to the bottom if it were not for the movements of the pelvic and tail fins which serve to maintain buoyancy in these fish. The early Osteichthyes gave rise to three main lines of bony fishes: the lobe-finned fish (Order Crossopterygii), the lungfish (Order Dipnoi), and the telecosts (Order Tele- ostei) which form the dominant group of marine and aquatic animals today). The lungfish and lobe-finned fish have functional lungs as well as gills. Teleosts posses a swim bladder, but no lungs.

Consider a unit cell of sodium chloride in the accompanying figure, (a) What fraction of a Cl^- at each corner of the cube is within the unit cell? (b) What fraction of a Cl^- appears in each face? (c) What fraction of each Na^+ is with-in the unit cell? (Note that one complete Na^+ is in the center.) (d) Add all the fractions for each ion within the unit cell. What is the ratio of Na^+ to Cl^-?

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Solution:

The following rules may be used to determine the number of unit particles (atoms) associated with each type of cubic unit cell: 1) An atom at a corner contributes 1/8 of its volume to each of 8 adjacent cubes. 2) An atom on an edge of a cube contributes 1/4 of its volume to each of 4 adjacent cubes. 3) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. 4) An atom completely within a cube contributes all of its volume to the unit cell. (a) According to rule 1 an atom at a corner contributes 1/8 of its volume to the unit cell. (b) From rule 3 one can see that an atom on the face of the unit cell contributes 1/2 of its volume to the cell. (c) From rule 2 one can see that atoms on the edge of the unit cell contribute 1/4 of their volume to the unit cell. Therefore the sodium ions on the edge of the cell contribute 1/4 of their volume and the one in the center of the cell contributes its entire volume. (d) There are 14 Cl^- ions contributing to the cell. 8 are at corners and 6 are on faces - using parts a + b of this problem one can calculate the number of atoms that the Cl^- ions contribute to the unit cell. total number of Cl^- ions = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/face = 4 atoms of Cl^- The Na^+ ions can be dealt with in a similar manner. There are 13 Na^+ ions, 12 on edges and 1 in the center - using part c of this problem one can find the number of Na^+ ions contributing to the unit cell. total number of Na^+ ions = 12 edges × 1/4 atom/edge + 1 atom in center = 4 atoms of Na^+ There are 4 atoms of Cl^- and 4 atoms of Na^+ contributing to the unit cell. The ratio will therefore be -Na^+ : Cl^- is 4 : 4 or 1 : 1. Na^+ and Cl^- contribute equally to the unit cell.

A single unit of core can take on a value of 1 or 0, depending upon whether the direction of its mag-netization is clockwise or counterclockwise. If all the cores were strung on a single wire each time current was passed down the wire, all the cores would become magne-tized in the same direction. This would render the use of core inefficient. To avoid this possibility, the method of coincident cur-rent is used. Three wires are strung through each core to mediate control of the direction of magnetization. One entire plane of cores resembles a grid with cores at each intersection. Two of the three wires are current wires, while the third is called a sense wire. Let us present an example of the procedure necessary to have a single core represent the binary digit 1. Each of the current wires sends only one-half the necessary cur-rent to make the core become magnetized in a clockwise direction. The core located where the current wires inter-sect will become magnetized in a clockwise direction; all other cores along the lines remain unaffected. The sense wire is a feedback device which can determine at any time if a particular core is representing a binary 1 or a binary 0.

Question:

At two points on a horizontal tube of varying circular cross section carrying water, the radii are 1 cm and 0.4 cm and the pressure difference between these points is 4.9 cm of water. How much liquid flows through the tube per second?

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Solution:

Since the tube is horizontal there is no pressure difference along the tube due to hydrostatic effects because the static pressure due to the weight of the fluid plays no part in the problem. Thus Bernoulli's equation is P_1 + (1/2) \rhov^2_1 = p_2 + (1/2) \rhov^2_2(1) where \rho is the density of the liquid, v its velocity, p its pressure, and the subscripts 1 and 2 refer to any two points along the tube. In time t, those particles of the fluid which were originally at point 1(see figure), move a distance v_1t. The total volume of fluid which moves past point 1 in time t is there-fore A_1V_1t. Its rate of flow per unit time is then A_1V_1. Similarly, the rate of flow past point 2 is A_2V_2. Assuming the fluid is incompressible, these two rates of flow must be equal. We have A_1V_1 = A_2V_2(2) This is the equation of continuity which states that the quantity of an incompressible liquid which flows through a tube per second is constant at all points. The pressure difference is given as 4.9 cm of water. This equals the pressure produced by the weight of 4.9 cm of water or P_1 - P_2 = \rhog x 4.9 cm(3) From equations (1) and (3), v^2_2 - v^2_1 = [2(p_1 - p_2)]/\rho = [2(\rhog) (4.9 cm)]/\rho = (2g)(4.9 cm) = (2)(980 cm/sec^2) (4.9 cm) andv^2_2 - v^2_1 = 98^2 cm^2/sec^2(4) Using equation (2), v_1/v_2 =A_2/A_1 = (\pi × 0.4^2 cm^2)/(\pi × 1^2 cm^2) = 0.16. Substituting v^2_1 = 0.16^2 v^2_2 in equation (4) yields v^2_2(1 - 0.16^2) = 98^2 cm^2 \bullet s^-2 orv_2 = \surd[(98^2 cm^2 \bullet s^-2)/(0.9744)] The quantity of water flowing through the tube per second is thus A_1V_1 = A_2V_2 = \pi × (0.4 cm)^2 × \surd[(98^2 cm^2 \bullet s^-2)/(0.9744)] = 50 cm^3 \bullet s^-1.

1) When the trolley moves with a uniform speed, the only external force acting on m is the gravitational force mg\ding{217}. It is balanced by the \ding{217} tension T\ding{217}in the rope, since m is in equilibrium (Fig. a). Hence \alpha = 0. 2) \ding{217} When the trolley has acceleration a, the effect of the acceleration is transmitted to the mass through the rope. We see (Fig. b), that the magnitudes of the gravitational and horizontal accelerations determine the magnitude of the angle \alpha. The tension in the rope provides the upward force -mg\ding{217}to hold the mass (since m is in vertical equilibrium) as well as \ding{217} the external force ma acting on the mass as a result of the motion of the trolley, (Fig. c). Hence, T sin \alpha = ma(1) T cos \alpha = mg(2) Dividing (1) by (2), we get [T sin \alpha] / [T cos \alpha]= (ma/mg) tan \alpha= (a/g) Therefore, \alpha is given by \alpha = tan^-1 (a/g).

During a single normal breath, the volume of air entering or leaving the lungs is called the tidal vol-ume. Under conditions of rest, this volume is approximately 500 ml. on average. The volume of air that can be in-spired over and above the resting tidal volume is called the inspiratory reserve volume, and amounts to about 3000 ml. of air. Similarly, the volume of air that can be expired below the resting tidal volume is called the expiratory reserve volume, and amounts to approximately 1000 ml. of air. Even after forced maximum expiration, some air (about 1000 ml.) still remains in the lungs, and is termed the residual volume. The vital capacity is the sum of the tidal volume and the inspiratory and expiratory reserve volumes. The vital capacity then represents the maximum amount of air that can be moved in and out during a single breath. The average vital capacity varies with sex, being 4.5 liters for the young adult male, and about 3.2 for the young adult female. During heavy work or exercise, a person uses part of both the inspira-tory and expiratory reserves, but rarely uses more than 50% of his total vital capacity. This is because deeper breaths than this would require exhaustive activities of the inspiratory and expiratory muscles. Vital capacity is higher in an individual who is tall and thin than in one who is obese. A well developed athlete may have a vital capacity up to 55% above average. In some diseases of the heart and lungs, the vital capacity may be reduced considerably.

Question:

An x-ray tube with a copper target is found to be emitting lines other than those due to copper. TheK_\alphaline of copper is known to have a wavelength of 1.5405 \AA, and the other two K_\alphalines observed have wavelengths of 0.7092 \AA and 1.6578 \AA. Identify the impurities.

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Solution:

One determines the total spin of an atom by calculating the sum of all the s quantum numbers of the electrons. There are four quantum numbers needed to describe each electron. They are the principal quantum number n, the azimuthal quantum number l, the magnetic quantum number m and the spin quantum number s. In each orbital of the atom two electrons can be accommodated. These electrons will have opposing spins of (+1/2) and (-1/2). The total spin is calculated by determining the total number of unpaired electrons and multiplying this number by (1/2). It is assumed that all unpaired electrons will have the same spin. Solving for the total spins: Figure A shows the total number of electrons that can occupy each orbital. (a)1s^2 2s^2. Filled s orbitals each contain 2 electrons. One writes this electronic configuration diagrammatically as shown in figure B. Because there are no unpaired electrons the total spin for this atom is 0. (b)1s^2 2s^2 2p^3. Here the two s orbitals are filled but the p orbital is only half filled. There are three subshells in the p orbital and there will be one electron occupying each of these. The configuration of lowest energy is assumed when each of these 3 electrons has the same spin. This configuration is written diagrammatically as shown in figure C. This configuration contains 3 unpaired electrons. The total spin is, thus 3 × (1/2) or (3/2). (c)1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2. From figure A one sees that the only unfilled orbital in this configuration is the 3d. Therefore, unpaired electrons can only occur in this shell. From figure A, one sees that the 3d orbital can contain 10 electrons. Because each subshell contains 2 electrons, there are 5 subshells in this orbital. In this configuration there are 5 electrons occupying the 3d orbital, therefore, each subshell con-tains 1 electron and there are 5 unpaired electrons in the configuration. The total spin of this configuration is, thus, 5 × (1/2) or (5/2).

The block diagram is shown in figure 1. The slave gets the inverted clock pulse. The input to the slave is taken from the x'-output of the master. b) The timing diagram is shown in figure 2. Normally a D flip-flop behaves simply as a delay line. However, here we have fed x' of the master as the input to the slave. Hence, our master-slave combination gives an inversion plus a delay (see D and Y). c) The Karnaugh Map is drawn as shown in figure 3. d) From the Karnaugh Map, we write down the excitation equation as: y\rightarrow= D. \rightarrow e) The state table can be prepared as shown in figure 4: Dy y\rightarrow \rightarrow REMARKS 00 01 0 0 Y\rightarrowIS SAME AS D \rightarrow 10 11 1 1 Y\rightarrowIS SAME AS D \rightarrow f) The state diagram is also plotted with the help of the Karnaugh Map as shown in figure 5.

Question:

Calculate B at the center o\pm a circular loop of wire (point P in the diagram).

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Solution:

For this problem we apply the Biot - Savart Law which gives us the magnitude and direction of the magnetic field at a certain point due to a current carrying wirer dB^\ding{217} = [(\mu_0 i)/(4\pi)] [(d^\ding{217}l × r^\ding{217})/(r^3)] where r^\ding{217} is a displacement vector from a current element to P, and dl^\ding{217} is the length of this current element in the direction of current flow. Since the contribution from any current element will be in the same direction (into the page), we may add (inte-grate) these contributions directly neglecting the vector nature. The magnitude, dB, is given by dB = [(\mu_0 i)/(4\pi)] [(dl sin \texttheta)/r_2] where \texttheta is the angle between dl^\ding{217} and r^\ding{217}. Since we are dealing with a circular current loop, \texttheta is 90\textdegree and sin \texttheta = 1. The magnetic field strength at point P is found by integrating dB over the entire loop B = \int dB = \int [(\mu_0 i)/(4\pi)] [(dl)/r^2] But [(\mu_0 i)/(4\pi)] and r are constants. Therefore, B = [(\mu_0 i)/(4\pir^2)] \int dl. The integral of dl is just 2\pir, the circumference. Therefore B = [(\mu_0 i)/(2r)] (directed into the page).

For this problem, you can enter any two numbers to represent the numerator and denominator of the fraction you choose. Then, by factoring the fraction continually, the program will print out the unit fraction's denominators. If the fraction has too many factors (i.e., more than 29), the program will print out a message telling you to try again. The flowchart will illustrate the procedure. Based on the flowchart, the program looks as follows: 10 REM UNIT FRACTIONS BY FACTORING 11 DIM M(29) 15 PRINT "TYPE N AND D, REMEMBERING THAT D MUST BE GREATER THAN N". 16 PRINT 18 INPUT N,D 19 FOR J = 0 TO 29 20 LET M(J) = 0 21 NEXT J 22 IF D < = N THEN 14 23 LET K = 0 24 LET P = INT(D/N) + 1 35 LET M(K) = P 38 LET K = K + 1 40 IF K < 30 THEN 50 44 PRINT "TOO MANY FACTORS: TRY AGAIN" 45 PRINT 48 GO TO 14 50 LET R = N\textasteriskcenteredP - D 55 IF R = 0 THEN 75 56 LET G = D/10 58 LET D = P\textasteriskcenteredD 60 LET N = R 62 LET W = D/N 64 LET Y = INT(W) 66 IF Y < > W THEN 24 70 LET M(K) = W 75 PRINT 76 PRINT "DENOMINATORS OF UNIT FRACTIONS ARE:" 77 PRINT 80 LET K0 84 IF M(K) = 0 THEN 96 86 PRINT 88 PRINT M(K) 92 LET K = K + 1 94 GO TO 84 96 PRINT"--------------------------" 97 PRINT 98 GO TO 15 99 END

Question:

A Venturi meter is a device by means of which the velocity of flow of a fluid can be measured. The diagram illustrates the operation of one type of such a meter. If the value of A is 10 times that of a, and colored water is used in the U-tube, what is the vel-ocity v_1 of flow of water when the difference in levels h is 6 in.? See Fig.

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Solution:

The operation of the meter, according to diagrams A and B is based upon Bernoulli's Principle. The indicator tube shows that the pressure at the constriction (P_2) is less than p_1. The pressure P_1 due to the flowing fluid in the wide tube must balance both p_2 as well as the pressure due to the weight of water in the column, hence it must be greater than p_2. By the law of continuity Av_1 = av_2 This result follows from the assumed incompressibility of the fluid. In the same time interval ∆t, the same volume of fluid which enters the constriction must leave it. Or (see figure C). (A∆x_1)/∆t = (a∆x_2)/∆t where A and a are the cross-sections indicated in figure B. By definition of velocity v_1 = (∆x_1)/(∆t) and (1/2) = (∆x_2)/(∆t). ThenAv_1 = av_2 We are given that A = 10a, then 10 av_1 = av_2 10 v_1= v_2(1) By Bernoulli's Principle (1/2) dv^2_1 + P_1 = (1/2) dv^2_2 + P_2 where d is the density of the fluid. This principle follows from the work energy theorem. Consider a volume of fluid (fig. C) which flows into the constriction. The total work done on this element of fluid, is equal to the work done on it by the fluid behind it in pushing it through the constriction minus the work it does on the fluid in front of it when it pushes this fluid forward. This total work is equal to the change in kinetic energy of this element of fluid. Or \intF_1^\ding{217} \bullet dx_1^\ding{217} - \intF_2^\ding{217} \bullet dx_2^\ding{217} = F_1∆x_1 - F_2∆x_2 = (1/2) mv^2_2 - (1/2) mv^2_1(2) Here F_1 is the constant force exerted on the element of fluid, by the fluid to the left of it, F_2 is the force it exerts on the fluid to the right of it, and v_1 and v_2 are the velocity of the fluid element before it enters the constriction, and after it enters the constriction, respectively. M is the mass of the fluid element. Equation (2) may be written as F_1∆x_1 + (1/2) mv^2_1 = F_2∆x_2 (1/2) mv^2_2(3) Using the fact that the volume of fluid element pushed to the right equals the volume of fluid element pushed to the left (i.e., A∆x_1 = a∆x_2) then F_1∆x_1 + (1/2) mv^2_1 = F_2(A/a)∆x_1 (1/2) mv^2_2 Dividing both sides by the volume A∆x_1 (F_1/A) + (1/2) (m/A∆x_1) v^2_1 = (F_2/a) + (1/2) (m/A∆x_1) v^2_2 orP_1 + (1/2) dv^2_1 = P_2 + (1/2) dv^2_2 since density = (Mass)/(volume)and Pressure = (Force)/(Area). d is the density of water. Then P_1 - P_2 = (1/2) d(v^2_2 - v^2_1) = (1/2)d (100v^2_1 - v^2_1) = (1/2)d (100v^2_1) (approx) where we used equation (1) But P_1 - P_2 = (F_1 - F_2)/A_x Where A_x is the cross sectional area of the indicator tube, and the difference of these forces (due to the fluid in motion) must equal the weight of water in the column of height h. Then F_1 - F_2 = mg = d(Ah)g where m is the mass of water in the column, d is the density of water, and Ah is the volume of the water. Upon division of both sides by A_x, P_1 - P_2 = (F_1 - F_2)/A_x = dhg Therefore hdg = (1/2)d (100)v^2_1 v^2_1 = [(2g)/(100)] h = [2(32)]/[100] h = .64h v_1 = .8\surdh Substituting h = 6 in. = .5 ft: v_1 = .8(.7 ft)_s = .56 ft/sec

As shown in the figure, before the collision the particles were moving along the x axis, so that the total y component of the momentum is zero. By momentum conservation the total y component of momentum must also be zero after the collision, so that M [ v_y(1) + v_y(2) ] = 0 where M is the mass of the particles, whence v_y(2) = - v_y(1) The Velocities (a) before and (b) after collision. We cannot calculate v_y(1) itself without specifying the initial trajectories and the details of the forces during the collision process.

Question:

If you hydrolyze a peptide bond in a protein molecule, what type of compounds are formed?

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Solution:

When proteins are boiled in dilute acids or bases they are hydrolyzed, i.e. degraded or broken down, to amino acids. Amino acids are compounds whose molecules possess both the amino (- NH_2) and the carboxyl (- CO_2H) functional groups. The general formula may be written as The union of amino acids, which result in the protein, is due to bonds that come about as a result of the elimi-nation of a hydrogen from the - NH_2 group and an OH from the - CO_2H group. This linkage is termed a peptide bond. It is depicted as When this substance is hydrolyzed, the following reaction occurs. In acid: (back) the amino and carboxylic group. Note: NH_2 is basic, so that in acid, you actually get - NH_3^+. Thus, you obtain a compound that possesses an amino and a carboxylic group, which is, as you recall, an amino acid. With base, the same result is obtained, but, you immediately obtain NH_2 and not NH_3^+. The carboxylic portion, which is acidic, reacts with the OH^- from base to yield COO^-.

Question:

The specific rotation [(\alpha)^25\textdegree_D] of L-alanine is + 1.8. The specific rotation [(\alpha)^25\textdegree_D] of L-alanine is + 1.8. Calculate the observed rotation of a 1.10 M solution in a 2.5-dm polarimeter tube at 25\textdegreeC.

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Solution:

All amino acids with the exception of glycine show optical activity, i.e., they can rotate the plane of plane-polarized light when examined in a polarimeter. Optical activity is shown by compounds capable of existing in two forms, whose structures are nonsuperimposable mirror images of each other. Optical activity is expressed quantitatively as the specific rotation [(\alpha)^25\textdegree_D] : [(\alpha)^25\textdegree_D] = (observed rotation ×100) / [optical path length (dm) × concentration (g/100 ml)] With this information, one can find the observed rotation of L-alanine after converting the concentration from (moles/liter) to (g/100 ml). The molecular weight of alanine is 89. [1.10 moles] / [1 liter (= 1000ml)] × 89 g/mole = (97.9 g) / (1000ml) = (9.79 g) / (100ml) Solving for the observed rotation: observed rotation = [(\alpha)^25\textdegree_D × optical path length (dm) × concentration (g/ml)] / [100] observed rotation =[+1.8 × 2.5dm × 9.79] / [100] = 0.44\textdegree.

Question:

Draw a typical bacterial growth curve and label the different phases. Discuss the factors for the existence of each phase.

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Solution:

If one inoculates a flask of nutrient broth with a given number of bacteria and follows the rate of growth during the incubation period, there is found a series of different growth rates. A plot of the logarithms of the number of cells versus time is used to illustrate the growth curve, which is shown in the following diagram: There is an initial period of no growth, followed by one of rapid growth, and then there is a leveling off. The last period is one of decline in the population. The curved portions designates the transitional period between phases. These represent periods where the bacteria enter a new phase. Following the addition of inoculum into the new medium, the bacterial population does not increase but each individual bacterium increases in size. During this period, called the lag phase, the bacteria are physiologically active and are adapting to the new environment, but the bacterial population number remains constant. In addition, the bacteria may alter the environment for their means (eg. exretion of CO_2 to lower pH, so that more favorable growth conditions are achieved.) At the end of the lag phase, the bacteria start to divide. The lag phase is followed by the logarithmic or exponen-tial phase where the cells divide at a constant and maximal rate according to their generation time. Most of the bacteria during this phase are uniform in terms of metabolic activity, unlike the other phases. After the log phase, growth begins to level off in the stationary phase. The population remains constant because of a cessation of reproduction or an equalization of growth and death rates. The decreased growth is usually due to the exhaustion of nutrients or the production of toxic or inhi-bitory products. The final phase is called the death phase. Here, the bacteria die faster than they are being produced, if any reproduction is occurring at all. Depletion of nutrients and accumulation of inhibitory products, such as acid, cause the increased death rate. The number of bacteria increases exponentially in the log phase, and the number of bacteria decreases exponentially in the death phase. Some species die rapidly, so that few living cells remain after about three days, while other species die after several months.

The device shown above is called a eudiometer. It is used to compare the pressures of several gases. The pressure of the confined gas is equal to the baro-metric pressure plus the pressure needed to depress the column of liquid (for P_1) or to the barometric pressure minus the pressure needed to support the column of liquid (for P_3). The pressure of the liquid column is given by pressure of liquid = height or depth of column (mm) × \rho (g/ml) (ml/cm^3) (cm/10mm) Hence: P_1 = barometric pressure + pressure needed to depress liquid = P_atm + h_1 × \rho (ml / cm^3 ) cm/10mm = P_atm + 0.1 h_1 (g/cm^2) \rho P_2 = barometric pressure + pressure needed to depress liquid (or - pressure needed to elevate liquid) = P_atm + 0 × \rho × m1/cm^3 × cm/l10mm = P_atm P_3 = barometric pressure - pressure needed to elevate liquid = P_atm - h_3 × \rho × ml/cm^3 × cm/10mm = P_atm - 0.1 h_3 (g/cm^2 ) = P_atm - 0.1 h_3 (g/cm^2 ) \rho

Question:

The half-life of radon is 3.80 days. After how many days will only one-sixteenth of a radon sample remain?

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Solution:

Question:

In the Cephalochordate, one of the three chordate subphyla, all three chordate characteristics are highly developed and retained in the adult. Outline the features of an adult Cephalochordate.

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Solution:

The cephalochordates are represented by the lancelets, most notablyAmphioxus. These small, marine chordates have been the subject of much study because of the almost diagrammatic way in which they illustrate the basic chordate characteristics. In Amphioxus, the notochord extends from the tip of the head to the tip of the tail, and aids the animal in burrowing through the mud. There are numerous gill slits in the pharynx. The dorsal hollow nerve cord exterids along the entire length of the body. Water enters the mouth (the opening into the pharynx), and passes through the gill slits into a chamber known as the atrium and leaves the body through the atriopore. Food is removed from the water in the pharynx with the aid of secretions from the endostyle in the ventral portion of the pharynx. The food is then sent posteriorly into the instestine; undigested food is eventually eli-minated via the anus. Metabolic wastes are excreted by segmentally arranged, ciliated protonephridia that open into the atrium. The muscles of the body are V shaped, segmentally arranged structures. A continuous dorsal fin, supported by fin rays, expands into a tail fin and then continues ventrally to the atriopore as the ventral fin. Although superficially similar to fishes, the lancelets are more primitive for they lack paired fins and jaws . Anterior to the atriopore, however, are two lateral folds that may have been the forerunners of the paired pectoral and pelvic fins of fishes. It hardly seems possible that any form living today is in the direct line of vertebrate ancestry, but the lancelet most closely represents the ancestral vertebrates. Its body plan, although relatively simple, resembles that of vertebrates, and the larva of the lamprey, one of the lowest vertebrates, is in many ways similar to the lancelets.

Question:

V \leftarrow 3 4 2 -1. Find the result of the following operations.

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Solution:

a) The symbol \vert, when written as a monadic function, as \vertV, specifies the absolute value of V. V is a vector, which consists of the elements 3 4 2 -1. The absolute value of V is 3 4 2 1 i.e., \vertV is equal to 3 4 2 1. b) The symbol \Gamma represents a maximum function. When used with the Reduction Operator /, the maximum function selects the largest from among all the elements of the vec-tor, V in this case. Thus, if V consists of 3 4 2 -1, then \Gamma/V is equivalent to the following: 3\Gamma4\Gamma2\Gamma - 1 i.e.,3\Gamma4\Gamma2 i.e.,3\Gamma4 i.e.,4 In the above, using the right to left rule of APL for eval-uating an expression, the elements of V are compared two by two from the right. The maximum is selected at each com-parison. Thus, finally when the whole string has been checked out, the largest element is obtained. c) The symbol L represents a minimum function. When used with the Reduction Operator / , the minimum function selects the least from among all the elements of the vec-tor (V in this case). Thus, if V consists of 3 4 2 -1, then L/V is equivalent to the following: 3L4L2L - 1 i.e.,3L4L-1 i.e.,3L-1 i.e.,-1 In the above, using the right to left rule, the elements of V are compared two by two, and the minimum is selected.

Question:

How does communication serve in predator avoidance?

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Solution:

Both recognition and avoidance of predators can be considered forms of communication between species. One method by which animals avoid detection is called cryptic appearance. In this method, the animal uses its color or shape to resemble objects in the environment in which it lives. Many insects are shaped to resemble inedible objects, such as twigs. Some animals can even change their colors to resemble the surface on which they are temporarily resting. Warning appearance is another form of inter-species communication. It is used by some animals to warn potential predators that they are dangerous or poisonous, thus serving to prevent attack. The bold red-and-black bands of the venomous coral snake and the black-and- yellow stripes of the hornet are familiar examples. Mimicry is another method used to avoid attack by predators. It is used by animals that are not naturally protected; that is, they do not actually possess any characteristics harmful to a predator. The mimic, however, resembles some other species that does have dangerous characteristics and warning signals. In the form of mimicry called Batesian mimicry, the predator is unable to differentiate between the dangerous model and the harmless mimic. It thus leaves both alone. For example, the Monarch butterfly is distasteful to birds, who therefore avoid it. It serves as the model for the Viceroy butterfly, the edible mimic who closely resembles the Monarch in color, shape and behavior. Having the same warning appearance as the Monarch the Viceroy butterfly is protected. (See Figure) A second form of mimicry, called Mullerian mimicry, involves the evolution of two or more inedible or unpleasant species to resemble each other. Since each species serves as both model and mimic, greater protec-tion is afforded to them because their repellent qualities are more frequently advertised.

The Paschen series illustrates the spectrum hydrogen exhibits in the infrared region. An example of a hydrogen line spectrum is shown in the accompanying figure. The series of lines of the Paschen series fit the equation ѵ = 3.29 × 10^15 [(1/9) - (1/b^2)], where v = frequency and b = 4, 5, 6, 7, 8, . . .. Thus, if you want the first four hydrogen spectral lines, you use b = 4, 5, 6, and 7. Your answer would be in terms of frequency, however, you want the wavelengths. To obtain this, employ the relation between frequency and wavelength, c = \lambdaѵ, where c = speed of light, \lambda = wavelength and ѵ = frequency, (c = 2.998 × 10^10 cm/sec.) Once you know ѵ, you can find \lambda. Therefore, 1st line:b = 4, ѵ = 3.290 × 10^15 [(1/9) - (1/16)] = 1.599 × 10^14 sec^-1. 2nd line:b = 5, ѵ = 3.290 × 10^15 [(1/9) - (1/25)] = 2.340 × 10^14 sec^-1. 3rd line:b = 6, ѵ = 3.290 × 10^15 [(1/9) - (1/36)] = 2.742 × 10^14 sec^-1. 4th line:b = 7, ѵ = 3.290 × 10^15 [(1/9) - (1/49)] = 2.984 × 10^14 sec^-1. The series limit is b = infinity. If B \rightarrow \infty, then (1/B^2) \rightarrow 0. Series limit: ѵ = 3.290 × 10^15 [(1/9) - 0] = 3.656 × 10^14 The wavelengths (\lambda) = [{speed of light (c)} / {frequency (ѵ)}] 1st line:\lambda = 1.87 × 10^-4 cm2nd line:\lambda = 1.28 × 10^-4 cm 3rd line:\lambda = 1.09 × 10^-4 cm4th line:\lambda = 1.004 × 10^-4 cm Series limit:.820 × 10^-4 cm.

The largest number to be squared is decimal seven or 111_2. With decimal seven as an input the output will be 49_10 or 110001_2 hence there will be three inputs, one for each bit, and six outputs. A truth table is made with inputs I_0, I_1, I_2 and outputs A_0, A_1, A_2, A_3, A_4, A_5. I_0 and A_0 are the most significant bits. mintermsInputs I_0I_1I_2 Outputs A_0A_1A_2A_3A_4A_5 0 000 000000 1 001 000001 2 010 000100 3 011 001001 4 100 010000 5 101 011001 6 110 100100 7 111 110001 From the truth table it is seen that A_0 (I_0 , I_1 , I_2) = \sum 6, 7 A_1 (I_0 , I_1 , I_2) = \sum 4, 5, 7 A_2 (I_0 , I_1 , I_2) = \sum 3, 5 A_3 (I_0 , I_1 , I_2) = \sum 2, 6 A_4 (I_0 , I_1 , I_2) = \sum none - connect to Logic 0 A_5 (I_0 , I_1 , I_2) = \sum 1, 3, 5, 7 Use 3-variable K-maps to simplify the expressions A_0 = I_0I_1 A_1 = I_0I_1 + I_0 I_2 A_2 = I_0I_1 I_2 +I_0 I_1 I_2 A_3 = I_1I_2 A_4 = 0 A_5 = I_2 A logical diagram is drawn from the minimized equations. Figure 1 shows the combinational circuit.

The vector diagram is shown in figure (a). The resultant displacement vector is labelled R\ding{217}. The magnitude of this vector is 2.8 km. The direction, as measured with a protractor, is 45\textdegree south of east, or the tangent may be used to find the direction, since a right triangle is formed. We shall also compute the solution analytically. In figure (b) a closeup of the resultant vector R\ding{217} is shown. We can see from the graph that side A and side B each equal 2 km. Thus, by the Pythagorean theorem: R^2 = A^2 + B^2 = (2 km)^2 + (2 km)^2 = 8 km^2 R = 2 \surd2 km = 2(1.4)km = 2.8 km tan \texttheta = (2 km/2 km) = 1, \texttheta = 45\textdegree R\ding{217} = 2.8 km, 45\textdegree south of east.

Question:

Calculate the pH at 25\textdegreeC of a solution containing 0.10 M sodium acetate and 0.03 M acetic acid. The apparentpKfor acetic acid at this ionic strength is 4.57.

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Solution:

This problem involves the calculation of [H^+] (pH = - log [H^+]) in a mixture of amonoproticacid (acetic acid) and its completely dissociated salt (sodium acetate). Because the addition of a salt represses the ionization of the acid, the concentration ofundissociatedacid is approximately equal to the molar concentration of added weak acid (C_a). The addition of the acid represses the hydrolysis of the salt so that the concentration of anions of the weak acid is approximately equal to the molar concentration of the salt C_s. Using several approximations, thepKfor a weak acid is the pH of a solution containingequimolarquan-tities of salt and acid. However, if the solution does not containequimolarquantities of salt and acid, then the following approximation holds: pH =pK+ log [C_s / C_a] , where C_s and C_a are the concentrations of the salt and the acid, respectively. Thus, in this problem, C_s = 0.10 MC_a = 0.03 M pK = 4.57 andpH = 4.57 + log [0.10 / 0.03] = 4.57 + 0.52 = 5.09.

Question:

A long, horizontal, rigidly supported wire carries a current of 50 A. Directly above it and parallel to it is a fine wire, the weight of which is 0.075 N per meter, which carries a current of 25 A. How far above the first wire should the second wire be strung in order for it to be supported by magnetic repulsion?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0726.htm

Solution:

If the upper wire is to be supported by magnetic repulsion, the magnetic force per unit length (F/l) must just equal the weight of a unit of length of the wire (mg/l). Further, the currents in the two wires must be in opposite directions in order for the force between the wires to be one of repulsion. Hence (mg/l) = (F/l) but(F/l) = (\mu_0 /2\pi) [(I I')/r](see the figure) \therefore(mg/l) = (\mu_0 /2\pi) [(I I')/r] r = [(\mu_0 l II')/(2\pimg)] = [(2 × 10^-7 N \bullet A^-2 × 50A × 25 A) / (0.075 N \bullet m^-1 )] = 0.33 × 10^-2 m = 0.33 cm. The wires must therefore be very thin in order to allow their centers to be so close together.

Question:

The earthworm has a central nerve cord running along the entire length of the body. When an earthworm is cut into several parts, severing its nerve cord in the process, each part will go on crawling around for some time. How can each fragment carry out crawling movement even after the nerve cord is severed?

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Solution:

The arrangement of the nervous system in lower invertebrates is quite different from that in higher vertebrates like man. In higher vertebrates, there is an expanded and highly developed anterior end of the spinal cord, forming the brain. The brain coordinates and regu-lates the activities of the entire body. When the brain is separated from the rest of the body, the animal cannot perform any complicated functions. However, in the earthworm, a higher invertebrate, coordinated activities such as crawling can still be ob-served when the body is cut into several transverse sections. This movement is possible because the earth-worm has more than one neural center controlling and coordinating its activities. The nervous system of the earthworm consists of a large, two-lobed aggregation of nerve cells, called the brain, located just above the pharynx in the third segment, and asubpharyngealganglion just below thepharnyxin the fourth segment. A nerve cord connects the brain to the subpharyngeal ganglion and extends from the anterior to the posterior end of the body. In each segment of the body there is a swelling of the nerve cord, called a segmental ganglion. Sensory and motor nerves arise from each segmental ganglion to supply the muscles and organs of that segment. The segmental ganglia coordinate the contraction of the longitudinal and circular muscles of the body wall, so that the worm is able to crawl. When the earthworm is cut into several pieces, thus sever-ing the connection to the brain (that is, the nerve cord), the resulting fragments still contain segmental ganglia which can fire impulses to the muscles of the body wall, resulting in crawling movement.

Question:

A long straight wire in a house carries an alternating current which has a maximum value of 20 amp. Calculate the maximum magnetic induction a distance lm away from the wire.

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0745.htm

Solution:

The magnetic induction at a distance R from a wire contain-ing a current i can be found by using Ampere's Law. Given a closed path c, and a current i through the closed curve, the following re-lation holds: \oint_c B^\ding{217} \textbullet dl^\ding{217} = \mu_0 i where the circle on the integral sign Indicates that c is closed, and the permeability constant \mu_0 = 4\pi × 10^-7 weber/amp-meter . Let the closed curve c of the above integral be in the shape of a circle of radius R, concentric with the current carrying wire as shown in the figure. Ampere's Law then becomes: \int_c B cos \texttheta dl = \int_c B dl Since B^\ding{217} and dl^\ding{217} are collinear and \texttheta = 0 o . B^\ding{217} is constant along the path c, therefore B \int dl = \mu_0 i B \bullet 2\piR = \mu_0 i orB = [(\mu_0 i)/(2\piR)] from which we have B_max = (\mu_0 /2\pi) (i_max /R) = 2 × 10^-7 (\omega/a-m) × (20a/1m) = 4 × 10^-6 (\omega/m^2) This can be compared with the earth's magnetic field, which has an Intensity of the order of 5 × 10^-5 (W/m^2) .

Question:

Write a BASIC program to solve Laplace's equation (\partial2V/\partialx2) + (\partial2V/\partialy2) = 0. Consider a rectangular 8 by 8 grid with boundary conditions such that 3 sides are assumed to be at a potential of 100. The remaining side is at zero potential as shown.

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Solution:

We use the replacement of partial derivatives by central dif-ferences (\partial2V/\partialx2) = (V_i-1,_j - 2V_i,_j + V_i+1,_j ) / (∆x2) and (\partial2V/\partialy2) = (V_i,_j-1 - 2V_i,_j + V_i, _j+1) / (∆y2) Thus [(V_i-1,_j - 2V_i,_j + V_i+1,_j ) / (∆x2)] + [(V_i,_j-1 - 2V_i,_j + V_i, _j+1) / (∆y2)] = 0. But, in the given rectangular grid of little squares, ∆x = ∆y and hence we obtain simply V_ij = (1/4) [(V_i-1,_j + V_i+1,_j + V_i,_j-1 + V_i, _j+1]. Thus when Laplace's equation is satisfied, the value of the function at each grid point is the average of the values of the function at the four closest neighbors. When exterior boundary conditions are specified, the solution, by repeated applications of this algorithm flows out from the boundary and stabilizes when the above condition for V_i,jis met. Initial values of the potential are given in the data statements within the program. Each line, represents one row of data in the grid. The problem specified the boundary values of the potential. Interior values of the potential are guesses. Convergence is studied by considering d = V_ij new -V_ij = = (1/4) [(V_i+1,_j + V_i, _j+1 + V_i-1,_j + V_i, _j-1 -4V_ij]. S is used to accumulate the sum of all changes as we sweep over the grid. In addition, the counter K is used to accumulate the number of passes over the grid. The process is repeated until the total change in any sweep (the variable S) is less than some arbitrary amount. 1\O\O REM ELLIPTIC PARTIAL DIFFERENTIAL EQUATION 1\O1 REM LAPLACE'S EQUATION 11\O FOR I = 1 TO 8 111 FOR J = 1 TO 8 112 READ V(I,J) 113 NEXT J 114 NEXT I 12\O LET K = \O 13\O LET S = \O 14\O FOR I = 2 TO 7 15\O FOR J = 2 TO 7 16\O LET D = (V(I+1,J) + V(I,J+1) + V(I-1,J) + V(I,J-1) -4\textasteriskcenteredV(I,J))/4 161 LET S = S + ABS(D) 162 LET V(I,J) = V(I,J) + D 163 NEXT J 164 NEXT I 17\O LET K = K + 1 18\O IF S > 2 THEN 13\O 2\O\O PRINT 2\O2 FOR I = 1 TO 8 2\O3 FOR J = 1 TO 8 2\O4 PRINT INT(V(I,J)); 2\O5 NEXT J 2\O6 PRINT 2\O8 NEXT I 2\O9 PRINT 212 PRINT "NUMBER OF PASSES FOR CONVERGENCE =\textquotedblright; K 8\O1 DATA 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O, 1\O\O 8\O2 DATA 1\O\O,9\O,9\O,9\O,9\O,9\O,9\O, 1\O\O 8\O3 DATA 1\O\O,8\O,8\O,75,75,8\O,8\O, 1\O\O 8\O4 DATA 1\O\O,75,6\O,5\O,5\O,6\O,75, 1\O\O 8\O5 DATA. 1\O\O,6\O,5\O,3\O,3\O,5\O,6\O, 1\O\O 8\O6 DATA. 1\O\O,5\O,3\O,15,15,3\O,5\O, 1\O\O 8\O7 DATA. 1\O\O,4\O,2\O,5,5,2\O,4\O, 1\O\O 8\O8 DATA 1\O\O,\O,\O,\O,\O,\O,\O, 1\O\O 999 END Sample output is given 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 1\O\O 97 95 94 94 95 97 1\O\O 1\O\O 94 9\O 88 88 9\O 94 1\O\O 1\O\O 9\O 83 8\O 8\O 83 9\O 1\O\O 1\O\O 84 73 68 68 74 84 1\O\O 1\O\O 73 58 52 52 58 73 1\O\O 1\O\O 52 34 29 29 35 52 1\O\O 1\O\O \O \O \O \O \O \O 1\O\O NUMBER OF PASSES FOR CONVERGENCE = 22

(a) For the car to just make it around the loop, its speed at the highest point of the loop must be such that the force of gravity on it is sufficient to provide the centripetal force needed to keep it in a circular path. For this to be the case, F_centripetal = (mv^2_c) / R = mg or the velocity must be given by v_c =\surdRg(1) at point C. Neglecting friction, we use conservation of energy and note that the velocity of the car at point D must be the same as at point c, since both correspond to the same net change in potential energy relative to the starting point. The change in potential energy equals the change in kinetic energy and is proportional to the square of the velocity. The change in potential energy from the starting point to point D equated to the corresponding change in kinetic energy yields. mgh = (1/2)m(vD- v_0)^2 = (1/2)mv^2_D v_D = \surd2gh(2) where h is the vertical height of the starting point above points C and D (as shown in figure), and V_0 is the initial velocity of the car, which is zero at the starting point. Equating equations (1) and (2), \surdRg = \surd2gh we haveh = R/2. This indicates that the starting point must be 2R + R/2 = (5/2)R above the bottom of the loop in order for the car to have just enough energy to make it around the loop. (b) Equating the change in potential energy to the kinetic energy of the car at the point in question, as we did in part (a), we find for point A mg(h + 2R) = (1/2)mv^2_A (5/2) Rg = (1/2) v^2_A v_A = \surd(5 gR) for point B,mg(h + R) = (1/2) mv^2_B (3/2) gR = (1/2)v^2_B v_B = \surd(3gR)

Question:

The concentration of sodium ions (Na+) inside most cells is lowerthan the concentration outside the cells. Why can't this phenomenonbe explained by simple diffusion across the membraneand what process is responsible for this concentrationdifference?

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Solution:

Since the cell membrane is somewhat permeable to sodium ions, simplediffusion would result in a net movement of sodium ions into the cell, until the concen-trations on the two sides of the membrane became equal. Sodium actually does diffuse into the cell rather freely, but as fast asit does so, the cell actively pumps it out again, against the concentrationdifference. The mechanism by which the cell pumps the sodium ions out is calledactive transport. Active transport requires the expenditure of energy forthe work done by the cell in moving molecules against a concentration gradient. Active transport enables a cell to maintain a lower concentration ofsodium inside the cell, and also enables a cell to accumulate certain nutrientsinside the cell at concentrations much higher than the extracellularconcentrations. The exact mechanism of active transport is not known. It has been proposedthat a carrier molecule is involved, which reacts chemically with themolecule that is to be ac-tively transported. This forms a compound whichis soluble in the lipid portion of the membrane and the carrier com- poundthen moves through the membrane against the concen-tration gradientto the other side. The transported molecule is then released, and thecarrier molecule diffuses back to the other side of the membrane whereit picks up another molecule. This process requires energy, since workmust be done in transporting the molecule against a diffusion gradient. The energy is supplied in the form of ATP. The carrier molecules are thought to be integral proteins; proteins whichspan the plasma membrane. These proteins are specific for the moleculesthey transport.

Question:

Explain how the changes in the phytochrome system are related to blooming.

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Solution:

Phytochrome is a pigment found in certain plants in very small amounts. The pigment is thought to be responsible for detecting the light environment of the plant. Phytochrome exists in two different forms - P_660 the inactive form, and P_730, the active form. P660 absorbs red light and is converted to P_730; P_730 absorbs far-red light and is converted back to P_660. In nature, P_660 is converted to P_730 in the daytime (when red wavelengths predominate over far-red). The reverse occurs in the night time. Hence, P_730 predominates in the plant during the day and P_660 predominates during the night. This is summarized in the figure below. This differ-ential response of phytochrome to light and dark is thought to be involved in the phenomenon of photoperiodism in plants. It has been found that red light inhibits flowering in short-night plants but promotes flowering in long-night plants, under conditions during which flowering normally takes place. This experimental observation led many investigators to hypothesize that the P_730 - P_660 inter-conversion might be the plant's time-regulator of flowering. According to this hypothesis, P_730, converted from P_660 by the absorption of red light, would inhibit flowering in short-day plants but promote flowering in long-day plants. Because P_730 accumulates in the day and diminishes at night, short-day plants could flower only if the nights were long enough, during which a great amount of P_730 would be inactivated. Long-day plants, on the contrary, would require short nights, during which the P_730 would not be completely inactivated, so that enough P_730 would remain at the end of the night to promote flowering. This hypothesis was highly regarded in the beginning. However, it was dealt a fatal blow by a later study showing that the reversion of all the P_730 to P_660 requires only three hours; this has been found to hold true in all plants, both long and short-day, that have been studied. What happens during the rest of the dark period? No sat-isfactory answer is as yet available. However, extended days provide P_730 for longer periods. It is generally agreed at present that the time-measuring phenomenon of flowering is not solely controlled by the interconversion of P_660 and P_730 . There seems to be two variables involved. Experimentally it has been found that one is the presence or absence of light, the other is the length of the dark or light period. Phytochromes seem to be responsible for the detection of either light or darkness, but the actual measuring of the time between the moment the plant senses the onset of darkness and the moment it senses the next exposure to light depends on some mechanism, not yet known. Further investigation is needed before we can fully understand the "biological clock\textquotedblright in flowering.

(a) The optical equation for the spherical end of the rod (which is a spherical refracting surface), is given by (n/p) + (n'/q) = {(n' - n)/R} where the object is in the region of refraction index n, the rays are going from the region of refractive index n to the region of refractive index n', and R is the curvature of the refractive surface. The radius R for the glass hemisphere has a positive sign because the center of curvature of the glass surface lies inside the glass, which is the real image region (see Fig. 1). R is negative when the curvature center lies in the virtual image region. As shown in Fig. 1 above, p = +8 cm, R = +2 cm. Therefore, the image distance q is obtained as (1.5/q) + (1/8)cm = {(1.5 - 1)/2 cm} or q = + 12 cm q is positive, i.e., the image is formed at the right of the vertex of the glass surface and is real. The magnification m for this problem is given by m = - (n/n),(q/p) = - (1/1.5) × (12 cm/8 cm) = - 1 that is, the image has the same height as the object, but is inverted, h_I = -h_0 = - 0.1 cm. (b) When the rod is in water, the optical equation yields for q (Fig. 2) (1.5/q) + (1.33/8 cm) = {(1.50 - 1.33)/2 cm} orq = - 18 cm. Since q is negative, the image is formed in the water and is imaginary. For magnification, we have m = - (1.33/1.5) × (-18 cm/8 cm) = 1.99, and the image is erect and has a height h_I = 1.99 h_0 = 0.99 cm.

Question:

Define a process, and describe its states during its life cycle.

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Solution:

The most important process in the growth of bacterial populations is binary (transverse) fission. This is a type of cell division in which two identical daughter cells are produced as a result of the division of the parent cell. New cell wall material begins to form on the inner surface of the wall of the parent cell at a point midway along its length and this new wall material invaginates, dividing the cellular material of the parent cell evenly into two halves. The cell is then, separated in two by the completion of the transverse wall. Each daughter cell possesses a complete set of genetic information. The genes, the units of inheritance, are arranged in sequence along a single, circular chromosome composed of DNA. During repro-duction, the DNA is replicated and the two chromosomes move apart into separate nuclear areas in each half of the parent cell before the completion of the transverse cell wall. Each daughter cell has genetic information identical to the parent cell. This process of giving rise to new individuals by cell division is termed asexual reproduction. Although binary fission is the major method of bacterial reproduction, another form of asexual reproduction budding is observed in some bacteria. Budding involves an outgrowth of the parental cell, which enlarges and separates to form a new cell.

Question:

Write a program which concatenates two strings "I LOVE" and "YOU.\textquotedblright, checks for presence of the character \textquotedblleft.\textquotedblright, replaces it with \textquotedblleft!\textquotedblright, and prints out the final string.

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Solution:

Concatenation is the basic operation for combining two strings to form a third. The general form of this operation is represented by the following statements: variable 1 = string 1 variable 2 = string 2 variable 3 = string 1 string 2 There must be at least one blank space on each side of the "=" sign (as well as any other arithmetic operator, except the unary minus) and between strings 1 and 2 in variable 3. The simple pattern matching statement has the form subject pattern where the two fields are separated by at least one blank. This statement tells the computer to check the string indicated by subject forthe occurrence of a string specified by the pattern. A replacement statement has the form subject pattern = object The left part of this statement operates identically with the pattern matching statement. If a match occurs, the object string replaces the pattern string in the subject string. The program follows: WORD 1 = 'I LOVE' WORD 2 = 'YOU.' WORDS = WORD 1 WORD 2 WORDS ' . ' = ' ! ' OUTPUT = WORDS END Note, that END is a label, indicating the end of program. Labels appear beforethe statements and consist of a letter or a digit followed by any number of other characters up to a blank. Also note, that single quotation marks can be used as well as double quotations.

Question:

Predict the reaction of propene with Br_2/CCI_4, H_2O/H_2SO_4, and HCI. Predict the reaction of toluene with these substances.

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Solution:

The procedure for solving this problem is as follows: write the structures of toluene and propene; classify these compounds from their structures; once these compounds are classified, you can predict the reaction that will occur with the given reagents Br_2/CCI_4, H_2O/H_2SO_4, and HCI. The structure of propene is H_2C = CH \textemdash CH_3. Notice, it has a double bond which indicates it is an alkene. Alkenes undergo the characteristic reaction of addition to the double bond. Thus, Note: Adding Br_2 breaks the double bond in a CCl_4 solvent. The structural formula of toluene is shown in figure A. The circle inside the ring indicates resonance. Its true structure is a hybrid (figure B) , which means it is stable to addition reactions. Thus, toluene will not undergo any reaction with these reactants.

Question:

Write a subroutine which computes the roots of the quadratic equation a1x2 + a2x + a3 = 0 according to the quadratic formula: x1,2 = (-a2/2a1) \pm \surd[(a2/2a1)2 - (a3/a1)] ( = [{a2 \pm \surd(a22 - 4a1a3)} / 2a1])

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Solution:

The only difficult part to this problem is taking into ac-count the fact that the character of the solution changes according to the sign of the discriminant, which we define as [a2/2a1]2 - a3/a1. If the discriminant is positive, then the solutions are real numbers given by the above formula. If the discriminant is negative, then the solution consists of the pair of complex conjugate numbers x1,2 = -(a2/2a1) \pm i │DISC│ where DISC is the discriminant. In this latter case, the subroutine should return the value of the real part and \vertDISC\vert for the absolute value of the imaginary part. The flowchart illustrates the required logic. CSOLUTION OF THE QUADRATIC EQUATION CA(1) × X × X + A (2) × X + A (3) = 0 SUBROUTINE QUAD (A, XR1, XR2, X1) DIMENSION A(3) X1 = -A(2)/(2.\textasteriskcenteredA(1)) DISC = X1\textasteriskcenteredX1 - A(3)/A(1) IF (DISC) 10, 20, 20 10X2 = SQRT (-DISC) XR1 = X1 XR2 = X1 XI = X2 GO TO 30 20X2 = SQRT (DISC) XR1 = X1 + X2 XR2 = X1 - X2 XI = 0.0 30RETURN END

The only force acting on the ball is the grav-itational force, which is directed downward, throughout its motion, therefore, the ball will be accelerated downward, at a = - g = - 32 ft/sec^2. We have chosen the positive direction to be up. We have two directions of motion to consider: first, the upward motion to the max-imum height and then the downward motion toward the ground. Therefore, we must be careful to use the proper signs in our equations. We choose the origin for distance (y=0) at the point from which the ball is thrown. The initial velocity of the ball is v_0 = +32 ft/sec. The ball grad-ually loses velocity until it reaches its maximum height. Then it falls down towards the ground. The equations for velocity and distance therefore become v = v_0 + at = (32 ft/sec) - (32 ft/sec^2) × t y = v_0t + 1/2 at^2 = (32 ft/sec) × t - (16 ft/sec^2) × t^2 From these equations we find t(sec) v(ft/sec) y(ft) 0 32 0 1 0 16 2 -32 0 3 -64 -48 4 -96 -128 After 1 sec, the velocity of the ball has become zero; that is, the maximum height has been reached (16 ft) and the sub-sequent motion is downward. All velocities for t > 1 sec are therefore negative. At t = 2 sec, the ball has returned to its starting point (y = 0) and for all subsequent times, y is negative. The diagrams below show the velocity and the distance as functions of the time.

Question:

10 cc of battery acid (density 1.21) is diluted to 100 cc. 30 cc of the dilute solution is neutralized by 45 cc of 0.5 N base. What is the normality of the diluted acid? What is its strength in grams per liter? What is the strength of the undiluted 1.21 density acid in g/l?

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Solution:

Question:

What general trends are observed in the evolution of the brain?

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Solution:

Generally, the higher an animal is on the ladder of evolution, the more complex and heavy is its brain (see Figure 1). In the earthworm, an invertebrate, the 'brain' is merely a ganglion (a clump of nerve cell bodies) of about the same size as the other ganglia present in each segment of the body. Even among the mammals, the size and complexity of the brain vary to a great extent. The weight of the brain of a full grown man is approximately three pounds whereas a gorilla, about four times heavier than man, has a brain that weighs only one and a half pounds. This discrepancy in brain weight can account part-ly for the difference in intellectual capabilities of the two mammals. Increasing size is not the only feature seen in the evolution of the brain. The brains of higher mammals dis-play increasing foldings of their surface. These foldings, or convolutions, are particularly pronounced in man. By stimulating various regions of the brain with an electrode, and observing the corresponding results in behavior, func-tional mapping of the brains of various mammals has been achieved. The mapping shows that different regions of the brain are associated with different functions. There are the areas for motor functions, somatic sensory functions, vision, hearing, olfaction, and so forth. The mapping also shows that the proportion of the total area of the cerebrum devoted to sensory and motor functions differs greatly among species. It has been found that, in general, the more convoluted the surface of the cerebrum (called the cerebral cortex), the smaller the proportion devoted ex-clusively to sensory and motor activities. Since, as mentioned, there is a trend toward increasing convolutions in the evolution of the brain, we can anticipate that higher animals have a more convoluted cerebral cortex, and that at the same time, a smaller proportion of their brain is involved in sensory and motor functions. This is supported by the finding that in the cat the sensory and motor functions occupy a major portion of the cortex, whereas in man, the area of the cortex devoted to those functions is relatively small (see Figure 2). The rest of the brain not associated with motor and sensory activities contain the so-called associative areas. These areas are made up of neurons that are not directly connected to sense organs or muscles but supply inter-connections between the other areas. They are believed to be responsible for the higher intellectual faculties of memory, reasoning, learning, imagination, and personality. They are, in short, responsible for the behavioral con-sequences that most clearly distinguish man from other animals.

#define NULL "\textbackslash 0' #define MAX_NAME 30 #define MAX_ID 10 structstudents { charname [MAX__NAME ]; charid [MAX_ID]; floatgpa; structstudents \textasteriskcenterednext; }; }; main ( ) { inti; structstudents \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr; char\textasteriskcenteredcalloc ( ); /\textasteriskcenteredroutine calloc returns a pointer to the location\textasteriskcentered/ /\textasteriskcenteredCreating first element in the list\textasteriskcentered/ if((ptr = (struct students \textasteriskcentered)calloc (1, sizeof (structstudents))) == NULL); { printf ("No memory available \textbackslashn"); exit (1); } ptr \rightarrow next = NULL; start = ptr; scanf ("%s%s%f\textbackslashn", ptr \rightarrow name, ptr \rightarrow id, ptr \rightarrow gpa); /\textasteriskcenteredcreating 9 more elements \textasteriskcentered/ for(i = 1; i <10; i++) { if((new _ ptr = (struct students \textasteriskcentered)calloc (1, sizeof (struct students))) == NULL) { printf ("No memory available \textbackslashn"); exit (1); } new _ ptr \rightarrow next = ptr \rightarrow next; ptr \rightarrow next = new _ ptr; ptr = new _ ptr; scanf ("%s%s%f\textbackslashn", ptr \rightarrow name, ptr \rightarrow id, ptr \rightarrow gpa); } } Explanation of Related Terms Linked List : A linked list is a list of structures. It util-izes the ability of a member of the structure to be a pointer to a structure of the same type. Structures where member points to a structure of the same type are known as self-referential structures. calloc ( ): This is a library routine that allocates space from the heap. The function calloc ( ) returns a pointer to a contiguous block of memory that is used for array or similar structures. It returns NULL value if insufficient space is available for allocation. sizeof ( ): It is an operator in C which returns the number of bytes required to store an object, eg. number _ of _ bytes = sizeof (structstudents); will give us the number of bytes required to store one struc-ture of student. exit ( ): It provides a method of terminating a program. It is one of the library routines i.e. it is a function in the li-brary. It returns several values but most common, are 0 and 1. Value 0 is for normal termination i.e. program completes execu-tion as required. Value 1 is for abnormal termination i.e. program terminates abnormally and does not execute as required. Explanation of Statements 1)char\textasteriskcenteredcalloc (); that has been allocated. It is mainly a function which returns a pointer of the typechar. 2)structstudents \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr This declaration signifies that \textasteriskcenteredstart, \textasteriskcenteredptr and \textasteriskcenterednew _ ptr are pointers to the structure students. 3) ptr = (structstudents \textasteriskcentered) calloc(1, sizeof (structstu-dents)); To follow the concept in this statement we take individual parts and analyze their meaning. a) sizeof (structstudents); will return the number of bytes required to store a struc-ture of students. b) calloc (1, sizeof (structstudents)); Arguments required by calloc ( ) are the number of storage elements to allocate and size of storage area in bytes. c) (struct students \textasteriskcentered) This statement assigns the value returned from the routine calloc ( ) to a pointer to the structure of type students. The asterisk specifies that the value is to be assigned to a pointer. This assigned pointer is to a structure of type students. d) ptr = (structstudents \textasteriskcentered) calloc (1, sizeof (structstudents)); The ptr is assigned to the pointer on the right side of the assignment. i.e. pointer ptr is assigned the value calcu-lated on the right side of the assignment. Note: Assignment is possible only when both the pointers are to the same structure. Here ptr is assigned to another pointer (calculated) to a structure of type students. 4) ptr \rightarrow next ptr \rightarrow name etc. Reference to the members of the structure is through dot notation, i.e. (\textasteriskcenteredptr).name (\textasteriskcenteredptr).next etc. C provides a unique notation for pointers to structures. The notation \rightarrow is used in place of dot notation with pointers to structures. To reference any member in a structure the \rightarrow is used. eg. ptr \rightarrow next ptr \rightarrow name ptr \rightarrow id etc. LINKED-LIST: Represented through diagrams. Note: In element 2 the next is occupied by NULL i.e. next of element 2 is pointing to NULL. Further expansion of this list according to the program will give the required structure. It is suggested that accord-ing to the program the whole list should be drawn to give a clear understanding of concepts involved.

Question:

What are some of the important properties and characteris-ticsofenyzmes?

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Question:

Which were the first vertebrates to exhibit a hinged jaw and paired fins? Describe the evolution of the hinged jaws of vertebrates.

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Solution:

Roughly four hundred million years ago, there evolved a group of fishlike organisms with an armor of bony scales; usually, but not always, a cartilagenous endoskeleton; both pectoral and pelvic fins, and most notably, hinged jaws. These organisms were grouped in the class Placodermi. The placoderms mark a notable advance in vertebrate evolution in that they possess hinged jaws. The acquisition of jaws was one of the most important events in the history of vertebrates, because it made possible a revolution in the method of feeding and hence in the entire mode of life of early fishes. Studies have indicated that the hinged jaws of the placoderms developed from a set of gill-support bars. Notice that hinged jaws arose independently in two important animal groups, the arthropods and the vertebrates, but that although they are functionally analogous structures, they arose in entirely different ways: from ancestral legs in the insects and from skeletal elements in the wall of the pharyngeal region in vertebrates. Though the placoderms themselves have been extinct for at least 230 million years, two other classes that arose from them are still important elements of our fauna. These are the Chondrichthyes (cartilaginous fishes) and the Osteichthyes (bony fishes).

All the information needed to write the program is given in the group of diagrams supplied with the problem. This program will read an input card file and prepare and print a report containing some of the data from the cards. ASSIGN clause . requirements vary with the system, the general format for the ASSIGN clause is: ASSIGNsystem-name An FD (FILE DESCRIPTION) entry must be included in the File section for each file. The entry must include FD with the file-name, a LABEL RECORDS clause, and may include a DATA RECORD clause. The characters in a file must all be accounted for in its record description entry. The FILLER item is thus used to refer to unused spaces in the printer file. For the PROCEDURE DIVISION new WRITE and MOVE statements are introduced. The WRITE statement accomplishes a MOVE and WRITE in a single COBOL operation. WRITE record-name FROM data-name. This format accomplishes both moving of data-name to record-name and writing of the record to its associated file. The MOVE CORRESPONDING statement; MOVE CORRESPONDING data-name-1 TO data-name-2 moves elementary data items from the data group of data-name-1 to elementary data items;with the exact same name, of the data group data-name-2. As an example, in this problem the statement: MOVE CORRESPONDING CARD- RECORD TO INFO-LINE, will move only the data items C- CODE, PAYCODE, PRESENT-DUE in CARD-RECORD to the corres-ponding items C-CODE, PAYCODE, PRESENT-DUE in INFO- LINE. In the following program, the new concepts introduced in this problem are demonstrated thoroughly

If the wall is frictionless, F_1\ding{217} is horizontal. The direction F_2\ding{217} is unknown (except in special cases, its direction does not lie along the ladder). Instead of considering its magnitude and direction as unknowns, it is simpler to resolve the force F_2\ding{217} into x- and y-components and solve for these. The magnitude and direction of F_2\ding{217} may then be computed. The first condition of equilibrium states that the net horizontal component of force on an object is zero. Similarly for the net vertical component. Hence, from the figure, \sumF_x = F_2 cos \texttheta - F_1 = 0 \sumF_x = F_2 cos \texttheta - F_1 = 0 (1st condition) \sumF_y = F_2 sin \texttheta - 80 lb. = 0. The second condition of equilibrium states that the net torque acting on a body is zero. If the body is in trans-lational equilibrium (F_net\ding{217} = 0) we may compute the torques about any axis. (Torques coming out of the plane of the figure will be considered positive) The resulting equation is simplest if one selects a point through which two or more forces pass, since these forces then do not appear in the equation. Let us therefore take moments about an axis through point A. \sum\cyrchar\cyrt_A = F_1 × 16 ft. - 80 lb. × 6 ft. = 0.(2nd condition) From the second equation, F_2 sin \texttheta = 80 lb., and from the third, F_1 = (480 lb \textbullet ft)/(16 ft) = 30 lb. Then from the first equation, F_2 cos \texttheta = 30 lb. Hence, F_2 = \surd[(80 lb)^2 + (30 lb)^2] = 85.5 lb. \texttheta = tan^-1 [(80 lb)/(30 lb) = 69.5\textdegree \texttheta = tan^-1 [(80 lb)/(30 lb) = 69.5\textdegree .

The general formula for the compound is B_a H_b where a and b are to be determined. To find these, we must first find the molecular weight of the compound by using the ideal gas law, PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature, n equals the mass, m, divided by the molecular weight MW,PV = nRT = (m/MW) RT,orMW = (mRT/PV) . MW = (mRT/PV) = [(1 g × 0.082 liter - atm/mole-\textdegreeK × 276\textdegreeK) / (1.66 atm × 0.820 liter)] = 27.6 g/mole. To find a and b, we use the relation MW_Ba _Hb = a × atomic weight of B plus b × atomic weight of H, or g/mole = a × 10.8 g/mole + b × 1.0 g/mole. By trial and error, we find that a = 2 and b = 6 (2 × 10.8 g/mole + 6 × 1.0 g/mole = 27.6 g/mole), so that the formula of our compound is B_2 H_6 (diborane) .

Question:

A suburban housewife, who lives at location A of the town street plan in Figure 1, drives to fill up her automobile at the local gas station, located at B. For some reason, all of the horizontal streets run from left to right, and all the vertical streets run from bottom to top. She is curious as to how many possible routes exist between her and the gas station. Write a FORTRAN program to solve her dilemma.

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Solution:

From the definition of the street plan, the direction of movement from each comer can be either up or to the right. Any parti-cular path can be written as a sequence of u's and r's (ups and rights), in which the total number of u's and r's can be written as n and m respectively. Hence, the total number of possible paths is (m + n) ! / n! m! We can write a program to generalize this problem for a street plan of n and m dimensions. The use of the FACT (factorial) function will make the program shorter and easier to understand. INTEGER N,M READ (5,{_\ast}) N,M CCOMPUTE NUMBER OF POSSIBLE PATHS CFROM CORNER TO CORNER PATHS = FACT (M+N) /FACT (N) {_\ast}FACT (M) WRITE (5,100) N,M, PATHS 100FORMAT (1X, 'NUMBER OF POSSIBLE PATHS 1FROM CORNER TO CORNER IN AN' ,I2, 'BY', 2I2, 'MATRIX =',I4) STOP END

Linear and angular velocity are related by the equation: v = \omegaR where v is linear velocity, \omega the angular velocity, and R the radius of the wheel. We will assume, in this problem, that the velocity increases uniformly since nothing to the contrary has been stated. By definition: a = (∆v)/(∆t)\alpha = (∆\omega)/(∆t) where a and \alpha are the linear and angular accelerations respectively. However: a = (∆v)/(∆t) = (∆\omega)/(∆t) - R = \alphaR We thus have a relationship between a and \alpha. a = (30 m/sec - 0 m/sec)/( 12 sec) = 2.5 m/sec^2 Therefore: \alpha = a/R = (2.5 m/sec^2)/(0.4 m) = 6.25 rad./sec^2 Note that although by dimensional analysis the units in a should reduce to 1/(sec.^2), we express angular acceleration in terms of rad./sec.^2. This is done because of the physical considerations of the problem. We consider the wheel as rotating through a discrete angle measured in radians, in a certain amount of time. However, our rigorous dimensional analysis has not really been violated since radians are dimensionless. As we know, radians are defined using a central angle of a circle. The magnitude of this angle in radians is calculated by dividing the length of the arc that the central angle subtends, by the length of the circle's radius. (See diagram.) \texttheta (in radians) = s/r Both s and r are measured in units of length which cancel in the ratio. We see, therefore, that radians are dimensionless. To calculate the angular velocity, we must remember the formula: \omega = \omega_0 + \alphat where \omega_0 is the initial angular velocity, and t is time. \omega = 0 + (6.25 rad./sec.^2)(t sec.) = 6.25 t rad./sec. The tangential velocity is: v = \omegaR = (6.25 t rad./sec.) (0.4 m) = 2.5 t m/sec. Note that we may now drop the radians from our units. To calculate the centripetal acceleration at the circumference we must remember the formula: a_c = (v^2)/(R) = (\omega^2R^2)/(R) = \omega^2R = (6.25 t rad./sec.)^2 (0.4 m) = 15.6 t2m/sec.^2 w, v, and \alpha are plotted against time in figures 1, 2 and 3.

we refer here to the relationship 1/0 + 1/i = 1/f Here, the image distance, i, is the unknown. The object distance, 0, is 8cm, while the focal length, f, is 16 cm. Thus, 1/0 + 1/i = 1/f 1/8cm + 1/i = 1/16cm 1/i = 1/16cm - 1/8cm 1/i =-(1/16cm) i= - 16cm This means that the image is 16 cm in front of the lens. To calculate the image size, we first calculate the magnification. Thus, m = -i/0 = +16/8 = 2 This means that the image is erect, therefore virtual, and twice as large as the object, or 16 cm high.

Question:

Explain what the computer takes as the attributes of the fol-lowingvariables by Default Rules of PL/I, if full explicit declarationsare not provided by the programmer. a) A variable named:PERCENT (1:8), b) A variable named:AMOUNT FIXED, c) A variable named:NUMBER COMPLEX, d) A variable named:TOTAL BINARY, e) A variable named:INCANT

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Solution:

a) First of all note that the name of the vari-able; PERCENT starts witha P. The letter P does notliebe-tween the letters I and N (inclusive). Hence, PERCENT is as-signed the following: PERCENT(1:8)DEC, REAL, FLOAT(6); b) The variable AMOUNT has the scale declared as FIXED. Hence, thecomputer completes the attributes as follows: AMOUNT FIXED (5) DEC, REAL. In the above, the precision is made equal to 5, i.e., (5,0). The Base and Mode are supplied as DEC and REAL. c) The variable NUMBER has the mode specified as COMPLEX. Hence, the following remaining attributes are supplied by the computer: Scale :FIXED Base :DEC Precision: (5), i.e., (5,0) Hence, the variable looks like that given below. NUMBERFIXED(5) DEC COMPLEX. d) The missing attributes are supplied for TOTAL BINARY as follows, according to the Default rules: TOTAL FLOAT (21) BINARY REAL; e) The missing attributes are supplied for INCANT as follows: INCANTFIXED(15) BIN, REAL; It should be noted in which order the attributes following a variable name aredeclared, especially the Scale and Preci-sion. The Precision attribute shouldalways follow the Scale attribute, and never vice versa.

Question:

A coil of wire is connected across one gap of a Wheatstone bridge and a temperature-controlled standard 1-\Omega resistor across the other. If the temperature of the coil is 0\textdegreeC, the other arms of the bridge have re-sistances in the ratio 0.923. If the temperature of the coil is 100\textdegreeC the ratio is 1.338. What is the temperature coefficient of resistance of the wire?

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Solution:

A Wheatstone bridge is shown in the figure. When the bridge is balanced, the current through galvanometer G is zero. In this case, then, the po-tential between points a and b is the same, whether we calculate it by path cab or path dab. Using Kirchoff 's Voltage Law around these 2 loops, we find - I_1 R_1 + I_2 R_4 = 0 andI_1 R_2 - I_2 R_3 = 0 Solving each equation separately for the ratio I_1 /I_2 (I_1 /I_2 ) = (R_4 /R_1 ) (I_1 /I_2 ) = (R_3 /R_2 ) and, therefore,(R_4 /R_1 ) = (R_3 /R_2 ) or(R_1 /R_2 ) = (R_4 /R_3 ) Let R_1 be the coil, and R_2 the temperature-controlled standard. In the first case, [{R_1 (0\textdegreeC)}/(1 \Omega)] = .923 orR_1 (0\textdegreeC) = .923\Omega In the second case R_1 (100\textdegreeC) = 1.338 \Omega But R_1 (100\textdegreeC) = R_1 (0\textdegreeC) (1 + \alpha t), where \alpha is the temperature coefficient of resistance of the wire and t is the difference in temperature experienced by the resistor (100\textdegreeC). Thus \alpha = [{R_1 (100\textdegreeC) - R_1 (0\textdegreeC)}/{t R_1 (0\textdegreeC)}] = [(1.338 \Omega - .923 \Omega)/{(100\textdegreeC) (.923 \Omega)}] \alpha = .0045 per \textdegreeC

Since the ray strikes the prism at grazing incidence, the refracted ray DE must enter at the critical angle \textphi_C. Now angles DAE, ADF, and AEF are all right angles. This follows because the prism is right-angled, and also since DF and FE are normals to the surfaces BA and AC. Thus DFE must be a right angle also, since the four angles of a quadrilateral (ADFE) add up to 360\textdegree. The angles of a triangle must add up to 180\textdegree, and therefore \angleDEF = 90\textdegree - \textphi_C.(1) Applying Snell's law to the refraction at face AC, we have n_1 sin \angleDEF = n_2 sin \texttheta, where n_1 and n_2 are the refractive indices of the prism and air, re-spectively. Since the medium to the right of face AC is air, n_2 = 1, whence n_1 sin \angle DEF = sin \texttheta.(2) At surface BA, we obtain n_2 sin 90\textdegree = n_1 sin \textphi_C or(1) (1) = n_1 sin \textphi_C Thensin \textphi_C = 1/n.(3) Using (3) and (1) in (2) (1/sin \textphi_C) [sin (90 - \textphi_C)] = sin \texttheta But sin (90 - \textphi_C) = cos \textphi_C, whence (cos \textphi_C/sin \textphi_C) = cot \textphi_C = sin \texttheta The ray will emerge from AC only if \angle DEF is less than the critical angle. If it is greater, total internal reflection occurs and the ray is directed along AC. Thus, for the ray to emerge, \angle DEF < \textphi_C. That is, 90\textdegree -\textphi_C < \textphi_C or \textphi_C > 45\textdegree. Using (3) \thereforesin \textphi_C = (1/n_1) > sin 45\textdegree = (1./\surd2)\therefore n_1 <\surd2.

Question:

What is meant by the autonomic nervous system? What are its subdivisions and to which of the large divisions of the entire nervous system does it belong?

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Solution:

Nerves from the central nervous system (CNS) that innervate the cardiac muscles, smooth muscles and secretory glands form the autonomic nervous system. Structural and physiological differences within the autonomic nervous system are the basis for its further subdivision into sympathetic and parasympathetic systems. Nerves of these two divisions leave the central nervous system at different regions. The sympathetic nerves emerge from the thoracic and lumbar regions of the spinal cord, and the parasympathetic nerves arise from the brainstem and sacral portions of the spinal cord. Thus the sympathetic system is known as the thoracolumbar outflow (originating in T_1 - T_12 and L_1 - L_12), while the parasympathetic system is referred to as the craniosacral outflow (originat-ing in cranial nerves III, VII, IX, and sacral nerves 2-4). The autonomic nervous system contains only motor nerves and is distinguished from the rest of the nervous system by several characteristics. There is no voluntary control by the cerebrum over these nerves. We cannot control our heart beat or the action of the muscles of the stomach or intestines. Another important charac-teristic of the autonomic nervous system is that most internal organs receive a double set of fibers, one set belonging to the sympathetic system and the other to the parasympathetic system. Impulses from the sympat-hetic and parasympathetic nerves always have antagonistic effects on the organs innervated. Thus if one functions to increase a certain activity, the other functions to decrease it. However, many tissues/cells do not receive parasympathetic innervation (i.e. fat cells, most arteri-oles, spleen). The autonomic nervous system is part of a larger unit called the peripheral nervous system. The peripheral nervous system contains two types of nerve fibers: the afferent fibers which convey information from receptors in the periphery to the CNS, and the efferent fibers which carry information from the CNS to the effectors. Effectors are tissues or organs which bring about appropriate responses to certain stimuli, both internal (such as from the brain) and external (such as from the environment). Some examples of effectors are the skeletal muscles, cardiac and smooth muscles, and secretory glands. The peripheral nervous system can be divided functionally into two parts. That part which innervates the skeletal muscles is known as the somatic nervous system, and that which innervates smooth muscle, cardiac muscle and glands is the autonomic nervous system.

The human eye consists of three layers or tu-nics. The fibrous tunic is the outermost layer of the eye: it consists of the sclera posteriorly and the cornea anteri-orly. The sclera is a protective coat and is the "white of the eye." The cornea is the transparent covering which func-tions in the refraction (bending) of light. The middle coat is the vascular layer. Posteriorly, it consists of the choroid coat. This pigmented layer absorbs excess light and nourishes the retina through its rich vas-culature. The choroid continues anteriorly as the ciliary body. The ciliary body consists of the ciliary muscle, a smooth muscle which functions in the accommodation reflex, and the ciliary process which secretes aqueous humor (fluid) into the anterior part of the eye. Finally, the ciliary body continues anteriorly as the iris. The iris is the colored part of the eye and also contains smooth muscle which func-tions in controlling the size of the pupil - the small open-ing through which light passes. The innermost layer is the nervous tunic. It consists solely of the retina in the back of the eye. The retina con-sists of photoreceptor cells called rods and cones. In the center of the retina is a small depressed area called the fovea centralis, the region of keenest vision. Medial to this is the optic disc. The optic nerve exits the back of the eye at this point. Because there are no photoreceptors at the disc, there is a blind spot in the peripheral field of vision. There is no anterior continuation of the retina. There are two fluid filled cavities inside the eye. The anterior cavity contains an aqueous humor (a watery fluid), and is located between the cornea and the lens. The poste-rior cavity is filled with vitreous humor (a gel-like fluid), and is located between the lens and the retina. The transparent lens which separates these two cavities is re-sponsible for focusing incoming light rays on the retina.

(a) Bose statistics determine the number of distinct ways of placing n_i identical particles in g_i states, where the number of particles in a state is not restricted. Consider the system under consideration to be a linear array of (n_i + g_i - 1) holes into which either particles or partitions that separate the states can be inserted. Note that there are (g_i - 1) such partitions to separate the entire cell into g_i states. The remaining n_i holes which are separated into groups by these partitions, represent the distribution of the particles among the various states (see Fig. a). The number of distinct permutations of n_i particles and (g_i - 1) partitions over (n_i + g_i - 1) holes is the same as the number of distinct arrangements of n_i particles among the g_i states. Now, the number of permutations of (n_i + g_i - 1) distinguishable objects is (n_i + g_i - 1) !. But,n_i particles and (g_i - 1) partitions actually are indistinguishable and we must divide the above number by the number of permutations of n_i particles and by the number of permutations of (g_i - 1) partitions. The result gives the number of distinct arrangements. N_BE = [{(n_i + g_i - 1) !} / {n! (g_i - 1) !}] Substituting the numerical values, N_BE = [6!/(2! 4!)] = 15 Two possible distributions are shown in Fig. a. (b) In Fermi statistics a quantum state can contain only one particle. As before, we first find the number of ways of distributing n_i in distinguishable particles over g_i states. The first particle can be put in any of the g_i states, and for each choice of a state, the second particle can be put in any of the (g_i - 1) remaining states, and so on. The last particle will have (g_i - n_i + 1) possible states to choose from. The total number of configurations is g_i (g_i - 1) .... (g_i - n_i + 1) = [(g_i !)/{(g_i - n_i) !}] However, the actual indistinguishability of the particles requires that we divide out the number of identical configurations from this number. n_i particles give rise to n_i! permutations of the part-icles among themselves. Thus, we have N_FD = [(g_i !)/{n_i (g_i - n_i) !} = [5!/(2! 3!)] = 10 Two possible distributions are shown in Fig. b.

The ovum must be fertilized while in the upper regions of the Fallopian tube or oviduct to develop suc-cessfully. Upon fertilization, it begins the series of cell divisions and differentiation which will eventually lead to development of the embryo (see Figure 1). The fertilized ovum continues its passage through the oviduct to the uterus, which requires three to four days. Once in the uterus, the egg floats free in the intrauterine fluid, from which it receives nutrients for several more days. During this time, it continues dividing and begins to differentiate into the blastocyst. The blastocyst stage (a fluid-filled sphere of cells) is attained by the time or shortly after the egg reaches the uterus. The egg would have slowly disintegrated during its passage to the uterus if fertilization had not occurred. While the egg had been traveling to the uterus, the uterus was being prepared under the combined stimula-tion of estrogen and progesterone to receive the egg. If the fertilized egg were to pass too quickly through the oviduct, it would reach the uterus prematurely, when both it and the uterus had not yet reached a suitable state for implantation. For the egg, the necessary event for implantation is the disintegration of the zona pellucida, a remnant of the Figure 1Schematic diagram of the maturation of an egg in a follicle in the ovary following its release (ovulation). a) fertilization in the upper part of the oviduct b) early cleavage as it desends c) late cleavage continuing desent d) blastocyst development in the uterus before implantation e) implantation of blastocyst in the wall of the uterus. follicle. This disintegration is completed at the blasto-cyst stage. The blastocyst is composed of an outer layer of specialized cells called trophoblasts, which are re-sponsible for implantation, and an inner cell mass destined to become the fetus itself. The trophoblast layer, upon disintegration of the zona pellucida, enlarges and adheres to the uterine wall. These trophoblasts secrete lytic en-zymes which will digest the uterine endometrium, allowing the blastocyst to completely embed itself in the uterine lining. This digestion of tissue continues for a period, with the breakdown products of the endometrium serving as a nutrient source for the developing embryo until both the fetal circulation and the placenta have developed. The placenta will form from an intermeshing of the trophoblast layer with the endometrium, and provides for contact and nutrient exchange between the mother and the fetus. In addition to enzymes, the trophoblast secrets hor-mones, including chorionic gonadotropin. This important hormone is necessary for retention of the corpus luteum, which would have involuted if its ovum had not been fert-ilized. The hormonal secretions of the corpus luteum and later the placenta, maintain the uterus in the proper condition for pregnancy and prevent menstruation. Detection of the presence of chorionic gonadotropin in the urine is the basis for most pregnancy tests. Because of the time required for the ovum to traverse the oviduct and for the uterus to be prepared, the total time that elapses from ovulation to implantation is from seven to eleven days.

Question:

Two wires are attached to a corner fence post with the wires making an angle of 90\textdegree with each other. If each wire pulls on the post with a force of 50 pounds, what is the resultant force acting on the post? See Figure.

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Solution:

As shown in the figure, we complete the par-allelogram. If we measure R and scale it, we find it is equal to about 71 pounds. The angle of the resultant is 45\textdegree from either of the component vectors. If we use the fact that the component vectors are at right angles to each other, we can write R^2 = 50^2 + 50^2 whence R = 71 pounds approximately at 45\textdegree to each wire.

The number of moles of gas present is the mass of gas present divided by the molecular mass of nitrogen, or n = (36/28) = 1.29 The pressure is p = 3 atm = 3 × 1.013 × 10^5 N/m^2 = 3.04 × 10^5 N/m^2 Thus the volume of the container is by the ideal gas law V = nRT/p = (1.29 moles × 8.313 joules/\textdegreek × 473\textdegree k)/(3.04 × 10^5 N/m^2) = (1.29 moles × 8.313 N-m/\textdegreek × 473\textdegree k)/(3.04 ×10^5 N/m^2) = 16.7 × 10^-3 m^3 That is, the volume is a cube with sides 25.6 cm. The total number of molecules is the number of moles of gas times Avogadro's number, or (1.29 moles)[6.02 × 10^23 (molecules/moles)] = 1.29 × 6.02 × 10^23 molecules so that the volume per molecule is (16.7 × 10^-3 m^3)/(1.29 × 6.02 × 10^23) = 2.15 ×10^-26 m^3 = 2.15 × 10^-20 cm^3 This corresponds to a cube with sides 2.8 × 10^-7 cm, which is the mean distance between molecules. The mean square speed is calculated from (1/2)mv^2= (3/2)kT where m is the mass of one molecule of nitrogen,v^2is the mean square speed of the molecule, k is Boltzmann's constant, and T is the absolute temperature. Furthermore, since k = R/N_0 where R is the gas constant and N_o is Avogadro's number, and mN_0 = M where M is the mass of one mole of gas, we have: v^2 = [(3kT)/m] = (3RT)/(N_0m) = (3RT)/M v^2 = (3 × 8.313 joule/\textdegreek mole × 473\textdegree k)/(2.8 × 10^-3 kg/mole) = (3 × 8.313 kg-m^2/s^2 \textdegreek mole × 473\textdegree k)/(28 × 10^-3 kg/mole) = 4.21 × 10^5 m^2/sec^2 Thus the average speed c of a molecule is c = \surd(v^2) = 6.5 × 10^2 m/sec. When two molecules of radius r collide, their centers are a distance 2r apart. Therefore, a molecule travelling along a straight line will collide with other molecules within a distance 2r of its center (we assume a simplified model in which one molecule moves and the others remain stationary and in which all collisions are elastic). During a time ∆t, the molecule which we are following travels a distance c∆t. Thus, the molecules with which our molecule will collide are all found within a cylinder of base radius 2r and height c∆t. This cylinder is called the collision cylinder. If N is the number of molecules in volume V, then the density of molecules is N/V. If we multiply the volume of the cylinder by the density of molecules we will have the number of molecules in the cylinder (4\pir^2c∆tN/V). Dividing this quantity by ∆t we have the number of collisions per unit time (4\pir^2cN/V) . The average time between collisions is the reciprocal of this, called the mean free time (V/4\pir^2cN). If we multiply this by the average velocity, we have the average distance a molecule travels before colliding with another (the mean free path): L = (cV)/(4\pir^2 cN) = V/(4\pir^2N) Substituting our values, we have (in the present case let r = 1.3 × 10^-10 m): L = [16.7 \pi 10^-3 m^3]/[4\pi(1.3 × 10^-10 m)^2(1.29 moles)(6.023 ×10^23 moles^-1) = 1.01 x 10^-7 m

Question:

Are the daughter nuclei formed during cleavage exactly equivalent , or is there some sort of parceling out of potentialities during cleavage?

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Solution:

There is evidence that there is no segregation of genetic potentialities during cleavage. Experiments have shown that isolated blastomeres (the cleavage cells) are capable of developing into an entire organism . In a classical experiment, the nucleus is removed from a ripe frog's egg with a micropipette. A cell from an advanced stage of embryonic development is then separated from its neighbors. The plasma membrane of the cell is ruptured in the process and the nucleus is injected into the enucleated frog egg. The egg is found to develop normally. When nuclei obtained from late blastula or early gastrula stages are transplanted into enucleated eggs, the eggs can develop normally. Even nuclei from later stages of development, such as from the neural plate (the precursor of the spinal cord and brain) or from cells lining the gut may lead to the development of normal embryos. In humans, there is no segregation of genetic potentialities during early cleavage. Identical twins are thus possible. More experimentation has to be done to determine just how far into development the "no segregation " rule will apply.

Question:

Two parallel plates of area A = 1 m^2 are given equal and opposite charges of Q = 30 microcoulombs each. A sheet of dielectric of permittivity \epsilon = 15 × 10^\rule{1em}{1pt}12 coul^2 / newton-m^2 occupies the space between the plates (see Fig.)A). Compute (a) the resultant electric in-tensity in the dielectric, (b) the induced charge per unit area on the faces of the dielectric.

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Solution:

(a) The charge per unit area on the plates is \sigma= q/A = 30 x 10 ^-6 coul/m2. The resulting electric field between the plates can be obtained by applying Gauss' Law to surface S in Fig. b. \int_s E^\ding{217} \bullet da^\ding{217} = Q / \epsilon In Fig. B, we see that on surface S_1 E^\ding{217} \textbullet da^\ding{217} = \rule{1em}{1pt}E da and on the rest of the surface E^\ding{217} \textbullet da^\ding{217} = 0 since the field through the sides S_2 of the surface is zero. The electric field is constant between the plates. Therefore \int_s E^\ding{217} \bullet da^\ding{217} = \rule{1em}{1pt}E \int_(s)1 da = \rule{1em}{1pt}EA, givingE = \rule{1em}{1pt}Q / \epsilonA = \rule{1em}{1pt}(\sigma / \epsilon ) │E│ = (30 × 10^\rule{1em}{1pt}6 coul / m^2) / (15 × 10^\rule{1em}{1pt}12 coul^2 / n-m^2) = 2 × 10^6 volts / m. (b) The dielectric coefficient is K = \epsilon / \epsilon_0 = (15 × 10^\rule{1em}{1pt}12) / (8.85 × 10^\rule{1em}{1pt}12) = 1.7. The induced charge on the dielectric of a capacitor is given by Q_ind = Q [(1) \rule{1em}{1pt} (1 / K)]. Then the induced charge density in the dielectric is \sigmaind= Q_ind / A = Q / A [(1) \rule{1em}{1pt} (1 / K)] = \sigma[(1) \rule{1em}{1pt} (1 / K)] or\sigma_ind = 30 × 10^\rule{1em}{1pt}6 (1 \rule{1em}{1pt} 1 / 1.7) coul / m^2 = 12.3 × 10^\rule{1em}{1pt}6 coul / m^2.

Question:

Using Positive Logic, show that the Diode circuit of fig.1 represents a three-input AND gate.

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Solution:

A diode is similar to a valve which lets current flow in only one direction. In fig. 1 it is seen that when all inputs are low (OV) all diodes conduct. This action brings V_out to the same voltage as Vin N, hence V_out is low. When one or two inputs are high (\geq 5V) , any diodes corresponding to the low inputs conduct and any diodes corresponding to the high inputs act as insulators. Hence the diodes that conduct bring V_out to the low voltage. The diodes that are insulators have no effect on the out-put voltage. When all inputs are high, all of the diodes act as insulators and only then will the output voltage be high. From this information, it is seen that the output is high only when all of the inputs are high. Using positive logic, this is seen to be an AND gate.

Col 1-8 Col 10-14 16-71 1 VARIANCE START 0 2 SAVE (14,12) SAVE REGISTERS 3 BALR 3,0 STORE BASE ADDRESS 4 USING \textasteriskcentered,3 USING REGISTER 3 AS BASE REGISTER 5 ST 13,SAVE+4 6 LA 13,SAVE STORE ADDRESS OF SAVE AREA 7 PRINTLIN HEADING,132 PRINTING THE LABELS 8 AP WRKGPNBR, = P '1' FIRST GROUP NUMBER 9 RDCD READCARD CARDINPA, CRDEOF READ IN A NUMBER, X; 10 CLI CRDATA, C 'E' CHECK TO SEE IF NUM-BER IS A DIVISOR 11 BE CALQUE IF 'E', THEN END OF A GROUP AP WRKGPCNT, = P '1' COUNT ELEMENTS OF GROUP 12 PACK WRKDATA, CRDATA CONVERT NUMBER TO PACKED DECIMAL 13 AP WRKSUM, WRKDATA FINDING SUM OF GROUP 14 ZAP WKDATASQ, WRKDATA PLACE NUMBER IN AN-OTHER AREA 15 MP WKDATASQ, WRKDATA SQUARE NUMBER 16 AP WRKSUMSQ, WKDATASQ FINDING SUM OF THE SQARES 17 ZAP WKDATASQ, ZERO ZERO AREA USED FOR SQUARING 18 CP WRKDATA, WRKMAX FINDING MAX 19 BH GPHIGH STORE MAX 20 CP WRKDATA, WRKMIN FINDING MIN 21 BL GPLOW STORE MIN 22 B RDCD GET NEXT NUMBER IN THE GROUP 23 GPLOW ZAP WRKMIN, WRKDATA STORE NEW MIN 1 - 8 10 - 14 16 - 71 23A B RDCD 24 CALQUE MVC PRTGPNBR, PATTI PREPARE GROUP 25 ED PRTGPNBR, WRKGPNBR NUMBER FOR THE PRINT LINE 26 MVC PRTGPCNT, PATTI PREPARE N FOR 27 ED PRTGPCNT , WRKGPCNT PRINT LINE 28 MVC PRTSUM, PATT2 PREPARE SUM FOR 29 ED PRTSUM, WRKSUM PRINT LINE 30 ZAP WRKSUMST, WRKSUM GET COPY SUM 31 MP WRKSUMST, WRKSUM FINDING THE SQUARE OF THE SUM 32 DP WRKSUMST, WRKGPCNT DIVIDE THE SQUARE OF THE SUM BY N 33 SP WRKSUMSQ, WRKSUMST(5) CALCULATE THE NUMERATOR OF VARIANCE 34 SP WRKGPCNT, = P'1' FIND DENOMINATOR OF VARIANCE 35 DP WRKSUM, WRKGPCNT FINDING VARIANCE 36 AP WRKGPCNT, = P'1' RESTORE N TO FIND AVERAGE 37 DP WRKSUM, WRKGPCNT FIND AVERAGE 38 MVC PRTMAX, PATT2 PREPARING THE 39 ZAP WRKSUM, WRKSUM (2) AVERAGE FOR THE 40 ED PRTAVER, WRKSUM PRINT LINE 41 MVC PRTVAR, PATT2 PREPARING VARIANCE 42 ZAP TEMPHOLD , WRKSUMS Q +V (4 ) FOR THE 43 ED PFTVAR, TEMPHOLD PRINT LINE 44 MVC PRTMAX, PATT2 PREPARING TO PRINT 45 ED PRTMAX, WRKMAX GROUP'SMAX 46 MVC PRTMIN, PATT2 PREPARING TO PRINT 47 ED PRTMIN, WRKMIN GROUP'S MIN 48 PRIMTLIN PRTDETL,133 PRINT OUTPUT LINE 49 AP WRKGPNBR, = P '1' INCREMENT GROUP NUMBER 50 ZAP WRKGPCNT, ZERO 51 ZAP WRKGPCNT, ZERO 52 ZAP WRKSUMSQ, ZERO 53 ZAP WRKMAX, = P '0000' 54 ZAP WRKMIN, = P '9999' 55 B RDCD 56 GPHIGH ZAP WRKMAX, WRKDATA STORE NEW MAX 57 CP WRKDATA, WRKMIN 58 BL GPLOW 59 B RDCD 60 CRDEOF L 13, SAVE + 4 60A RETURN (14, 12) RESTORE REGISTERS \textasteriskcenteredDATA DEFINITION FOR THE CARD INPUT AREA 62 CRDINPA DS 0CL80 63 CRDDATA DS CL3 64 DS CL77 \textasteriskcenteredDATA DEFINITION FOR THE WORK AREAS 66 SAVE DS 18F USE TO SAVE THE CONTENTS REGISTERS 67 WRKDATA DS PL3 68 ZERO DC PL4 '0' 69 WRKSUM DC PL '3' 70 WRKDATASQ DS PL6 71 WRKSUMSQ DC PL6'0' 72 WRKSUMST DC PL6 73 WRKGPNBR DC P '0' 74 WRKGPCNT DC P '0' 75 WRKMAX DC PL3 '0' 76 WRKMIN DC PL3 '9' 77 TEMPHOLD DS PL4 78 PATT1 DC X '40202020' 79 PATT2 DC X '40202021 2020202020' \textasteriskcenteredDATA DEFINITION FOR THE HEADING 81 HEADING DS \varphiCL133 \varphi 1 - 8 10 - 14 16 - 71 72 82 PRTCC1 DC C '1' 82A DC 19C ' ' 83 DC C'# N SUM AVG VAR X 83A MAX MIN' 84 DC 57C ' ' \textasteriskcentered DATA DEFINITION FOR THE PRINTER WORK AREA 86 PRTDETL DS \varphiCL133 \varphi 87 PRTCC2 DC C'-' SKIP TWO SPACES 88 DC 19C ' ' 89 PRTGPNBR DS CL 2 90 DC 3C ' ' 91 PRTGPCNT DS CL2 92 DC 4C '' 93 PRTSUM DS CL 6 94 DC 4C '' 95 PRTAVER DS CL6 96 DC 4C '' 97 PRTVAR DS CL8 98 DC 6C'' 99 PRTMAX DS CL6 100 DC 6C'' 101 PRTMIN DS CL 6 102 DC 50C'' 103 END VARIANCE Comments: For the beginning assembler language programmer, the calculations in his program will generally be done in either of two forms of arithmetic, packed decimal or fixed point binary. Our VARIANCE program written above is an example of an assembler language program which uses only packed de-cimal arithmetic. It should be understood that both forms of arithmetic can be done simultaneously, and, in fact, this is what is done on most occasions. We must first point out that the numbers, as written in the leftmost edge of the program, are simply placed there for the purposes of reference, and are not in any way re-lated to the program itself. The calculations involved in our program have been done in an order corresponding to the listing of our question, and as such, we do not think it necessary to explain the logic of the program. In our explanation, we will discuss only those instructions we think would pose problems for the student. The first six instructions of our program constitute the housekeeping procedures required. The label in Columns 1-8 of Line 1 is the name of the program. The other in-structions we will consider are: 7. PRINTLIN HEADING, 132 - The label 'HEADING\textasteriskcentered is the name of a most cases, specified as the length of a print line (132 or 133 spaces) on an output page. This print area usually contains carriage controls (Column 1), output information (whether defined as constants or moved into the area), and blanks. It is good assembler programming practice to fill up unused portions of a print area with blanks since the computer has a tendency to print undesired characters in 'unblanked' spaces. The above discription given for the "heading' print area applies, in general, to all print areas defined by a data definition. Variations of these areas lie only in their names (chosen by the programmer) and in the fact that data to be printed may be explicitly defined in the area (DC statement), moved into it (MVC statement) or a combination of both. The coding procedures for the way data is to be read from a card are quite similar to those of printing. The only differences are that the length of the input area is 80 bytes (80 columns on a card), and instead of placing blanks in the unused fields, we just define them (DS state-ment - Line 64). 10. CLI CRDATA,C 'E' - This instruction determines if the card read in has an 'E' in column 1. If so, we are at the beginning of a new group, but we must first process (find the variance of) the old group before commencing this one. Lines 14 - 15 and 30 - 31 illustrate the technique used in multiplying (and squaring) two packed decimal numbers. The multiplicand is placed in a field twice as long as that of the multiplier. When the calculation is done, the result will replace the multiplicand. Thus, in making its area twice as long as that of the multiplier, we quard against overflow. In general, when calculations are done in assembler language, Operand - 1 is replaced by the result, so one should always check to see that this area is large enough to re-ceive the largest (most spacious) possible result. For ex-ample, in the DP instructions of Lines 32 and 35, Operand -1 was made large enough to store the resulting quotients and their remainders. The length of these fields can be checked by inspecting them in their various data definitions. In-cidentally, it is good programming practice to somehow in-dicate in which data definition a particular symbol maybe found. This is usually achieved by allowing the first (or last) two or three letters of the symbol to correspond to the name of the data definition. For example, 'CRDDATA' is to be found in the data definition for the card input area, and 'PRTSUM' is located in the definition for the print area. The only other instructions we think necessary to discuss are those involved in the EDITING (ED) of packed de-cimal results to be printed. It is imperative that we edit packed decimal results prior to printing, since our output will not be what we expect unless this is done. Our prob-lem here is one of translation. When the computer reads in our data, it does so in EBCDIC code. This process involves replacing each character of our data with two EBCDIC digits. For example, it reads 1 (data) as F1(ECDIC), E as C5, and so forth. Now, in printing it does the reverse; the informa-tion to be printed is taken as EBCDIC and translated back to 'readable' code. For example, the packed decimal number 406F, taken as EBCDIC, would be translated as Having pointed out the importance of editing, we will now illustrate the way in which it is done. A packed deci-mal number has the form: This is a representation of the number in memory. Two deci-mal digits are stored per byte, except for the rightmost byte, which has a sign factor in its last four bits. The integers are those of our actual answer. As you can see, the storage space alotted a pack decimal number is usually such that unused high order locations are packed with zeros. By first placing a lexadecimal pattern in a print area, a packed number will be translated to the correct EBCDIC code before printing is done. The first byte of this pattern usually contains a hexadecimal 40, which causes suppression of high order zeros. In general, the length of the pattern field is twice that of the packed decimal field. The afore-mentioned 40 of the pattern field may be said to replace the sign factor of the packed decimal field, but it will be in the first byte instead of the last. Each integer in the packed field will then be replaced by a DIGIT SELECTOR (HEX 20) in the pattern field. For example, let us consider the following instructions: MVCPRTSUM,PATTERN EDPRTSUM,WRKSUM Before MVC and ED instructions After instructions:1. MVC PRTSUM, PATTERN The main idea behind editing is to place a hexadeci-mal pattern in the print field, to ensure correctness of the packed decimal data to be printed. The pattern used can be longer than the print field since it will be truncated on the right. Although this means less coding and simple edit-ing, it should not be done unless the portion to be trun-cated is of no importance. One should also note that since there are two digits per byte for a packed decimal field the print field should be at least twice as long. As long as our intentions are to print only the inte-gers from the packed field, it is okay to make the pattern, and hence the printfield, twice as long. But if we intend to place decimal points, commas, and/or any special char-acters in our output, we must use a pattern and a print-field that are more than twice the length of the packed field. Lines 83 and 83A of our program illustrate how a literal is continued from one card to another in a data definition. The X in column 72 of the card is a continu-ation character. Our variance program is one coded to completion. It lacks only the JCL cards and the actual data. But after studying this program, the student should not find it dif-ficult to write a program which uses the packed decimal in-structions.

Question:

Does a yeast cell metabolize more efficiently in the presence or in the absence of oxygen? Explain.

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Solution:

Under aerobic conditions (in the presence of oxygen), glucose undergoes glycolysis to form pyruvate, which after conversion to acetyl CoA, enters the TCA cycle. The NADH produced is oxidized in the electron transport system provided that oxygen is available as the ultimate electron acceptor. Let us calculate the maximum amount of ATP produced from the oxidation of one mole of glucose. A net gain of 2 ATP is obtained directly in glycolysis (see figure). In addition, 2 NADH are genera-ted. This occurs in the cytoplasm. To enter the electron transport system, the NADH must be transported into the mitochondrion, a process which requires 1 molecule of ATP for every NADH which crosses the mitochondrial membrane. So intead of three ATP being produced per NADH, each cytoplasmic NADH yields only 2 net ATP; that is, a total of 4 ATP are produced by the 2 NADH generated in glycolysis. Recall that two 3-carbon molecules are obtained in glycolysis. Each of these two 3-carbon compounds can eventually enter the TCA cycle as acetyl CoA, and later, their electrons enter the electron transport system. Thus ATP production from the TCA cycle is twice that of a single turn. Two ATP are produced in the cycle via the formation of 2 molecules of GTP from 2 molecules of succinyl CoA. The conversion of pyruvate to acetyl CoA yields one NADH while the Krebs cycle yields 3 NADH. Thus 8 (4 × 2) NADH are produced per glucose at this point. The 8 (4 per turn) NADH formed in the cycle yield 24 ATP (3 ATP per NADH) by means of the electron transport chain. The 2 FADH_2 (1 per turn) produce 2 ATP each via electron transport, giving a total of 4 ATP. Summation of all the ATP from glycolysis, the citric acid cycle, and electron transport leads to a net production of 36 ATP per mole of glucose fully oxi-dized. Under anaerobic conditions, oxygen is not available as a terminal electron acceptor, and the reactions of the electron transport system cease as the reduced intermediates build up (refer to questions 3-20 and 3-26). This leads to an accumulation of the TCA cycle intermediates and, since it cannot enter the cycle, an accumulation of pyruvate.Under these conditions in yeast cells, pyruvate is then decarboxylated to acetaldehyde, which is in turn reduced to ethyl alcohol (ethanol), regenerating oxidized NAD+ for further use in glycolysis. This process is called alcoholic fermentation conditions, this too regenerates NAD+ in order to continue glycolysis for energy production: Pyruvic acid gets reduced to lactic acid as NADH gets oxidized back to NAD+ . (see next question). Under anaerobic conditions then, only 2 ATP molecules are produced, as opposed to the 36 ATP produced when glycolysis is supplemented with aerobic respiration Thus, a yeast cell metabolizes 18 times more efficiently in the presence of oxygen than in the absence of oxygen. ATP yield from the complete oxidation of glucose Reaction sequence ATP yield per Glucose GLYCOLYSIS: GLUCOSE TO PYRUVATE (in the cytoplasm) Phosphorylation of glucose \Elzbar1 Phosphorylation of fructose 6-phosphate \Elzbar1 Dephosphorylation of 2 molecules of 1, 3-DPC +2 Dephosphorylation of 2 molecules of phosphoenolpyruvate +2 2 NADH are formed in the oxidation of 2 molecules of glyceraldehyde 3-phosphate CONVERSION OF PYRUVATE TO ACETYL CoA (inside mitochondria) 2 NADH are formed CITRIC ACID CYCLE (inside mitochondria) Formation of 2 molecules of guanosine triphosphate from 2 molecules of succinyl CoA +2 6 NADH are formed in three oxidation steps. 2 FADH_2 are formed in 1 oxidation step. OXIDATION PHOSPHORYLATION (inside mitochondria) 2 NADH formed in glycolysis; each yields 2 ATP (not 3 ATP each, because of the cost of the shuttle) +4 2 NADH formed in the oxidative decarboxylation of pyruvate; each yields 3 ATP +6 2 FADH formed in the citric acid cycle; each yields 2 ATP +4 6 NADH formed in the citric acid cycle; each yields 3 ATP +18 36 ATP/glucose

Question:

A woman opens a can of ham and beans and suspects food poisoning . She boils the contents for half an hour. A few hours after eating the ham and beans, nausea, vomit-ing, and diarrhea set in. Explain.

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Solution:

The woman probably suspected that the can of ham and beans was contaminated withClostridiumbotulinum, the cause of the microbial food poisoning known as botulism. These bacteria sometimes, but not always, produce foul odor.Clostridiumbotulinumare usually found in canned, low-acid products since they are anaerobic, growing without the presence of oxygen. The poisoning is due to an extremely potent toxin produced by the bac-teria. The toxin acts upon nerves and causes the paralysis of the pharynx and diaphragm, thus causing respiratory failure and sometimes death (65% of all cases are fatal). The toxin is heat-labile (unstable) and is destroyed by boiling for 15 minutes. Therefore, the woman denatured thebotulinustoxin and her food poisoning cannot have been botulism. She also did not show the symptoms of botulism which affects the neuromuscular junction. There is a flaccid paralysis which may precede cardiac or respiratory failure. Other symptoms include diffi-culty in swallowing and speech. Double vision is often present as well. The woman was probably poisoned by the toxin produced by Staphylococcusaureus , the most common agent causing food intoxication in the U.S. today. The bacteria may have come from food handlers who harbored the organism in the nose, throat, or sores on the skin (boils). The organism is a facultative anaerobe, able to live in the anaerobic environment of the can. Staphylococcal food poisoning usually involves food such as ham and dairy products.The important point in this type of poisoning is that the toxin is heat-stable and would be unaffected by boiling for 30 minutes. Therefore, the woman's precautionary action was not successful. Fortunately, death is rarely caused by this toxemia.

Question:

A slit of width a is placed in front of a lens of focal length 50cm and is illuminated normally with light of wavelength 5.89 × 10^-5 cm. The first minima on either side of the central maximum of the diffraction pattern observed in the focal plane of the lens are separated by 0.20 cm. What is the value of a?

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Solution:

This is an example of Fraunhofer diffraction (see figure). Parallel rays of light are incident on slit NN' from the left. The rays are diffracted diffracted and encounter the lens of focal length f. The rays are then focused on a screen (SS') lying in the lens' focal plane, and a diffraction pattern is observed. The minima of this diffraction pattern are described by the formula sin \texttheta = m\lambda/a where \texttheta locates the minima fringes, a is the slit width, m is the minima number, and \lambda is the wavelength of the light used. In this problem, the first minima on either side of the central maximum is found at angle \texttheta, such that \texttheta = sin^-1 \lambda/a. The angular separation of the 2 minima is then (see figure) \delta = 20 = 2 sin^-1 \lambda/a(1) But, we may write 2 \texttheta \approx DB /f(2) where DB is the linear separation of the 2 minima. Here, we have approximated the arc DB (shown dotted in the figure) by the linear distanceDB. Using (2) in (1) DB/f \approx 2 sin^-1 \lambda/a orsin (DB/f) \approx \lambda/a anda \approx [\lambda/{sin(DB/2f)}] Hencea \approx [{5.89 × 10-5cm} / {sin(.20 cm/100 cm)}] = [{5.89 ×10^-5 cm} / {sin(.0020)}] Since.0020 rad is a very small angle, we may write sin (.0020) \approx .0020 = 2× 10^-3 whencea \approx (5.89 ×10-5cm)/(2 × 10^-3) = 2.495 × 10^-2 cm.

The string is stretched under the action of the radial centripetal force which keeps the mass M on its circular path. When we pull in the string we shorten r_0 by increasing the radial force F_centrip^\ding{217} on M. As we know, a force can only produce a torque about the axis of rotation if it has a component perpendicular to the radius which locates the mass M. A purely radial force like F_centrip^\ding{217} has no such component, therefore the angular momentum must remain constant as the string is shortened. Mv_0r_0 = Mvr(1) The kinetic energy at r_0 is (1/2) Mv_0^2; at r it has been increased to (1/2) Mv^2 = (1/2) Mv_0^2 [r_0/r]^2 because v = v_0r_0/r from above. It follows that the work W done from outside in shortening the string from r_0 to r is W = (1/2) Mv^2 - (1/2) Mv_0^2 = (1/2)Mv_0^2 [(r_0/r)^2 - 1](2) This can also be calculated directly as the work done by F_centrip^\ding{217} along the distance r_0 - r; W = ^r\int_r0 F_centrip^\ding{217} \bullet dr^\ding{217} = - ^r\int_r0 F_centrip dr W = -^r\int_r0 dr(Mv^2/r) = - ^r\int_r0 dr(M/r) [(v_0^2r_0^2)/(r^2)] where we have used (1). Hence W = -Mv_0^2r_0^2^r\int_r0 [(dr)/(r^3)] W = (Mv_0^2r_0^2)/(2r^2) \vert^r_r0 = (Mv_0^2r_0^2)/(2) [(1/r^2) - (1/r_0^2)] W = (1/2) Mv_0^2 [(r_0/r)^2 - 1] which is (2). We see that the angular momentum acts on the radial motion as an effective repulsive force. We have to do extra work on the particle on bringing it from large distances to small distances if we require that the angular momentum be conserved in the process.

The Bohr model of the hydrogen atom consists of a single electron of charge -e revolving in a circular orbit about a single proton of charge +e. The electrostatic force of attraction between electron and proton provides the centripetal force that retains the electron in its orbit (see the figure). Bohr's model assumes that the angular momentum of each orbit is re-stricted to integral multiples of Planck's constant divided by 2\pi, L_n = n(h/2\pi). This can be seen by considering the wave nature of the electron. In order to have a stable electronic orbit, the matter wave of the electron must be stationary around the orbit, as shown in the figure. Therefore, the orbit is an integral multiple of the wavelength (de Broglie wavelength) \lambda = h/mv , 2\pir_n = n\lambda or r_n = n(h/2\pimv) where v is the orbital velocity of the electron. This equation can also be written as r_n = (L_n/mv). The electrostatic force on the electron as it moves in the nth circular orbit, can be written as F = (mv^2/r_n) = k(e^2/r^2_n), ormv^2 r_n = ke^2. This is the centripetal force needed to keep the electron in its cir-cular orbit. Bohr's postulate for the quantization of the orbital angular momentum is L_n = mvr_n = n (h/2\pi). Squaring both sides, we have m^2 v^2 r^2_n = n^2 (h^2/4\pi^2). Now, we form the ratio (m^2 v^2 r^2_n/mv^2 r_n) = n^2 [h^2/(4\pi^2 ke^2)] we getmr_n = n^2 [h^2/(4\pi^2 ke^2)] r_n = [h^2/(4\pi^2 ke^2m)] n^2 The smallest orbit will have a radius r_1 = [h^2/(4\pi^2 ke^2m)] When the numerical values are substituted, we find for the first Bohr Radius r_1 = [{(6.63 × 10^-34 J-sec)^2} / {4\pi^2 × (9 × 10^9m/farad) × (1.6 × 10^-19 coul)^2 × (9.11 × 10^-31kg)}] = 5.29 × 10^-11 m. In Bohr's theory, the electron may revolve only in some one of a number of specified orbits.

The initial velocity of the car is zero. Since there is no friction, the change in potential energy of the car equals its increase in kinetic energy. On the smaller hill of height h, the change in potential energy with respect to the starting point is PE = mg (200 - h) Its kinetic energy is given as kE = (1/2)mv^2 = (1/2)m(32)^2 Equating the two, mg(200 - h)= (1/2)m(32)^2 g(200 - h)= 32(200 - h) = (1/2)(32)^2 200 - h= (1/2)(32) h= 200 - 16 = 184 ft. Therefore in order for the car to have speed 32 ft/sec, the lower hill must be at a height of 184 ft above the lowest point on the track.

This problem can be solved by two methods: the mole and the equivalent methods. The mole method requires consideration of the balanced equation that illustrates the reaction. From the data provided, this equation becomes 5Fe^2+ + MnO^-_4+ 8H_3O^+ \rightarrow 5Fe^3+ + Mn^2+ + 12H_2O. Now 25.0 ml of .40 M FeSO_4 furnishes (.025 liters)(.40 mol / liter) = .010 moles of Fe^2+, since the definition of molarity is molarity is M = (number of moles of solute number of liters) / (number of liters). The balanced equation indicates that the number of moles of MnO^-_4 will be 1/5 that of Fe^2+. As such, the number of moles of MnO_4^- = (.010)(1/5) = .002 moles. Since the KMnO_4 solution has a concentration of .2 M, then the numberof liters required is [.002 mole of MnO^-_4] / [.2 mol/liter)] = .01 liters, which equals 10 ml [.002 mole of MnO^-_4] / [.2 mol/liter)] = .01 liters, which equals 10 ml . The equivalent method functions differently. An equi-valent is defined as that mass of oxidizing or reducing agent that picks up or releases the Avogadro number of electrons. Normality is defined as the number of equi-valents per liter. Since, in going from Fe^2+ to Fe^3+, you lose 1 electron, .40 M FeSO_4 is equal to .40 N FeSO_4. Recalling the definition of normality, you have (.025 liter) (.40 equiv/liter) = .01 equiv. of Fe^2+, the reducing agent. In an oxidation-reduction reaction, the number of equivalents of oxidizing agent must equal that of the reducing agent. This means you must have .01 equiv. of MnO^-_4. You know that for KMnO_4, there exists 1 equiv/liter. Therefore, the number of liters equals (.01equiv.)/(1.0 equiv./liter) = .01 liters or 10 ml.

Question:

Name each of the organic compounds shown in figure A below.

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Solution:

A system of nomenclature called the IUPAC system has been formulated so that all organic molecules may have their structures defined adequately. In this system for hydrocarbons (i.e. compounds containing only hydrogen and carbon atoms) the longest chain is taken as the basic structure, and the carbon atoms are numbered from the end of the chain closest to a branch chain or other modification of simple alkane structure. The position of substituents in the chain are denoted by the number of carbon atom or atoms to which they are attached. In the problem, you are not given alkanes, you are given alkenes, compounds that contain a double bond between a pair of carbon atoms (i.e. unsaturated), and a ring compound. In unsaturated compounds, the rules are the same, except that the position of the double bond is indicated; the numbering starts at the end of the chain nearest the double bond. For rings, name the ring and any substituent present based on which carbon is located. Thus, you proceed as follows: (a) There is only one chain, and it possesses three carbon atoms. The double bond is located on the first carbon, not the second, since you want to use the lowest possible number. As such, you form the molecule propene. Three carbons suggest propane. It's an alkene, so that you change ane to ene. Therefore, propane becomes propene. You have no need to name the position of the double bond here; the double bond can only be in two po-sitions and both yield the same molecule. That is,CH2 = CH - CH_3 is equivalent to CH_3 - CH = CH_2. = CH - CH_3 is equivalent to CH_3 - CH = CH_2. (b) The longest chain containing the double bond has 7 carbon atoms, so that "hep" prefix is suggested. Because it is a double bond, you add the suffix ene to obtain heptene. The double bond is located on the second carbon, not the third, since you want the lowest number. You have, therefore, the 2-heptene. Using this numbering system, you see that CH_3, a methyl group, is located on the fifth carbon. Thus, the name of the molecule is 5-metyl 2 heptene. (c) Has a six carbon chain, which suggests the pre-fix ''hex\textquotedblright. There is a double bond so that the molecule has the suffix ene. Thus, you obtain hexene. The double bond is located on the second carbon, which means the name of the molecule is 2-hexene. (d) This ring compound has a methyl group (CH_3) positioned on a benzene ring. The compound can be called methyl benzene. It is also given the special name of toluene.

The change in a body's gravitational potential energy is the negative of the work done by gravity on the object in displacing it. By definition, this is W = - \intFg^\ding{217} \bullet dr^\ding{217} where Fg^\ding{217} , the force of gravity, is Fg^\ding{217} = -mg(1) The symbolis a unit vector in the positive y direction (see figure). Now dr^\ding{217} = dx+ dy whereis a unit vector in the positive x direction. Then Fg^\ding{217} \bullet dr^\ding{217} = -mg\bullet (dx+ dy) = - mgdy andW = -\int-mhdy(2) We evaluate (2) over the path of motion of the block. If we take the origin of our coordinate system at the foot of the plane, y varies from 0 to 100 ft. Therefore W = mg^100 ft\int_0 dy = mg(100 ft) W = (40 lb)(100 ft) = 4000 ft. lb.

Question:

In addition to its function in secreting female sex hormones, the human ovary serves as the organ of egg production. Explain the process of egg formation in the mammalian ovary. What happens to the follicle after ovulation?

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Solution:

The mammalian ovary (see figure) consists of two types of cells: the germ cells and the germinal epithelial cells. The germ cells are the oogonia, which will eventually become the eggs. The germinal epithelial cells line the walls of the ovary and will eventually form part of the follicle. At birth, the human ovaries contain an estimated 400,000 immature eggs or oogonia. Thus, in marked contrast to the male, the newborn female already has all the germ cells she will ever have: only a few, perhaps 400, are destined to reach full maturity during her reproductive life. The process of maturation of an oogonium is called oogeneis. Its beginning is marked by the enlargement of the oogonium to form a primary oocyte. Following the formation of the primary oocyte, the prospective egg undergoes the first meiotic division. Although the nuclear division proceeds normally, there is unequal division of the cytoplasm, resulting in one large secondary oocyte and a smaller cell made up almost entirely of nuclear material, called the first polar body. The secondary oocyte also divides unequally in the second meiotic division, producing a large ootid and a small second polar body. The first polar body may or may not divide; in any event, the polar bodies all soon disintegrate. The ootid becomes the mature egg or ovum. The single haploid ovum that results from the meiotic divisions of the primary oocyte contains most of the original cytoplasm and also some stored food. During oogenesis, the developing egg becomes in-creasingly surrounded by a layer of germinal epithelial cells, and by the secondary oocyte stage, it is well enveloped by these cells. The oocyte and its surrounding cells are together termed the follicle. After secondary oocyte formation is completed, about two weeks after the onset of oogenesis, the oocyte alone is released from the follicle in a process called ovulation. This involves the rupture of part of the ovarian wall and the subsequent expulsion of the oocyte from the ovary. The secondary oocyte then enters the Fallopian tube, and remains at the secondary oocyte stage until after fertilization. It is important to realize that the ovary is not directly connected to the Fallopian tube; during ovulation the secondary oocyte is expelled into the body cavity. However, because of the close proximity of the ovary and Fallopian tube and the presence of finger-like projections from the mouth of the Fallopian tube which help direct the oocyte into it, almost all oocytes that are ovulated do end up in the Fallopian tube. The follicular cells which stay behind in the ovary following ovulation enlarge greatly and become filled with a lipid substance; the entire gland-like yellowish structure is known as the corpus luteum (meaning "yellow body"). Under hormonal influence, the corpus luteum secretes progesterone and estrogen, two female sex hormones. These hormones stimulate the proliferation and vascularization of the uterine lining, preparing it for implantation of the embryo. If the ovum is fertilized, the corpus luteum enlarges to form a corpus luteum of pregnancy, which increases its hormone production to maintain pregnancy. If the ovum is not fertilized, the corpus luteum begins to degenerate about nine days after ovulation and is called a corpus luteum of menstruation.

Alkenes readily react with a number of oxidizing agents. A test for the presence of an olefin is the reaction with an acidic solution of permanganate ion (MnO-_4 ). The purple color of permanganate ion disappears as the olefin is oxidized. The course of such a reaction is that the olefin cleaves into two oxidized fragments. A carbon atom with two alkyl groups attached is converted to the carbonyl group of ketone, while a carbon atom with one attached hydrogen becomes the carboxyl group of an acid. This is summarized by the following To solve this problem, write the structural formula of each hydrocarbon, find which alkyl groups correspond to R, R', and R" in the general formula, and then substitute into the general formula. The product of this reaction is acetic acid. The product of this reaction is methyl ethyl ketone. The products of this reaction are acetic acid and acetone.

Question:

A radio capacitor consists of a stack of five equally spaced plates each of area 0.01 m^2, the separation between neighbors being 2.0 mm. Calculate the capacit-ance (a) if the top and bottom plates are connected to form one conductor and the center three are connected to form the other, and (b) if the top, center, and bottom plates are connected to form one conductor and the other two plates are connected to form the other conductor. (See figs. (a) and (b)).

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(a)Since the middle three plates are at the same potential, no field exists between them. The capacitor essentially consists of two capacitors of equal capacitance in parallel, one formed from the top two plates, the other from the bottom two plates (see figure (a)). The capacitance of a parallel plate capacitor is C = (\epsilon_0 A/d) where \epsilon_0 is the permittivity of free space, d is the plate separation, and A is the area of 1 capacitor plate. Since the equivalent capacitance of 2 capacitors in parallel is equal to the sum of their individual capacitances, we obtain C_1 = 2\epsilon_0 (A/d) = [(2 × 8.85 × 10^-12 C^2 \bullet N^-1 \bullet m^-2 × 0.01 m^2 )/(0.002 m)] = 8.85 × 10^-11 F. (b)Here a field exists between any pair of plates. Each of the two connected plates acts as the positive plate for each of two capacitors formed between itself and the outer and middle plates. There are effectively four equal capacitors all connected in parallel. Hence C_2 = 4\epsilon_0 (A/d) = 17.7 × 10^-11 F, which is twice the capacitance of the previous arrangement.

i)2110101 + 7+ 00111 2811100 ii)45101101 - 33-100001 12001100 Now to do binary subtraction using 1's complement, take the complement of 33: 100001 = 011110. Next add it to 45 Similarly, for the 2's complement method, take the complement of 33 and add binary one to it. 33 = 100001 binary equivalent 011110\leftarrow 1's complement of it + 1 011111\leftarrow 2's complement Then (+ 45)\equiv101101 +(- 33)011111 121001100= 12 drop last digit iii)- 15 +6 - 9 15 = 1111 binary equivalent 00001's complement + 1 00012's complement 6 = 0110binary equivalent - 1510000 ^++ 6001101's complement - 90110 Now as no carry is generated the answer is negative and is in 1's complement form. Taking back its complement, we get the answer in binary equi-valent form. 0110 = 1001 \equiv 9 the correct answer. - 15- 10001 +6+ 001102's complement - 9- 10111 Again, as no carry is generated at the most significant bit position, the answer is negative and is in 2's complement form. Complement and add binary one to obtain the answer in binary equivalent form. 0111 = 1000 + 1 1001= - 9.

Let R be the extra resistance needed in the circuit. The 50 cells have a total emf of 50 cells × 2.3 V/cell = 115 V and a total internal resistance of 50 cells × 0.1\Omega/cell = 5 \Omega. Let us then represent the battery by an emf source \epsilon_b in series with a resist-ance r. Since the battery is being charged, its polarity will be in the direction opposite to that of the dc supply source. The net emf \epsilon of the circuit is then the difference between the 230 volt dc supply voltage and \epsilon_b = 115 V. The charging current I = 6 A. We apply Ohm's law \epsilon = I(R + r) to obtain (230 - 115)V = 6 A(R + 5\Omega). \thereforeR = [(115 V)/(6 A)] - 5 \Omega = 14.2 \Omega .

Electrical power is the rate at which electrical energy is produced or consumed. It is equal to the product of the current in a device and the fall of potential across its terminals or, for the motor, P = VI = (1.2 × 10^1V)(2 × 10^2A) = 2.4 × 10^3W. Since the motor only has an efficiency of 80%, 20% of the available power goes into losses such as heat. The other 80% is available to the motor to do mechanical work. The mechanical power delivered by the motor, is (2.4 × 10^3W)(8 × ^.10^\rule{1em}{1pt}1) = 1.92 × 10^3W.

Question:

A sample of bristle cone pine wood of age 7,000 \pm 100 years, known by counting growth rings, possesses an activity of 6.6 d(disintegrations) / minute-g of carbon. Calculate the exact age of the wood sample from radiochemical evidence.

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Solution:

Radioactive decay is related to the rate of disintegration and the half-life by the following equation: q_t = q_0 (0.5) q_t = q_0 (0.5) t/(t)1/2 where q_t is the rate of disintegration of the substance con-taining carbon, q_0 is the rate of disintegration of carbon (14 d / min-g), t is the time elapsed during the reaction, and t_1/2 is the half-life of the element concerned. For carbon, t_1/2 = 5730 yrs. Here one needs to solve for t. q_t = q_0 (0.5)^t/t(1/2)q_t= 6.6 d / min - g q_0= 14 d / min - g t= ? t_1/2= 5730 yrs 6.6 d / min - g = 14 d / min - g × (0.5)^t/5730 ^yrs [(6.6 d / min - g) / (14 d / min - g)]= 0.5^t/5730 ^yrs .471= 0.5^t/5730 ^yrs Log .471= (t / 5730 yrs) × log 0.5 - .327 = (t / 5730 yrs) × (- .301) (- .327 / - .301)= (t/ 5730 yrs) t = [(.327 × 5730 yrs) / .301]= 6225.0 yrs. Therefore, the tree is 6,225 years old.

Question:

City A is 100 miles north and 200 miles west of city B. An airplane flies in a direct line between the cities in a time of one hour. What are the vectors that describe the distance of A from B, and the velocity of the airplane?

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Solution:

We will define first a coordinate system with B at the origin (see the figure below). The x-direction is east and the y-direction is north. The vector BA is specified by its coordinates x = -200 mi y = 100 mi or by its magnitude and direction (BA)^2= x^2 + y^2 = [(-200)^2 + (100)^2]mi 2 BA = 100\surd5 mi sin \texttheta = (CA/BA) = (100/100)\surd5\approx(1/\surd5) \approx \texttheta = 26.5\textdegree \textphi = 180\textdegree - \texttheta = 153.5\textdegree The velocities are given in a similar way. Since they are constant V_x = (x/1 hr) = (-200 mi/1 hr) = - 200 mi/hr V_y = (y/1 hr) = (100 mi/1 hr) = 100 mi/hr V^2 = V^2_x + V^2_y = [(-200)^2 + (100)^2]mi^2/hr^2 V = 100\surd5 mi/hr \textphi = 153.5\textdegree

The height of the mercury column of the barometer depends on density \rho and g as well as on the atmospheric pressure. Hence both the density of mercury and the local acceleration of gravity must be known. The density varies with the temperature, and g with the latitude and elevation above sea level. All accurate barometers are provided with a thermometer and with a table or chart from which cor-rections for temperature and elevation can be found. Let us assume g = 980 cm/sec^2 and p = 13.6 gm/cm^3. The pressure due to the atmos-phere supports the weight of mercury in the column of the barometer (see the figure). If the cross sectional area of the column is A, then the weight of mercury in the column is W = mg where m is the mass of the mercury. Since Density (\rho) = mass/volume thenW = \rho(Ah)g where V = Ah, h being the height of mercury in the column. There-fore W/A = \rhogh is the pressure due to the weight of the mercury acting downward. It must equal p_a for equilibrium to be maintained in the fluid, (see figure). Hence p_a = pgh = 13.6 gm/cm^3× 980 cm/sec^2× 76 cm = 1,013,000 dynes/cm^2 (about a million dynes per square centimeter). In British engineering units, 76 cm = 30 in. = 2.5 ft, \rhog = 850 lb/ft^3 p_a = 2120 lb/ft^2 = 14.7 lb/in^2.

Zero offsets for X and Y mean that the tools are at the origin 0 at the start. A Z-offset of 2" means that the tools are 2" above the workpiece in their idle positions. From the data given, obviously, hole #1 requires spindle T01 carrying the 1" drill. We select a slower speed for this large hole, say S02, i.e., 375 rpm. The feed rate is taken as F02, i.e., 1" per min. The hole #2 requires spindle T04 carrying a 1/4" drill. The speed can be faster, say, S03, i.e., 440 rpm., and the spindle feed rate can be F01, i.e., 2". The hole #3 requires spindle T02 first, and then spindle T03 for the tap. While tapping, we will also require a coolant fluid feature, which is obtained by specifying M06. The program for the machining of the workpiece can be written as follows: OP.NO. `G' `X' `Y' `R' `Z' Auxiliaries N001 G80 X02000 Y01000 S02 F02 T01 G82 R01750 Z02250 N002 G80 X04000 Y01000 S01 F03 T04 G81 R01750 Z03500 N003 G80 X06000 Y02000 S02 F04 T02 G81 R01750 Z03500 N004 G80 X06000 Y02000 S03 F03 T03 M06 G84 R01750 Z02500 G80 X00000 Y00000 M30 In step No. 001, the G80 clears any stray instructions lying about in the storage. The spindle T01, holding a 1" drill moves to the X and Y coordinates. The G82 command makes the spindle rotate at the specified speed S and the spindle descends at a rapid rate by 1-75", so that it is still 1/4" above the workpiece. The spindle then descends at a programmed feed rate F02, for a distance of 1/2" (Z-R = 2.25 - 1.75 = 0.5"). In operation #N002, we perform a G80, change the spindle to T04, move to the X and Y specified, and lower the spindle as specified by R and Z. We perform a G81 drilling operation which differs from the G82 in that in G81 the spindle is withdrawn immediately rather than dwell-ing at the end of the drilling as in G82. In N003, we again do a drilling operation, but by us-ing the T02 spindle which carries a 41/64" drill. Note that as Z = 3.5", the drill goes completely through the 1" thick workpiece and comes out 0.5" on the other side. In N004 we keep X and Y same, and so, perform a tapp-ing operation using spindle T03. Note that the M06 starts the coolant during the tapping operation. Finally, the G80 in the last step clears all previous commands, and moves the spindle to the edge of the work-piece so as to facilitate removal of the workpiece. The M30 command signifies an end of the tape on which the program will be punched, and is also a command to the tape reader on which the tape will be later mounted, to rewind the tape to the starting position so that it is ready to be re-run for the machining of the next workpiece.

A taxis is a directed type of orientation in which an animal moves toward or away from a stimulus. But exactly how does the animal detect which direction to move in relation to the stimulus? One simple way is by testing the intensity of the stimulus at different short intervals of time: The animal simply takes momen-tary readings of the stimulus and directs itself toward the strongest reading. If it accidentally moves away from the source, it changes direction until a higher intensity of the stimulus is sensed. For example, dogs locate the source of an odor by repeatedly sniffing as they move. They periodically change their direction so as to continue smelling a high intensity of the odor. A second, more complex taxis guidance system is the instantaneous reading of the stimulus using two receptor organs. Many animals move in response to a stimulus by positioning themselves so that equal intensities of the stimulus are sensed by the two organs. They are then precisely aligned to move either toward or away from the stimulus, depending on whether they are attracted or repelled by it. For example, an animal can orient toward a light source by moving its body, or simply its head, until an equal intensity of illumination is sensed in both eyes. This quidance system can be demonstrated by covering one eye, for instance, the right eye, of a pos-itively phototactic animal. Since the animal senses all the light in its left eye, it moves toward the left. However, its next instantaneous reading directs the animal to the left again, since a balance of illumination was made impossible. This causes the animal to turn to the left indefinitely, resulting in movement in conterclockwise circles! Another example is the phototaxis of a planarian. If two equally bright lights are located at equal distances from the worm, it will move toward a pt. midway between the two lights (see Figure 1) The orientation of the honeybee toward an odor is also one involuing two simultaneous readings of a stimuli. The honeybee tests the strength of an odor simultaneously with both antennae, and then turns in the direction of the antenna that sensed the stronger odor. If an investigator crosses the antennae and glues them into this position, the bee reads the odor as coming from the opposite direction. Should such a bee be placed in a Y-maze with an odor source in one channel, the bee will detect the source as originating from the other channel and move in that direction. (See Figure 2)

Question:

How much energy in electron volts must be expended to separate the atoms in the potassium iodide molecule, 39_k 127_I, if the ions are originally separated by a distance R = 3.04 × 10^\rule{1em}{1pt}10 m? The potential energy for coulomb's law of force is, E_P = \rule{1em}{1pt}ke^2 / d where e is the electric charge of each ion and d is the distance between the charges. K = 9 × 10^9 N \textbullet m^2 / c^2

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Solution:

When the ions are bonded together their separation distance is R, so the potential energy is, E_P(R) = \rule{1em}{1pt}ke^2 /R = \rule{1em}{1pt}{[{9 × 109N(m^2/ C^2)}(1.6 × 10^\rule{1em}{1pt}19c) 2 ] / (3.04 × 10^\rule{1em}{1pt}10m)} =\rule{1em}{1pt}7.6 × 10^\rule{1em}{1pt}19 J When the ions are very far apart, the separation distance d becomes very large and may be assumed to be infinite. The potential energy for infinite separation distance is zero: E_P (\infty) = \rule{1em}{1pt}ke^2 / \infty = 0 The binding energy of the molecule is the difference between these two potential energies (the amount of energy needed to breakup the molecule). Binding energy = Ep (\infty) \rule{1em}{1pt}E_P(R) = 7.5 × 10^\rule{1em}{1pt}19 J Since I eV = 1.6 × 10^\rule{1em}{1pt}19 J, the energy to separate the atoms may be written, (7.6 × 10^\rule{1em}{1pt}19J)/ (1.6 × 10^\rule{1em}{1pt}19J/ eV) =4.8 eV

Question:

Explain the terms primary, secondary, and tertiary in regards to covalent bonding in organic compounds.

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Solution:

If a carbon atom is bound to only one other carbon atom, then the former carbon atom is called primary. If a carbon atom is bonded to two other carbon atoms, then that carbon atom is called secondary. If a carbon atom is bonded to three other carbon atoms, then that carbon atom is called tertiary. Any group that is attached to a primary, secondary, or tertiary carbon is called a primary, secondary, or tertiary group. For example, Carbons 1, 2, and 5 are primary, carbon 4 is secondary, and carbon 3 is tertiary. The hydrogen atoms attached to carbons 1, 2, and 5 are called primary hydrogens, those attached to carbon 4 are called second-ary hydrogens, and those attached to carbon 3 are called tertiary hydrogens. This same principle applies to alcohols, and depending upon where the hydroxyl group (-OH) is attached (that is, primary, secondary, or tertiary carbon), the alcohol is called primary, secondary, or tertiary, respectively.

Question:

Are all antibodies structurally and functionally similar? Explain.

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Solution:

All antibodies are proteins and are referred to as immunoglobulins. Five major classes of immunoglobulins can be distinguished based on their character-istic structures and separate functional roles: IgG, IgM, IgA, IgD and IgE. IgG immunoglobulins constitute about 80% of all the antibodies in the blood serum. These are the major anti-bodies involved in the destruction of invading micro-organisms. IgM immunoglobulins appear first in response to an antigen; they are later replaced by IgG. IgA im immunoglobulins are found in secretions such as saliva and tears and are involved in the defense of body cavities such as the mouth and vagina. The function of IgD is still unknown, while IgE is thought to be involved in allergic reactions. There are thus five basic types of immunoglobulins. But we are certainly exposed to more than five different antigens. What distinguishes one immunoglobulin molecule from the next? Let us look at the structure of an IgG molecule. It is composed of four polypeptide chains; two heavy chains (mol. wt. \sim 53,000) and two light chains (mol. wt. \sim 22,500). In a given IgG molecule, both heavy chains are identical and both light chains are identical. The polypeptide chains are held together by disulfide (\rule{1em}{1pt}S\rule{1em}{1pt}S\rule{1em}{1pt}) linkages to form a Y-shaped molecule: The ends of each arm of the IgG molecule contain an active site which can combine with a specific antigen. Thus, two identical binding sites permit each antibody molecule to form a complex with two antigens. Both light and heavy chains contain two regions: a "constant" region which has very similar amino acid sequences in all immuno-globulins, and a "variable" region which has different amino acid sequences for each type of antibody. This variability accounts for the specificity of antibodies- that is, one particular antibody molecule can bind only one particular kind of antigen. The specific sequence of amino acids in the variable region at the end of each arm of the antibody molecule determines the spatial configuration of the active site which will only bind one particular antigen.

In naming complex open chain organic compounds, first find the longest continuous chain containing the functional groups. Write down the parent name. The parent for this particular compound is heptene. Number the chain starting from the end that will give the smallest prefix numbers for the functional groups. This is shown in figure B. Add the suffix functionality with the appropriate numbering: 4- hepten-2-ol. Add the prefix functionality, remembering to group together like prefixes. Then, double check to make sure a substituent has not been forgotten or one substituent has not been included twice. By following these steps, one arrives at the name for the structure: 4-n-butyl-3, 5-diethyl-4-hepten-2-ol.

Question:

Calculate themolarityof a solution containing 10.0 grams of sulfuricacid in 500 ml of solution. (MW of H_2SO_4 = 98.1.)

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Solution:

Themolarityof a compound in a solution is defined as the number ofmoles of the compound in one liter of the solution. In this problem, one istold that there are 10.0 grams of H_2SO_4, present. One should first calculatethe number of moles that 10.0 g represents. This can be done by dividing10.0 g by the molecular weight of H_2SO_4 . numberof moles= (amount present in grams/molecular weight) numberof moles of H_2SO_4 = [10.0 g/(98.1 g/mole)] = 0.102 moles Sincemolarityis defined as the number of moles in one liter of solution, and since, one is told that there is 0.102molesin 500 ml (1/2 of a liter), one should multiply the number of moles present by 2. This determinesthe number of moles in H_2SO_4 present in 1000 ml. Number of moles in 1000 ml = 2 × 0.102 = 0.204. Becausemolarityis defined as the number of moles in 1 liter, the molarity(M) here is 0.204 M.

Question:

(a) What force T, at an angle of 30\textdegree above the horizontal is required to drag a block of weight w = 20 lb to the right at constant speed along a level surface if the co-efficient of sliding friction between block and surface is 0.20? (b)Determine the line of action of the normal force N\ding{217} exerted on the block by the surface. The block is 1 ft high, 2 ft long, and its center of gravity is at its center.

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Solution:

(a) There can be no net vertical force because there is no accelerated motion upwards; similarly, there is no net horizontal force because the block moves with constant velocity. If T is the rope tension, N, the normal force, and f_k the force of friction, we have using Newton's Second Law, \sumF_x = T cos 30\textdegree - f_k = 0, \sumF_y = T sin 30\textdegree + N - 20 lb = 0.(1) A body may experience translational equilibrium without experiencing rotational equilibrium. The general condition for rotational equilibrium is that the sum of torques taken about the center of mass of the body be zero. However, if no net external force acts, this con-dition is less stringent. The condition in this case is that the sum of torques taken about any point (i.e. C) be zero. Let x represent the distance from point 0 (see diagram) to the line of action of N\ding{217} , and take moments about an axis through 0. Then, from the condition of rotational equilibrium, the net torque about 0 must be zero, or T sin 30\textdegree × 2 ft - T cos 30\textdegree × 1 ft + N × x - 20 lb × 1 ft = 0.(2) We may now solve for N. From (1) and (2) T sin 30\textdegree = 20 lb - N(3) T cos 30\textdegree = f_k.(4) (2 ft) (T sin 30\textdegree) - (1 ft) (T cos 30\textdegree) + (X) (N) = 20 ft \bullet lb(5) But f_k = \mu_kN, where \mu_k is the coefficient of static fraction, Hence T cos 30\textdegree = \mu_kN = .2 N(5) Dividing (3) by(5) tan 30\textdegree = (20 lb - N)/(.2 N) \surd3/3= (20 lb - n)/(.2 N) [(.2\surd3)/3] N + N =20 lb N = (20 lb)/[{(.2\surd3)/3} + 1] = 17. Lb(6) Substituting (6) into (5), we obtain T(\surd3/2) = .2(17.9 lb) orT =(.4/\surd3) (17.9 lb) = 4.15 lb(7) Using (6) and (7) in (5) (2 ft)(4.15 lb)(1/2) - (1 ft)(4.15 lb)(\surd3/2) + x(17.9 lb) = 20 ft \bullet lb x(17.9 lb) = (20 ft \textbullet lb) - (4.15 ft \textbullet lb) + (3.6 ft \textbullet lb) x = (20 ft \bullet lb - 4.15 ft \bullet lb + 3.6 ft \bullet lb)/(17.9 lb) x = 1.08 ft Therefore, the line of action of N must be .08 ft to the right of the center of mass if the block is to maintain its rotational equilibrium.

The extremely thin hairlike appendages that protrude through the cell wall of some bacterial cells are called flagella. The length of a flagellum is usually several times the length of the cell, but the diameter of a flagellum is only a fraction of the diameter (the width) of the cell. Baccilli and spirilla usually have flagella, which are rarely found on cocci. Flagella occur singly, in tufts, or equally distributed around the periphery of the cell. Flagella are responsible for the motility of bacteria. Not all bacteria have flagella, and thus not all are motile. The mechanism by which the flagella move the bacterial cell is not completely understood; however, it is proposed that movement requires rotation of the semi-stiff flagellum. Whatever the mechanism, the flagellum moves the cell at a very high speed and it enables the bacterium to travel many times its length per second. Bacterial flagella are structurally different from the flagella Qfi eucaryotic cells. This difference can be used to distinguish the bacterial (procaryotic) flagella from the protozoan (eucaryotic) flagella. By examining a cross section of the flagellum under an electron microscope, one can conclusively determine its identity. The eucaryotic flagellum contains a cytoplasmic matrix, with ten groups of tubular fibrils embedded in the matrix: nine pairs of fibrils around the periphery of the eucaryotic flagellum surround two central single fibrils. (see below). Bacterial flagella consist of only a single fibril and lack the ("9 + 2") structural organization of the eucaryotic flagellum. Most bacterial flagella have the same diameter as a single fibril from a eucaryotic flagellum and hence they are usually thinner.

Question:

Show how i ) the compression function ii) the expansion function iii) Take and Drop functions are used in APL.

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Solution:

Assume the following matrix has been stored in memory \mid61220144153\mid \mid282284441370\mid M =\mid4961451184169474\mid \mid1729405851210961208\mid i) Compression is a dyadic function used to select rows or columns from a given matrix. Its general form is L/ [K]M. L is a binary vector composed of ones and zeros. M is a matrix, K is an index indicating which dimension of M is to be compressed and / separates the two arguments. Suppose it is desired to select the first and third rows only of (1). The function 1 0 1 0 / [1]M gives 61220144153 4961451184169474 The K index indicates the dimension of compression, i.e., [1] = row compression and [2] = column compression. The command 0 1 0 1/[2]M yields, when applied to (1) : 2441 145169 40585961 (The first column here represents all numbers between 1 and 50,000 that are equal to the sum of the factorials of their digits. For example, 145 = 1! + 4! + 5!. The second column gives two pairs of numbers that are the reverse of one another and whose product is a square.) ii) Expansion is a dyadic function used to insert rows and columns of zeros into numeric matrices and rows and columns of blanks into character matrices. Its general form is L/ [K]M LetG = B S H A H T W A D E Q B A R T This is a matrix of names; the first character is the initi-al and the remainder is the last name. To make M more legible, perform the function: 1 0 1111\textbackslash [2]G, which yields B S H A H T W A D E Q B A R T iii) Take and drop: These are dyadic functions used for selecting a submatrix from a given matrix. Their gen-eral form is L+M and L+M. L is a vector containing two components. The first component represents the number of rows to be taken or dropped while the second component gives the number of columns taken or dropped. Suppose we want to select Ramajunam's number (1729) from M. This is a 1×1 sub-matrix. The command 3 - 4\downarrowM causes the top three rows (3) and the right four columns (-4) to be dropped (\downarrow). The result is 1729.

Question:

Given the setup in the figure, what would be the pressure of the gas (in atm) if P_atm is 745 Torr and P_1iq equivalent of a mercury column 3.0 cm high?

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Solution:

A manometer is used to measure the pressure of a trapped sample of gas. If the right hand tube is open to the atmosphere, the pressure which is exerted in the right-hand surface is atmospheric pressure, P_atm. If the liquid level is the same in both arms of the tube, the pressures must be equal; otherwise, there would be a flow of liquid from one arm to the other. At the level indicated by the dashed line in the figure, the pressure in the left arm is equal to the pressure of the trapped gas, P_gas , plus the pressure of the column of liquid above the dashed line, P_1iq . One can write p_gas = p_atm - p_1iq Here, one is given that P_atm . is 745 Torr and that P_1iq is equivalent to a mercury column 3.0 cm high. One wishes to find P_gas in atm, thus, one must convert 745 Torr to atm and find the P_1iq . There are 760 Torr in 1 atm, which means no. of atm in P_atm = [(745 Torr) / (760 Torr/1 atm)] = .98 atm One atmosphere pressure supports 76 cm of mercury, thus 3 cm of mercury supports 3/76 atm. Therefore, P_1iq = 3 cm/76 cm/atm = .039 atm. Solving for P_gas P_gas = .980 atm - .039 atm = .941 atm.

Question:

The passenger compartment of a jet transport is essentially a cylindrical tube of diameter 3 m and length 20 m. It is lined with 3 cm of insulating material of thermal conductivity 10^-4cal \bullet cm^-1 \bullet C deg^-1 \bullet s^-1, and must be maintained at 20\textdegreeC for passenger comfort, although the average outside temperature is - 30\textdegreeC at its operating height. What rate of heating is required in the compartment neglecting the end effects?

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Solution:

The hull of the aircraft is a good conductor of heat and may be considered to be at the outside temperature. The circular cylinder of insulat-ing material has thus a temperature of 20\textdegreeC inside and - 30\textdegreeC outside (see figure). Consider a cylinder of the material at distance R from the center of the craft and of thickness dR. By the normal equation of conductivity, the flow of heat across this infinitesimal cylinder per second is H = KA (dt/dR) where dt/dR is the variation of temperature with R (the temperature gradient) and A is the area of the surface across which the heat flow occurs. Hence A = 2\pi R L andH = 2\pi R L K dt/dR ThereforedR = (2\pi R L K dt)/H where L is the length of the cylinder. Thus dR/R = (2\piKL/H) dt. The quantity H is a constant since the system is in equilibrium. If the same quantity of heat did not pass over every cross section of the insulating mater-ial, heat would build up somewhere and the temperature would rise. This is contrary to the condition that equilibrium shall have been attained. For the whole cylinder of insulating material, then ^R2\int_R1 dR/R =(2\piKL/H) ^t2\int_t1 dt where the temperature at R_1 is t_1, and the temperature at R_2 = t_2 (see where the temperature at R_1 is t_1, and the temperature at R_2 = t_2 (see figure). Then ln (R_2/R_1) = (2\piKL/H)(t_2 - t_1) Hence H = [(2\piKL/H)(t_2 - t_1)]/[ln (R_2/R_1)] = {2\pi × 10^-4 cal \bullet cm^-1 \bullet Cdeg^-1 \bullet s^-1 × 20 × 10^2 cm[20\rule{1em}{1pt} (-30)]\textdegreeC}/{1n (300/294)} = 3100 cal \bullet s^-1 = 12,980 J\bullets^-1 \approx 13.0 kW. Here we have used the fact that 1 cal/s = 4.186 J/s This is the heat which flows through the walls. In order to keep the temperature of the cabin constant, an equal amount of heat power must be supplied to the cabin by external means.

The expansion of air during pumping is an isothermal (constant temperature) process since, as a result of thermal contact with the surroundings, the air in the vessel will have the same temperature as that of the air outside. The process is shown in Fig. A, where expansion proceeds from the initial state A to the final state B. It is governed by the ideal gas law PV = nRT_1 = Constant. If the volume is increased by an infinitesimal amount dV; then the pressure change can be obtained from d(PV) = 0 PdV + VdP = 0, ordP/P = - dV/V.(1) The volume changes by an amount v during one cycle. This The volume changes by an amount v during one cycle. This corresponds to a decrease ∆P in the pressure of the vessel. Integrating (1) from initial to final parameters of the system during one cycle, we get ^Pi-∆P\int_ Pi dP/P = -^V+v\intVdV/V or1n P ]_PiPi - ∆P= - 1n V]V^V+v 1n [(P_i - ∆P)/P_i] = - 1n [(V + v)/v](2) When the next cycle starts, the pressure of the vessel goes down to P_i - ∆P, but the volume of the vessel is again V. Therefore at the end of the second cycle the pressure becomes P_i - 2∆P and (2) for this case is 1n[(P_i - 2∆P)/(P_i - ∆P)] = - 1n[(V + v)/V](3) Rearranging (3), we have 1n {[(P_i - 2∆P)/P_i]\bullet[P_i/(P_i - ∆P)]}= - 1n [(V + v)/V] 1n [(P_i - 2∆P)/P_i]} - {1n [(P_i - ∆P)/P_i]} = - 1n [(V + v)/V] 1n [(P_i - 2∆P)/P_i] = {- 1n [(V+v)/V]} + {1n [(P_i - ∆P)/P_i]}.(4) Now, we substitute (2) in (4), 1n [(P_i - 2∆P)/P_i] = {- 1n [(V + v)/V]} - {1n[(V + v)/V]} = - 2 1n[(V + v)/V] If we repeat the cycle n times; 1n [(P_i - n∆P)/P_i] = - n 1n[(V + v)/V]. If P_i is the original pressure P_1 of the vessel, then (P_1 - n∆P) is the final pressure P_2, 1n P_2/P_1 = - n 1n[(V + v)/V]. Thus, the number of pumping cycles required to achieve the pressure P_2 is n = [1n p_1/p_2] / [1n (V + v)/V]

Applying Boyle's law to the first part of the problem, we obtain p_0V_0 = p × V_f = p × (5/6) V_0 where p_0 and V_0 are the initial pressure and volume of the air inside the diving bell, p and V_f are the final pressure and volume of the air in the bell. p = (6/5) p_0 Now, the pressure change that the air experiences can only be due to a change in pressure of the water with depth. The relation between pressure (p) and depth (h) in a fluid is given by p = p_0 + \rhogh(1) where p_0 is the ambient pressure at the surface of the fluid (see figure) and \rho is the fluid density. Now, the change in pressure caused by submerging the bell a distance h in the water is p - p_0 = \rhogh(2) But, the change in pressure experienced by the air in the bell is ∆p = (6/5) p_0 - p_0 = (1/5) p_0(3) and this must be caused by the submerging of the bell a distance h in the fluid. Therefore, using (3) and (2) (1/5) p_0 = \rhogh orh = p_0/5\rhog(4) The ambient pressure is given as 30 in. of mercury. This is to be interpreted as meaning that the ambient pressure is equal to the pressure exerted by a column of mercury 30 in. long. Thus p_0 = (30 in.) \rho' g(5) where \rho' is mercury's density. Thus, inserting (5) in (4) h = [(30 in) \rho' g]/[5 \rhog] = [(30 in) \rho']/[5 \rho] Since the relative density of mercury is defined as R = \rho'/ \rho we obtain h = [(30 in) R]/5 = (6 in.) R h = [(30 in) R]/5 = (6 in.) R But R = 13.6, and h = (6 in)(13.6) = (1/2 ft)(13.6) h = 6.8 ft. The water level in the bell is thus at a depth of 6.80 ft below the surface of the fluid. But 1/6 of the bell's volume, V_0, is water. Noting that V_0 = A × 8 ft where A is the bell's cross-sectional area, we find that (1/6)V_0 = [(8 ft) / 6 (1/6)V_0 = [(8 ft) / 6 ] × A is water. Hence, the height of water in the bell is (8/6) ft. (See figure). The depth of the foot of the bell is thus 6.80 ft + 8/6 ft = 8.13 ft. The bell has thus been lowered 8.13 ft into the water. (b) If air filled the whole jar at this depth, the pressure on it would be that due to atmospheric pressure plus 8.13 ft of water. Using (1) p = p_0 + \rhog h p = 1 atm + (1.95 sl/ft^3) (32 ft/s^2) (8.13 ft) p = 1 atm + (62.4 lb/ft^3)(8.13 ft) p = 1 atm + 507.312 lb/ft^2 Since atmospheres can't be added to lb/ft^2, we note that 1 lb/ft^2 = 4.725 × 10^-4 atm Thenp = 1 atm + (507.312 lb/ft^2) (4.725 × 10^-4 atm/lb/ft^2) p = 1.239 atm. The pressure acting on the air in the bell is thus 1.239 atm. The bell's volume, V_0, is constant at V_0 = 8 ft × A = 8 ft × 100 ft^2 V_0 = 800 ft^3 The number of moles in this volume is obtained from the gas equation p V_0 = n R T where R is the gas constant, T is the gas temperature in Kelvin degrees, and n is the number of moles of gas. Hence, at its submerged position n = pV_0/RT = [(1.239 atm) (800 ft^2)]/[.0821 {(liter\bulletatm)/(mole\bullet\textdegreeK)}(283\textdegreek)] where we used the fact that 10\textdegree C = (273 + 10)\textdegreeK = 283\textdegreeK In order to be consistent, we transform 800 ft^3 to liters by noting that 1 ft^3 = 28.32 liters. Then n = (pV_0)/(RT) = [(1.239 atm × 800 × 28.32 liters)/(0.0821 liter\textbulletatm\bulletmole^-1 \bulletK deg^-1 × 283\textdegreeK)] = 1208 moles. On the surface of the water the number of moles in the diving bell was n_0 = p_0V_0/RT = [(1 atm × 800 × 28.32 liters)/(0.0821 liter\textbulletatm\bulletmole^-1 \bullet k deg^-1 × 283\textdegreeK)] = 975 moles. The number of moles that have passed through the pump is thus 1208 - 975 = 233.

The electric field ,E^\ding{217}, experienced by the electron is radial and has a magnitude of E = e / r^2 where e is the electronic charge of a proton, and r is the dist-ance between electron and proton. E =(4.8 × 10^\rule{1em}{1pt}10 statC) / (0.53 × 10^\rule{1em}{1pt}8 cm)^2 = 1.7 X 10^7 statV / cm which is a very large field indeed. Sparking usually occurs in air when a field strength of 100 statV/cm is reached. The electric poten-tial of the electron is \varphi_E = e / r = (4.8 × 10\rule{1em}{1pt}10 statC) / (0.53 × 10^\rule{1em}{1pt}8 cm) = 0.09 statV which is rather small. The potential difference between the terminals of a flashlight is 1.5 V or 0.005 statV. The electric field E depends on 1/r^2 , whereas the potential \varphi_E depends on 1/r; since r is extremely small (0.5 X 10^\rule{1em}{1pt}8 cm) in the case of the hydrogen atom, the field strength is large while the potential is small.

The structure of the IgG molecule has al-ready been visualized in the electron microscope. It is a Y-shaped molecule with one binding site, or valency, at the end of each arm. Each IgG molecule is thus divalent, having two binding sites capable of combining with two antigens. Each antigen molecule also has more than one binding site; that is, it is divalent or multivalent. The reason that both antibodies and antigens are multivalent is to permit a high degree of agglutination (clumping). This allows precipitation of the antibody- antigen complex to occur. If only one binding site existed on each antibody and antigen molecule, then only one molecule of antibody could bind to one molecule of antigen: Precipitation would not occur if the complex were this small. Even if the antibody were divalent, as we know to be so, complex size would still be limited to 3 molecules if the antigen were univalent: Therefore, the antigen molecule must also be at least divalent for there to be extensive bridging of antigenantibody complexes. This large complex of antigens and antibodies, called a lattice, is now able to settle out of the body fluid. It then tends to be engulfed and destroyed by macrophages.

In figure 1, both particles are moving towards each other with the same velocity in our frame of reference. In figure 2 they have collided and stuck together. The velocity of the joined body would seem to be zero, but we must prove that this is so. By the law of conservation of momentum, the total momentum before and after collision must be conserved. We are told that the velocities of the two particles are equal but opposite and we assume that they are collinear so that the problem is one dimensional making the velocities v and - v respectively. There-fore, the momentum before the collision is mv + m(- v). Since this is equal to 0, the momentum after the collision must also equal 0, or m_final v_final = 0. Since no energy is produced in the collision, no mass can be lost (it can only be gained) so that m_final isnot 0, meaning that v_final is 0. Further, we note that the two particles had total kinetic energy 2 \textbullet 1/2 mv^2 before the collision, and 0 kinetic energy after the collision. Therefore, all this energy must have been converted to heat. Since E = mc^2 in the rest frame, this heat energy has an equivalent mass given by \DeltaE = \DeltaMc^2. Hence, \Deltam =(mv^2 - 0) / c^2 \Deltam = { 1gm (105cm/sec )^2 } / ( 3 ×105cm/sec )^2 } = 1.1 × 10^-11 gm

Question:

Write a program in FORTRAN to prepare a payroll for a business employing N workers. The following computations must be performed: 1) regular hours and overtime must be totaled. 2) wages equal to regular rate times regular hours plus overtime rate times overtime hours. 3) Deduc-tions must be made for insurance taxes, and government income tax. 4) Net wages are printed for each worker. Assume that tax schedules can be accessed from secondary memory.

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In preparing a payroll register, the first step is collecting and processing the time cards. The items 1) - 3) are entered by reading a card. This infor-mation is then used in calculating gross and net pay. Since the wages values require the accuracy of only two decimal places (cents), while the FORTRAN calculations are much more accurate, the rounding should be made by asking for only two places in the FORMAT statement. If the computer uses truncation as the approximation routine, .005 should be added to each value before truncating. To calculate the taxes, a table look-up routine must be performed. Assume that Table 1 has already been loaded into an array. Each time a card is read it will give the salary and number of exemptions. 10FORMAT (F 6.2, I2) 15DIMENSION A (28, 8) CREAD SALARY AND EXEMPTIONS 20READ (3, 10) S,E CEX INDICATES WHERE THE TAX WILL BE EX = E + 2 CC INDICATES ROW OP THE TAX C = 1 CR IS THE LOWEST TAXABLE SALARY R = 80 CTAX NOT IN TABLE IF (S.LT.78) GO TO 140 CCHECKING FOR PROPER ROW 100IF (S.LT.R) GO TO 110 CADVANCING TO NEXT SALARY STEP R = R + 2 CADVANCING TO NEXT ROW IN TABLE C = C + 1 GO TO 100 CPRINTS SALARY AND TAX 110WRITE (4,120) S, A (C,EX) 120FORMAT (F6.2, 5X, F 5.2) GO TO 20 140WRITE (4,150) 150FORMAT ("TAX NOT IN TABLE") GO TO 20 STOP END Note that FORTRAN is not the best choice of language for this type of problem as evidenced by the numerous uncon-ditional GO TO statements scattered throughout the program. If the program had been written in COBOL, the structure would have been easier to follow. Married personWeekly payroll period The wages are the number of withholding exemptions claimed. At Least But less than 0 1 2 3 4 5 6 The amount of tax to be withheld shall be. $ 78 80 82 84 86 88 90 92 94 96 98 100 105 110 115 120 125 130 135 140 145 150 160 170 180 190 200 $80 82 84 86 88 90 92 94 96 98 100 105 110 115 120 125 130 135 140 145 150 160 170 180 190 200 210 $11.10 11.40 11.70 12.00 12.30 12.70 13.00 13.30 13.70 14.00 14.40 15.00 15.80 16.70 17.50 18.40 19.20 20.10 20.90 21.80 22.60 23.90 25.60 27.50 29.50 31.50 33.50 $9.10 9.40 9.70 10.00 10.30 10.60 10.90 11.20 11.50 11.80 12.10 12.70 13.50 14.40 15.20 16.10 16.90 17.80 18.60 19.50 20.30 21.60 23.30 25.00 26.80 28.80 30.80 $7.00 7.30 7.60 7.90 8.20 8.50 8.80 9.10 9.40 9.70 10.00 10.60 11.30 12.10 12.90 13.80 14.60 15.50 16.30 17.20 18.00 19.30 21.00 22.70 24.40 26.10 28.10 $5.00 5.30 5.60 5.90 6.20 6.50 6.80 7.10 7.40 7.70 8.00 8.50 9.30 10.00 10.80 11.50 12.30 13.20 14.00 14.90 15.70 17.00 18.70 20.40 22.10 23.80 25.50 $3.00 3.40 3.60 3.90 4.20 4.50 4.80 5.10 5.40 5.70 6.00 6.50 7.30 8.00 8.80 9.50 10.30 11.00 11.80 12.60 13.50 14.70 16.40 18.10 19.80 21.50 23.20 $1.10 1.40 1.70 1.90 2.20 2.50 2.80 3.10 3.40 3.70 4.00 4.50 5.30 6.00 6.80 7.50 8.30 9.00 9.80 10.50 11.30 12.40 14.10 15.80 17.50 19.20 20.90 $0.00 0.00 0.00 0.10 0.30 0.60 0.90 1.20 1.50 1.70 2.00 2.50 3.20 4.00 4.70 5.50 6.20 7.00 7.70 8.50 9.20 10.40 11.90 13.60 15.30 17.00 18.70

The forces acting on the yo-yo are the horizontal pull F and the frictional force f f = \muW where \mu is the coefficient of friction. Instantaneous rotation takes place about an axis through the point of contact P (not about the center of the yo-yo, although it might appear so) since the instan-taneous velocity of the contact point is zero. Therefore, the yo-yo rolls in the direction of the pull and its rota-tion is determined by the torque about P, \cyrchar\cyrt = Fh. It should be observed that the frictional force does not contribute to this torque, since f is acting at P.

An engine works by accepting heat at a high temperature, converting part of it into work, and rejecting the remainder at a lower temperature. A re-frigerating machine works the opposite way. By supply-ing to it an amount of work W, it will accept heat Q_1 at a low temperature and reject heat Q_2 = Q_1 + W at a higher temperature. The efficiency \eta of a reversible heat engine is given by \eta = W/Q_1 = (Q_1 - Q_2)/Q_1 = (T_1 - T_2)/T1 where W is the net work done by the machine in one cycle and Q_1 is the heat absorbed in one cycle. Since a refrigerator and a heat engine operate in opposite ways, and Q_2 > Q_1, its efficiency can be defined as \eta = W/Q_2 = (Q_2 - Q_1)/Q_2 = (T_2 - T_1)/T_2 = 1 - (T_1/T_2) where W is the work done on the machine. The sequence of events is shown in the diagram. The machine operates between reservoirs at 280\textdegreeK. and 320\textdegreeK. If Q_1 = 5000 joules, we have \eta= 1 - (280/320) = 0.125 = (Q_2 - Q_1)/Q_2 = (Q_2 - 5000 J)/Q_2 Q_2= 5715 J The work done can be found from \eta = W/Q_2 orW = \etaQ_2 = 0.125 × 5715 J \approx 715 J Irreversible transfers of heat occur when the entropy of the system increases. This occurs for a transfer of heat from a high temperature reservoir to a low temperature reservoir. In this problem, there are two irreversible transfers of heat, one between the inside of the house and the low temperature reservoir, for which entropy,∆S= Q_1[{1/(Texp coil)} - {1/(T_inside)}] = Q_1[(1/280) - (1/290)] = 0.62 J/\textdegreeK. and one between the high temperature reservoir and the outside: ∆S= Q_2[{1/(T_out)} - {1/(T_cylinder)}] = Q_2[(1/305) - (1/320)] = 0.88 J/\textdegreeK.

Question:

The lymphatic system in man constitutes an extensive network of thin vessels resembling veins. What are the functions of the lymphatic system?

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The lymphatic are not part of the circulatory system per se, but constitute a one-way route from interstitial fluid to the blood. The lymphatic system in man constitutes an extensive network of thin-walled vessels resembling the veins. These vessels ultimately drain into veins in the lower neck. One of the function of the lymphatic system is to transfer excess interstitial fluid back to the blood. There is a net movement of plasma out of the capillaries and into the tissues; the lymphatics restore the blood volume by returning this fluid that has filtered out. In addition, the lymphatic system serves to return proteins to the blood. Since the capillaries have a slight permeability to plasma proteins, and since the concentration of proteins in the blood plasma is greater than that in the interstitial fluid, there is a small but steady loss of protein from the blood into the interstitial fluid. This protein returns to the blood via the lymphatics. Should there be a malfunction in the lymphatic system, the interstitial fluid protein concentration would increase to that of the plasma. This eliminates the protein concentration difference between the plasma and the interstitial fluid, thus eliminating the plasma colloid osmotic pressure. Only blood pressure and interstitial fluid pressure remain, and permit the net movement of fluid out of the capillary into the interstitial space (edema). Another function of the lymphatic system is to provide the pathway by which fat and other substance absorbed from the gut reach the blood. It is also believed that certain high-molecular weight hormones reach the blood via the lymphatics. In addition to its function in transport, the lymphatic system plays a critical role in the body's defense mechanism against disease. The lymph nodes, which are found at that junctions of lymph vessels, act as filters and are sites of formation of certain types of white blood cells. Lymph, the fluid in the lymph capillaries, flows slowly through the nodes where invading bacteria are phagocytosed by the cells of the lymph node. Indigestible particles such as dust and soot, which the phagocytic cells cannot destroy, are stored in the nodes. Since the nodes are particularly active during an infection, they often become swollen and sore, as the lymph nodes at the base of the jaw are apt to become during a throat infection. It should be noted that since the lymphatic system is not connected to the arterial protion of the blood circulatory system, lymph is not moved by the hydrostatic pressure developed by the heart. Lymph flow, like that in the veins, depends primarily upon forces external to the vessels. These forces include the contractile action of the skeletal muscle (through which the lymphatics flow) and the effects of respiration on the pressure in the chest cavity. Since the lymphatic have valves similar to those in veins, external pressure would permit only unidirectional flow.

Question:

a) log_1010b) log_10100c) log_101d) log_10 0.1e) log_10 0.01

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Question:

Translate the following flowchart into a pseudocoded program and a BASIC program:

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You probably don't know what this flowchart means at this point. Actually, if you can understand the flowchart symbols alone, you could render the flowchart into pseudocode without understanding the meaning of the problem. This is not advisable, but sometimes programmers are asked to write programs without being given the theory behind the algorithm. If you are familiar with the three basic control constructs - sequencing, looping, decision-making - you can design programs to execute many kinds of tasks. In this example, the logic is straightforward, but the idea behind it may not be so. This flowchart illustrates a program that determines how many of each part is needed for a specific purpose. The process uses the term explosion matrix to find the expense of materials and labor if a contractor plans to build 6 ranches, 11 split-levels, and 15 Tudors in the next year. Assume the average building costs are as given in the table: ExpensesRanchesSplit-LevelsTudors material175001620012400 labor620055004900 We can write the table above and the desired number of homes as the product of two matrices. Thus: \mid175001620012400\mid\mid6\mid\midmaterials\mid \mid620055004900\mid\mid11\mid=\midlabor\mid \mid15 \mid The program outputs the answers in the following format: Ranches Splits Tudors Materials Labor 1 2 3 \bullet \bullet \bullet 10 1 3 5 \bullet \bullet \bullet 19 5 7 9 \bullet \bullet \bullet 23 TOTAL COSTS First, we present the pseudocoded version. Notice that we have referenced array elements with zero subscripts. This was done to ac-commodate this convention in the BASIC language. real A(1,2), B(2,0), C(1,0) output headings: 'RANCHES SPLITS TUDORS MATERIALS LABOR' input data into A M \leftarrow 0.0 L \leftarrow 0.0 do for R = 0 to 9 B(0,0) \leftarrow R + 1 . B(1,0) \leftarrow 2. \textasteriskcentered R + 1 . B(2,0) \leftarrow 2. \textasteriskcentered R + 5. C \leftarrow A \textasteriskcentered B (A, B, and C refer to entire matrices) output B(0,0), B(1,0), B(2,0), C(0,0), C(1,0) m \leftarrow m + C(0,0) L \leftarrow L + C(1,0) end do for output 'TOTAL COSTS', M,L end program In BASIC, we add DATA statements and MAT instructions to calculate the product. The BASIC program is given below: 10DIM A(1, 2), B(2, 0), C(1, 0) 12PRINT 14PRINT "RANCHES\textquotedblright, "SPLITS", "TUDORS", "MATERIALS", "LABOR" 15PRINT 20MAT READ A 25LET M = 0 26LET L = 0 30FOR R = 0 TO 9 34LET B(0,0) = R + 1. 36LET B(1,0) = 2. \textasteriskcentered R + 1. 38LET B(2,0) = 2. \textasteriskcentered R + 5. 40MAT C = A \textasteriskcentered B 45PRINT 50PRINT B(0,0), B(1,0), B(2,0), C(0,0), C(1,0) 54LET M = M + C(0,0) 58LET L = L + C(1,0) 60NEXT R 62PRINT 70PRINT "TOTAL COST = ", M,L 80STOP 82DATA 17500, 16200, 12400 84DATA 6200, 5500, 4900 99END

In birds, the cells from the epiblast (the upper of two embryonic cell layers) migrate down-ward. Initially, a strip of epiblast extending forward in the midline of the embryo from the posterior edge of the area pellucid becomes thickened as the primitive streak, with a narrow furrow, the primitive groove, in its center. Hensen's node, a thickened knot of cells, is present at the anterior end of the primitive streak. It is by the migration of cells from the lateral portion of the epiblast that the primitive streak is formed. The cells of the epiblast invaginate at the primitive streak, move into the space between the epiblast and hypoblast and reach the latter, forming a mass of moving cells. The cells continue to migrate, moving laterally and anteriorly from the primitive streak area. Though cells are migrating, the shape of the primitive streak remains intact. The primitive streak, where invagination occurs, is considered to be homologous to the amphibian blastopore. The cells which form the notochord arise from Henson's node. The notochord is the rod-shaped structure that forms a skeletal axis in chordates which is re-placed in the vertebrates by vertebrae (in frogs, a special area of mesoderm develops into the notochord). Cells which grow laterally and anteriorly from the primitive streak between the epiblast and the hypoblast become the mesoderm. The original hypoblast cells plus other cells that migrate into the lower layer from the primitive streak form the endoderm, which gives rise to the digestive tract and yolk sac. The epiblast forms the ectoderm.

Question:

A mass M is connected to a linear spring of stiffness K. The masspring system is initially at rest. At time t = t_0, the system is displaced a distance ofx_0 , and released. It can be shown that the motion of the system is described by the following differential equa-tion: Mẍ + Kx(t) = 0 where x(t) is the displacement of the mass and ẍ is the second derivative of x with respect to time. If v_0 is the initial velo-city, construct a model of this system and devise a digital program which simulates the motion of the system from time t = t_0 to time t = t_f.

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Solution:

The general procedure for developing a model for simulation purposes can be summarized as follows: 1.Definition of the problem and the system that is being studied. 2.Making simplifying assumptions concerning the components of the system. 3.Determining the factors outside the system's environment (exogenous factors) that will affect it. 4.Formulation of a mathematical model (if possible). 5.Compiling the relevant data. 6.Construction of a computer program of the model. Applying these steps to this particular problem: 1.The system consists of the mass and the spring. The problem of following the motion of the system can be viewed through the change in displacement, x. So essentially, the problem is to solve the given differential equation and record samples of the values x assumes from t = t_0 to t = t_f . 2.The differential equation given, results from assuming that M and K are time invariant (constant). Also, the mass of M is considered "lumped" at a single point. 3.The differential equation does not include the effect of air resistance for simplicity. 4.The differential equation is a dynamic mathematical model of the system. 5.The data available is, t_0, t_f, x(t_0) = x_0 ; v(t_0) = v_0, K and M. 6.Although analytical methods exist for solving second-order linear differential equations with constant coefficients such as the one given, digital simulation is applicable even when no analytical method applies. The numerical method used here is called the Euler predictor-corrector method (slow, but straightforward and very general). To use it we must transform the equation to a system of simultaneous first-order equations; using the substitution v = x: ẋ = v(t)(1) v̇ = -(K/M) \textasteriskcentered x(t)(2) The flow-chart for the subroutine is as follows. The program closely follows the flow chart: (I/O statements can be appropriately formatted): REAL K,M,N READ K,M,N,T,TFIN, XCOR, VCOR, ACCVR CN IS THE NUMBER OF SAMPLES TO BE CHOSEN DT = (TFIN - T)/N CDT IS THE INCREMENT SIZE 50X = XCOR V = VCOR A = -(K/M){_\ast}X PRINT T,X T = T + DT IF (T.GT.TFIN) STOP XPRED = X +.V {_\ast} DT VPRED = V + A {_\ast} DT 100XDOT = VPRED CEQUATION (1) VDOT = -(K/M) {_\ast} XPRED CEQUATION (2) XCOR = X + 0.5 {_\ast} (V + XDOT) {_\ast} DT VCOR = V + 0.5 {_\ast} (A + VDOT) {_\ast} DT XDIF = ABS (XCOR - XPRED) VDIF = ABS (VCOR - VPRED) IF = (XDIF. LE. ACCUR. AND. VDIF. LE. ACCUR) GO TO 50 XPRED = XCOR VPRED = VCOR GO TO 100 END

A hydrogen bond is a molecular force in which a hydrogen atom is shared between two atoms. Hydrogen bonds occur as a result of the uneven distri-bution of electrons in a polar bond, such as an O-H bond. Here, the bonding electrons are more attracted to the highly electronegative oxygen atom, resulting in a slight positive charge (\delta+) on the hydrogen and a slight negative charge (\delta-) on the oxygen. A hydrogen bond is formed when the relatively positive hydrogen is attracted to a relatively negative atom of some other polarized bond. For example: The atom to which the hydrogen is more tightly linked or specifically the atom with which it forms the polar bond, is called the hydrogen donor, while the other atom is the hydrogen acceptor. In this sense, the hydrogen bond can be thought of as an intermediate type of acid-base reaction. Note, however, that the bond is an electro-static one - no electrons are shared or exchanged, between the hydrogen and the negative dipole of the other molecule of the bond. Hydrogen bonds are highly directional (note the arrows in the figure), and are strongest when all three atoms are colinear. Bond energies of hydrogen bonds are in the range of about 3 to 7 kcal/mole. This is intermediate between the energy of a covalent bond and a van der Waals bond. How-ever, only when the electronegative atoms are either F, O, or N, is the energy of the bond enough to make it important. Only these three atoms are electronegative enough for the necessary attraction to exist. Hydrogen bonds are responsible for the structure of water and its special properties as a biological solvent. There is extensive hydrogen bonding between water molecules, forming what has been called the water matrix. This structure has profound effects on the freezing and boiling points of water, and its solubility properties. Any molecule capable of forming a hydrogen bond can do so with water, and thus a variety of molecules will dissociate and be soluble in water. Hydrogen bonds are also most responsible for the maintenance of the shape of proteins. Since shape is crucial to their function (as both enzymes and struc-tural components), this bonding is extremely important. For example, hydrogen bonds maintain the helical shape of keratin and collagen molecules and gives them their characteristic strength and flexibility. It is hydrogen bonds which hold together the two helices of DNA. Bonding occurs between the base pairs. The intermediate bond strength of the hydrogen bond is ideal for the function of DNA - it is strong enough to give the molecule stability, yet weak enough to be broken with sufficient ease for replication and RNA synthesis.

Question:

Explain how you would prepare benzoic acid from benzene. Use any inorganic reagents.

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Solution:

One of the most common ways of preparing benzoic acid from benzene is through an intermediate grignard. To plan this synthesis, first write your starting material and desired product. Then, think of inorganic reagents that lead you step by step through intermediates until you obtain the product. A grignard reagent can be represented as R-MgX, where R is any aryl or alkyl group and X is a halogen. To obtain this grignard intermediate, you must halogenate the benzene ring. This can be accomplished by using a solution of halogen and FeX_3, where X is the halogen. Let the halogen be Br. Thus, you have the reaction in Figure B. This is electrophilic substitution of a hydrogen in the ring by a bromine atom. By electrophilic, you mean electron seeking. By addition of Mg, in dry ether, you obtain the grignard reagent. See figure C. Now that you have the grignard, add carbon dioxide. The reaction proceeds as shown in figure D. With the addition of the acid, HBr, the reaction in figure E proceeds to the desired product benzoic acid.

Question:

In the Ostwald process for the commercial preparation of nitric acid, ammonia gas is burned in oxygen in the presence of a Pt catalyst. The balanced equation is: 4NH_3 + 5O_2^Pt\rightarrow4NO + 6H_2 O What volume of O_2 and what volume of NO is formed in the combustion of 500 liters of NH_3. All gases are under the same conditions of temperature and pressure.

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Solution:

In order to solve this volume-volume problem set up a ratio between the volumes involved. No molecular weights are needed in volume-volume problems. First, one wants to find out how much O_2 is required to react with 500 liters of NH_3. As can be seen from the stoichiometry of the reaction, the ratio of the mole volumes (i.e. how much NH_3 reacts with O_2) is 4 : 5. Using the equation [(volume NH_3 present)/(mole volume NH_3 )] = [(volume O_2 required)/(mole volume O_2 )] [(500 liters) / (4 liters)] = [(volume O_2) / (5 liters)] volume O_2 = 625 liters. To find the amount of NO produced, set up a similar equation. In this case, 4 mole volumes of NH_3 produce 4 mole volumes of NO as indicated by thestoichiometryof the reaction [(volume NH_3 present) / ( mole volume NH_3)] = [(volume NO produced) / (mole volume NO)] [(500 liters) / (4 liters)] = [(volume NO produced) / (4 liters) volume NO produced = 500 liters. It was mentioned in the problem that a Pt catalyst was used. A catalyst is a substance which speeds up the rate at which a reaction occurs by lowering the amount of energy needed to start a reaction (i.e. lowering the activation energy). Catalysts have no effect on the final concentrations of the compounds involved in the reaction. Therefore, the presence of a catalyst does not change the values calculated in this question.

Question:

Distinguish a metallic bond from an ionic bond and from a covalent bond.

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Solution:

The best way to distinguish between these bonds is to define each and provide an illustrative example of each. When an actual transfer of electrons results in the formation of a bond, it can be said that an ionic bond is present. For example, Potassiumsulfurpotassiumsulfur ionionic bond atomsatomionsdue to the (unlike ions due transferattraction of electron from potassiumof unlike to sulfur)ions When a chemical bond is the result of the sharing of electrons, a covalent bond is present. For example: A pure crystal of elemental metal consists of millions of atoms held together by metallic bonds. Metals possess electrons that can easily ionize, i.e., they can be easily freed from the individual metal atoms. This free state of electrons in metals binds all the atoms together in a crystal. The free electrons extend over all the atoms in the crystal and the bonds formed between the electrons and positive nucleus are electrostatic in nature. The electrons can be pictured as a "cloud" that surrounds and engulfs the metal atoms.

The relativistic Doppler shift is given by the expression v՛ = v\surd[(1 - u / c) / (1 + u / c)] Multiplying both the numerator and denominator by the factor \surd(1 - u / c), one obtains the expression v՛ = v[(1 - u / c) / \surd{1 - (u / c)^2 } ] where v' is the observed frequency of light of frequency v in the frame of reference of the source; u is the relative source-observer velocity, and c is the speed of light. For case (a), the light is travelling parallel to the earth's motion, and u is given by u = R\Omega , with R = 30,000 light-years, and \Omega = [2\pi / {200(10^6)}](rad / year). Thus, u = [{30,000(2 \pi)} / {200(10^6)}] light- year(rad / year) = 9.425(10^-4)c (where c represents the speed of light) or (u / c) = 9.425(10^-4). Now since u << c, v' \approx v {1 - (u / c)}, or, since v' = c / \lambda' and v = c / \lambda (\lambda = the wavelength of light), 1 / \lambda' \approx 1 / \lambda {1 - (u / c)}, with \lambda = 6563\AA. Thus, \lambda' = 6569.2 \AA, or ∆\lambda = \lambda' - \lambda = 6.2 \AA. b) Where the line of observation is perpendicular to the direction of the earth's motion, the observed frequency shift, which is purely a relativistic effect, is called the transverse Doppler effect. The frequency shift is given by the expression v = v'[1 - (u / c)^2)1 / 2 The derivation of this effect is Indicated in the figure. The generalized Doppler shift Is given by v = v'[{1 + (u / c)cos \texttheta՛ } / \surd{1 -(u/c)^2} ] where, as indicated, the observer (on the earth) is located at point P at rest in the unprimed coordinate system, and the source of light is at rest in the primed system. For transverse observation, \texttheta = \pi / 2, cos \texttheta՛ = -(u / c) Thus,v = v'[{1 - (u / c)^2} / \surd{1 -(u/c)^2} ] = v' [1 - (u / c)^2]1 / 2 or 1 / \lambda = 1 / \lambda՛[1 -(u / c)^2]1 / 2], with (u / c) = 9.425(10^-4) . Thus, \lambda = 6563.003 \AA ,or∆\lambda = 0.003 \AA.

The activities of the various part of the body of higher animals are integrated by the nervous and endocrine systems. The endocrine system consists of a number of ductless glands which secrete hormones. The swift responses of muscles and glands, measured in milliseconds, are typically under nervous control. Nerve impulses are transmitted along pathways consisting of neurons. The hormones secreted by the endocrine glands are transported by the bloodstream to other cells of the body in order to control and regulate their activities. Nervous stimulation is required by some endocrine glands to release their hormones, particularly the posterior pitui-tary gland. The responses controlled by hormones are in gen-eral somewhat slower (measured in minutes, hours, or even weeks), but of longer duration than those under nervous control. The long term adjustments of metabolism, growth and reproduction are typically under endocrine regulation. We already mentioned that hormones travel in the . blood and are therefore able to reach all tissues. This is very different from the nervous system, which can send messages selectively to specific organs. However, the body's response to hormones is highly specific. Despite the ubiquitous distribution of a particular hormone via the blood, only certain types of cells may respond to that hormone. These cells are known as target-organ cells. The central nervous system, particularly the hypo-thalamus plays a critical role in controlling hormone secretion; conversely hormones may markedly influence neural function and behavior as well.

The displacementy of the medium due to wave motion at a positionxand at a timetis (see the figure) y = A sin(2\pi/\lambda) (x - vt) where A is the amplitude, \lambda is the wavelength, andv is the velo-city with which the wave is traveling along the x-axis. If we compare this equation with the expression given in the question, we see that (a)A = 0.20 cm (b)(2\pi/\lambda) = 0.40\pi \lambda = (2/0.40) cm = 5.0 cm (c)v = 60 cm/sec (d)f = (v/\lambda) = {(60 cm/sec)/5.0 cm} (e)y = (0.20 cm) sin [0.40\pi(5.5 - 60 × 0.020)] = (0.20 cm) sin [0.40\pi(5.5 - 1.2)] = (0.20 cm) sin (0.40 × 4.3\pi) = (0.20 cm) sin 1.72\pi = (0.20 cm) (\rule{1em}{1pt}0.77) = \rule{1em}{1pt}0.15 cm

Question:

A chemist forms magnesium oxide by burning magnesium in oxygen. The oxide obtained weighed 1.2096 grams. It was formed from .7296 g of magnesium. Determine the mass equivalent of magnesium in this reaction.

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Solution:

An equivalent is defined as that mass of oxidizing or reducing agent that picks up or releases the Avogadro number of electrons in a particular reaction. One equivalent of any reducing agent reacts with one equivalent of any oxidizing agent. In this problem, the key is to determine the number of equivalents of oxygen involved. Once this is known, you also know the number of equivalents of magnesium. Since the oxide weighed 1.2096 g and the magnesium weighed .7296 g, the mass of the combined oxygen must be 1.2096 - .7296 = .4800 g. Before the oxygen reacted, its oxidation state was zero. After the reaction, however, it was - 2. As such, each oxygen atom gained 2 electrons. Therefore, the Avogadro number of electrons will be taken up by one half of a mole of O atoms. It follows, therefore, that there are 8.000 g per equivalent for oxygen, since 1 mole of oxygen atoms weighs 16 grams. It was found, however, that there were .4800 g of oxygen. As such (.4800 g) /(8.00g/equiv)= .0600 equiv of oxygen . This means that magnesium also has .06 equiv. 0.7296 g of Mg participated in the reaction. Therefore, the grams per equivalent of Mg = (.7296 gMg) / (.060 equiv) = 12.16 g / equiv .

Question:

A 3000-lb car traveling with a speed of 30 mi/hr strikes an obstruction and is brought to rest in 0.10 sec. What is the average force on the car?

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Solution:

The momentum of a body is defined as p = mv. Since F^\ding{217} = ma^\ding{217} = m (dv^\ding{217}/dt) anddp^\ding{217} = d(mv^\ding{217}) = mdv^\ding{217} we have F^\ding{217} dt = dp^\ding{217} . Consider a collision between two masses m_1 and m_2 as shown in the figure. During the collision, the two objects exert forces on each other where F_1 is the force on m_1 due to m_2, and F_2 is the force on m_2 due to m_1. By Newton's third law, these forces at any instant are equal in magnitude but opposite in direction. The change in momentum of m1 resulting from the collision is \Deltap_1^\ding{217} =^t1\int_t0 F_1^\ding{217} dt =F_1 ∆t = mv_1 - mv_0 whereF_1 is the average value of the force F_1^\ding{217} during the time interval of the collision \Deltat = t_1 - t_0. For the car colliding with an obstruction, we can say F\Deltat = mv_1 - mv0 m = W/g = (3000 Ib) / (32 ft/sec^2) = 94 slugs v0= 30 mi/hr = 44 ft/sec F= [m(v_1 - v_0)] / \Deltat = [94 slugs (0 - 44 ft/sec)] / [0.10 sec] = - 4.1 × 10^4 lb.

The best way to approach this problem is to consider the geometry of the carbon-carbon bonding. To do this, consider its hybridization. Benzene can be written as shown in figure A. Notice that each carbon has 4 bonds, and is bonded to three other elements (2 other carbons and 1 hydrogen). This suggests sp^2 hybridization, sp^2 hybridization allows bonding to three other elements, and leaves a \textquotedblleftfree" p orbital. This p orbital can overlap with another p orbital to give a pi bond, which serves to form the double bond. With sp^2 hybridization, the molecule must assume a geometry where the orbitals are 120\textdegree apart. This is the only way that the repulsion which occurs when like charges are brought to-gether can be avoided. See figure B. An angle of 120\textdegree creates a planar geometry. This is true because a planar molecule with bond angles of 120\textdegree is a configuration where the 3 atoms involved are the farthest apart. Consider cyclohexane. It can be written as shown in figure C. Each carbon atom, is bound to four other atoms (2 other carbons and 2 hydrogens). This suggests sp^3 hybrodization. With sp^3 hybridization, a molecule must assume a geometry that permits 109\textdegree28' bond angles. The atoms of the bond are arranged in a tetrahedral formation, this allows for the greatest distance between the atoms. The way cyclohexane can accomplish this is to assume a puckered ring of six carbon atoms, as can be seen in figure D.

Question:

Although it is often possible to see unicellular organisms if a dropof pond water is placed on a microscope slide, it is impossibleto seecuboidalepithelium if a rat kidney is placedunder the microscope. Describe the steps necessary forthe preparation of tissues for light microscopy.

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Solution:

There are four basic steps in the preparation of tissues for light microscopy. (1) Fixing. The tissue is killed and stabilized with chemical agents, sothat structures are preserved with a life-like appearance. Fixing preventspost-mortem changes of tissue structure caused by autolysis (self--digestion) . (2) Embedding. The fixed tissue is embedded in a hard material thatprovides firm support. This material is often paraffin or wax. The tissueis dehydrated and soaked in molten wax which penetrates the tissuesand then is cooled to form a hard matrix. (3) Sectioning. The support provided by the em-bedding material enablesthe tissue to be cut into very thin slices without being crushed by theknife edge. Sectioning tissue permits study of complex structures by providingan unimpeded view of deep layers. (4) Staining. The cells are stained with dyes that color only certain organelles. This provides contrast between different cellular structures, for example, be-tween nucleus and cytoplasm or between mitochondria and othercytoplasmicstructures. Most cellular structures are colorless. If unstained, it is difficult and often impossible to distinguish different cellular regionsby light microscopy. Cytochemistryrefers to staining procedures which specifically color certaincellular substances. For ex-ample, theFeulgenprocedure specificallydyes DNA. DNA reacts with dilute hydrochloric acid to give aldehydegroups; these then react with the colorless Schiff reagent to changethe dye to red. No other cellular components react in the right way withHClto cause the Schiff re-agent to become colored, so only DNA is stained. Othercytochemicaltechniques permit visualization of RNA, of the acidhydrolasesoflysosomes, and of specific lipids, proteins, and carbohydrates.

Question:

Two circular cylinders have the same mass and dimensions, but one is solid while the other is a thin hollow shell. If they are released together to roll without slipping down a plane inclined at 30\textdegree to the horizontal, how far apart will they be after 10 s?

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Solution:

In either case the forces acting on the cylinder are as shown in the diagram. The weight mg^\ding{217} acting downward is split up into its components parallel to and perpendicular to the plane. The other two forces acting on the cylinder are the normal force N^\ding{217} exerted by the plane and the frictional force attempting to prevent motion, F^\ding{217}, which has magnitude \muN. Since the cylinder does not lift from the plane, N = mg cos 30\textdegree. Further, by Newton's second law, mg sin 30\textdegree - \muN = mg(sin 30 - \mu cos 30) = ma, where m is the mass of the cylinder and a the acceleration produced. Rotation about the center of the cylinder also takes place. Since we are dealing with rotations of dis-tributed masses, a relation involving the moment of iner-tia must be used. That relation is the rigid body analogue of Newton's second law, F = ma, or \cyrchar\CYRG = I\alpha where the torque \cyrchar\CYRG corresponds to F, the moment of inertia I corresponds to m, and the angular acceleration \alpha of the cylinder corresponds to a. If an axis is taken about the center of the sphere, then the only torque acting on the sphere is \muNr due to the frictional force \muN. Then \muNr = \mur mg cos 30\textdegree = l\alpha, Since no slipping takes place, the point A is instantan-eously at rest. Hence a = r\alpha, and mg (sin 30\textdegree - \mu cos 30\textdegree) = ma,(1) \mur mg cos 30\textdegree = Ia/r.(2) or, multiplying (1) by r^2 , and (2) by r mgr^2 sin 30 - \mumgr^2 cos 30 = mar^2(3) and\mur^2mg cos 30 = Ia(4) Substituting (4) in (3) mg r^2 sin \texttheta - Ia = mar^2 or a = (mg r^2 sin \texttheta)/(I + mr^2) For the solid cylinder, I = 1/2mr^2 a_1 = (mg r^2 sin \texttheta)/(1/2mr^2 + mr^2 ) = (2/3)g sin \texttheta, and for the hollow cylinder, I = mr^2 a_2 = (mg r^2 sin \texttheta)/(mr^2 + mr^2) = (1/2)g sin \texttheta. The distances traveled in 10 s from rest are, found by using the kinematic equations for constant acceleration. If the top of the inclined plane is taken as the initial position, then s_0 = 0. Also the initial velocity v_0 = 0. ThenS = s_0 + v_0t + (1/2)at^2 . Hence S_1 = 0 + (1/2)a_1t^2 = (1/2) × (2/3) × 32 ft\bullets^-2 ×(1/2) × 100 s^2 = (1600)/3 ft andS_2 = (1/2)a_2t^2 = (1/2) × (1/2) × 32 ft\bullets^-2 ×(1/2) × 100 s^2 = 400 ft. = 400 ft. \therefore S_1 - S_2 = (533.3 - 400)ft = 133.3 ft.

Figure 1b defines the positions relative to the fixed pulley A needed to solve this problem. The lengths of the cord are assumed to be constant and the section of cord over the pulleys to remain constant, thus yielding the following constraint equations y_1 + y_B = K_1,(a) (y_2 \rule{1em}{1pt} y_B) + (y_3 \rule{1em}{1pt} y_B) = k_2, ory_2 + y_3 \rule{1em}{1pt} 2_(y)B = K_2.(b) Differentiating each equation (a) and (b) twice yields ẏ_1 + ẏ_B = 0(c) ẏ ̇_1 + ẏ ̇_B = 0(d) ẏ_2 + ẏ_3 \rule{1em}{1pt} 2ẏ_B = 0(e) ẏ ̇_2 + ẏ ̇_3 \rule{1em}{1pt}2ẏ ̇_B = 0(f) These equations lead directly to the desired solutions upon substitution of the information from Figure 1a. If it is supposed that down is positive, then v_1 = ẏ_1 = \rule{1em}{1pt}2 ft / sec,v_2 = ẏ_2 = 3 ft/sec,(g) a_1 = ẏ ̇_1 = 1 ft / sec^2,a_2 = ẏ ̇_2 = \rule{1em}{1pt}2 ft / sec^2. First solve (c) and (d) to find the motion of pulley B. \rule{1em}{1pt}2 ft / sec + ẏ_B = 0 ẏ_B = 2 ft / sec(h) 1 ft / sec^2 + ẏ ̇_B = 0 ẏ ̇_B = \rule{1em}{1pt}1 ft / sec^2.(i) Now using (g), (h) and (i) in equations (e) and (f), it is possible to obtain the desired results: 3ft / sec + ẏ_3 \rule{1em}{1pt} 2 (2 ft / sec) = 0 ẏ_3 = 1 ft / sec (\rule{1em}{1pt}2 ft / sec^2) + ẏ ̇_3 \rule{1em}{1pt} 2 (\rule{1em}{1pt}1 ft / sec^2) = 0 ẏ ̇_3 = 0 Thus, block 3 is moving down with a constant velocity v_3 = 1 ft / sec.

Question:

Explain briefly the background and applications of COBOL.

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Solution:

COBOL, an acronym forCOmmonBusiness-Oriented Lan-guage, is a high-level, procedural language for business appli-cations. The language was first introduced in 1959 by a commit-tee named Conference on Data System Languages (CODASYL) .This committee, sponsored by the Department of DefenseDoD), issued a report in 1960 specifying the first version of COBOL, and the first compiler appeared shortly after. Subsequent revision is-sued in 1961 became the cornerstone of further developments including standard versions by American National Standards In-stitute (ANSI). The following is a quick chronicle of COBOL's background : 1959-CODASYL sponsored byDoD 1960-First COBOL compiler 1961-COBOL-61(First Revision) 1968-ANSI COBOL-68(First Standard) 1974-ANSI COBOL-74(Widely Used Revised Standard) 1985-ANSI COBOL-85(New Standard) COBOL applications are mostly commercial rather than scien-tific. It is still the most widely used language for business applications (accounting, banking, organizational, management information, etc.) accounting for 60% to 80% of the programs written for this purpose.

Question:

Compute the resistance across the terminals in figure A. Compute the resistance across the terminals in figure A.

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Solution:

First we regroup the resistors to simplify our calculations. We are allowed to do this as long as we do not change the orientation of the resistors with respect to each other and to the terminals. An equivalent arrangement is shown in figure B, In figure C we sum the resistors in series. We are left with two re-sistances in parallel which we find from the formula: (1/R_T ) = (1/R_1 ) + (1/R_2 ) = (1/105) + (1/75) = (12/525) = (4/173) where R_T is the total resistance across the terminals. Thus: R_T = (173/4) = 43.25

Question:

Monochromatic light of wavelength 400m\mu from a distant point source falls on an opaque plate in which there is a small circular opening. As a screen is moved toward the plate from a large distance away, the Fresnel diffraction pattern on the screen first has a dark center when the distance from plate to screen is 160cm. Find the diameter of the central disk in the diffraction pattern if a lens of focal length 160cm is placed just to the right of the circular opening.

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Solution:

We have a circular aperture illuminated by coherent light from a source infinitely distant from the aperture. This implies that all points on the aperture are in phase with all other points. We now place a screen an infinite distance from the aperture and slowly move it towards the aperture observing the center spot of the screen. At some distance the path length from the light at the circumference of the aperture will be just one wavelength out of phase with light from the center of the aperture, which can be represented algebraically as R_1 = R_0 + \lambda in figure (2). We have learned from our study of diffraction that this is just the condition for destructive interference. Now if we move the screen closer to the aperture, the center of the screen will have a bright spot when the path difference between a circum-ferential ray and a central ray is (3\lambda / 2) and dark when there is a 2\lambda path difference, etc. Each of the path differences defines the edges of Fresnel zones. Looking at figure (2) for our problem when R_O is 160cm, r will just be the outer radius of the second Fresnel zone. From figure (2), R^2 _1 = R2_O + r^2(1) andR_1 = R_O + \lambda(because there is destructive interference at R_O = 160 cm.)(2) Substituting for R_1 in equation (1) yields R^2 _O + 2\lambdaR_O + \lambda^2 = R2_O + r^2 The R2_O terms cancel, and so, r^2 = 2\lambdaR_O + \lambda^2 or r =\surd(\lambda^2 + 2\lambdaR_O) = [(400 × 10^-9m)^2 + 2 × 400 × 10^-9m × 1.6m]1 / 2 = 1.13 × 10^-3m,(3) which is the radius of the aperture. If we now place a 160cm focal length lens on the screen side of the aperture, the lens will convert an infinite distance on the screen side to a point 160cm from the aperture (as in figure (1)). Stating it another way, the screen at 160cm from the aperture with a 160cm focal length lens will produce a Fraunhofer diffraction pattern of the source. Now from any standard optics text Fraunhofer diffraction of a circular aperture is (d / 2) / l =1.22\lambda/ 2r(4) where d is the diameter of the central disk; l is the dis-tance from the aperture; \lambda is the wavelength of the light used and r is the radius of the aperture. Now solving for the diameter of the central disk we have d = (l × 1.22 × \lambda) / r where l = 1.6m, \lambda = 400 × 10^-9m, and r = 1.13 × 10^-3m. Henced = (1.6m × 1.22 × 400 × 10^-9m) / (1.13 × 10^-3m) = 6.91 × 10^-4m.

Figure (a) shows the cabin. Axes have been set up with the x, y and z directions coinciding with the length, breadth, and height of the room. The astronaut must get from point A to point B. The vector A\ding{217} going from the origin 0 to point A is A\ding{217} = (15 ft, 0, 0) The vector from 0 to point B is: B\ding{217} = (0, 15 ft, 15 ft) The vector going from A to B is then: B\ding{217} - A\ding{217} = (0, 15 ft, 15 ft) - (15 ft, 0, 0) = (- 15 ft, 15 ft, 15 ft) Its length is: \vertB\ding{217} - A\ding{217}\vert = \surd[(- 15ft)^2 + (15 ft^2) + (15 ft)] = 15\surd3 ft = 26 ft This is the distance the astronaut must glide. The angle to the floor at which he launches him-self is \texttheta, where tan \texttheta = BC/AC. Point C has coordinates (0, 15 ft, 0). Thus BC has length 15 ft and AC has length \surd(15^2 + 15^2 ft) = 15 \surd2 ft = 21.2 ft. \therefore tan \texttheta = [15 ft/(15 \surd2 ft)] = (1/\surd2) = 0.707or \texttheta = 35.25\textdegree . Figure (b) shows the cabin minus the roof in an exploded diagram; points A, B, and C are again marked in. For convenience a new set of coordinate axes has been chosen. The astronaut walks the same distance along the walls from A to B by any particular route whether the walls are upright or flat as in the ex-ploded diagram. But in the diagram it is much easier to see that the minimum distance from A to B is the straight-line path between the two points. The vector B\ding{217} - A\ding{217} has components 15 ft in the x-direction and 30 ft in the y-direction. Distance \vert B\ding{217} - A\ding{217} \vert therefore equals\surd(15^2 + 30^2 ft) = 33 ft 6(1/2) in. Note that B' is the same point as B in the exploded diagram; AB' is thus an alternative route. There is a further alternative route AB'' which can be seen most clearly in figure (c) in which the wall marked I in figure

Question:

In a double-slit experiment, D = 0.1 mm and L = 1 m. If yellow light is used, what will be the spacing between adjacent bright lines?

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Solution:

The wavelength of yellow light is approximately 6 × 10^-5 cm. Let the separation between the (n+1)th and m' th maxima be ∆x, as shown in the figure. Then, sin \texttheta_m = [(m)/(D)] \lambda sin \texttheta_m+1 = [(m+1)/D]\lambda The angles \texttheta_m and \texttheta_m+1 are related to the positions h_m ,and h_m+1 of these maxima on the screen; sin \texttheta_m \approx [(h_m+1)/L] sin \texttheta_m+1 \approx h_m/L , hence , sin \texttheta_m+1 - sin \texttheta_m = \lambda/D [(hm+1-h_m)/L] \approx \lambda/D ∆x/L \cong L/D = [{(6 ×10^-5 cm) × (100 cm)} / {10^-2 cm}] = 0.6 cm Thus , the spacing between lines is about 6 mm or (1/2) of an inch.

Question:

Draw the general structure of an amino acid and discuss why it is both an acid and a base.

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Solution:

The general structure of an amino acid can be drawn as: where R represents a side group which can be changed to create any amino acid. Note that the molecule contains two polar groups(COO^- and NH_3^+) and for this reason, amino acids are called dipolar ions or zwitterions. It is these two polar groups which give amino acids the ability to function as both acids and bases. Depending on the pH, amino acids can exist in one of three ionic forms: The pH at which most of the amino acids exist in the zwitterion (dipolar) form is called its isoelectricion point. When the pH is made more acidic, the dipolar ion is converted to the cation I. In this form, the amino acid will act as an acid with the carboxyl group capable of donating its proton. When the pH is made alkaline, the anion III is formed and predominates in solution. In this form, the amino acid will act as a base, with the amino group as the proton acceptor. At its isoelectric point, the amino acid is capable of acting as both an acid and a base, with both ionic forms existing in exact balance with each other. The relative acidity and basicity of an amino acid is strongly influenced by the structure and properties of its particular R group. In addition, the R group can itself act as an acid or base. Thus, each amino acid will have its own characteristic isoelectric point, and its own particular K_a and K_b for its amino and carboxylic acid groups. At a given pH, some amino acids will be acidic (protonated or cationic), while others will be basic (anionic), and others will be neutral (zwitterionic). At physiological pH, most amino acids exist in the dipolar form, with acidity and basicity being determined by each particular R group. In the dipolar form, however the amino acids can react with each other, via their polar groups, to form a covalent peptide linkage: This peptide bond is the means by which large protein molecules are constructed from amino acids.

The NADH and FADH2 formed in glycolysis, fatty acid oxidation and the citric acid (TCA) cycle are energy-rich molecules. When each of these molecules transfers its pair of electrons to molecular oxygen, a large amount of energy is released. This released energy can be used to generate ATP in oxidative phosphorylation. As electrons flow through the electron transport chain from NADH to O2, ATP is formed at three sites along the chain. How do we know that three separate sites exist instead of one site where 3 ATP are produced? The first evidence is the ATP yield from the oxidation of different substrates. The oxidation of NADH yields 3 ATP but the oxidation of succinate, which forms FADH2, only yields 2 ATP. The electrons from FADH2 enter the chain at coenzyme Q, which is past the first phosphory-lation site. Only 1 ATP is formed when ascorbate (an artificial substrate also called Vitamin C) is oxidized, since its electrons enter at cytochrome c, which is past the first two phosphorylation sites. The P:O ratio (the ratio of organically bound phosphorous produced per mole of oxygen, which is often used as an index of oxida-tive phosphorylation) is 3:2:1 for NADH : succinate : ascorbate, respectively. Hence, by knowing the P:O ratios and that each substrate enters the process at different steps,we see that one mole of ATP is produced at each site. The steps involved can be determined by the free energy change, ∆G, released during electron transfer. The ∆G for electron transfer from NADH to FMN is -12 kcal/mole, from cytochrome b to c, -10 kcal/mole, and from cytochrome a to O2, -24 kcal/mole. These reactions are sufficiently exergonic to drive the synthesis of ATP, whereas the other electron transfer reactions do not release sufficient energy to allow ATP synthesis to take place. There also exist inhibitors which specifically block certain steps in the chain. Rotenone specifically inhibits electron transfer from NADH to coenzyme Q, pre-venting ATP synthesis at the first site. Antimycin A inhibits electron flow at the second site, while cyanide and carbon monoxide block electron transfer at the third site. Each inhibitor has been observed to block all successive steps of the electron transport system. However it is possible to bypass the action of the inhi-bitor. For instance, the action of rotenone can be avoided through the addition of either succinate or ascorbate. Similarly, the action of antimycin A on the second site can be bypassed through the addition of ascorbate. By inhibiting the various sites and measuring the resulting P:O ratios it can be deduced that each site in the electron transport chain is responsible for the production of one molecule of ATP (see figure). It should be noted that the addition of antimycin A will result in the blockage of site 1 as well as site 2. This is due to a backflow of electrons which results in the reduction of site 1, thus inactivating it. Likewise, the addition of cyanide results in the reduction of sites 1 and 2 as well as 3, thus rendering them all incapable of ATP formation. Hence cyanide has the well deserved reputation as one of the world's strongest toxins. The advantage of the sequential transfer over one single transfer is that the sequential reactions divide up the free energy change of the oxidation of NADH, which is highly exergonic. The free energy of oxidation of NADH + (1/2) O2 + H+ \rightleftharpoons H2O + NAD+ is about - 53 kcal/mole. If all this energy were released at once, much of it would be wasted as heat since only 7 kcal are needed to phosphorylate ADP to form ATP. If all the energy from the oxidation of NADH is released at once to form 1 ATP, the efficiency of this system is 7/53 × 100% = 13%. However, if the same energy is released in a stepwise process to form 3 ATP, the efficiency is 21/53 × 100% = 40%. Thus, stepwise reactions increase the efficiency of oxidative phosphorylation from 13% to 40%. In addition, the large amount of heat liberated would very easily destroy enzyme activity and thus is harmful to the cell if released in one burst. The three reactions involved are "coupled" with the phosphorylation of ADP so that the liberated energy is used immediately to drive this otherwise non-spontane- ous reaction. The exact mechanism of oxidative phosphorylation, that is, how the energy liberated is transferred so as to form the high energy phosphate bond of ATP, is un-known. The exact mechanism of oxidative phosphorylation, that is, how the energy liberated is transferred as so to form the high energy phosphate bond of ATP, is unknown. Mitchell proposed a chemiosmotic hypothesis in 1961. The three sites of "ATP synthesis" along the respiratory chain are actually sites where, at each, a pair of protons (2 H+) are pumped from the matrix to the intermembrane space. This creates a proton gradient. The protons flow back onto the matrix through special "lollypop" enzymes. As a pair of H+ flow back in down the proton gradient, one ATP is synthe-sized. Three pairs of H+ are extruded into the intermem-brane space per NADH, so three pairs of H+ can re-enter and thus produce 3 ATPs. The mechanisms is similar for FADH2, except since FADH2 donates their electrons at the second site, only two pairs reenter and two ATPs are synthesized.

The best or the most direct method of preparing aspirin from phenol is by use of the Kolbe reaction, which requires a knowledge of the chemistry of phenols. To plan this synthesis, first write your starting material, phenol, and end product, aspirin, and then think of reactions that direct you in a step by step manner to the desired end. These structures are shown in figure A. As mentioned, you want to employ the Kolbe reaction; it produces the carboxylate salt shown in figure B from phenol, which begins to look like aspirin. The procedure is shown in figure C. The hydrogen of the OH group in phenol is acidic, due to the fact that the ring delocalizes the unpaired electrons of oxygen in resonance structures. Thus, when base is present, you have a neutralization reaction to produce sodium phenoxide, a salt, and water. Now, that you have this material, perform the a salt, and water. Now, that you have this material, perform the Kolbe reaction. You have the reaction shown in figure D. If you acidify (by adding a mineral acid like HCI) , you obtain the reaction in figure E. Salicylic acid is produced. You are told that acetic acid is available. The only differ-ence between salicylic acid and aspirin is an ester linkage. Thus, if you perform an esterification reaction with acetic acid on the alcohol portion of salicylic acid, you obtain aspirin. By esterification you mean the reaction of a carboxylic acid (R-COOH) and an alcohol (R-OH) to produce an ester and water. Thus, you perform the reaction in figure F. The acid (H+) is a catalyst, which speeds up the process.

The father does work in pushing the swing. The work done increases the total energy of the swing and child as the swing moves from its initial position, \texttheta_0 = 1 radian, to the vertical position where the father stops pushing. From that point on, total energy of the swing and child re-mains constant. Work done by the father is found from W = \int F^\ding{217} \textbullet dr^\ding{217} = \int F dr cos \texttheta(1) where r is the distance along the path. W =^l(\texttheta)0\int_0 mg cos (0\textdegree) dr = mgl\texttheta_0 .(2) (a) Now we can use conservation of energy to find how high the swing goes. (Let the potential of the swing and child be O at the vertical position where the father lets go.) (PE at \texttheta_0) + W = PE at \texttheta = E_total = constant(3) mg (l \rule{1em}{1pt} l cos \texttheta_0) + mgl\texttheta_0 = mg (l \rule{1em}{1pt} l cos \texttheta) mgl (1 \rule{1em}{1pt} cos \texttheta_0 + \texttheta_0) = mgl (1 \rule{1em}{1pt} cos \texttheta) 1 \rule{1em}{1pt} 0.54 + 1 = 1 \rule{1em}{1pt} cos \texttheta cos \texttheta = 0.46 \texttheta = 63\textdegree above the horizon. (b)We need to find the tangential acceleration of the velocity of the swing at the bottom of its arc, which can be found by a second application of the conservation of energy principle as follows: PEat \texttheta+ W = KE_at _bottom = E_total(4) mgl (1 \rule{1em}{1pt} cos \texttheta) + mgl\texttheta_0 = 1/2 mv^2 mg l (1.46) + mgl = 1/2 mv^2 v^2 =2 gl (2.46) = 4.92 gl(5) Application of Newton's Law allows us to find the tangential acceleration, a_T . \sumF_T = ma_T mg + mg sin \texttheta_0 = ma_T(6) But mg sin \texttheta_0 the component of gravity in the tangen-tial direction, does not stay constant. It gets smaller as the angle decreases. Thus, the tangential acceleration is not constant. We can obtain an approximate solution by integrating the gravity force between O and \texttheta_0, and using that equivalent force. ^(\texttheta)0\int_O mg sin \texttheta d \texttheta = \rule{1em}{1pt}mg cos \texttheta │(\texttheta)0_O = 1 rad = [\rule{1em}{1pt}0.54 + 1] mg = 0.46 mg(7) Substituting this value for mg sin \texttheta_0 in Equation (6) yields mg + 0.46 mg = ma_T a_T = 1.46 g. Now to find how long the father pushed write a_T = dv / dt = 1.46 g ^v\int_O dV = 1.46 g^t\int_O dt t = (v/1.46g). Using the value of v from Equation (5) yields t = \surd[(4.92 gl) / (1.46 g)] t = 1.52 \surd(l/g) . The force decelerating the child will be the gravity force. This force will be of the magnitude given in equa-tion (7). Writing F = ma, 0.46 mg = ma a = 0.46 g. The time will again be given by t = v/a = (v/0.46g).(8) If there is no push, conservation of energy yields a velocity of \surd[2 gl (1 \rule{1em}{1pt} cos \texttheta_0)]. Substituting this in equa-tion (8) yields t = \surd[{2 gl (0.46)} / (0.46 g)] t = 2.09 \surd(l / g).

Question:

Are there comparable changes, in the male, to female menopause?

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Solution:

The changes that occur in the male due to aging are much less drastic than those which occur in the female. The testosterone and pituitary secretions which are initiated when the male reaches puberty continue throughout his adult life. Should the testosterone level decrease, this would result, through the switching-off of the negative-feedback inhibition on the pituitary, in a rise in the pituitary secretions of FSH and LH. Increased levels of FSH and LH stimulate testosterone production. Despite this, there is a steady decrease in testosterone secretion in later life, but this is probably caused by the slow deterioration of testicular function. Despite this decrease, testosterone secretion remains high enough in most men to maintain sexual vigor throughout life, and fertility has been documented in men in their seventies and eighties. Thus, there is usually no complete cessation of reproductive function in males analogous to female menopause.

Question:

A 2-ft-long string which can just support a weight of 16 lb is fixed at one end to a peg on a smooth horizontal surface. The other end is fixed to a mass of 1/2 slug. With what maximum constant speed can the mass rotate about the peg? (See figure).

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Solution:

If the tension in the string exceeds 16 lb, the string will break. Thus the maximum centripetal force that can be exerted on the mass is 16 lb. But if the mass is circling the peg with a velocity v, the centripetal force necessary to keep it in the circle is mv^2/R, where m is 1/2 slug and R is the length of the string, 2 ft. Thus 16 lb = T_max = (mv^2_max) / R = [(1/2) slug × v^2_max]/(2ft) Thereforev^2_max = 64 ft \bullet lb \bullet slug^-1 = 64 ft^2 \bullet s^-2 orv_max= 8 ft \bullet s^-1 .

Since the beam of light passes obliquely from an optically less' dense medium, air, to an optically denser medium, glass, the angle of incidence is larger than the angle of refraction. The equation for indices of refraction is (sin \texttheta_1/sin \texttheta_2) = (n_2/n_1) Since the index of refraction for air is approximately one, this reduces to n_2 = (sin \texttheta_1/sin \texttheta_2) = (sin 20.0\textdegree/sin 12.4\textdegree) = (0.342/0.215) = 1.58.

The point of junction between two bones is called a joint. Some joints , such as those between the bones of the skull, are immovable and extremely strong, owing to an intricate intermeshing of the edges of the bones . The truly movable joints of the skeleton are those that give the skeleton its importance in the totaleffectormechanism of locomotion. Some are ball and socket joints, such as the joint where the femur joins the pelvis, or where thehumerusjoins the pectoral girdle. These joints allow free movement in several directions. Both the pelvis and the pectoral girdle contain rounded, concave depressions toaccomodatethe rounded convex heads of the femur andhumerus, respectively. Hinge joints, such as that of the human knee, permit movement in one direction only. The pivot joints at the wrists and ankles allow free-dom of movement intermediate between that of the hinge and the ball and socket types. (Refer to Fig. 1.) The different bones of a joint are held together by connective tissue strands called ligaments. Skeletal muscles, attached to the bones by means of another type of connective tissue strand known as a tendon, produce their effects by bending the skeleton at the movable joints. The ends of each bone at a movable joint are covered with a layer of smooth cartilage . These bearing surfaces are completely enclosed in a liquid-tight capsule , called the bursa. The joint cavity is filled with a liquid lubricant, called the synovial fluid , which is secreted by the syn-ovial membrane lining the cavity. (Refer to Fig. 2) During youth and early maturity the lubricant is replaced as needed , but in middle and old age the supply is often decreased, resulting in joint stiffness and difficulty of movement. A common disability known as bursitis is due to the inflammation of cells lining the bursa, and also results in restrained movement.

The concentration of a solution is given in terms of molarity, i.e., moles/liter. This means that the number of moles of H_2 O_2 must be determined, since you already know the volume. To determine the number of moles of H_2 O_2, you know that, when oxidized, 1 mole of H_2 O_2 yields 1 mole of O_2. Thus, if you know the number of moles of O_2 produced, you also know the moles of H_2 O_2 originally present. To find the number of moles of O_2, you must use the equation of state. This tells us PV = nRT, where P = pressure, V = volume, n = number of moles, R = universal gas constant, (.082057 1-atm mol^-1 deg^-1)/ and T = temperature in degrees Kelvin (degree Celsius plus 273\textdegree). Hence, n_O(2) = (PV/RT). All the values needed to find n_O(2) are known or can be calculated. The pressure of O_2 will not be .941 atm, since this figure also includes vapor pressure from water. Therefore, the actual pressure of O_2 is .941 - .0373. That is, the difference between barometric and vapor pressure. Substituting, you obtain n_O(2) = [{(.941 - .0373) (3.94 × 10^-3)} / {(.082) (301)}] = 1.44 × 10^-4 moles of O_2. This means there were originally 1.44 × 10^-4 moles. Now, recalling the definition of molarity, which represents the concentration of a solution, you have [(1.44 × 10^-4 moles)/(.350 liters)] = 4.12 × 10^-4 M which is the concentration of the H_2 O_2 solution.

Question:

The volume of hydrogen evolved during the course of a reaction is measured by the displacement of water as shown in the diagram below. Hydrogen is evolved in flask A and displaces water from flask B into beaker C. If, during a particular run of this experiment in which atmospheric pressure is 765 torr and the water temperature is 293.15\textdegreeK, 65.0 ml of water is displaced, how much water would be displaced at 760 torr and 298.15\textdegreeK? (The equi-librium vapor pressure of water at 293.15\textdegreeK is 17.5 torr and at 298.15\textdegreeK is 23,8 torr.)

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Solution:

We can convert from the volume under non- standard conditions in the problem to the volume under standard conditions by means of the ideal gas equation, PV = nRT (where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature) . For our nonstandard conditions we may writeP_nV_n = nRT_n and for final conditions (760 torr and 298.15\textdegreeK) we may write P_sV_s = nRT_s Note that n and R are the same in both cases. Dividing the second equation by the first we obtain P_sV_s / P_nV_n = nRT_s / nRT_n = T_s / T_norV_s = [P_nV_n / V_s] × [T_s / T_n] . The pressure experienced by the hydrogen under non-standard conditions is equal to the pressure exerted by the atmosphere (765 torr) plus that from water vapor pressure at 20\textdegreeC (17,5 torr), or P_n = 765 torr + 17.5 torr = 782.5 torr. Under the final conditions, the pressure experienced by hydrogen is equal to the sum of the atmospheric pressure (760 torr) and the vapor press-ure of water at 298.15\textdegreeK (23.8 torr), or P_s = 760 torr + 23.8 torr = 783.8 torr. We can now calculate the volume of hydrogen under the final conditions. Thus, V_s = [P_nV_n / P_s] × [T_s / T_n] = [(765 torr × 65.0 ml) / 783.8 torr] × [298.15\textdegreeK / 293.15\textdegreeK] = 64.5 ml.

The small intestine is that region of the digestive tract between the stomach and the cecum. Its long, convoluted structure is an adaptation for absorption of nutrients. Structural modifications of the internal surfaces of the small intestine act to increase its surface area for absorption. First, the mucosa lining the intestine is thrown into numerous folds and ridges. Second, small fingerlike projections called villi cover the entire surface of the mucosa (see figure). These villi are richly supplied with blood capillaries and lacteals (for absorption of fats) in order to facilitate absorption of nutrients. Third, individual epithelial cells lining the folds and villi have a "brush-border" on the surface facing the lumen, consisting of countless, closely-packed, cylindrical processes known as microvilli. These microvilli add an enormous amount of surface area to that already present. The total internal surface area of the small intestine is thus incredibly large; this is advantageous for the purpose of absorption. The large intestine also has villi to increase its surface area. However, the number of villi present in the small intestine far exceeds that found in the large intestine. The main function of the large intestine is to absorb water from the undigested food substances and reduce the remains to a semi-solid state before it is expelled through the anus.

a) (i) The conversion for integer part consists of dividing by the desired base (e.g. 2) repeatedly and saving the remainder: Writing the remainders from right to left we have 10001 17_10 = 10001_2 Another format is the top-down division method as shown below: (ii) Proceeding in the same way for octal conversion As the binary, octaland hexadecimal systems can easily be converted from one form to another, it is only necessary to convert in any one of them and use the triad or quartet methods to obtain the results for remaining bases. Thus, 1321_10 = 2451_8 = 010 100 101 001_2 1321_10 = 010 100 101 001_2 = 0101 0010 1001_2 = 529_16 Thus 360_10 = 550_8 \textbullet Now 550_8 = 101 101 000_2 and 101 101 000_2 = 0001 0110 1000_2 = 168_16 d) (i) The fraction conversion procedure is exactly the opposite of the integer procedure .75_10 to octal .75 ×8 6.00 Take the digit 6, and save it; it's the first digit of the octal num-ber. Now take the decimal part .00 and again multiply it by 8. We get all 0's and hence 0.75_10 = 0.60_8 = 0.110_2 (ii) The binary conversion can be obtained directly or from converted octal numbers as shown above. (iii).75_10 = .110_2 and.110_2 = .1100_2 = .c_16 e) (i).390625_10 = X_8 Thus.390625_10= .31_8 .31_8 = .011 001_2 (ii).390625_10 = .011001_2 (iii).390625_10 = ?_16 Observe that the arrow connecting the 7.2, in the third line and the 3.2 in the eleventh line points out that after multiplying by 8 we shall have a repetition pattern of 1463. Hence, to convert .9_10 to X_8 , the precision must be known, say, up to the 9^th digit after the deci-mal point. Hence, .9_10 = (.7 1463 1463 ...)_8 . ii) The same will be the case for binary conversion. We convert to binary from the obtained octal number .9_10 = (.71463...)8 = (.111 001 100 110 011 ...)_2 For direct conversion iii) Now, to convert 0.9_10 into hexadecimal base, use the binary form already obtained. 0.9_10 = 0.1 1100 1100 1100 11... _2 = 0.1110 0110 0110 0110 ..._2 (grouping by four bits) = 0.E666 ...._16 = 0.E666_16 (up to 4_10 places after the point)

This example is used to illustrate the basic concepts of genetics and the methods of solving genetic problems. First one must understand the common nomencla-ture used in the study of inheritance: nucleus which contain the hereditary units, the genes. b) gene - the part of a chromosome which codes for a certain hereditary trait. c) genotype - the genetic makeup of an organism, or the set of genes which it possesses. d) phenotype - the outward, visible expression of the hereditary constitution of an organism. e) homologous chromosomes - chromosomes bearing genes for the same characters. f) homozygote - an organism possessing an identical pair of alleles on homologous chromosomes for a given character or for all given characters. g) heterozygote - an organism possessing different alleles on homologous chromosomes for a given character or for all given characters. In solving genetic problems, one uses letters to represent the genotype of the organism. For example, "a" represents the gene for blue color and "A\textquotedblright represents the gene for red color. A capital letter is used for a dominant gene; that is, the phenotype of that gene will be expressed in a heterozygous state. For example, if the genotype Aa is expressed as red, then A is the dominant gene. Small letter a represents the recessive gene; that is, one whose phenotype will be expressed only in the homozygous state. Therefore, aa would be expressed as blue. To solve a genetic problem, one writes down the genotypes of the two parents in the cross. Mendel's First Law tells us what to do. This law, also known as the Law of Independent Segregation, states that genes, the units of heredity, exist in individuals in pairs. In the formation of sex cells or gametes, the two genes separate and pass independent of one another into different gametes, so that each gamete has one, and only one, member of each pair. During fertilization, the gamete of one parent fuses with that of the other parent. Fusion brings the genes from each parent together, giving rise to offspring with paired genes. Now we will illustrate the Law of Segregation as it applies to the problem given. Let G represent the gene for green pod color and g rep-resent the gene for yellow pod color. Since the parents come from pure lines, meaning that the two members of each gene pair are identical, we write the genotype of the parent plant with green pods as GG, and that of the parent plant with yellow pods as gg. Each gamete from the first parent will have one G and each gamete from the second will have one g. (Recall, the Law of Independent Segregati-on). Writing out a schematic cross, we have: The genotype of the F_1 generation is written as Gg because it results from the fusion of the two gametes. We are told that all the F_1 generation from the above cross are green, and we observe that they are all heterozygous. Therefore G, which stands for the gene for green pods, must be dominant (by definition), and g, which stands for the gene for yellow pods, must be reces-sive. To determine the possible genotypes of the F_1 plants, or the second generation offspring, we mate two F_2 plants. The possible gametes derived from the parents are again obtained using Mendel's First Law. But now we obtain two types of gametes, G and g, from each parent, because either gene can come from the parental genotype of Gg. Schematically, then: It is easier to determine the F_2 generation using the Punnet square. The square is constructed as follows: It gives all possible combinations of the parental gametes, so in F_2 we have genotypically: 1 GG:2 Gg:1 gg, Phenotypically: GG is homozygous dominant and green because G is dominant; Gg is hetorozygous and green because G is dominant; and gg is homozygous recessive and yellow because g is recessive. It is important to note that we cannot observe the genotype itself because it lies in the constitution of the gene. Our observations come only from what we actually see, that is, the phenotypic differences. We know from the Punnet square that in the F_2 gene-ration, the ratio of homozygous dominant to the entire progeny is (1/4); the ratio of heterozygous is 2/4 or (1/2) and the ratio of homozygous recessive is Therefore, the number of homozygous dominant (GG) plants is 1/4 × 580or145, the number of heterozygote (Gg) plants is 1/2 × 580or290 and the number of homozygous recessive (gg) plants is 1/4 × 580or145. When added together, they total 580, the number of F_2 plants.

In order to trace this path, we must first outline some aspects of the human male and female repro-ductive systems, illustrated in Figures 1 and 2, respec-tively. In normal males, the testes lie in the scrotum, a sac attached to the lower anterior wall of the abdomen. The testes reside in the body cavity during early embryonic development, but before birth they descend into the cavities of the scrotum. The inguinal canals are connections be-tween the scrotum and body cavity; these canals are blocked off by connective tissue after the testes descend. The testes are located in the scrotal sacs because the sperm within them require cooler temperatures than the internal body temperature in order to survive and mature. Each testis consists of roughly 1,000 coiled tubules called seminiferous tubules. It is in these tubules that the germ cells produce the sperm. Sertoli cells are also present and nourish the developing sperm. Connected to the seminiferous tubules via fine tubes called vasa efferentia is the epididymis, which is a single, complexly coiled tube in which sperm are stored and mature. The epididymis, which is derived from the embryonic kidney, empties into the vas deferens. This duct passes from the scrotum through the inguinal canal into the abdominal cav-ity, over the urainary bladder to a point where it opens into the ejaculatory duct, which empties into the urethra, the duct that leads from the urinary bladder to the exterior. Thus in the male, the urethra serves as a common passageway for both reproductive and excretory functions. The urethra passes through the penis and is flanked by three columns of erectile tissue. This tissue becomes enlarged with blood during peroids of sexual excitement, causing the penis to become erect. The engorgement of the penis is caused by arterial dilation and increased blood flow at unchanged arterial pressure. Prior to ejaculation, sperm from the testes pass through the vasa efferentia into the epididymis. During ejaculation, sperm from the epididymis are moved through the vas deferens by peristaltic contractions of its walls. Fluids are added to the sperm at the time of ejaculation. These fluids come from three pairs of glands; the seminal vesicles, the prostate glands (which in the human are fused into a single gland), and Cowper's bulbour-ethral glands, The mixture of secretions from these three sets of glands is termed seminal fluid. The semen consists of the sperm and seminal fluid. The seminal fluid consists of mucus (from seminal and bulbourethral secretions) and nutrients (seminal secretions), in a milky alkaline fluid (prostatic secretions). During copulation, the male's penis is inserted into the female vagina and the semen is released there. Sur-rounding the vagina are the labia majora, two folds of fatty tissue covered by skin richly endowed with hair and sebaceous glands. These folds extend dorsally and down, enclosing the openings of the urethra and the vagina and merging behind them. The labia minora - thin folds of tissue devoid of hair - lie within the folds of the labia majora and are usually concealed by them. At the ventral junction of these two is the clitoris, a sensitive, erectile organ, which is the major site of female sexual excitement. The external female sex organs are collectively known as the vulva. The vagina is a single, muscular tube which extends from the exterior to the uterus. From the vagina, the sperm swim, by motion of their flagella, through the cervix, the muscular ring of the uterus which projects into the vagina and pass into the uterus. From there, they enter the Fallopian tubes (also known as the oviducts) where one may fertilize the secondary oocyte, if it is present. If fertilization does occur, the oocyte completes the second meiotic division and the zygote (fertilized egg) is formed. Cleavage of the zygote begins in the Fallopian tube and will have proceeded to a multicellular state by the time the egg enters into the uterus and is implanted.

Question:

A flexible wire 80 cm long has a mass of 0.40 gm. It is stretched across stops that are 50 cm apart by a force of 500 nt. Find the frequencies with which the wire may vi-brate.

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Solution:

The speed of a wave through a medium is determined by its elasticity and inertia. The elasticity is what causes a restoring force to act on any part of the medium displaced from its equilibrium position. The reaction of the displaced portion of the medium to the re-storing force depends on its inertia. for a stretched wire, the tension T is a measure of its elasticity; the greater the tension, the greater is the elastic restoring force on the displaced portion of wire. The inertia is measured by m, the mass per unit length of the wire. The speed of the wave has been found both analytically and experimentally to be, v = \surd(T/\mu) = \surd[(500 nt)/(0.40 × 10^\rule{1em}{1pt}3 kg/0.80 m)] = \surd[(10^6 m^2)/(sec^2)] = 1000m/sec The wire can only vibrate in a standing wave pattern, since it is stopped at two points (see figure). The incident wave is reflected at points A and B and rein-forces the later incident waves, producing the pattern shown. Hence, \lambda is related to L by, L = (n\lambda)/2(1) Hence, \lambda = (2L)/n Since the velocity of a sound wave is related to its fre-quency f and wavelength by v = f we obtain, (v/f) = (2L)/n or, f = (nv/2L) = [n(10^3 m/s)]/[(2) (50 × 10^\rule{1em}{1pt}2 m)] f = n × 10^3 s^\rule{1em}{1pt}1 Putting in values of n, we find, f_1 = 10^3 s^-1 f_2 = 2 × 10^3 s^\rule{1em}{1pt}1 f_3 = 3 × 10^3 s^-1 The other possible frequencies are the integral multiples of 1000 vibrations/sec, that is, 2000, 3000, 4000,...., vibrations/sec.

The total volume of liquid Q which flows across the entire cross section of a cylindrical tube in time t, is given by Poiseuille's law, Q = (\pi/8) [(R^4)/(\etaL)]∆p t where R is the radius, \eta is the viscosity of the liquid, and ∆p is the pressure difference between the two cross- sectional surfaces separated by a distance L. The water in the capillary flows as a result of the pressures due to the water in the cylinder and its own weight, as shown in the figure. Under its own weight, it would have a rate of flow given by, Q_1/t = (\pi/8) [(R^4)/(\etaL_2)] (\rhogL_2) where \rho is the density of water and g is the gravitational acceleration. The weight of the water exerts a pressure \rho gL, on the upper cross-section of the capillary and gives rise to another pressure difference between the two ends of the capillary. The rate of flow due to this pres-sure difference is, Q_2/t = (\pi/8) [(R^4)/(\etaL_2)] (\rhogL_1) The total rate of flow is therefore Q/t = Q_1/t + Q_2/t = (\pi/8) [(R^4)/(\etaL_2)] \rhog(L_1 + L_2) The quantity of water, Q, flowing from the capillary in time t causes a drop in the level of the cylinder, h. The area of the cylinder is A, and thus Q/t = Ah/t. \thereforeh/t = (\pi/8A) (R^4/\eta) [\rhog(L_1 + L_2)]/L_2 When the cylinder is full, L_1 = 25 cm, and h/t = [\pi × (0.25 × 10^-3m)^4 × 10^3 kg/m^3 × 9.8m/s^2 × 0.65m]/[8 × \pi/4 × 10^-2m^2 × 10^-3Ns/m^2 × 0.40m] = 3.11 x 10^-6 m/s = 1.12cm/hr When the cylinder is empty,L_1 = 0 cm, and, h'/t = [(\piR^4)/(8A\etaL_2)] \rhogL_2 = [(\piR^4)/(8A\etaL_2)] (L_1 + L_2) \bullet (L_2)/(L_1 + L_2) = (h/t) (L_2)/(L_1 + L_2) = (h/t) (0.40/0.65) = 1.12 cm/hr \bullet(0.40/0.65) = 0.69 cm/hr Thus hour divisions are separated by 1.12 cm at the top and 0.69 cm at the bottom. Note that L_1 varies slightly during the hour and, to be quite exact, an integration ought to be performed. The error involved is, however, slight, since the variation in L_1 is very small in compari-son with L_1 + L_2.

Since this problem deals with the pressure and volume of a gas at a constant temperature, Boyle's law can be used. Boyle's law states that the volume, V, of a given mass of gas, at constant temperature, varies inversely with the pressure, P. It can be stated as V = k × 1/p, where k is a constant. Hence,k = PV For a particular system, at constant temperature, k is constant. Therefore, if either the pressure or the volume is changed, the other must adjust accordingly. Here,P = 760 TorrandV = 100 m1so k = 760 × 100 = 76000Torr-m1. If P is doubled to 1520 Torr, then k = 76000 Torr-ml = 1520 Torr × V v = [(76000 Torr-ml)/(1520 Torr)] = 50 m Since k is a constant for a given system, another form of Boyle's Law can be expressed as P_1 V_1 = P_2 V_2 This says that the pressure of the original system times the volume of the original system is equal to the new pressure times the new volume. Here, 760 torr × 100 m1 = 1520 torr × V_2 50 m1 = V_2.

Let the cylinder have a cross-sectional area A, length 1, and density \rho , and let it float in the mercury of density \rho' immersed to a height y. In this position, shown by the solid line in the diagram, the cylinder is in equilibrium. Hence, the net force on the cylinder, composed of its weight (F_G) and the buoyant force (F_B) is zero. Then, taking the positive direction downward, F_G - F_B = 0 orF_G = F_B(1) ButF_G = m_c g where m_c is the cylinder mass. Since \rho = m_c / [Volume of cylinder (V)] m_c = \rhoV = \rhoAl andF_G = \rhoAlg The buoyant force is equal to the weight of fluid displaced by the object. Hence F_B = m_mg where m_m is the mass of mercury displaced. But m_m = \rho' V' where V' is the fluid volume displaced. Hence, m_m = \rho' A(l - y) andF_B = \rho' A(l - y) g(3) Using (3) and (2) in (1) \rhoAlg = \rho' Ayg If the cylinder is pushed in a further distance x, the up thrust is greater than the weight, and there is a restoring force attempting to return the cylinder to its original position. If a is the downward acceleration, we find, from Newton's Second Law F_net = m_ca Again,F_net = F_G - F_B = m_ca or\rhoAlg - \rho'A(y + x)g = \rhoAla Since\rhoAlg = \rho'Ayg -\rho' Axg = \rhoAla anda = - (\rho'g)/(\rhol) x Comparing this with the equation of motion of a simple harmonic oscillator, we realize that \omega^2 = (\rho'g)/(\rhol) since T = 2\pi/\omega, where T and \omega are the period and angular frequency of the motion, we note that \omega = 2\pi/T and(4\pi^2)/(T^2) = (\rho'g)/(\rhol) \rho = (\rho'gT^2)/(4\pi^2l) = [(13,600 kg \bullet m^-3) (\pi^2 m \bullet s^-2)(.56)^2 s^2]/[(4\pi^2) (.14 m) \rho = 7616 kg \textbullet m^-3.

Question:

Calculate the uncertainty in the location of a 100 g turtle crossing a road at a velocity of 1 mm/sec, the uncertainty of the latter being 10^-4 mm/sec. (h = 6.626 × 10^-27 erg-sec and 10 mm = 1 cm.)

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Solution:

This problem demands the use of the Heisenberg Uncertainty Principle. This principle states that one can-not simultaneously locate both the position of a particle and its momentum with absolute precision. Whenever one quantity, either position or momentum, is known precisely, the other is known less precisely. Heisenberg showed that the uncertainty in the determination of momentum of a particle, ∆p, and the simultaneous determination of its location, ∆x is related by the equation ∆p∆x\geq h, where h is Planck's constant. Momentum p is equal tomv, where m is mass and v the velocity of the mass, thus ∆p =m∆v. One substitutes this into the Heisenberg inequality attainingm∆v∆x\geq h. One can now solve for ∆x. 1 erg = g cm^2/sec^2, 1 mm = .1 cm. ∆x = h/(m∆v) = [(6.626 × 10^-27 erg-sec) / {(100 g) (10^-4 mm/sec)}] \bullet [(1mm)/(.1 cm)] = 6.626 × 10^-24 cm.

Question:

Water flows into a tank at a constant rate Q. Q is determined by the position of a valve connected to the inflow pipe. Write a digital computer program to simulate this system from time t = 0 to t = t_f if V(O) = V_O . (Consider the walls to be of infinite height, so that overflow is not possible.)

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Solution:

Following the steps outlined in the previous problems: 1.Changes in this system's state can be viewed through changes in the volume V of water in the tank. 2 & 3. Once the valve position is set, Q is determined and will remain unchanged. Thus, it can be considered external to the system consisting of water in the tank and water flowing into the tank. Q is changed only when a new simulation run is being made with a new parameter Q. 4. Using the notation of the preceding problems, the behavior of the system can be diagrammed as shown in Fig. 2 [Q] \rightarrow [V] Fig. 2 V is the accumulation of the rate of change Q (when Q has units of Volume/time). To conform more closely with notation previously developed, let V̇ = the constant Q, and since accumulation of rates of change is mathematically expressed as integration, the complete diagram is shown in Fig. 3 From this, the equation can be written: V̇ = Q or dV/dt = Q(1) 5. Since the solution of (1) can easily be seen to be: V(t) = Q {_\ast} T + V_0(2) ȧ simple program would just substitute values of t into (2) and output V. A more general program would use the Euler predictor-corrector algorithm to solve the first-order differential equation (1). But since the rate of change is constant, the corrected value would always equal the predicted value. The method used here directly simulates the behavior of the system according to observed changes in V for small, discrete changes in t: V_new= V_old + V_added(3) whereVadded= Q {_\ast} change in time = Q {_\ast} (t_new - t_old) = Q {_\ast} ∆t This is the original Euler's method (generally, it is less accurate than This is the original Euler's method (generally, it is less accurate than predictor-corrector method).Thus V_Nis calculated from V_N-1. If V_0 is predictor-corrector method). is calculated from V_N-1. If V_0 is always used for Voldand t_0 = 0 is always used for t_old equation (3) would be reduced to equation (2). The program is as follows: DATA T/0.0/ READ N, TFIN,Q,VOL PRINT T, VOL REALN = N CUSING REALN AVOIDS MIXED-MODE ARITHMETIC DT = (TFIN - T)/REALN DO 50 I = 1,N T = T + DT VOL = VOL + Q {_\ast} DT PRINT T, VOL 50CONTINUE STOP END

The Golgi apparatus is composed of several membrane-bounded, flattened sacs or cisternae arranged in parallel array about 300 \AA apart (see Figure 1) . The sacs are disc-like and often slightly curved. Note the concavity on the trans face near the plasma membrane and the convexity of the cis face are thinner (more like reticulum membrane than like plasma membrane). The function of the Golgi apparatus is best understood in cells involved in protein synthesis and secretion. The protein to be secreted is synthesised on the rough endoplasmic reticulum. Vesicles containing small quan-tities of the synthesized protein bud off from the endplasmic reticulum. These vesicles carry the protein to the convex face of the Goldi complex. In the Goldi appar-atus, the protein is concentrated by the removal of water. In addition, chemical modifications of the protein, such as glycosylation (addition of sugar) occur. The modified protein is released from the concave surface in the form of secretory granules. The secretory granules containing the protein are from the cytoplasm by a membrane that can fuse with the plasma membrane, its content (protein in this case) is expelled from the cell, a process known as excytosis. This process is outlined in Figure 2. Most of the cell organelles are found in a specific arrangement within the cell to complement their function. For example, the Golgi apparatus is usually found near the cell membrane and is associated with the endoplasmic reticulum. Since they are relatively close to each other, transport of materials between them is considerably efficient.

Question:

What are gallstones and how are they formed?

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Solution:

Gallstones contain about 80% cholesterol. Cholesterol is a steroid molecule that is absorbed from certain digested fats, such as butter, and is also produced in the liver. It is a precursor of many of the body's important steroid hormones (see fig.) Even though cholesterol is vital for proper body functioning, it is often not handled well by our bodies. This is because it is almost completely insoluble in body fluids. Excess cholesterol in the blood may be deposited along the arterial walls and in the heart. This can lead to arteriosclerosis and ultimately, to heart disease. The body's major mechanism for the removal of cholesterol from the blood takes place in the liver, which releases cholesterol into the intestine as part of the bile. Bile contains bile salts and lecithin that emulsify lipids such as cholesterol. In the gallbladder, bile is stored and slowly concentrated by the removal of water. As this happens, the concentration of cholesterol increases to a level near the saturation point. If there is an excess of cholesterol, or insufficient amounts of bile salts or lecithin to keep it soluble, the cholesterol will precipitate, forming a hard solid mass called a gallstone. If a gallstone is small, it will pass through the bile duct and out into the intestine, causing no complications. A larger stone, however, may lodge in the neck of the gall-bladder. This leads to pain and contractile spasms of the gallbladder wall. Since the gallbladder is off to the side of the bile duct, bile can still flow from the liver to the intestine for digestion. If the spasms of the gall-bladder are powerful enough, the stone may be forced out of the gallbladder and become lodged in the bile duct, obstructing it. Complete blockage may lead to a condition known as jaundice, in which the skin yellows due to an accumulation of bile pigments in the body. Since no bile reaches the intestine, there is also decreased emulsification and absorp-tion of fats. Gallstones may sometimes lodge at the point where the bile duct and pancreatic duct join before they enter the duodenum. When this happens, both bile and pancreatic juice are blocked in their passages, resulting in severe impairment of both digestion and absoprtion in the intestinal tract. Gallstones may be removed surgically if they are not dissolved by the body. If necessary, the gallbladder may be removed without any great effect on digestion.

Question:

A newly discovered planet has twice the density of the earth, but the acceleration due to gravity on its surface is exactly the same as on the surface of the earth. What is its radius?

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Solution:

This problem must be approached carefully. We must express the acceleration due to gravity in terms of the density and the radius of the planet. If the radius is R and the mass of the planet M, then the acceleration due to gravity on its surface is found from Newton's Second Law, F = ma. Consider an object of mass m on the surface of the planet. Then the only force on m is the gravitational force F, and F =GMm/R^2 But a is the acceleration of m due to the planet's gravitational field, org_p. Then g_v= GM/R^2 Assuming the planet is spherical, its volume is the volume of a sphere of radius R: V = (4/3)\piR^3 Since Mass = Volume × Density. M = (4\piR^3\rho/3) where \rho (the Greek letter rho) is the density of the planet. Therefore g_v= {G(4/3)\piR^3\rho}/R^2 = (4\pi/3)GR\rho Similarly, the acceleration due to gravity on the surface of the earth is g= (4/3)\piGR_e_(\rho)e where\rho_eis the density of the earth, and R_e is its radius. If g_v= g Then (4/3)\piGR\rho= (4/3)\piGR_e_(\rho)e Canceling (4/3)\piGon both sides R\rho= R_e(\rho)e If the density of the planet is twice that of the earth, \rho = 2\rho_e So R2\rho_e = R_e(\rho)e Whence R = (1/2)R_e = (1/2) × 6.38 × 10^6 m = 3.19 × 10^6 m The radius of the planet is one half of the radius of the earth, or 3.19 × 10^6 meters.

Question:

What is the radius of the orbit of a 1-MeV proton in a 10^4 - gauss field?

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Solution:

An electron which is moving in a magnetic field B^\ding{217} with velocity v^\ding{217}, experiences a force F^\ding{217} = - (1/c) ev^\ding{217} × B^\ding{217} . As a result of this force, the electron moves on a circular path in the plane of v^\ding{217}. The radius of the orbit is obtained from the expression for the centripetal force F_center = F = [(mv^2 ) / R] or(1/c) evB = [(mv^2 ) / R] R = [(mvc) / eB] . The velocity of a 1- MeV proton is obtained as follows KE = total energy - rest energy = [(m_0 c^2)/{\surd(1 - \beta^2 )}] - m_0 c^2 = [1/{\surd(1 - \beta^2)}] m_0c^2 = 1 MeV = 1.6 × 10^-6ergs The rest energy of a proton is m_0c^2 = 1.67 × 10^-24 gr × (3 × 10^10 cm/ sec)^2 = 1.5 × 10^-3ergs Since KE << m_0 c^2 , we have a non - relativistic motion. In this case, \beta = (v/c) << 1, hence we can use the approximation [1/{\surd(1 - \beta^2 )}] \approx 1 + (1/2) \beta^2 and (1/2) \beta^2 \approx (KE)/(m_0 c^2 ) =[(1.6 × 10^-6 ergs)/(1.5 × 10^-3ergs)] = 1.07 × 10^-3 \beta = \surd(2.14 × 10^-3 ) = 0.046 v = \betac = (0.046 × 3 × 10^10 cm/sec = 1.38 (10^9 cm/sec) In the expression for R we can use the nonrelativistic (rest mass) m^0 for m.Therefore R \approx (m_0 vc)/(eB) Since m_0 = 1.67 × 10^-24ge = 4.8 × 10^-8stat c or R = [(1.67 × 10^-24 g) × (1.38 × 10^9 cm/sec) × (3 × 10^10 cm/sec)] / [(4.8 × 10^-10 statC) × 10^4 gauss] = 14.4 cm

As in the previous problem, a similarity can be noticed between the given differential equation and that of a differential equation describing harmonic motion (damped). Using the same equi-valences with the addition of . R\rightarrow D, the equation of the damped system can be constructed from the equation of this problem. The programmer needs only to substitute the following as input data in the previous program: USE THE VALUER as D USE THE VALUE(1/C) as K USE THE VALUE(q_0) as XCOR USE THE VALUE (i_0) as VCOR

Question:

Algae may be microscopic single-celled organisms or enormousmulticellularseaweeds hundreds of feet long. The body of one of these organisms is termed athallus. How does athallusdiffer from a plant body?

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Solution:

A plant body shows differentiation of parts, with its roots, stems, and leaves all varying greatly in structure and in function.Multicellular algae show very little, if any, tissue differentiation. For this reason, the body is termed athallus(from the Greekthallos: a young, undifferentiated shoot or sprout). There is no anatomical basis for distinguishing the leaves or roots ofmulticellularalga, though in some species there may be a functional basis. The plant kingdom has traditionally been divided into two groups: theThallophyta(algae) and theEmbryophyta(land plants). The Embryophyta , in addition to showing tis-sue differentiation, have a life cycle different from that of theThallophyta. The reproductive structures of theThallophytesare usually unicellular, and the reproductive cells lack a protective surrounding wall or jacket of non-reproductive cells.Thallophytezygotes do not develop into embryos until after they are released from the parentthallus. In Embryophyta , the reproductive structures aremulti-cellularand are surrounded by non-reproductive cells. The early stages of embryonic development occur while the zygote is still contained within the parent plant. The body of a fungus is also termed athallus. Fungi, like algae, show little differentiation of parts, and lackmulticellularsex organs with jacket cells. The embryo of a fungus develops outside the parent fungus.

Question:

Show that the principle of conservation of momentum results Show that the principle of conservation of momentum results from the principle of conservation of kinetic energy and the Galilean trans-formation

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Solution:

Suppose we observe a collision from two separate frames of Suppose we observe a collision from two separate frames of reference, S and S', moving with a relative velocity V^\ding{217}see figure (B). see figure (B). The position of a particle relative to S is related to its position relative to S' by see figure (A) . r^\ding{217} = r'^\ding{217} + R^\ding{217}(1) Differentiating (1) with respect to time, v^\ding{217} = v'^\ding{217} + V^\ding{217}(2) where v'^\ding{217} and v^\ding{217} are the velocity of the particle relative to S' and S respectively. Equations (1) and (2) constitute the Galilean trans-formation. We apply the principle of conservation of kinetic energy in frame S see figure (B),or (1/2)M_aV^\ding{217}_aiV^\ding{217}_ai + (1/2)M_bV^\ding{217}_bi \bulletV^\ding{217}_bi = (1/2)M_aV^\ding{217}_afV^\ding{217}_af + (1/2)M_bV^\ding{217}_bf.V^\ding{217}_bf(3) But, from (2) v^\ding{217}_ai = v'^\ding{217}ai+V^\ding{217}v^\ding{217}_af = v ׳ \ding{217} af +V^\ding{217} v^\ding{217}_bi = v ׳ \ding{217} bi +V^\ding{217}v^\ding{217}bf= v'^\ding{217}bf+V^\ding{217}(4) Substituting (4) in (3) (1/2)M_a(v'^\ding{217}_ai + V^\ding{217})\bullet (v'^\ding{217}_ai + V^\ding{217}) + (1/2)M_b(v'^\ding{217}_bi -+ V^\ding{217})\bullet (v'^\ding{217}_bi + V^\ding{217}) = (1/2)M_a(v'^\ding{217}_af + V^\ding{217})\bullet (v'^\ding{217}_af + V^\ding{217}) + (1/2)M_b(v'^\ding{217}_bf + V^\ding{217})\bullet (v'^\ding{217}_bf + V^\ding{217}) or(1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi + (1/2)(M_a + M_b)(V^\ding{217}\bulletV^\ding{217}) + M_av'^\ding{217}_ai\bulletV^\ding{217} + M_bv'^\ding{217}_bi \bulletV^\ding{217} = (1/2)M_av'^\ding{217}_af \bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf\bulletv'^\ding{217}_bf + (1/2)(M_a+M_b)(V^\ding{217}\bulletV^\ding{217}) + M_av'^\ding{217}_af \bulletV^\ding{217} + M_bv'^\ding{217}_bf \bulletV^\ding{217} Rewriting this last equation: (1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi + (M_av'^\ding{217}_ai + M_bv'^\ding{217}_bi)\bulletV^\ding{217} = (1/2)M_av'^\ding{217}_af \bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf \bulletv'^\ding{217}_bf + (M_av'^\ding{217}_af + M_bv'^\ding{217}_bf)\bulletV^\ding{217}(5) Now, if conservation of kinetic energy holds (1/2)M_av'^\ding{217}_ai \bulletv'^\ding{217}_ai + (1/2)M_bv'^\ding{217}_bi \bulletv'^\ding{217}_bi = (1/2)M_av'^\ding{217}af\bulletv'^\ding{217}_af + (1/2)M_bv'^\ding{217}_bf\bulletv'^\ding{217}bf and (5) becomes (M_av'^\ding{217}_ai + M_bv'^\ding{217}_bi) \bulletV^\ding{217}_ = (M_av'^\ding{217}_af + M_bv'^\ding{217}_bf) \bulletV^\ding{217}(6) But, sinceV^\ding{217}is an arbitrary vector (6) holds for allV^\ding{217}and we obtain the V^\ding{217} V^\ding{217} principle of conservation of momentum in frame S' M_av'^\ding{217}_ai + M_bv'^\ding{217}bi= M_av'^\ding{217}_af + M_bv'^\ding{217}_bf(7) M_av'^\ding{217}_ai + M_bv'^\ding{217}bi= M_av'^\ding{217}_af + M_bv'^\ding{217}_bf(7) Note also that S' is an arbitrary frame, and, therefore, (7) holds in all frames.k

Because there are two possible ways in which the chromosomes can line up and segregate in meiosis I, there are two possible sets of gametes that could be formed by meiosis. Each set has 2 pairs of identical gametes. There are thus 4 types of gametes possible. We know this from looking at the genotypes of the parental cell (Aa Bb) . Four combinations of these genes are possible:AB, Ab, aB, and ab. All these gametes have been produced in the two possible meiotic divisions. It is important to note, however, that each separate meiosis produced only two types of gametes - AB and ab or Ab and aB - and only in these combinations . Gametic genotype combinations such as AB and Ab or aB and ab could not be produced in the same meiotic division. Although there are eight chromosomes in total, the separation of homologs occurs as a function of duplicated chromosomes. Thus, if in meiosis I, AA is paired with BB, aa must be paired with bb. If AA is paired with bb, then aa must be paired with BB. The genotypes of the gametes produced in a meiotic division are determined in meiosis I, and meiosis II is merely a mitotic type division resulting in the separation of identical daughter chromosomes. In meiotic gamete formation, the randomness of the way the chromosomes line up is one method that ensures genetic variation. Diverse gamates, differing in their combinations of chromosomes, are produced by the independent assortment of chromosomes when they first segregate in meiosis. In a cell with many chromosomes, such as a human cell with 23 pairs, the number of possible gametes that could be produced is very large indeed.

Question:

Which of the following compounds are saturated? Which are unsaturated? (a) CH_3CH_2CH_2CH_3, (b) CH_2 = CHCH_2CH_3, (c) cyclohexane, (d) cyclohexene, (e) benzene.

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Solution:

The most fundamental class of organic compounds is the saturated hydrocarbons. A hydrocarbon is a compound that contains only carbon and hydrogen. Saturated is used to describe the absence of double bonds and / or triple bonds. The term comes from the fact that these compounds do not react with hydrogen because they are saturated with hydrogen. When a double and/or triple bond is present, the compound is termed unsaturated. Thus, compounds a and c are saturated and compounds b, d, and e are unsaturated.

To solve this problem you need to write out the structures of the initial and final compounds. In any synthesis, think of reactions that lead you step by step to the desired result. If you had nitrobenzene, you could go directly to aniline because nitrobenzene can be reduced with metal and acid to aniline. To obtain nitrobenzene, perform a nitration reaction, as shown below in figure B. (2) Ethyl amine \equiv C_2H_5NH_2, Ethane \equiv C_2H_6. Now, if you had ethyl chloride, C_2H_5CI, you could obtain ethyl amine directly, since upon the addition of ammonia (NH_3) ethyl chloride is converted to the corresponding amine. To obtain ethyl chloride, halogenate ethane with Cl_2 and light. C_2H_6 + Cl_2C_2H_5CI + HCI C_2H_5CI + NH_3 \rightarrow C_2H_5NH_2 + HCI .

Question:

A workman sitting on top of the roof of a house drops his hammer. The roof is smooth and slopes at an angle of 30\textdegree to the horizontal. It is 32 ft long and its lowest point is 32 ft from the ground. How far from the house wall is the hammer when it hits the ground?

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Solution:

Figure A illustrates the first part of the motion. Two forces are acting on the hammer as it slides down the roof; the weight mg\ding{217} acting downward, one component of which, mg cos \texttheta, balances the second force, the normal force exerted by the roof. At the same time, the component parallel to the roof, mg sin \texttheta, is unbalanced and produces the acceleration on the hammer. Apply Newton's second law to the unbalanced force to obtain mg sin \texttheta = ma. Thus the hammer accelerates down the roof with acceleration a = g sin \texttheta . In this case sin \texttheta = sin 30\textdegree = (1/2). The kinematic relation for constant acceleration which does not involve time is used to find the velocity with which the hammer leaves the roof. It is v^2 = v^2_0 + 2as, where v_0,the initial velocity; is 0 and s is the distance the hammer moves on the roof (= 25 ft). Hence, v is obtained from v^2 = 2 × (32/2) ft/sec^2 × 32 ft; that is, v = 32 ft/sec. In the second stage of the fall, the hammer undergoes projectile motion. It drops 32 ft in time t while traveling a distance x horizontally. Let the positive direction of y be taken as downward, and resolve v into its vertical and its horizontal com-ponents: v sin \texttheta and v cos \texttheta, respectively (see fig. B). \texttheta is the same as the angle of the slope of the roof. Since there is no horizontal component of force acting on the hammer when it leaves the roof, there is then no horizontal acceleration. The kinematic equation for constant velocity is then x = (v cos \texttheta)t. The vertical acceleration is the constant acceleration of gravity g. Therefore y = (v sin \texttheta)t + (1/2) gt^2 where t = x/v cos \texttheta. y = v sin \texttheta [x/(v cos \texttheta)] + (g/2) × [(x^2)/(v^2 cos^2 \texttheta)]. [{(x^2 × 32 ft)/sec^2}/{2 × (32 ft/sec)^2 × (3/4)}] + (x/\surd3) - 32 ft = 0 orx^2 + 16 \surd3 × ft - 1536 ft^2= 0 (x + 32 \surd3 ft) (x - 16 \surd3 ft) = 0 x = - 32 \surd3 ftor+16 \surd3 ft. The negative answer is clearly inadmissible. It is the answer that would result if the direction of projection were reversed. Hence the correct answer is x = 16 \surd3 ft = 27.7 ft from the house.

A first representation of the system might be the flow graph on figure #1. The rates of flow out of the sources and into the sinks are determined by natural "valves". The rates of flow between N_1 and N2 are determined by valves which are controlled by density levels. Using this information, a detailed diagram of the system is shown on figure #2. The equations for the natural valves are equations (1)-(6). The equation for the inter-region population flow is Q = C (N_1/A - N_2/B)(7) where C is a proportionality constant. Density at A > Density at B \Rightarrow Q > 0. Therefore, the arbitrarily assumed direction of Q is toward B, away from A, and the complete state equations are : Ṅ_1^\textasteriskcentered = Ṅ_1 - Q(8) N_2^\textasteriskcentered= N_2 + Q(9) Case (1):Substitution of equations (1) and (2) into equations (8) and (9), respectively, yields: Ṅ_1^\textasteriskcentered= rN_1 - Q Ṅ_2^\textasteriskcentered = (b - mN_2/2N_20)N_2 + Q Substituting the formula for Q given by equation (7): Ṅ_1\textasteriskcentered = rN_1 - C(N_1/A - N_2/B) = (r - C/A)N_1 + (C/B)N_2 = C_11N_1 + C_12N_2(10) where C_11 = r - C/A and C_12 = C/B can be evaluated outside of the simulation loop, reducing the number of operations per iteration by three: Ṅ_2^\textasteriskcentered = (b - mN_2/2N_20)N_2 + C(N_1/A - N_2/B) = (C/A)N_1 + (b - C/B)N_2 - (m/2N_20)N^2_2 = C_21N_1 + C_22N_2 - C_3N^2_2(11) where C21= C/A, C_22 = b - C/B, C_3 = m/2N_20 . For this coupled, non-linear first-order system, the modified method is first applied to equation (10) and then to equation (11). A logical flag, TWO, is used to determine which equation is being integrated. The variable BIRTH is used for b: REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2, N10,N20,A,B,BIRTH,C,C11,C12,C21 C22,C3,M,R INTEGER I,N LOGICAL TWO/.FALSE./ COMMON N1,N2,TW0,Cll,C12,C21,C22,C3 READ, N, TFIN, ACCUR, N10,N20, A, B, BIRTH, C,M,R PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = R - C/A C12 = C/B C21 = C/A C22 = BIRTH - C/B C3 = M/(2.\textasteriskcenteredN20) N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1,ACCUR,DT) TWO = .TRUE. CALL MEULER(T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,Cll,C12,C21,C22,C3 LOGICAL TWO COMMON N1,N2,TWO,Cll,C12,C21,C22,C3 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 IF (TWO) G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW - C3\textasteriskcenteredW\textasteriskcentered\textasteriskcentered2 RETURN END Case (2): Substituting equations (3) and (4) into equations (8) and (9), respectively, yield: Ṅ_1^\textasteriskcentered = K(S - N_1) - Q Ṅ_2^\textasteriskcentered = r_2N_2 + Q Substituting the formula for Q gives: Ṅ_1^\textasteriskcentered = K(S - N_1) - C(N_1/A - N_2/B) = -(K + C/A)N_1 + (C/B)N_2 + KS = C_11N_1 + C_12N_2 + C_3 where C_11 = -(K + C/A), C_12 = C/B , C_3 = KS Ṅ_2^\textasteriskcentered = r_2N_2 + C(N_1/A - N_2/B) = (C/A)N_1 + (r - C/B)N_2 = C_21N_1 + C_22N_2 where C_21 = C/A, C22= r - C/B. Using the same procedure as used in Case(1): REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2, 1N10,N20,A,B,C,K, R, S,C11,C12,C21 1C22,C3 INTEGER I,N LOGICAL TWO/.FALSE./ COMMON Nl,N2,TWO,Cll,C12,C21,C22,C3 READ,N,TFIN,ACCUR,N10,N20,A,B,C,K,R,S PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = -(K + C/A) C12 = C/B C3 = K\textasteriskcenteredS C21 = C/A C22 = R - C/B N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER (T,N1,ACCUR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT, T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,C11,C12,C21,C22,C3 LOGICAL TWO COMMONN1,N2,TWO,C11,C12,C21,C22,C3 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 + C3 IF (TWO)G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW RETURN END Case (3):Substituting equations (5) and (6) into equations (8) and (9), respectively , yield: Ṅ_1^\textasteriskcentered = K(S - N_1 ) - Q = -(K + C/A)N_1 + (C/B)N_2 + KS = C_11N_1 + C_12N_2 + C_3 where C_11 = -(K + C/A), C_12 = C/B, C_3 = KS Ṅ_1^\textasteriskcentered = (b - mN_2/2N_20)N_2 + Q = (C/A)N_1 + (b - C/B)N_2 - (m/2N_20)N^2_2 = C_21N_1 + C_22N_2 - C_4N^2_2 where C_21 = C/A, C_22 =b - C/B, C_4 = m/2N_20 . Using the same procedure as used in Case (1), and using the variable BIRTH for b : REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1,N2,N10, 1N20,A,B,BIRTH,C,C11,C12,C21,C22,C3,C4, 1K,M, S INTERGER I,N LOGICAL TWO/.FALSE./ COMMON N1, N2, TWO, C11, C12, C21, C22, C3, C4 READ, N, TFIN, ACCUR, N10, N20, A, B, BIRTH, C, K, M, S PRINT,T,N10,N20 REALN = N DT = TFIN/REALN C11 = -(K + C/A) C12 = C/B C3= K\textasteriskcenteredS C21 = C/A C22 = BIRTH - C/B C4 = M/(2\textasteriskcenteredN20) N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER (T,N1,ACCUR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,C11,C12,C21,C22,C3,C4 LOGICAL TWO COMMON Nl,N2,TWO,C11,C12,C21,C22,C3,C4 G = C11\textasteriskcenteredW + C12\textasteriskcenteredN2 + C3 IF (TWO) G = C21\textasteriskcenteredN1 + C22\textasteriskcenteredW - C4\textasteriskcenteredW\textasteriskcentered\textasteriskcentered2 RETURN END

Question:

In corn, the gene R for red color is dominant over the gene r for green color. The gene N for normal seed is dominant over the gene n for tassel seed. A fully heterozygous red plant with normal seeds was crossed with a green plant with tassel seeds, and the following ratios were obtained in the offspring: 124 red, normal: 77 red, tassel: 126 green, tassel: 72 green, normal. Does this indicate linkage? If so, what is the map distance between the two loci?

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Solution:

Let us look at the cross, F RN RN rN rn rn RrNn Rrnn rrNn rrnn 124 red, normal (RrNn) - parental 77 red, tassel (Rrnn) - recombinant 73 green, normal (rrNn) - recombinant 126 green, tassel (rrnn) - parental The phenotypic ratio is indeed indicative of linkage. The two parental types (RrNn and rrnn) are found in much greater numbers than the recombinant types (Rrnn and rrNn). The recombination frequency (RF) is: RF = [(total recombitants) / (total offspring)] = (73 + 77) / (124 + 77 + 126 + 73) = (150) / (400) = .375 An RF of .375 would give a map distance of 0.375 × 100 or 37.5 map units. The two loci are thus located on the same chromosome and are 37.5 map units apart.

In 7 min 100 g of water are raised in temperature by (100 -16) \textdegreeC = 84\textdegreeC. The amount of heat provided by the heater to the water is, Q_W = C_Wm_W∆T where C_W is water's specific heat, and M_W is the mass of water. Q_W = (1 cal/gr.\textdegreeC) (100gr) (100\textdegreeC - 16\textdegreeC) = 8.4 × 10^3 cal = (8.4 × 10^3 cal) × (4.186 J/cal) = 3.52 × 10^4 J The rate of delivery of the heat energy is, P = Q_W / t =Q_W / [(7 min) (60 sec/min)]= [(8.4 × 10^3)/(420)] cal/s = 20 cal/s = [(3.52 × 10^4) / 420] J/s = 83.7 W Therefore the power of the heater is 83.7 W, With 200 gr. of alcohol in the vessel, the temperature rises from 16\textdegreeC to 78\textdegreeC in 6 min 12 s. If C_a is the specific heat of alco-hol and M_a is the alcohol's mass, it absorbs, Q_a=C_aM_a(78\textdegreeC - 16\textdegreeC) = C_a (200gr) (62\textdegreeC) amount of heat during the temperature rise. The power of the heater remains the same while heating the water or the alcohol, hence, in calories per second, we have, P =Q_a/ [(6 min) (60s/min) + 12 s] =Q_a/ [372 s] 20 cal/s = [C_a × (200gr) × (620c)] / [372 s] giving c_a = [(20 cal/s) (372 s)] / [200gr(\textasteriskcentered620c)] = 0.6 cal/C\textdegree.gr 30grof alcohol are vaporized in 5 min 6 s = 306 s, hence the amount of heat, Q', required to vaporize it at 78\textdegreeC is, Q ' = p × (306 s) = (20 cal/s) (306 s) = 6.12 × 10^3 cal. If L is the heat of vaporization of alcohol, Q' is given by, Q ' = L (30gr) Then,L = (Q' / 30gr) = [(6.12 × 10^3 cal) / 30gr] = 204 cal/gr

Question:

Obtain the 7's and 8's complement of the following octal numbers: 770, 1263, 00010 and 0000.

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Solution:

In the octal system, 8 = 10, 8^2 = 100, 8^3 = 1000, etc. Thus, the octal system contains only the numbers 0,1,2,3,4,5,6,7. a) The seven's complement of 770 is 777 -770 007 The eight's complement is then 007 + 1 = 010 This could also have been obtained directly: 1000 - 770 010 b) The seven's complement is: 7777 -1263 6514 The eight's complement is: 6514 + 1 = 6515 according to the relationship Radix complement (X) = Radix complement minus one (X) + 1 (In base eight arithmetic, 8 is the radix and the radix complement of a number equals c) The radix minus one complement of 00010 is: 77777 - 00010 77767 and the radix complement is: 77770 Note that although 00010 = 10, the radix complements are different. In the computer, numbers are stored in words of length 12 - 36 bits. In a 12-bit word, the number 10 would be stored as 000000000010 The radix complement is: 777777777770 d) The radix complement of 0000 is: 10000 - 0000 10000 and the radix minus one complement is: 10000 - 1 = 7777.

Question:

What is the total binding energy of _6C^12 and the average binding energy per nucleon?

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Solution:

The mass of an atom, in general, is not equal to the sum of its component masses. The mass of the com-ponent parts (protons, neutrons, and electrons) is slightly greater than the mass of the atom. This difference in mass has an energy equivalent (E = mc^2), which is called the binding energy of the nucleus. Although binding energy refers to the nucleus, it is more convenient to use the mass of the whole atom in calcu-lations. Then,M_n= M -ZM_e, whereM_n, M, and M_e are the nuclear, atomic, and electron masses, respectively, and Z is the atomic number. Since a carbon 6 atom, C_6^12, is made up of 6 protons and 6 electrons (or 6 H^1 atoms) plus 6 neutrons, then the binding energy, (b.e.) can be represented as follows: b.e. =M_n+ZM_e- M b.e. for _6C^12 = 6 [mass of electron and proton (H_1) + mass of neutron] - atomic mass of _6C^12. In other words, a mass difference equation can be written in terms of whole atom masses. Mass of 6 H^1 atoms = 6 × 1.0078= 6.0468 Mass of 6 neutrons = 6 × 1.0087=6.0522 Total mass of component particles= 12.0990 Atomic mass of _6C^12=12.0000 Loss in mass of formation of _6C^12= 0.0990 Binding energy (931.5MeV/ ∆mass) (.0990 ∆mass) = 92.22MeV Since there are 12 nucleons (protons plus neutrons), the average binding energy per nucleon is [(92.22MeV) / (12 nucleons)] = 7.68MeV.

Question:

What is structure of a typical neuron?

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Solution:

Before we get to the answer, it is important to recognize the fact that the term 'typical neuron' is rather vague. Neurons, which are the basic structural and functional units of the nervous systems of multicellular animals, show a great diversity in types. In humans, many types of neurons are present (Figure 1). Nevertheless, often three parts of a neuron can be distinguished: a cell body, an axon, and a group of processes called dendrites. Dendrites are usually rather short and numerous extensions from the cell body. They frequently branch profusely, and their many short terminals may give them a spiny appearance. When stained, they ordinarily show many dark granules. There is usually only one axon per neuron (very rarely two), and it is frequently longer than the dendrites. It may branch extensively, but unlike dendrites it does not have a spiny appearance and does not show dark granules when stained. The most fundamental distinction between dendrites and axons is that dendrites receive excitation from other cells whereas axons generally do not, and that axons can stimulate other cells whereas dendrites cannot. Thus, den-drites carry information to the cell body while axons carry information away from the cell body. Axons may be several feet long in some neurons. A bundle of many axons wrapped together by a sheath of connective tissue is what we commonly call a nerve. Each vertebrate axon is usually enveloped in a myelin sheath formed by special cells, the Schwann cells, that almost com-pletely encircle the axon. Schwann cells play a role in the nutrition of the axons, and provide a conduit within which damaged axons can grow from the cell body back to their original position. The myelin sheath is interrupted at regular intervals; the interruptions are called nodes of Ranvier. At these nodes, the myelin sheath disappears. The myelin functions in speeding up the transmission of impulses in the axon it envelopes. It is crucial to note that the myelin sheath is not a separate layer by itself. Through electron microscopy, it has been proved that the sheath is not a secretion product of the axon or Schwann cells, as was once believed, but a tightly packed spiral of the cell mem-brane of the Schwann cells. Thus the sheath is composed of the lipid from the membrane's bilayer. The nucleus and cyto-plasm of the Schwann cell are pushed aside to form the neu-rolemma. The nodes in the myelin sheath are simply the points at which one Schwann cell ends and another begins (see Figure 2). The neuronal cell body, or perikaryon, contains the nucleus and cytoplasmic organelles distributed around the nucleus. The perikaryon has well-developed endoplasmic reticulum and Golgi apparatus for manufacturing all the substances needed for the maintenance and functioning of the axon and dendrites.

The blood flow in the veins is partially aided by the movements of skeletal muscles near the veins. The middle layer of the venous walls has much less smooth muscle than does the middle layer of the arterial walls. Veins are therefore much (8 X) more distensible and easily collapsible. Most of the veins are surrounded by skeletal muscles, which contract as the body moves. This contraction exerts a pressure on the veins, com-pressing them. The compression of the veins forces the blood to move. As the muscle relaxes, the compressed section of vein fills with the blood, which is again pushed along by the next contraction. This "milking" action of skeletal muscle is shown in the accompanying figure. Why does the blood only move in the direction of the heart? At numerous points along the interior of the vein, there are valves which allow blood to flow in only one direction - towards the heart. Thus when contracting muscle forces the blood to move in the veins, the valves act to ensure unidirectional movement of blood by preventing backflow. When one stands perfectly still for a long period, the milk-ing action of the skeletal muscles is decreased, venous pressure increases in the lower parts of the legs increasing capillary pressure. This forces fluid from the circulatory system into the tissue spaces, causing swelling (edema). The movement of the chest during breathing also aids the flow of blood to the heart. As the diaphragm and chest muscles contract during inspiration, the chest cavity increases in size. The in-creased volume causes a decrease in pressure within the cavity to below atmospheric pressure. This pressure gradient causes air to flow into the lungs. But this decrease in pressure also lowers venous pressure in the chest cavity. During inhalation, blood thus tends to be drawn into the veins in the chest cavity. During exhalation, blood cannot be forced back because of the venous valves.

Question:

The hump on a camel's back is mostly depot fat. Assuming complete hydrolysis of the fat, and its subsequent oxi-dation to CO_2 and water, calculate the number of kilocalories that would be released from 25 kg of camel fat.

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Solution:

Depot fat is usually made up of a mixture of three triglycerides - triolein, tristearin and tripalmitin. These fats are produced by a reaction of three moles of their respective fatty acids (oleic, stearic or palmitic) and one mole of glycerol. The structure of a fatty acid is where R is a long chain saturated or unsaturated alkyl group. The equation for this reaction is It has been found that 9 Kcal of energy is released by each gram of fat hydrolyzed and then oxidized. In this problem, one is asked to find the amount of energy re-leased when 25 kg of fat is oxidized. 25 kg is equal to 25000 g. Therefore, the number of kilocalories produced in this reaction is 25000 g × 9. Kcal/g = 2.25 × 10^5 Kcal.

The electrons of an atom are distributed so that they occupy concentric atomic orbitals with each orbital containing two electrons. With increasing distance from the nucleus, there is a corresponding increase in the energy level of the orbital. Each atomic orbital has a characteristic energy level, hence the orbitals are said to be quantized. The quantum theory describes the characteristics of electrons in an atom by assigning four quantum numbers to each electron. The principal quantum number, n, denotes the main energy level or shell in which the electron resides. The main energy levels of an atom are assigned integral values, starting with 1 as the shell closest to the nucleus, and continually increasing as the main energy levels become more distant from the nucleus. Hence, n = 1, 2, 3, ... . The angular momentum quantum number, l, determines the shape of the volume in which the electron is most likely to be found. The shape of this volume can be illustrated with electron cloud diagrams, and depends upon the magnetic and electrostatic inter-actions of the electron with the nucleus and other electrons in neighboring orbitals. These interactions will be dif-ferent for varying distances of the orbital to the nucleus. Consequently, the angular momentum of the electron depends upon the main energy level in which it resides. It has been determined that the quantum numbers l and n are related as follows: l = n - 1. Therefore, l = 0, 1, 2, 3, ..., depending on n. A moving charge generates a magnetic field. Since an electron is an elementary particle with a negative charge, and it moves within some volume described by l, it will propagate a magnetic field. This is accounted for in quantum theory by the magnetic quantum number m, which depends upon l in the following way: m = -l, 0, l, I.e., when l = 0, m = 0; when l = 1, m = -1, 0,1; when l = 2, m = - 2, - 1, 0, 1, 2. One final characteristic of an electron is its rotational motion. An electron spins on its own axis. There are only two possible directions (clockwise and counterclockwise) that an electron can spin about its axis. The spin quantum number, s, accounts for this phenomenon, and the arbitrary values + (1/2) and -(1/2) are used to describe the spin of an electron. It is important to realize that no two paired electrons in the same atomic orbital can have the same spin quantum numbers. This would be a violation of the Pauli Exclusion Principle, which states that no two paired electrons in an atom may have identical sets of quantum numbers. Electrons in the same orbital have the same principal, angular momentum, and magnetic quantum numbers. Therefore their spin quantum number must differ to be in accordance with the Pauli Exclusion Principle. The various atomic orbitals are labeled s, p, d and f. When l = 0, the orbitals are of the s variety; when l = 1 they are p orbitals; when l = 2 they are d orbitals; and when l = 3 they are f orbitals. The number of values of m will determine how many orbitals of each kind will be contained in each main energy level. For l = 0, m = 0 and this means that there is only one s orbital contained in each main energy level. For l, = 1, m = - 1, 0, 1 and hence there are three p orbitals contained in the main energy levels starting with n = 2. It is similarly found that there are five d orbitals and seven f orbitals in the main energy orbitals starting with n = 3 and n = 4, respectively. With this basic knowledge of quantum theory, we can now assign an electron configuration to an atom. Electron configurations show how the electrons within an atom are distributed. From the electron configuration we can deduce the valence of the atom, and hence determine its bonding capacity. An important note should be made here: in determining the electron configuration of an atom, one uses the Aufbau process of filling atomic orbitals. By this process, the electrons fill the orbitals in order of increasing energy, and when there is more than one orbital in an energy sublevel (e.g. p orbitals - there are three of them in a sublevel), the electrons will fill the orbitals so that each one will have one electron before any orbital receives a pair of electrons. The relative energies of the orbitals in a main shell is given as s < p < d < f. We will now assign an electron configuration for oxygen (atomic number = 8) and deduce its bonding capacity from its valence. There are eight electrons in the oxygen atom. In the first main energy level, n = 1, l = 0 and m = 0. There is just one s orbital in this level. Since an orbital can accommodate two electrons, the first main energy level will hold two electrons, leaving us with six electrons to account for. In the second main energy level, n = 2, l = 0, 1 and m = - 1, 0, 1. There is one s orbital (l = 0) and three p orbitals (l = 1) in this level. Since s orbitals are lower in energy than p orbitals, they will be filled first, by two electrons. There are four electrons remaining to be placed in orbitals. Because there are three p orbitals in this level, the electrons will fill them so that, each orbital has one electron before any orbital has a complete pair. Hence the four electrons will occupy the three p orbitals so that one orbital will have a pair of electrons and the other two orbitals will each have one electron. The electron configuration of oxygen can be represented as: Above: The numbers represent the principal quantum number n. Electrons with opposite spins are depicted as arrows (\upharpoonleftand \downharpoonright) . The three p orbitals are labeled p_x, p_y, and p_z for differentiation. The valence of an atom is the number of electrons that can be accommodated via bonding by the atom. The Octet Rule states that all atoms that can obtain an octet of electrons in the valence shell (outermost shell), will try to do so. By looking at the electron configuration of oxygen, we can see that its valence (the number of electrons needed to fill the valence shell)is two. Oxygen complies with the Octet Rule in that it seeks to obtain two electrons to fill up its second main energy level. Hence the bonding capacity (or the unpaired electrons available for bonding) of oxygen is two. Let's now look at the electron configuration of carbon (atomic number = 6) which is one of the most biologically important elements. There are six electrons in a carbon atom. As in oxygen, two electrons will occupy the lone s orbital in the first main energy level. This leaves us with four electrons to account for it. If these remaining electrons filled the orbitals in the second main energy level as was true for oxygen, we would get the electron configuration for carbon: In this case the valence of carbon is four, and the bonding capacity is two. This is because there are two unpaired electrons in the 2p orbitals. It should be noted, however, that this representation of carbon's electron configuration does not comply with the Octet Rule. In this case, a fully bonded carbon atom would contain only six electrons in its valence shell, and has one empty orbital (2p_z). This is an energetically unstable condition. If it is at all possible for carbon to complete its valence shell, it will do so. It so happens that carbon does comply with the Octet Rule, and has all its orbitals in the first two main energy levels filled to capacity when it bonds with other atoms. This is facilitated by the hybridization of atomic orbitals. The electrons in the second main energy level occupy an orbital that is intermediate in energy between the 2s and 2p orbitals. This orbital is called an sp^3 hybrid orbital because it is an orbital formed by one s and three p orbitals. There are four sp^3 orbitals, each containing an unpaired electron. The electron configuration of carbon is shown as: Now it can be seen that carbon has a valence of four, and a bonding capacity of four. In this case the Octet Rule is satisfied.

a) Using NAND gates: The equivalence function can be written as: A \equiv B= A\bulletB + A'\bulletB'[Using the Sum of Products form] = [(A\bulletB + A'\bulletB') ' ]'[Complementing twice] = [(A\bulletB)' \textbullet (A'\bulletB')']'[Applying DeMorgan's Rules] = (A\uparrowB)\uparrow(A'\uparrowB')[Using the Shaeffer stroke notation] \textasciigraveThe circuit realization is shown in fig. 1. b)Using NOR gates: We can write A \equiv B= (A+B')\bullet (A'+B)[using the Product of Sums form] = [ ( (A+B') \textbullet (A'+B) )' ]'[Complementing twice ] = [(A+B')'; + (A'+B)']'[Applying DeMorgan's rule to the inner primed bracket] = [(A\downarrowB')\downarrow (A' \downarrowB)] . The circuit realization is shown in figure 2.

Question:

Define the following terms as they relate to Polish string notation: a) infix b) prefix c) postfix Then, convert the following tree into the three types of expressions just defined.

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Solution:

Infix notation refers to the way we normally write expressions, namely, with the operators between the two operands. This is also known as algebraic notation, which is different from Polish notation. Prefix notation refers to the fact that the operator precedes the expression in the string. So, if we have the string \div XY, it means that the division operator acts to divide X by Y. One inconvenience with prefix notation is that to understand the string, you must read from end to beginning. Postfix notation avoids this difficulty. For the string above, postfix notation would be XY \div. The meaning is the same as above, but the operator comes after the two operands. The infix translation of the tree diagram is then (P + Q) × (R - S). This is obvious if you follow the tree up from the terminal nodes to the root node. Prefix translation yields x + PQ - RS. By simply looking at the numbering of the individual nodes, you can see that they correspond with the order of the string. In other words, you can write the following to highlight the correspondence: X + PQ - RS 1234567 Postfix notation looks somewhat different. For this tree, we get PQ + RS - x. This may be read from be-ginning to end as "add P and Q, then subtract S from R, then multiply the sum and the difference together."

Let subscript A refer to the aircraft, and subscript F to the moving air. Subscript E refers to the earth. We have given v_AF = 120 mi/hr, due north v\ding{217}_FE = 50 mi/hr, due east, and we wish to find the magnitude and direction of v\ding{217}_AE . By the law of addition of velocities v\ding{217}_AE = v\ding{217}_AF + v\ding{217}_FE . The three relative velocities are shown in the figure. It follows from this diagram that \mid v\ding{217}_AE \mid = 130 mi/hr. Furthermore, tan \varphi = [(50 mi/hr)/(120 mi/hr)] tan \varphi = .4167 and \varphi = 22.5\textdegree The airplane travels at a speed of 130 mi/hr at an angle of 22.5\textdegree east of due north.

Question:

A mixture of nitrogen and oxygen gas is collected by displacement of water at 30\textdegreeC and 700torrHg press-ure. if the partial pressure of nitrogen is 550torrHg, what is the partial pressure of oxygen? (Vapor pressure of H_2O at 30\textdegreeC = 32torrHg.)

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Question:

The vertebrate skeleton maybe divided into two general parts, the axial skeleton and the appendicular skeleton. Whatbones constitute these in man?

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The axial skeleton consists of the skull, ver-tebral column, ribs, sternum and hyoid bone. The primary function of the vertebrate skull is the protection of the brain. The part of the skull that serves this function is the cranium. The rest of the skull is made up of the bones of the face. In all, the human skull is composed of twenty-eight bones, six of which are very small and located in the middle ear. At the time of birth, several of the bones of the cranium are not com-pletely formed, leaving five membraneous regions called fontanelles. These regions are somewhat flexible and can undergo changes in shape as necessary for safe passage of the infant through the birth canal. The human vertebral column, or spine, is made up of 33 separate bones known as vertebrae, which differ in size and shape in different regions of the spine. In the neck region there are 7 cervical vertebrae; in the thorax there are 12 thoracic vertebrae; in the lower back region there are 5 lumbar vertebrae, in the sacral or hip region, 5 fused vertebrae form the sacrum to which the pelvic girdle is attached; and at the end of the vertebral column is the coccyx or tailbone, which consists of four, or possibly five, small fused vertebrae. The vertebrae forming the sacrum and coccyx are separate in childhood, with fusion occurring by adulthood. The coccyx is man's vestige of a tail. A typical vertebra consists of a basal portion, the centrum, and a dorsal ring of bone, the neural arch, which surrounds and protects the delicate spinal cord which runs through it. Each vertebra has projections for the attachment of ribs or muscles or both, and for articulating (joining) with neighboring vertebrae. The first vertebra, the atlas, has rounded depressions on its upper surface into which fit two projections from the base of the skull. This articulation allows for up and down movements of the head. The second vertebra, called the axis, has a pointed projection which fits into the atlas. This type of ar-ticulation allows for the rotation of the head. In man there are 12 pairs of ribs, one pair articu-lating with each of the thoracic vertebrae. These ribs support the chest wall and keep it from collapsing as the diaphragm contracts. Of the twelve pairs of ribs, the first seven are attached ventrally to the breastbone, the next three are attached indirectly by cartilage, and the last two, called "floating ribs", have no attach-ments to the breastbone. The sternum or breastbone consists of three bones - the manubrium, body and xiphoid process - which usually fuse by middle-age. The sternum is the site for the anterior attach-ment of most of the ribs. The ribs and sternum together make up the thoracic cage which functions to protect the heart and lungs. The hyoid bone supports the tongue and its muscles. It has no articulation with other bones, but is held in place by muscles and ligaments. The bones of the girdles and their appendages make up the appendicular skeleton. In the shoulder region the pecto-ral girdle, which is generally larger in males than in fe-males, serves for the attachment of the forelimbs. The pec-toral girdle consists of two collarbones, or clavicles, and two shoulder blades, or scapulas. In the hip region, the pelvic girdle serves for the attachment of the hindlimbs. The pelvic girdle, which is wider in females so as to allow room for fetal development, consists of three fused hip-bones, called the ilium, ischum and pubis, which are at-tached to the sacrum. Collectively, the "hip bone" is called the oscoxae or innominate bone. Articulating with the scapula is the single bone of the upper arm, called the humerus. Articulating with the other end of the humerus are the two bones of the forearm called the radius and the ulna. The radius and ulna permit rotation of the forearm. The ulna has on its end next to the humerus a process often referred to as the "funny bone." The wrist is composed of eight small bones called the carpals. The ar-rangement of these bones permits the rotating movements of the wrist. The palm of the hand consists of 5 bones, known as the metacarpals, each of which articulates with a bone of the finger, called a phalanx. Each finger has three phalanges, with the exception of the thumb, which has two. The pattern of bones in the leg and foot is similar to that in the arm and hand. The upper leg bone, called the femur, articulates with the pelvic girdle. The two lower leg bones are the tibia (shinbone) and fibula, cor-responding to the radius and ulna of the arm, respectively. These two bones are responsible for rotation of the lower leg. Ventral to the joint between the upper and lower leg bones is another bone, the patella or knee cap, which serves as a point of muscle attachment for upper and lower leg muscles. This bone has no counter part in the arm. The ankle contains seven irregularly shaped bones, the tarsals, corresponding to the carpals of the wrist. The foot proper contains five metatarsals, corresponding to the metacarpals of the hand, and the bones in the toes are the phalanges, two in the big toe and three in each of the others.

Question:

A 200-kg satellite is lifted to an orbit of 2.20 × 10^4 mi radius. How much additional potential energy does it acquire relative to the surface of the earth?

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As in the diagram, R is the earth's radius, M is the earth's mass, m is the satellite mass, and r is the distance between the earth's center and the satellite. R = 6.37 × 10^6 m r = 2.20 × 10^4 mi = 3.54 × 10^7m M = 5.98 × 10^24 kg m = 200 kg The additional potential energy is equal to the work done against the earth's gravitational field. At a distance R from the earth's center, that is, on the earth's surface, the satellite has a potential energy, U_surface U_surface = - (GMm/R) In orbit of radius r the potential is U_orbit = - (GMm/r) Then the additional potential energy involved in launching the rocket to its orbit, ∆U, is given by ∆U = U_orbit - U_surface = - (GMm/r) - {- (GMm/R)} = GMm (1/R - 1/r) = (6.67 × 10^-11 nt - m^2/kg^2) (5.98 × 10^24 kg)(200kg) × [1/(6.37 × 10^6 m) - 1/(3.54 × 10^7 m)] = 1.0310^10 joules This is about equal to the work needed to lift an object weighing 3800 tons to a height of 1000 ft above the earth. Note that the change in potential energy of the satellite cannot be found by using U = mgh. This formula applies only to objects near the earth's surface, where g is approximately constant.

a) The block diagram is as shown below. The slave gets the inverted clock pulse of what the master gets. The J-K flip-flop differs from the SC flip-flop in that the J-K flip-flop does not have the restriction that the S-C type flip-flop has, i.e., both the inputs of the S-C type flip-flop cannot simultaneously be equal to one. In the J-K type flip-flop, it is permissible to have both the J and K inputs 1. b) The timing diagram is given in Figure 1 for 5 clock pulses up to t_5. Note that y does at the trailing edge of the clock pulse exactly what x does at the leading edge of the clock pulse. Also note that x' and y' are just the reciprocals of x and y respectively. c) The Karnaugh Map for the J-K flip-flop is plotted in Figure 2. The entries in the Map represent the next state y\rightarrowfor a given combination \rightarrow of y, i.e., the present state, and the input J and K. d) The excitation equation is obtained from the Karnaugh map of figure 2, as follows, for the next state y\rightarrow \rightarrow y\rightarrow= Jy' + K'y. \rightarrow e) The Karnaugh Map also helps us to prepare the state table as given in figure 3: JKy y\rightarrow \rightarrow REMARKS 000 001 0 1 NO CHANGE 010 011 0 0 CLEARS 100 101 1 1 SETS 110 111 1 0 FLIPS f) The state diagram is drawn, in figure 4.

The dielectric is polarized when inserted between the plates and the polarization charges (or bound charges) effectively neutralize some of the surface charges on the plates, as shown in figure 2. This then reduces the field strength in the dielectric, thus lower-ing the voltage difference across the plates, since the latter is proportional to the electric field (V = Ed). What remains unchanged in this case is the true charge density on the plates, i.e. the total charge on the plates. The ratio of the field in the dielectric to that in the vacuum gives the relative permittivity of the dielectric (a) K = (E / E_0) = (V / V_0) = (3000 / 1000) = 3 (b) \epsilon = K\epsilon_0 = 3\epsilon_0 (c) X = \epsilon_ \rule{1em}{1pt} \epsilon_0 = 2\epsilon0 (d) E_0 = V_0 / լ= (3 × 10^3 volts) / (10^\rule{1em}{1pt}2 m) = 3 × 10^5 (volts / m) (e) E = V / լ = (1 × 10^3 volts) / (10^\rule{1em}{1pt}2 m) = 10^5 (volts / m) The bound charges of the dielectric set up a new electric field E_b in the slab which opposes the electric field E0due to the plate charges. The new field E is the resultant of these two (f)E_b = E_0 \rule{1em}{1pt} E = 2 × 10^5 (volts/m). The surface charge density \sigmais given by (g)\sigma = \epsilon_0E_0 = 3 × 10^5 \epsilon_0 (coul / m^2) The bound charges on the surface of the slab have a density \sigma_b = \epsilon_0E_b = \epsilon_0(E_0 \rule{1em}{1pt} E) = \epsilon_0E_0(k\rule{1em}{1pt}1) = XE = 2 × 10^5 \epsilon_0 (coul / m^2), and we get [(\sigma_b) / (\sigma_f)] = (2 / 3).

Question:

When light is incident on a thin slit, a diffraction pattern, shown in figure (a), is produced. Find an expression which give the location of the minima of diffraction pattern in terms of the angle \psi.(see figure (b)).

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Solution:

In this problem, we consider only Fraunhoferdiffraction. (See figure (b).) This type of diffraction is characterized by parallel, incident light rays encountering a slit, and being diffracted, again parallel to one another. In practice, this restriction may be effected in 2 ways. The light source is placed very far away from the slit, and the screen is far from the slit. Another method employs a converging lens to the left of the slit, and a converging lens to the right. Light rays incident upon the first converging lens leave par-allel to one another, encounter the slit, and are diffracted parallel to one another (see figure (c)). They then are focused by the second con-verging lens on screen F, and the diffraction pattern results. Using figure (c), we can derive an equation locating the minima of the diffraction pattern. Let us focus on rays coming from points A and B. Since the incident light is in phase, the only phase differencebetween r_1 and r_2 must occur because r_2 travels a larger distance r_1. (Here, we assume that the rays are effectively parallel). If this distance, BB', is equal to (1/2) \lambda, where \lambda is the wavelength of light, the rays r_1 and r_2 will interfere destructively and p_1 will be a minimum. Since BB' = a/2 sin \psi, minima will be observed when a/2 sin \psi = \lambda/2 Hence, a sin \psi = \lambdaminima. This formula gives the 1st order minima of the diffraction pattern. Now, examine figure (d), where the slit has been divided into 4 equal portions of length a/4. Looking at rays r_1, and r_2, we note that the only phase difference between them occurs because they travel unequal distances. Ray r_2 travels a distance BB' greater than r_1. If BB' = \lambda/2, r1and r_2 will destructively interfere. Furthermore, r_3 and r_4 will also destructively interfere since DH = \lambda/2. Hence, each pair of similar rays will destructively interfere, and p_2 will be a minimum of the diffraction pattern. Therefore, since BB' = a/4 sin \psi the second order minimum is described by Thena/4 sin \psi = \lambda/2minimum a sin \psi = 2\lambdaminimum. In general, if we divide the slit into n equal segments, where n is an even integer, we will find anda/n sin \psi =\lambda/2minima a sin \psi = n\lambda/2miniman even sincen = 2m wherem = 1 , 2 ,... we obtain a sin \psi = m \lambdaminima m = 1 , 2 ,... Note that n must be even because each pair of rays emanating from the slit interfere.

In this problem momentum must be conserved in both the horizontal and vertical directions. Let us refer to the bullet by using the subscript 1, and the gun by the subscript 2. We can state this conser-vation of momentum in the horizontal direction as follows: P_1x = p2x or m_1v_1x = m_2v2x but V_1x = v1cos 45\textdegree Therefore, we may express p_1x as, P_1x = m_1v_1 cos 45\textdegree = (500 kg) × (200 m/sec) × 0.707 = 7.07 × 10^4 kg-m/sec This must equal (except for the sign) the recoil momen-tum of the gun which moves only horizontally: P_2 = m_2v_2 = -7.07 × 10^4 kg-m/sec Therefore, v^2 = - [ {7.07 × (10^4 kg-m/sec)} / {7 × 104kg} ] \cong - 1m/sec or, approximately 2 mi/rh. What has happened to the vertical component of the recoil momentum, P_1y = 7.07 x 10^4 kg-m/sec ? Since the railway platform is in contact with the Earth, the Earth absorbs the ver-tical momentum. The Earth does recoil, but because of the extremely large value of the Earth's mass compared to that of the railway gun, the recoil velocity cannot be measured.

Question:

An electron with angular momentum L = 6 × 10^-35 N-sec and total energy H_0 = -4 × 10^-18 J is in orbit around a proton. What are the distance of nearest approach and farthest separation?

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The electrostatic potential is given by V_E = -k_E(e^2/R) where the negative sign indicates its attractive nature. The kinetic energy of the electron is k.E. = (1/2)mv^2 = [(m^2u^2R^2)/(2mR^2)] since, by definition of angular momentum, L =mvR. The total energy of the electron in its orbit is H_0 =k.E. + V_E = (L^2/2mR^2) - [k_E(e^2/R)] orH_0 = (L^2/2mR^2) - (k_Ee^2)/R R^2H_0 = (L^2/2m) - k_Ee^2R R^2 = (L^2/2mH_0) - [(k_Ee^2R)/(H_0)] R^2 = [(k_Ee^2R)/H_0] - (L^2/2mH_0) = 0 Solving for R via the quadratic equation R = - [(k_Ee^2/H_0) \pm \surd{(k^2_Ee^4/H^2_0) + (2L^2/mH_0)}] / 2 or2R = -[(k_Ee^2/H_0) \pm {(k^2_Ee^4/H^2_0) + (2L^2/mH_0)}^1/2] (k_Ee^2/H_0) = [{9 × 10^9 (N\bulletm^2/c)}(2.56 × 10^-30 c^2)] / [ -4 × 10^-18 J] (k_Ee^2/H_0) = -5.76 × 10^-11 m 2L^2/mH_0= [2(36 × 10^-70 N^2 \bullet s^2)] / [(9.1 × 10^-31 kg) (.4 × 10^-18 J)] 2L^2/mH_0= -1.98 × 10^-21 m^2 Thus 2R = 5.76 × 10^-11 m \pm (3.32 × 10^-21 m^2 - 1.98 × 10^-21 m^2)^1/2 = 5.76 × 10^-11 m \pm 3.66 × 10^-11 Thus R = 1.05 × 10^-11 m or 4.71 × 10^-11 m Which gives the nearest approach and farthest separation.

Question:

What is an operon? With the aid of diagrams, show how a known operon works.

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Solution:

A controllable unit of transcription is called an operon. An operon consists of a binding site for RNA polymerase (a promoter), a binding site for a specific repressor (an operator), and one or more structural genes (see figure). These structural genes are usually associated in their functions, and their coordinated synthesis is de-sirable. Thus they are linked together in the operon, and when they are transcribed, a single mRNA molecule is produced. The promotor and operator act as controls for this trans-cription, and the entire system is controlled by an associated regulatory gene, which codes for the repressor molecule of the operon. A known example of an operon is the lactose (lac) operon, present in E. coli. This operon contains the structural genes for \beta-galactosidase, an enzyme which cleaves lactose into glucose and galactose; galactoside permease, which allows for the increased rate of entry of lactose into the bacteria; and galactoside acetylase, whose function is not known. When lactose is absent from the medium, these enzymes are not needed, and so their levels are low. But when lactose is added to the medium containing these bacteria, the production of the enzymes increases tremendously, almost a thousand fold. How is the manufacture of this prodigious quantity of enzyme initiated and controlled? The regulatory gene (1) for the lac operon is transcribed constitutively, that is, its product, the repressor, is constantly being synthesized. This repressor binds to the operator(o) and blocks the transcription of the structural genes. That is why the levels of the three enzymes are normally so low. The situation is altered when lactose is introduced into the medium. Lactose molecules move into the nucleus of the bacterium, and bind to any repressor molecules in such a way that the repressors are inactivated. An inactivated rep-ressor cannot bind to the operator. It therefore cannot block the transcription of the structural genes by RNApoly-merase which binds to the promotor. Messenger RNA now transcribed from the genes of the operon is translated on the ribosomes into the three enzymes of lactose metabolism. As long as the inducer (lactose) is present, the synthesis of the enzymes can occur. But when the inducer is depleted by metabolism, repressor proteins (remember that synthesis of the lac repressor isconstitutive) become free to again bind to the operator and effectively turn off the synthesis of the enzymes. This model implies that the usual condition of the lactose operon is to be turned off. This is accomplished through the control of the regulator gene, which constitu-tively produces the repressor. In other words, the system is under negative control. Because the operon can be turned on by the presence of lactose (the inducer), it is said to be inducible. An important implication is that some genes, such as (1), produce proteins whose function is to regulate the expression of other genes, while certain other genes, namely operators and promoters, very probably do not code for any proteins at all.

Question:

Figure A shows a strut AB, of length l , pivoted at end A, attached to a wall by a cable, and carrying a load w at end B. The weights of the strut and of the cable are negligible. Suppose the weight w, and the angles \texttheta_1 and \texttheta_2 are known. What is the direction of C\ding{217}?

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Solution:

The system shown in figure A is in equilibrium. Hence, the net external force on the system must be zero. Also, the net external torque acting on the system about any point must be zero. Then, using figures (a) and (b) T_1\ding{217} + T_2\ding{217} + C\ding{217} = 0(1) and, taking torques about point A l(T_2 sin \texttheta_1) -l(T_1 sin \texttheta_2) = 0(2) Changing (1) into 2 scalar equations using figure B, C sin \varphi + T_2 sin \texttheta_1 = T_1 sin \texttheta_2(3) C cos \varphi = T_2 cos \texttheta_1 + T_1 cos \texttheta_2 To find the direction of C\ding{217}, we must solve for \varphi. Using (3) C sin \varphi = T_1 sin \texttheta_2 - T_2 sin \texttheta_1 C cos \varphi = T_1 cos \texttheta_2 + T_2 cos \texttheta_1 Dividing these last 2 equations tan \varphi = (T_1 sin \texttheta_2 - T_2 sin \texttheta_1)/(T_1 cos \texttheta_2 + T_2 cos \texttheta_1)(4) Solving (2) for T_1 T_1 = (T_2 sin \texttheta_1)/(sin \texttheta_2) Substituting this equation in (4) tan \varphi = [{(T_2 sin \texttheta_1)/(sin \texttheta_2)} sin \texttheta_2 - T_2 sin \texttheta_1]/ [{(T_2 sin \texttheta_1)/(sin \texttheta_2)} cos \texttheta_2 + T_2 cos \texttheta_1] andtan \varphi = 0 \varphi = 0\textdegree Hence, C\ding{217} is directed along the strut AB.

This problem involves a photon moving with the velocity of light in a gravitational field. We do not get the correct answer without doing a careful calculation using special relativity, but we can get the order of magnitude of the correct answer by a naive calculation. Suppose that the photon has an effective mass M_L; it will turn out that M_L drops out of the calculation of the deflection and thus we do not have to know what it is. Let the light beam pass the sun at a distance of closest approach, r_0, as measured from the center of the sun. We suppose that the deflection will turn out to be very small, so that r_0 is essentially the same as if the light beam were not deflected. The force F on the photon at the position (r_0, y) is F = -[(G M_S M_L)/(r^2 _0 + y^2)] The transverse component, F_X, is F_X = - [(GM_SM_L)/(r^2 _0 + y^2)] cos \texttheta = - [(GM_SM_L)/(r^2 _0 + y^2)] [r /(r^2 _0 + y^2)]1/2 = - GM_SM_L [r_0/(r^2 _0 + y^2)^3/2] where y is measured from the point P ,as in the figure The final value of the transverse velocity component v_x of the photon has the value given by M_LV_X = \int F_X dt = \int F_X (dy/v_y ) \cong 1/c \int F_X dy so that v_x \cong - [(2GM_Sr_0)/C ]^\infty\int_0[dy/(r^2 _0 + y^2)^3/2 ] \cong - [(2GM_S)/(cr_0)]

The radius of gyration is the distance from the axis of rotation to the radius at which we may consider all the object's mass to be concentrated in a thin hoop, as the diagram shows. For a thin hoop of radius k_0, the moment of inertia, I_0, is mk_0^2, where m is the mass of the hoop. To find k_0, we set I of the object equal to I_0 the moment of inertia of the equivalent hoop I = I_0 or(1/12)m1^2 = mk_0^2 Hence k_0 = [\surd{(1/12)mL^2}] / m = L/(2\surd3) = 0.289 L. The radius of gyration, like the moment of inertia, depends on the location of the axis. Note carefully that, in general, the mass of a body can not be considered as concentrated at its center of gravity for the purpose of computing its moment of inertia. For example, when a rod is pivoted about its center, the distance from the axis to the center of gravity is zero, although the radius of gyration is L/(2\surd3).

Question:

How can radioactive isotopes be used in biological research such as tracing the pathways of compounds synthesized in the living cell?

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Solution:

Labeling substance with radioactive isotopes is a powerful method for studying specific metabolic pathways, as well as being applicable to many other areas of biological research. Radioactive labeling has been used, for example, in determining the rates of metabolic processes in intact organisms, and also in determining whether a given metabolic pathway is the predominant route, for a given metabolite. It is used in establishing whether a pathway which has been studied in a test tube is the same reaction that occurs in vivo. Labeling is done by taking the compound in question and substituting in a radioactive isotope for one of the naturallyoccuringelements comprising the compound.(see Table). The radioactive isotope, when incorporated into a biological system will undergo spontaneousemmisionof radiation while taking part in the same biological acti-vities as its non-radioactive counterpart. Although this is not observable by eye, there are many known methods by which these emissions can be detected. The methods commonly employed include Geiger counters, photographic methods, fluorescent methods, and cloud chambers. In this way, the course of the reaction can be followed. The value of radioactive tracers in biological research can be seen by considering the photosynthetic equation: 6 CO_2 + 6 H_2O(sunlight/chlorophyll)\rightarrowC_6H_12O_6 + 6 O_2. Although in the equation the photosynthetic process appears to be a simple reaction, it actually follows a very complicated course. This complex reaction, like most others, proceeds through a series of changes involving one or two molecules at a time. Researchers were able to use 14 6 C, which was incorporated into CO_2, to determine the course of this reaction. They monitored the radio-activity through the use of a Geiger counter. Some isotopes useful as radioactive tracers in biological system. naturallyoccuringelement radioactive isotopes half - life 1 1 H 3 1 H 12.1 years 12 6 C 14 6 C 5,700 years 23 11 Na 24 11 Na 15 hours 31 15 P 32 15 P 14.3 days 32 16 S 35 16 S 87.1 days 35 17 Cl 36 17 Cl 3.1 × 10^5 years 39 19 K 42 19 K 12.5hours 40 20 Ca 45 20 Ca 152 days 56 26 Fe 59 26 Fe 45 days 127 53 I 131 53 I 8 days

Question:

Minimize the following minterm function containing 'don't cares' using the Karnaugh-Map. f(A,B,C,D) = \summ(5,6,7,8,9) + d(10,11,12,13,14,15)

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Solution:

f(A,B,C,D) = \summ(5,6,7,8,9) + d(10,11,12,13,14,15) . In the design of digital circuits one often encounters cases in which the switching function is not completely specified. In other words, a function may be required to contain certain minterms, omit certain minterms, with the remaining minterms being optional; that is certain minterms may be included in the logic design if they help simplify the logic circuit. A minterm which is optional is called a don't care minterm. The plotting of the given function f(A,B,C,D) is shown in fig. 1 . In the use of don't cares there is one additional rule which may be used in mapping. Recall that the don't cares by definition can be either 0 or 1. Hence in minimizing terms in Sum of Product form, the don't cares may be chosen to be 1 if in doing so the set of blocks on the map which can be grouped together is larger than would other-wise be possible without including the don't cares. In other words, with regard to don't cares one can take it or leave it, depending on whether they do or do not aid in the simplification of the function. In the map of fig. 1 the following minterms and don't cares may be grouped. m_8 = AB C D m_9 = AB CD d_11 = ABC D d_10 = ABCD d_12 = A BC D d_13 = A BCD d_15 = A B C D d_14 = A B CD and by the use of Boolean theorem A\bulletA= 0, terms B,C, and D are eliminated. The result from this group would be A. The next group is minterms 5 and 7, and don't cares 13 and 15. m_5 =ABCD m_7 =AB C D d_13 = A BCD d_15 = A B C D With the same argument, the eliminated terms are A and C. Leaving the resulting terms BD. The third and the last group consists of minterms 7 and 6, and don't cares 15 and 14. m_6 =AB CD m_7 =AB C D d14= A B CD d_15 = A B C D Giving the result BC. The resulting total function can now be written as the sum reduced terms that are found, namely; f(A, B, C, D) = A + BD + BC Note that this function is much simpler to deal with than it would have been without the inclusion of don't cares.

Question:

Explain the events which take place during glycolysis.

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Solution:

Glycolysis is the series of matabolic reactions by which glucose is converted to pyruvate, with the concur-rent formation of ATP. Glycolysis occurs in the cytoplasm of the cell, and the presence of oxygen is unnecessary. Glucose is first "activated," or phosphorylated by a high- energy phosphate from ATP (See reaction 1). The product, glucose-6- phosphate, undergoes rearrangement to fructose- 6-phospate, which is subsequently phosphorylated by another ATP to yield fructose-1, 6- diphosphate (reactions 2 and 3). This hexose is then split into two three- carbon sugars, glyceraldehyde-3-phosphate (also called PGAL) and dihydroxyacetone, phosphate (see reaction 4). Only PGAL can be directly degraded in glycolysis; dihydroxyacetone phosphate is reversibly converted into PGAL by enzyme action (reaction 4a). Since two molecules of PGAL are thus produced per molecule of glucose oxidized, the products of the subsequent reactions can be considered "doubled" in amount 1 glucose \textemdash\textemdash\ding{217} 2 PGAL PGAL gets oxidized and phosphorylated. NAD+, the coenzyme in the dehydrogenase enzyme which catalyzed this step, gets reduced to NADH, and 1,3-diphosphoglycerate is formed (reac-tion 5). The energy-rich phosphate at carbon 1 of 1,3 diphosphoglycerate reacts with ADP to form ATP and 3-phospho- glycerate. This undergoes rearrangement to 2- phosphoglycerate, which is subsequently dehydrated, forming an energy- rich phosphate: phosphoenolpyruvate (PEP) (reaction 8). Finally, this phosphate group is transferred to ADP, yield-ing ATP and pyruvate, (reaction 9) Since two molecules of PGAL are formed per molecule of glucose, four ATP molecules are produced during glycolysis. The net yield of ATP is only 2, since 2 ATP were utilized in initiating glycolysis (reactions 1 and 3 ). Pyruvate is then converted to acetyl coenzyme A which enters the Krebs cycle. In addition, two molecules of NADH are pro-duced per molecule of glucose. Hence, the net result of glycolysis is that glucose is degraded to pyruvate with the net formation of of 2 ATP and 2 NADH. The process of glycolysis can be summarized as follows: glucose + 2ADP + 2Pi + 2NAD+ \textemdash\textemdash> 2 pyruvate + 2ATP + 2NADH + 2H+

The feeding organelle in Paramecium begins with an oral groove. This oral groove is a ciliated channel located on one side of the organism. Food particles are swept into the oral groove by water currents pro-duced by the beating cilia, and are carried down the groove to a point called the cytostome, which can be thought of as the mouth. Food is carried through the cytostome into the cytopharynx, which is also lined with cilia. As food accumulates at the lower end of the cytopharynx, a food vacuole forms around it. The vacuole eventually breaks away and begins to move toward the anterior of the cell. Lysosomes fuse with the food vacuole and secrete digestive enzymes into the vacuole. As digestion proceeds, the end products, such as sugars and amino acids, diffuse across the membrane of the vacuole into the surrounding cytoplasm, and the vacuole begins to move toward the posterior end of the cell. When at the posterior end of the cell, the vacuole fuses with a struc-ture there called the anal pore. Undigested material in the vacuole is expelled to the outside through the anal pore. In addition to serving as the digestive organ, the food vacuole also serves, by its movement, to distribute the products of digestion to all parts of the cell.

Question:

Develop a BASIC program to use Euclid's Algorithm to find the greatest common divisor of the two positive integers A and B.

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Solution:

Briefly, Euclid's Algorithm is a method which takes the smaller of two numbers and divides it into the larger number. The remainder from this division then becomes the smaller number, and the process is repeated. When the remainder reaches zero, the last smaller number is the greatest common divisor. A flowchart of the program's logic is given below. The program is given below: 100 REM GREATEST COMMON DIVISOR 110 READ A,B 120 IF A < = B THEN 130 125 GO TO 140 130 IF A > 0 THEN 150 131 GO TO 160 140 LET C = B 141 LET B = A 142 LET A = C 143 GO TO 130 150 LET B = B - A 151 GO TO 120 160 PRINT "THE GREATEST COMMON DIVISOR IS", B 800 DATA 16,36 999 END

Acceleration is defined as the time rate of change of velo-city. Since we are concerned with motion along the x-axis, we may neglect the vector nature of acceleration (a\ding{217}), velocity (v\ding{217}) and position (r\ding{217}) and write a = dv/dt dv = adt Because a is constant ^v2\int_v1 dv = at2\int_t1 dt As the limits of integration, we that v_1 = v_0 at t_1 = 0, and v_2 = v at t_2 = t. Then ^v\int_v0 dv=at\int_0 dt v - v_0= at v= v_0 + at Butv= dx/dt = v_0 + at where x is the position of the body along the x-axis. Then dx= v_0dt + at dt ^x1\int_x2 dx= v_0\int^t2_t1 dt + a\int^t2_t1 t dt(1) We take x_1 = x_0 at t_1 = 0, and x_2 = x att_2 = t, whence ^x\int_x0 dx= v_0t\int_0 dt + at\int_0 t dt x - x_0= v_0t + (at^2)/2 x= x_0 + v_0t + (1/2)at^2 We can also obtain this result by noting that evaluating the integral in (1) is equivalent to finding the area under the velocity vs. time curve shown in the figure.

Question:

Design the block diagram of the accumulator and code the micro-operations that go into it.

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Solution:

Each micro-operation and its associated control function describes what logic operations need to be car-ried out. For example, the statement: C_0t_2 :OPR\leftarrow[MBR]_1-3 implies that a two-input AND gate and a three bit register (called OPR) are needed. Figure 1 shows the block diagram implementation of this statement. To design the block diagram of the accumulator, a list is made of all the micro-operations whose results go into the AC and their associated control functions. Control function Micro-operation Definition q_0c_2t_2 : AC\leftarrow[AC] \wedge [MBR] AND with MBR q_1c_2t_2 + q_2c_2t_2 : AC\leftarrow[AC] + [MBR] Add q_2c_2t_1 + q_7I ̅c_2t_3 [MBR]_4 : AC\leftarrow0 Clear q_7I ̅c_2t_3 [MBR]_6 : AC\leftarrow[AC] Complement q_7I ̅c_2t_3 [MBR]_8 : AC_n+1\leftarrow[AC]_n, AC_0\leftarrow[E] Shift-right q_7I ̅c_2t_3 [MBR]_9 : AC_15\leftarrow[E]; AC_n-1\leftarrow[AC]_n Shift-left q_7I ̅c_2t_3 [MBR]_10 : AC\leftarrow[AC]+1 Increment q_7Ic_2t_3 [MBR]_4 : AC_8-15 \leftarrow[INPR] Transfer INPR Figures 2 and 3 show the block diagram of the accumulator. Two figures are used to simplify the circuit, however, the total circuit is the combination of Figures 2 and 3. The accumulator register is an integrated circuit package capable of storing, shifting, complementing, incrementing and clearing a sixteen bit number. The add and transfer operations are shown in Figure 3.

Animals such as the South American llama live at high altitudes where the partial pressure of oxygen is significantly lower than at sea level. For example, at an altitude of 10,000 feet, the partial pressure of oxygen drops to 110 torrs (from 159 torrs at sea level). These animals have evolved a different kind of hemoglobin that has a higher affinity for oxygen than does the hemoglobin of the average mammal living at sea level. Their hemo-globin picks up oxygen more easily so that sufficient oxygen can be obtained at high altitudes. Human fetal hemoglobin is similar to llama hemoglobin in that it is also adapted to pick up oxygen in a medium which has a partial pressure of oxygen lower than normal. Fetal hemoglobin thus has a higher affinity for oxygen than adult hemoglobin. This adaptation is significant since the fetus obtains its oxygen from the mother's blood. For the hemoglobin of the fetus to be able to take oxygen from the hemoglobin of the mother, it must have a higher affinity for the oxygen than does the mother's hemoglobin. Fetal hemoglobin is therefore oxygenated at the expense of adult hemoglobin in the maternal circulation. What accounts for the higher oxygen affinity? The structure of fetal hemoglobin is slightly different from that of adult hemoglobin. This altered structure binds an organic phosphate called 2,3\rule{1em}{1pt} diphosphoglycerate (DPG) less strongly than does adult hemoglobin. DPG, which is present in human erythrocytes, reduces the oxygen affinity of hemoglobin: DPG reduces the oxygen affinity by stabilizing the deoxy-hemoglobin structure; the binding of oxygen is thus less favored. Since fetal hemoglobin binds DPG more weakly than does adult hemoglobin, its deoxygenated form does not undergo this degree of stabilization; its oxygen affin-ity is therefore enhanced. The oxygen dissociation curves of both fetal and maternal are shown in the following diagram: For fetal hemoglobin, the oxygen dissociation curve is shifted to the left, indicating a higher oxygen affinity. When an adult human adapts to high altitude, there is an increase in DPG in the red blood cells which shifts the curve to the right, favoring unloading of oxygen from hemoglobin at any given partial pressure.

Since the process of heat (Q) addition occurs at constant pressure, we may write Q = \mu C_p(T_2 - T_1) where \mu is the number of moles of gas in the system, C_p is the molar specific heat at constant pressure, and T_2 - T_1 is the temperature difference between the 2 equilibrium states [see figures (a) and (b)]. Now, Q is given (10,000 Joules), as is \mu and T. Hence, we may solve for T_2 as a function of C_p. If we can calcul-ate the value of C_p from kinetic theory, we will have obtained T_2. We now perform the appropriate calculation. Consider a gas moving from one equilibrium state to another, via some thermodynamic process. We assume that the process occurs at constant volume. Using the First Law of Thermodynamics, we obtain ∆U = Q - W(1) Here, ∆U is the change in internal energy of the gas during the process, and Q and W are the heat added to and the work done by the gas, respectively, during the process. Writing (1) in differential form dU = dQ - dW(2) In general, the element of work done by the gas in an expansion is, by definition dW = F^\ding{217} \textbullet ds^\ding{217} where F^\ding{217} is the force the gas exerts on the piston [see figure (a)] and ds^\ding{217} is the element of distance the piston moves during the expansion. Since F^\ding{217} acts perpendi-cular to the face of the piston (that is, F^\ding{217} and ds^\ding{217} are parallel), we obtain dW = F ds(3) But F may be written in terms of the pressure p, that the gas exerts on the piston face of area A F = pA Using this in (3) dW = p ds A = p dV where dV is the differential change in volume of the gas during the expansion. Substituting this in (2), we find dU = dQ - p dV(4) Applying this equation to the above-mentioned isovolumic process, we obtain dU = dQ since dV = 0. By defintion, however dQ = \mu C_v dT where C_v is the molar specific heat of the gas at constant volume, and dT is the differential temperature change the gas experiences due to the addition of heat dQ. Then dU = \mu C_v dT(5) We now assume that the change in internal energy of a gas is only a function of the temperature differ-ence experienced by the gas. Then, no matter what thermodynamic process the gas experiences, (5) holds. Consider next an isobaric thermodynamic process. Again, we apply (4) dU = dQ - p dV(4) Since the process occurs at constant pressure, dQ = \mu C_p dT(6) where C_p is the molar heat capacity at constant pressure. Furthermore, from the ideal gas law pV = \muRT If p is constant, p (dV/dT) = \muR orp dV = \mu R dT(7) Using (6), (7), and (5) in (4) \mu C_v dT = \mu C_p dT - \mu R dT orC_p = C_v + R(8) Equation (8) relates the molar specific heat at constant pressure to the molar specific heat at constant volume. All the derivations up to now have been necessary in order to obtain certain relations involving molar specific heats, namely equations (5) and (8). We now turn to an examination of the internal energy of a diatomic gas. Each method of energy storage of a diatomic molecule is called a degree of freedom. If we view a diatomic molecule as being dumbell-shaped, then it has 5 degrees of freedom [see figures (c) and (d)] . The molecule may move translationally in 3 directions (x, y, z) with 3 kinetic energies [(1/2) mv^2_x, (1/2) mv^2_y, (1/2) mv^2_z). Furthermore, it may rotate about 3 axes (x, y, z), again, with 3 kinetic energies [(1/2)I_x\omega^2_x, (1/2)I_y\omega^2_y, (1/2)I_z\omega^2_z]. However, the rotational kinetic energy about the y axis is negligible [see figure (d)] because I_y < < I_x , I_z . Hence a diatomic molecule has 5 independent methods of energy absorption, or 5 degrees of freedom. Notice one important fact: each of these kinetic energy terms has the same form, mathematical-ly. That is, they are all of the form of a positive constant times the square of a variable which has a domain extending from - \infty to + \infty. The theorem of equipartition of energy tells us that, when Newtonian mechanics holds, and the number of gas particles is large, each term of this form has the same average value per molecule, namely, (1/2) kT. In other words, each degree of freedom of a gas molecule contributes an amount (1/2) kT to the internal energy of the gas. For a diatomic gas, then, the internal energy per molecule is U_i = 5 (1/2)(kT) = (5/2) kT The internal energy for \mu moles of molecules is U = \muN_0U_i = (5/2) \mukN_0T = (5/2) \muRT Using this in (5), we may solve for C_v (1/\mu)[d{(5/2) \muRT}/{dT}] = C_v C_v = (5/2)R Using this fact in (8) C_p = C_v + R = [(5 R)/2] + R = (7/2)R Getting back to the original problem, use this value of C_p in the first equation Q = \mu C_p(T_2 - T_1) = (7/2)\muR(T_2 - T_-1) orT_2 = T_1 + Q/[(7/2)\muR] = 350\textdegreeK + [10,000 Joules] / [(7/2)(3 moles)(8.31Joules/mole \textdegreeK)] T_2 = 350\textdegreeK + 114\textdegreeK = 464\textdegreeK T_2 = 350\textdegreeK + 114\textdegreeK = 464\textdegreeK Using the ideal gas law for the 2 equilibrium states pV_1 = \muRT_1 pV_2 = \muRT_2 orV_2/V_1 = T_2/T_1 whence V_2 = (T_2/T_1)V_1 = (464/350)1.3 m^3 = 1.72 m^3.

Let the incident proton have mass m and velocity u^\ding{217}, the velocity becoming v^\ding{217} after scatter. Let the struck particle of mass M acquire velocity V^\ding{217} after the collision. Then, by the principle of conservation of energy, (1/2)mu^2 = (1/2)mv^2 +(1/2)MV^2 and since momentum is conserved both parallel and perpendicular to the original direction of travel of the proton mu = mv cos 60\textdegree + MV cos 30\textdegree and mv sin 60\textdegree = MV sin 30\textdegree . Thus V = (m/M) v [(sin 60\textdegree)/(sin 30\textdegree)] = \surd3 (m/M) v. Substituting into the other two equations gives (1/2) mu^2 = (1/2) mv^2 + (3/2) (m^2/M)v^2 and mu = (1/2) mv + (3/2) mv u^2 = v^2 + 3 (m/M) v^2 = v^2[1+ (3m/M)](1) u = (1/2) v + (3/2) v = v[(1/2) + (3/2)] or u^2 = v^2[(1/2) + (3/2)]^2(2) Equating (1) and (2) v^2 [(1/2) + (3/2)]^2 = v^2[1+ (3m/M)] or 1/4 + 9/4 + 3/2 = 1 + (3m/M) m/M = 1 and the struck particle must have been a hydrogen nucleus (a proton).

The man can only see an image of his eyes if light leaves them, strikes the mirror, and is re-flected back along the same path. The central image is thus formed by light traversing the perpendicular from his eyes to the mirror. The outer images are formed by light striking the beveled edges at the point A [see figure (b)] at an angle of incidence \textphi such that the angle of refraction \textphi' makes the refracted ray strike the silvered surface normally. This must be the case if the ray of light is to leave the beveled edge by the same path with which it arrived. The angle \textphi' lies between the normal to the beveled edge and the normal to the back surface. Since \angleSAX = 90\textdegree [see figure (b)] \textphi' + \angle DAX = 90\textdegree But \angle DAX = 90\textdegree - \texttheta Hence \textphi' = 90\textdegree - 90\textdegree + \texttheta = \texttheta Draw BA, a construction line at A parallel to the back of the mirror. Angle BAC is also equal to \texttheta. But by Snell's Law n_1 sin \textphi = n sin \textphi', where n_1 is the refractive index of air (n_1 = 1) and n is that of glass. Then sin \textphi = n sin \textphi' = n sin \texttheta. Also \alpha = \texttheta +(90 - \textphi). [See figure (b)]. sin[90 - (\alpha - \texttheta)] = n sin \texttheta. But sin (90 - \psi) = cos \psi and cos (\alpha - \texttheta) = n sin \texttheta. By the trigonometric relation for double angles cos \alpha cos \texttheta + sin \alpha sin \texttheta = n sin \texttheta. cos \alpha + sin \alpha tan \texttheta = n tan \texttheta cos \alpha = tan \texttheta [n - sin \alpha] tan \texttheta = {cos \alpha/(n - sin \alpha} Looking at figure (a) cos \alpha = {1(1/4) ft}/{\surd[1 (1/4) ft]^2 + (3 ft)^2} = (5/13) sin \alpha = [(3 ft)/{[1 (1/4) ft]^2 + (3 ft)^2}] = (12/13) whencetan \texttheta = [(5/13)/{1.54 - (12/13)}] = 0.625. \texttheta = 32\textdegree.

Question:

Write the structural formulas of the hydrogenation products of the following olefins: (a) propylene; (b) 2-methy1-2- pentene; (c) 1-methyl cyclohexene.

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Solution:

The most important reactions of alkenes (olefins) involve addition of reagents to the double bonds. An alkene can be converted to an alkane by addition of hydrogen to the double bond. These reactions are usually carried out by using a high pressure of hydrogen gas in the presence of a catalyst such as finely divided platinum, palladium, or nickel. To solve this problem, draw the structural formulas of each alkene. The addition of hydrogen breaks the double bond, and the H_2 molecule attaches itself to the carbon atoms involved. Thus,

Let A be the dominant gene and a be the recessive gene. Since the flies are heterozygous, each must therefore be Aa. a)In the cross, we have: homozygoushetero-homozygous dominantzygousrecessive The proportion of offspring expressing the dominant phenotype is 1/4 + 2/4 = 3/4. Therefore, the expected number with the dominant phenotype is 3/4 × 200 or 150. b) To establish which of the dominant phenotypes are homozygous and which are heterozygous, we have to perform a test cross; i.e., crossing one fly whose genotype is to be determined with a homozygous recessive fly. If the test fly is homozygous dominant: All the progeny will show the dominant phenotype. This implies that the test fly is homozygous dominant. If the test fly is heterozygous, then in the back cross: So half the progeny will show dominant phenotype and the other half will show recessive phenotype. This is indicative that the test fly was heterozygous. Therefore, by performing a test cross between a fly whose genotype we do not know and a homozygous recessive fly, we can determine if the fly is homozygous or heter-ozygous by looking at the phenotypes of the progeny.

An acid-base reaction involves proton transfer between two species: the proton donor and the proton acceptor. The proton donor is better known as a Bronsted acid and the proton acceptor as a Bronsted base. The general equation for an acid-base reaction can be written as: Proton donor \textemdash\textemdash\textemdash\textemdash\rightarrow H^+ + proton acceptor . For example, the following reaction occurs between acetic acid (CH_3 COOH) and water (H_2O): CH_3COOH + H_2O \textemdash\textemdash\textemdash\textemdash\rightarrow CH_3COO^- + H_3O^+ Acetic acid is the Bronsted acid because it donates a proton to become the acetate anion, and water is the Bronsted base since it accepts a proton to form the hydronium ion. The acetate anion now has the capacity to accept a proton, and therefore is a Bronsted base by definition. The acetic acid and the acetate anion together constitute a conjugate acid- base pair. The generalized formula of these types of reactions would be: Bronsted acid_J + Bronsted base_K \textemdash\textemdash\rightarrow conjugate Bronsted base_J + conjugate Bronsted acid_K where the compounds written with the subscript, J are on conjugate acid-base pair and the compounds written with the subscript K are the other conjugate acid-base pair. An oxidation-reduction reaction includes the transfer of electron(s), rather than proton(s), between two species. We can write the following general equation for oxidation- reduction (also known as redox) reactions, analogous to the one we wrote for acid-base reactions: Electron donor \textemdash\textemdash\rightarrow e^- + electron acceptor The electron donor is the species that is undergoing oxidation (loss of electrons), and is also called the reducing agent (i.e., one that reduces another species).The electron acceptor is the species undergoing reduction (gain of electrons), and is also known as the oxidizing agent (i.e., one species that oxidizes another) . The electron donor and acceptor together constitute a conjugate redox pair or couple. In some oxidation- reduction reactions, the transfer of one or more electrons is made via the trans-fer of hydrogen. Dehydrogenation is thus equivalent to oxidation. One example of a redox reaction is the one be-tween NAD^+ and NADH + H^+: (NAD^+, nicotinamide adenine dinucleotide, will be discussed at length in a later chapter.) As different acids and bases differ in their tendency respectively to dissociate and accept protons, different reducing and oxidizing agents differ in their tendency to lose and gain electrons or hydrogen, respectively. The acidity of a species is measured by its pH, while the electron-donating ability of a reducing agent is measured by its standard oxidation-reduction potential.

Question:

What is meant by evolution? Explain briefly the concepts of inorganicand organic evolutions.

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Solution:

Evolution, in the broadest sense, means a gradual, orderly change fromone condition to another. Evolution may be either inorganic or organic. The planets and stars, the earth's topography, the chemical compoundsof the universe, and even the chemical elements and their subatomicparticles have undergone gradual, orderly changes. This is referredto as inorganic evolution since these changes do not involve living matter. Evolution that involves changes in the substance of living organismsis known as organic (meaning "living") evolution. The principle oforganic evolution states that all the various plants and animals existing atthe present time have descended from other, usually simpler, organismsby gradual modifications accumulated in successive generations. These modifications may be anatomical, physiological, biochemical, reproductive, or any other aspects of a species, that could havebeen instrumental in the emergence, proliferation, or complete extinctionof a species.

Question:

A variable to be used in a program is called BIRTH-DATE. This variable is not an independent variable, that is to say it has subdivisions such as MONTH, which is three char-acters long, DAY, which is two characters long, and YEAR, which is four characters long. Write group and elementary items to describe this variable.

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Solution:

The variable BIRTH-DATE will be described in DATA DIVISION with level number 01, which refers to the highest level of a data item, and coded with margin A. Following this, using margin B and level number 02 (actually any level number from 02-49 could be used) the subdivisions are defined as shown in fig. 1. PIC stands for Picture and describes the length and type of the item. Each X corresponds to one character in the ele-mentary item descriptions, thus YEAR is 4 characters long. The use of the X in the item description means that the item is alphanumeric, it can be any combination of characters, numbers and special characters.

Bacteria are microorganisms found everywhere around us. They are responsible for turning food sour either by way of fermentation or putrefaction. Fermentation is the decomposition of carbohydrates by bacteria while putrefaction is the decay of proteins by the same agents. Although the result in tube A suggests that meat soup spontaneously generates bacteria, the outcome in tube D shows this to be incorrect because it shows no sign of decay even though it contains boiled meat soup. The process of boiling meat soup does not prevent the soup from spoiling as shown by tube B. The soup in tube B does spoil, disproving statement (b). The result in tube C disproves the claim that boiling the soup alters the air because although the tube is sealed after boiling, the content spoils after a week. It should be noted that the process of boiling kills the bacteria already present in the soup. Tube A spoils in a day because of the activity of the bacteria already present in it. Tube B shows that although the soup is free from bacteria after boiling, it still spoils because it is exposed to the bacteria in the air. Tube C, with a cotton seal across its mouth, restricts the exposure of the soup to the air, thus reducing the chances of the bacteria getting into the soup and causing it to decay. But the cotton seal is not airtight and air does get into the soup gradually. That is why the soup spoils in a week. Tube D has a special "S" tube attached to the mouth of it. Although the tube has direct access to the air, all the bacteria present in the air are trapped in the bend of the "S" tube. In this way, tube D can remain fresh inde-finitely, because bacteria cannot get into the soup. By this series of experiments, we conclude that ex-posed food does not spoil spontaneously. It decays because the bacteria that are present in the atmosphere have settled on and interacted with the food and produced decay.

In the position stated in the problem, three forces are acting on the drawbridge: the weight acting downward, the tension in the chain, and the reaction at the hinge. The direction of W\ding{217} is as shown in the figure, and must act through the center point of the drawbridge, since the drawbridge is uniform. The tension T\ding{217} acts along the chain which makes an angle \texttheta with the drawbridge, where tan \texttheta is the height of the chain's pulley above the hinge divided by the length of the drawbridge: tan = \texttheta (34.6 ft)/(20 ft) = 1.73 Thus:\texttheta =60\textdegree Since the three forces are in equilibrium, the horizontal component of F\ding{217} must be equal and opposite to that of T\ding{217} (since W\ding{217} has no horizontal component) and the sum of the vertical components of T\ding{217} and F\ding{217} must equal W\ding{217}. The latter bit of information alone doesn't help us to solve for the vertical component of F\ding{217}. However, we know that the moments about any point in the drawbridge must cancel. Taking moments about point C, we note that the vertical components of T\ding{217} and F\ding{217} both act over moment arms of the same length (10 ft). This being the case we know that the vertical components of T\ding{217} and F\ding{217} must be equal. Since both the horizontal and vertical components of F\ding{217} are equal in magnitude to those of T\ding{217}, F\ding{217} must also make an angle \texttheta = 60\textdegree with the drawbridge. Thus, since we know that the vertical component of F\ding{217} equals half the weight of the drawbridge: F sin \texttheta= (1/2) W F= W/(2 sin \texttheta) = (3,600 lb)/(2 sin 60\textdegree) =(3,600 lb)/{2 (\surd3/2)} = 2,079 lb.

The figure shows the situation before and after the collision. The momentum of the first star is p_1 = m_1v_1 = (2 × 10^30 kg) (2 × 10^4 m/sec) = 4 × 10^34 kg-m/sec and is depicted in a vertical direction. The momentum of the second star is p^2 = m_2v_2 = (5 × 10^30 kg) (3 × 10^4 m/sec) = 15 × 10^34 kg-m/sec and is depicted in a horizontal direction. The vector diagram shows how to add the initial moment of the two stars. The total momentum of the system of the two stars before the collision is represented by the re-sultant in the triangle of vectors. According to the law of conservation of momentum this also represents the momentum of the single star resulting from the coa-lescence. Applying Pythagoras' theorem to the triangle of vectors, the magnitude of the total momentum is [\surd(4^2 + 15^2)] × 10^34 = 15.52 × 10^34 kg-m/sec. Since the combined mass is 7 × 10^30 kg, the velocity after coalescence is (15.52 × 10^34) / (7 x 10^30) = 2.22 × 10^4 m/sec. To obtain the angle \texttheta, notice that tan \texttheta = [(4 × 10^34) / (15 x 10^34)] = 0.2667 Whence\texttheta = 14.93\textdegree. Therefore, the single star resulting from the coa-lescence moves with a velocity of 2.22 × 10^4 m/sec in a direction making an angle of 14.93\textdegree with the direc-tion in which the second star was initially moving.

Question:

Consider two large parallel plates of area 1m^2 separated by 1mm. (a) What is the capacitance of the system? (b) Suppose that the capacitor is charged so that there is a charge +q on the upper surfaceand \rule{1em}{1pt}q on the lower where q = 10^\rule{1em}{1pt}3 coul. How much work must be done to charge the capacitor? (c) What is the force between the plates?

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Solution:

(a) The capacitance of a parallel plate capacitor is given by: C = (\epsilon_0A) / d = A / (4\pik_Ed) where A is the area of each of the plates, d is the separation of the plates, and k_E = 1 / (4\pi\epsilon_0 ) (c)1m^2 / [4(3.14)(9 ×10^9 nt \rule{1em}{1pt} m^2 coul^2) (10^\rule{1em}{1pt}3 m)] = 8.05 × 10^\rule{1em}{1pt}9 (coul^2 \rule{1em}{1pt}sec^2/kg\rule{1em}{1pt}m^2 ) = 8.05 × 10\rule{1em}{1pt}9farads (b) The work done in charging the capacitor equals the energy stored in it: W = E = (1/2) (q^2/c) = (1/2)[(10^\rule{1em}{1pt}3 coul)^2 /{8.05×10^\rule{1em}{1pt}9(coul^2\rule{1em}{1pt}sec^2/kg\rule{1em}{1pt}m^2)}] = 62.1(kg\rule{1em}{1pt}m^2/sec^2) = 62.1 joules (c)To find the force between the plates, we must first calculate the field between the plates. We construct as a Gaussian surface, a rec-tangular box with one face within the top plate and another between the plates, as in the diagram (a). Since the plates are separated by a distance which is small compared to their length, the field between them is uniform. Therefore: \intE^\ding{217} \bullet ds^\ding{217} = q / \epsilon_0 = \sigmaA / \epsilon_0, E = \sigmaA /A\epsilon0= \sigma / \epsilon_0 where A is the area of the face of the Gaussian surface between the plates. We see that since the vertical sides are parallel to the field, they make no contribution to the flux. The face within the top plate also makes no contribution to the flux since the field within a conductor is zero. Next, we must calculate the force on an infinitesimal charge element dq on the bottom plate. Part of the field between the plates is due to charge elements such as this one. Therefore, in order to obtain the net external field acting on dq (which we use to calculate the force on dq) we must subtract the field due to dq, from the field between the plates. We construct as a Gaussian surface a rectangular box (see figure (b)) with horizontal faces very close to the surface of dq. Close to dq, the field is vertical and uniform. This occurs because, at this distance, dq ap-pears to be a long sheet of charge. Since dq has an almost infinitesimal width, the flux through the vertical faces is negligible. By Gauss's law: \intE^\ding{217} \bullet dS^\ding{217} = 2dAE_1 = dq / \epsilon_0 = \sigmadA /\epsilon_0, E_1 = \sigma / 2\epsilon_0 Thus, the net external field acting on dq is: E' = E \rule{1em}{1pt} E_1 = (\sigma / \epsilon_0) \rule{1em}{1pt} (\sigma / 2\epsilon_0) = (\sigma / 2\epsilon_0) The force on dq then is: dF = E'dq = (\sigma / 2\epsilon_0)dq Therefore the total force on a plate is F = \intdF = \surd(\sigma / 2\epsilon_0)dq = (\sigma / 2\epsilon_0)Q = [(Q/A ) / (2\epsilon_0)]Q = q^2 / (2\epsilon_0A) F = (10^\rule{1em}{1pt}3 coul)^2 / [2(8.85 × 10^\rule{1em}{1pt}12 farads/m)( 1m^2)] = (5.65 × 10^4 [coul^2/{( coul^2 \rule{1em}{1pt}sec^2 / kg\rule{1em}{1pt}m^2)\rule{1em}{1pt}m}] = 5.65 × 10^4 (kg\rule{1em}{1pt}m / sec^2) = 5.65 × 10^4 N

The age distribution of a population is the proportion of individuals belonging to each age group. When a population is allowed to exist in a stable environment for several generations, so that its birth and death rates become balanced, its age distribution becomes stable. The stable age distribution differs greatly from species to species, and is expressed by a survivor-ship curve. The oyster is an example of a species in which large numbers of young are produced, but the majority die in a short time. Only a tiny fraction succeeds in attaching themselves to a rock or to some other support which is the necessary step for continuing the life cycle. The survival rate among oysters after reaching this point is much higher. The hydra exemplifies species with constant mortality rates. In this case, an individual is just as likely to die when it is one year old as when it is one day old. Man and fruit flies, in contrast to the hydra and oyster, are species with a definite period of senescence. If provided with a good environment, most individuals live to a certain age in reasonably sound health. Then diseases and infirmities of old age begin to set in, and death becomes increasingly probable with each passing year. Changes in environmental conditions may radically alter the shape of the survivorship curve for any given population, and the altered mortality rates in turn have profound effects on the age distributions and on its future size. For example, the chief cause for the enormous increase in the population size of human beings has been a great reduction in mortality during the early life stages as a result of improvements in sanitation, nutrition and medical care. These improvements have caused a shift in the human survivorship curve from one intermediate between the curves C and D in primitive societies to one approaching A in most advanced societies.

Let us represent the dominant allele for the trait of tongue-rolling by R, and the recessive allele by r. We know that if a gene is dominant for a trait, it will always be expressed if it is present. Since the son cannot roll his tongue, he cannot possess the dominant gene for this trait, and his genotype must be rr. This means that, each parent must have at least one recessive allele to donate to their son. Also, each parent must have a dominant allele for this trait because they have the ability to roll their tongues. Thus the genotype of parents for this trait must be Rr. We can illustrate this by looking at the mating of two such parents, The offspring will be obtained in the following ratios F1/4 RR,Homozygous dominant;tongue roller. 1/2 Rr,Heterozygous;tongue roller. 1/4rr,Homozygous recessive;non-tongue roller. We see then that there is a one-in-four chance that two parents who are heterozygous for a dominant trait will produce offspring without that trait. It is not unlikely then, that the parents in this problem could have had a son who does not have the ability to roll his tongue.

Question:

Write a program to create, initialize, insert, delete,search, etc., a Hash Table, using PL / I.

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Solution:

PL / Ihasno Hash Table as a standard data struc-ture type. Hence, we shall use a two-dimensional array to simulate a Hash Table. Creation of Hash Table: DCLHASHTAB(N, 2)CHAR (6) VAR, MARKER FIXED (3); / \textasteriskcenteredWehave created an array named HASHTAB of N, say, 180 data items each of which can be a character variable, ofupto6 characters. This is our Hash Table. \textasteriskcentered / / \textasteriskcenteredThevariable MARKER will be useful later as a pointer \textasteriskcentered / / \textasteriskcenteredForany location J of the Table, the first element HASH-TAB (J, 1) will store a data item and the second element HASHTAB(J, 2) will store a pointer , if necessary, to the next linked element having the same HASH address . \textasteriskcentered / Search and Insert: / \textasteriskcenteredAssumethat the value of the HASHING function has turned out to be J \textasteriskcentered / IF J<1\mid >N THEN GO TO ('recalculate' error routine); / \textasteriskcenteredWecheck if the Hashed address is within bounds of our table \textasteriskcentered / DO WHILE HASHTAB (J, 1)\rceil= 0; / \textasteriskcentered We check if location is empty \textasteriskcentered / MARKER = J J = J + 1; / \textasteriskcentered If not empty, look at the next location \textasteriskcentered / IF J = N THEN J = 1; / \textasteriskcentered If we reach the end then we go round to the top of the array \textasteriskcentered / If J = MARKER THEN GO TO (table full error routine); / \textasteriskcenteredWecome back around to the starting point again, which indicates the Table has no empty spaces left \textasteriskcentered / HASHTAB( MARKER, 2) = J; / \textasteriskcenteredPointer of location calculated by the hashing function points to the new location J, \textasteriskcentered / End; / \textasteriskcentered End of DO WHILE \textasteriskcentered / HASHTAB (J, 1) = DATA; / \textasteriskcentered If we find the new location HASHTAB (J, 1) to be empty we do not enter the DO WHILE block, and jump to this statement, where we insert the data item. \textasteriskcentered / Search and Read Out; / \textasteriskcenteredAssumethat we are looking for a data item denoted by KEY, and whose HASH function value has been calculated as J \textasteriskcentered / DO WHILE HASHTAB (J, 1)\rceil= KEY; MARKER = J / \textasteriskcenteredWestore the original value of J in the variable MARKER \textasteriskcentered / J = HASHTAB (J, 2); / \textasteriskcenteredWechange the value of J to that pointed to by the pointer part of J \textasteriskcentered / If j = 0 THEN GO TO ('not available' error routine); / \textasteriskcenteredThismeans there are no more linkages to search, and the data item is not in the table \textasteriskcentered /. End; / \textasteriskcentered End of DO WHILE \textasteriskcentered / Y = HASHTAB (J, 1); / \textasteriskcentered If the position denoted by J is equal to the Key, we do not enter the DO WHILE, but come to this statement, and read out the value written in the location into some required location Y \textasteriskcentered / HASHTAB (MARKER, 2) = HASHTAB (J, 2); / \textasteriskcenteredWenow complete the pointer linkages of the chain by making the pointer of the previous element , stored in Marker, to point to the next element to the deleted element \textasteriskcentered / End;

Question:

Differentiate between sapwood and heartwood and between spring wood and summer wood. What are the vascular rays?

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Solution:

The sapwood is the youngest, outermost portion of the xylem in a woody plant. It contains tracheids and vessels which function in carrying water and minerals from the root to the leaves. The heartwood is the older, inner portion of the xylem whose tracheids and vessels have lost the ability of conduction. The heartwood serves to increase the strength of the stem, and to support the increasing load of foliage as the tree grows. The spring wood and summer wood together make up the secondary xylem ring (see figure) of each growing season. They account chiefly for the increase in diameter of a woody tree trunk. They are derived from the vascular cambium, which also gives rise to the secondary phloem. At the beginning of each growing season, the cambial cells divide to produce relatively large, thin-walled xylem cells, which make up the spring wood (early wood). Toward the end of each season, the xylem cells produced are smaller and have thicker walls. These smaller cells constitute the summer wood (late wood). Ordinarily, there is only one growth season per year and abrupt change in cell size between the spring wood and summer wood clearly marks off distinct alternate circular zones called annual rings. The number of annual rings found in the crosssection of a woody trunk is now used as an indicator of the age of the tree. In addition, because the width of the annual rings varies according to the climatic conditions prevailing when ring was formed, it is possible to infer what the climate was at a particular time by examining the rings of old trees. The vascular rays are rows of small, parenchymal cells that lie at right angles to the long axis of the stem or root, reaching radially from the secondary xylem to the secondary phloem. They are, like the secondary xylem and phloem, derived from the vascular cambium, and extend in length as the plant increases in girth. The vascular rays conduct water and minerals from the secon-dary xylem to the vascular cambium, the secondary phloem, and the cortical cells. Nutrients form the secondary phloem are channeled to the vascular cambium, the living cells of the secondary xylem, and the cells of the pith (if living).

Question:

The time required for a volume of gas, X, to effuse through a small hole was 112.2 sec. The time required for the same volume of oxygen was 84.7 sec. Calculate the molecular weight of gas X.

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Human skin is composed of a comparatively thin, outer layer, the epidermis, which is free of blood vessels, and an inner, thick layer, the dermis, which is packed with blood vessels and nerve endings. The epidermis is a stratified epithelium whose thickness varies in different parts of the body. It is thickest on the soles of the feet and the palms of the hands. The epidermis of the palms and fingers has numerous ridges, forming whorls and loops in very specific patterns. These unique finger-prints and palmprints are determined genetically, and result primarily from the orientation of the underlying fibers in the dermis. The outermost layers of the epidermis are composed of dead cells which are constantly being sloughed off and replaced by cells from beneath. As each cell is pushed outward by active cell division in the deeper layers of the epidermis, it is compressed into a flat (squamous), scalelike epithelial cell. Such cells synthesize large amounts of the fibrous protein,keratin, which serves to toughen the epidermis and make it more durable. Scattered at the juncture between the deeper layers of the epidermis and the dermis are melanocytes, cells that produce the pigment melanin. Melanin serves as a protective device for the body by absorbing ultraviolet rays from the sun. Tanning results from an increase in melanin production as a result of exposure to ultraviolet radiation. All humans have about the same number of melanocytes in their skin. The difference between light and dark races is under genetic control and is due to the fact that mela-nocytes of dark races produce more melanin. The juncture of the dermis with the epidermis is uneven. The dermis throws projections called papillae into the epidermis. The dermis is much thicker than the epider-mis, and is composed largely of connective tissue. The lower level of the dermis, called the subcutaneous layer, is connected with the underlying muscle and is composed of many fat cells and a more loosely woven network of fibers. This part of the dermis is one of the principle sites of body fat deposits, which help preserve body heat. The subcutaneous layer also determines the amount of possible skin movement. The hair and nails are derivatives of skin, and develop from inpocketings of cells from the inner layer of the epidermis. Hair follicles are found throughout the entire dermal layer, except on the palms, soles, and a few other regions. Individual hairs are formed in the hair follicles, which have their roots deep within the dermis. At the bottom of each follicle, a papilla of connective tissue projects into the follicle. The epithelial cells above this papilla constitute the hair root and, by cell division form the shaft of the hair, which ultimately extends beyond the surface of the skin. The hair cells of the shaft secrete keratin, then die and form a compact mass that becomes the hair. Growth occurs at the bottom of the follicle only. Associated with each hair follicle is one or more sebaceous glands, the secretions of which make the surface of the skin and hair more pliable. Like the sweat glands, the sebaceous glands are derived from the embryonic epidermis but are located in the dermis. To each hair follicle is attached smooth muscle called arrector pili, which pulls the hair erect lipon contraction. Nails grow in a manner similar to hair. Both hair follicles and nails develop from inpocketings of cells from the inner layer of the epidermis. The translucent, densely packed, dead cells of the nails allow the underlying capillaries to show through and give the nails their normal pink color. Like the hair and nails found in man, the feathers, hoofs, claws, scales, and horns found in other vertebrates are also derivatives of the skin.

First, apply Dalton's Law and then make use of the Combined Gas Law to obtain the desired answer. Dalton's Law states that the total pressure of a mixture of gases, P_total is equal to the sum of the individual partial pressures, P_(H)2 and P_(H)2O In this example the individual gases are hydrogen and water vapor. Therefore, the total pressure is the sum of these two gases. At 35\textdegreeC, the partial pressure of H_2O vapor is 40 mm, P_total = P_(H)2+ P(H)2O P_(H)2 =P_total -P_(H)2O = 753 - 40 = 713 mm This means that if no water vapor was present the H_2 gas would fill the container at 713 mm of pressure. Since pressure is indirectly proportional to the volume of H_2 gas, the effect of changing the pressure of 713 mm to STP pressure (760 mm) is to decrease the volume. In addition, however, the temperature of the H_2 gas is directly proportional to the volume of the gas, so that the effect of changing the temperature of (34\textdegreeC + 273\textdegree =) 307\textdegreeK to STP temperature (273\textdegreeK) is also to de-crease the volume of the gas. Therefore, we can predict that the net effect of changing the pressure and tem-perature of the gas to STP conditions is to decrease the volume. Using the combined gas law equation to solve for the dry H_2 gas volume, V_2 , [P_1V_1] / T_1 = [P_2V_2] / T_2 whereP_1 = 713 mmP_2 = 760 mm V_1 = 425 mlV_2 = volume at STP T_1 = 3070KT_2 = 2730K Thus, substituting, V_2 =[P_1V_1T_2] / [P_2T_1] = [(713 mm) (425 ml) (273\textdegreeK)] / [(760 mm) (307\textdegreeK)] = 355 ml. We see that this answer is in total agreement with our prediction that the volume at STP, V_2 = 355 ml, is less than the volume at the initial conditions, V_1 = 425 ml.

The diagram is labeled and current values have been inserted in each part of the circuit. Apply-ing Kirchhoff's current law to point A, we have I_1 + I_2 = I_3(1) Note that in the figure, boxes have been drawn representing the 2 batteries in the circuit. The given voltages, \epsilon_1 and \epsilon_2, are the voltages of the batteries assuming zero internal resistance. (The terminal volt-ages are V_BC and V_FE , not \epsilon_1 and \epsilon_2 .) Applying Kirchhoff's voltage law to the closed circuit containing both cells, and then to the closed circuit through the lower cell and the external re-sistance (see figure for the directions in which the loops are traversed) we have + \epsilon_2 - \epsilon_1 + (.5 \Omega) I_1 - (.1 \Omega) (I_2 ) = 0 and+ \epsilon_2 - (5 \Omega) (I_3) - (.1 \Omega) (I_2 ) = 0 Hence,\epsilon_2 - \epsilon_1 = (2 - 1.2)V = (.1 \Omega) (I_2 ) - (.5 \Omega) (I_1 ) \epsilon_2 = 2V = (5 \Omega) (I_3 ) + (.1 \Omega) (I_2 ) or, upon multiplication of both sides of each equation by 10, 8 V = (1 \Omega) (I_2 ) - (5 \Omega) (I_1 )(2) 20 V = (1 \Omega) (I_2 ) + (50 \Omega) (I_3 )(3) Substituting equation (1) in (3) 20 V = (1 \Omega) (I_2 ) + (50 \Omega) (I_1 + I_2) 20 V = (51 \Omega) (I_2 ) + (50 \Omega) I_1(4) Multiplying (2) by 10, 80 V = (10 \Omega) (I_2 ) - (50 \Omega) (I_1 )(5) Adding (4) and (5), we may solve for I_2 100 V = (61 \Omega) I_2 andI_2 = [(100 V)/(61 \Omega)] = 1.64 A(6) since 1 ampere = 1 volt/ohm. Substituting (6) in (3), we obtain I_3, 20 V = (1 \Omega) (1.64 A) + (50 \Omega) (I_3 ) [(18.46 V)/(50 \Omega)] = I_3 orI_3 \approx .37 A .

Question:

Find the first five energyeigenvaluesfor a hydrogen atom. (An equation in which an operation on a function gives a constant times the function is called aneigenvalueequation. The function is called aneigenfunctionof the operator and the constant is called aneigenvalue. TheSchroedinger equation haseigenfunctionswhich are the possible spatial portions of the wave functions andeigenvalueswhich are the possible values of the energies involved.)

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Solution:

Question:

Complete the program defined earlier in the previous three problems by writing the PROCEDURE DIVISION. The trans-action records for this program each contain a TRANSACTION- CODE based on the type of transaction. A code of 1 indi-cates a new customer's record is to be added to the file. A code of 2 indicates a charge to the customer's account, and a 3 indicates a credit, either from payment or return. The program must specify different paths for each trans-action. In coding your solution, write the three routines ADD-RECORD, ADD-CHARGE, and SUBTRACT-CREDIT first. Then write the basic program branching to and from routines as necessary. Any time an INVALID KEY clause is activated while a record is being replaced in the file, display "INVALID KEY", C-NR, then transfer control to READ- UPDATE- INFO. Refer back to the Data Division as needed.

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Solution:

The complete procedure division is shown in fig. 1.In the OPENING paragraph Indexed-file is opened as I/O , allowing us to write data to and from it. The statement is a conditional Go To statement. In this pro-gram paragraph 1 (ADD RECORD) is specified when TRANSACTION- CODE is a 1, paragraph 2 (ADD-CHARGE) is specified when TRANSACTION-CODE is a 2, etc. The statements after INVALID KEY are executed whenever there is an error while writing or reading from an indexed file. The error messages are printed using the ERROR-LISTING PARAGRAPH, which displays the error message and the record which contained the error.

During an isothermal change, T is constant. The work done on the gas in such a change is W = -^Vf\intVip dV. This is negative because the pressure on v_i the gas is in opposition to the volume change. From the ideal gas law, pV = nRT, we get W = -^Vf\int_Vi (nRT/V) dV = -nRT ^Vf\intVidV/V = -nRT ln (V_f/V_i) = -p_iV_i 1n(V_f/V_i) = -p_iV_i 1n(p_i/p_f). where the subscripts i and f refer to initial and final states respectively. Thus the work done on the gas in the first change is [from (1) to (2), as shown in the figure] W_1 = -p_1V11n(V_2/V_1) = -(1 atm ×1.013 × 10^6 dynes/cm^2 \bullet atm) × (1 liter × 10^3 cm^3/lit)×1n 2 = - 1.013 × 10^6 dynes/cm^2 × 10^3 cm^3 × 1n 2 = - 7.022 × 10^8 ergs = - 70.22 J. Further, since the volume is doubled, by the application of Boyle's law p_1V_1 =p_2V_2 we see that the pressure is halved at (2) p_2 = p_1 (V_1/V_2) = (1/2)p_1. The work done on the gas in the second change is [from (2) to (3)] W_2 = -p_2 ^V3\int_V2 dV = -p_2(V_3 - V_2) = (1/2)p_1(V_1 - 2V_1) = (1/2)p_1V_1 = (1.013 × 10^6 dynes/cm^2 × 10^3 cm^3) / (2 x 10^7ergs/J) = 50.65 J. The work done on the gas in the final change from (3) to (4) is W_3 = -p_3V_3 1n (p_3/p_4) = -(1/2)p_1v_1 1n(P_2/P_1) = (1/2)p_1V_1 1n (1/2) = (1/2) p_1V_1 1n 2 = [1.013 × 10^6 dynes/cm^2 × 10^3 cm^3 × in(2)] / [2 x 10^7ergs/J] = 35.11 J The total work done on the gas is thus W1+ W_2 + W_3 = (50.65 + 35.11 - 70.22)J = 15.54 J. In the first and third processes the temperature does not change. In an ideal gas the internal energy depends only on the temperature, so that no change of internal energy takes place in the first and third processes. Any work done on the gas in these changes is equal to the heat transfer taking place. The second process is isobaric. The change in internal energy during the process is given by the first law of thermodynamics as ∆U = Q - W, where Q is the heat energy added to the system and W the work done by the system. Hence ∆U = -50 J - (-50.65 J) = +0.65 J. The internal energy thus increases by 0.65 J during the three processes.

Question:

Trace the evolution of man using the existing classifications and characteristics of fossil men.

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Solution:

The fossils of the Dryopithecine date back to about 30 million years ago during the oligocene era. It is believed that there once existed several species of Dryopithecus: Dryopithecus africanus which was thought to be ancestral to the chimpanzee, Dryopithecus major - thought to be ancestral to the gorilla, and Dryopithecus sivolensis thought to be ancestral to the present day orangutan. Although not enough information has been collected to tell exactly when and where the human lineage first started, it is believed that it may have had its roots in the Dryopithe-cine line. The apelike Dryopithecus was probably the ancestor of man. By the middle of Miocene (about 20 million years ago), the evolutionary lines leading to the various types of modern anthropoids were distinct. The first set of fossils found to be considered in the hominid line, representing Ramapithecus, dates back to the upper Miocene epoch (about 10 to 14 million years ago). It must be pointed out, however that this is supported only by vague evidence since only the jaws and teeth from Ramapithecus have been found. The canine and incisor teeth are relatively small in man, and this may be an important distinction from the apes. If Ramapithecus did not use his teeth for defense and his diet changed to one that required more chewing, then this change may have been correlated with the increased use of hands. Since no fossil skull of Ramapithecus has been found, little can be learned about his intelligence. If he did reply more on his hands, there is the implication that he might stand at least semi-erect and use bipedalism at times. A gap of about five million years in man's evolutio-nary history is found between Ramapithecus and the next distinct homonid form, called Australapithecus, who appeared about 5.5 to 1.2 million years ago. Relatively large portions and numbers of the Australopithecine skeleton have been found scattered throughout South and East Africa. An important advance toward the modern human condition shown by the Australopithecines is the acquisition of some degree of bipedalism. They might have stood and walked completely erect or perhaps they were fully erect only while running. From fossil evidence it is deduced that Australopithecus was relatively small, an adult weighing about 75 pounds. His dental anatomy was humanlike, especially that of the canines. His cranial capacity was very small and comparable to that of a present day gorilla. It must however be noted here that although the Australopithecine cranial capacity is comparable to a gorilla's, the ratio of his brain size to body weight is much higher that the gorilla's proportion. Australopithecus probably was not very intelligent but was obviously developing human characteristics and some form of culture. He probably lived in a group and had a daily hunting-gathering mode of living. This is supported by the finding of bones of small animals and evidence of gathered fruits, nuts, and shoots at or around excavated sites. Stone tools are found also at these sites and are of a style either known as pebble tools or as the Oldowan industry. Therefore, Australopithecus was a toolmaker as well as a tool user, a feature much less common among animals other than man himself. A later stage in human evolution is represented by a group of fossil men, including Java and Peking man, that can be classified as Homo erectus. This species first appeared about 1.2 million years ago and existed till about years ago. As compared to Australopithecus, Homo erectus had greater body size, increased cranial capacity, reduced facial shape and dental structure, and the beginning of a human type culture. A major feature found in Homo erectus is the elaboration of cultural adaptations that originated in the earlier Australopithecine stage. The cultural associations of Homo erectus, in the making of tools (called Acheulian tools) and in social activities, are more complex than those associated with the Australopithecines. By the Homo erectus stage of human evolution, adap-tations based on culture are seemingly well established. If culture is the basis of human adaptation, then natural selection would favor a more efficient culture. In these fossil men, an enlarged brain is related to an increased capacity for more complex cultural skills and behavior patterns. Reduction in facial and dental structure also may be linked to more efficient use of tools and cooking of food. Homo erectus appear to have led the same kind of life, including the hunting and gathering that characterizes some groups of modern man. Because of their similarities to modern men, these fossil men are included within the genus Homo. The ancestral line, or lines, leading to modern man becomes hazy approximately 300,000 years ago. From around to 150,000 years ago, the fossil record is rather scanty. However, a collection of fossils found in Terra Amata, France, which is about 200,000 to 35,000 years old, gives us a decent picture of man's transition from Homo erectus to Homo sapiens. The structures of these skeletal remains clearly show an increase in brain size, as well as the increasing likeness of facial features to those of modern man. The Homo sapiens neanderthalensis, known as the cavemen, is a hominid group with skulls that were large and massive with a thick, bony ridge over the eyes and a receeding forehead. Their nose was broad and short and there was almost no chin at all. Their cranial capacity was as large as, or larger than that of modern man. In general, these characteristics are intermediate between those of Homo erectus and those of Homo sapiens. The population of Neanderthal men that inhabited regions that are now Europe, North Africa and the Middle East some 75,000 to 35,000 years ago seemed to be both culturally and physically modern. They were a widely spreading species - their remains have been found over a wide area including parts of western Europe, Asia and Africa, indicating that they were able to adapt to various climates. They were a society of small family groups who lived primarily in caves, used fire, made flint weapons of the Mousterian tool tradition, hunted a variety of game ranging in size up to the mammoth and rhinoceros, and buried their dead re-verently with food, flowers and ornaments. This last mentio-ned fact indicates that they were capable of abstract thinking including the concept of a life after death. Reconstruction of one Neanderthal skeleton caused these men to be depicted as standing stooped, with their knees bent forward. However, it has been determined later that this portrayal of Neanderthal man is incorrect. The skeleton used to form this illustration actually come from a man who died at a very old age with rickets, which accounts for the skeleton's deformation. It is now known through examination of many other skeletons that Neanderthal man stood as erect as modern man and was probably just as efficient a bipedal walker. Presently, Neanderthal man is considered to belong to a sub-species of Homo sapiens.

Question:

In a pound-for-pound comparison, which would you prefer as a rocket fuel, hydrogen (H2) or dimethylhydrazine 2 (CH_3)_2NNH2? H_2(g) + (1/2)O_2 (g) \rightarrow H2O(l), 2 2 O(l), ∆H= - 68.3Kcal/mole (CH_3)2NNH_2 + 4O_2 (g)\rightarrow N_2(g) + H 2 \rightarrow N_2 4H_2O(g) + 2CO_2(g), ∆H = - 404.9 Kcal/mole.

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Solution:

The material that is the better fuel will release more energy per unit weight in its reaction with oxygen. Calculate the energy released per unit weight for H_2 and (CH_3)_2NNH_2. You can do this by using the heats of formation and enthalpy changes given. Calculate the total mass or weight of materials from moles that react and divide the ∆H by it to obtain the amount of energy re-leased per unit weight. For hydrogen: 1 mole of H_2 reacts with (1/2) mole of oxygen (1 mole H_2) (1.008 g H_2/mole H_2) + ((1/2) mole O_2) (16.00 g O_2/mole O_2) = 1.008 g + 8.00 g = 9.01 g. ∆H = - 68.3 Kcal. Thus, (-68.3) / (9.01) = - 7.58 Kcal/g. For dimethylhydrazine: [1 mole (CH_3)_2NNH_2] [60.10 g (CH_3)_2NNH_2/mole (CH_3)_2NNH_2] + (4 moles O_2)(16.00 g. O_2 /mole O_2) = 60.10 g + 64.00 g = 124.10 g. ∆H = - 404.9. The amount of heat released per unit weight with (CH_3)_2NNH_2 + O_2 = (- 404.9 Kcal) / (124.10 g) = - 3.26 kcal/g. Hydrogen yields over twice as much energy per gram of fuel mass, and is thus preferable in this respect.

Question:

Solution:

Until the beginning of the twentieth century, large amounts of KNO3 (saltpeter) were used in the pre-paration of black gunpowder. This powder as prepared by mixing powdered saltpeter, carbon, and sulfur. When ignited, sulfur and carbon are very rapidly oxidized by KNO_3, pro-ducing several gaseous products. 4KNO_3 + S + 4C \ding{217} 2K_2O + 2N_2 + SO_2 + 4CO_2 + heat This equation can be applied to solve this problem. The only difference is that a red flare usesSr(NO_3)_2 to give it a distinctive color and is mixed in different proportions with carbon and sulfur.Sr(NO_3)_2 also oxidizes carbon and sulfur to form the same gaseous products as black gunpowder. Thus, the products for the oxidation of carbon and sulfur are CO2 and SO_2 and the product of reduction is N_2 (going from N^+5 in NO^-_3 to N0 in N_2). The complete reaction is: 2Sr (NO_3)_2 + 3C + 2S \ding{217} 2SrO + 2N_2 + 3CO_2 + 2SO_2.

Question:

Find the pH of a 0.2 M solution of formic acid. K_a = 1.76 × 10^-4.

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Solution:

To solve the problem, first note that the acid dissociates according to the general equation HA+H_2O\rightleftarrowsH^++A^-, where HA is the acid and H^+ and A^- are the dissociation products. To find the pH, calculate [H^+]. One can see from the equation that equal con-centrations of H^+ and A^- form when HA dissociates. At equilibrium, x moles of HA dissociate into x moles of H^+ and x moles of A^-. HA+H_2O\rightleftarrowsH^++A^- Before:0.2OO After:0.2-xxx Now, set up the equilibrium constant equation, which is K_a = [products] / [reactants]= {[H+][A^-]} / [HA] = 1.76 × 10^-4 Before substituting in the values, simplify the calculation by assuming that 0.2 - x \allequal 0.2. This is a valid assumption since the acid is very weak and only a very small percentage dissociates. Thus, one has K_a = {(x)(x)} / (0.2) = 1.76 × 10^-4 . Solving: x = [H^+] = 5.93 × 10^-3. H^+ concentration is then converted to pH according to the equation, pH = - log [H^+] = - log 5.93 ×10^-3 = - [- 3 + log 5.93] = 3 - 0.77 = 2.23.

Question:

What are the properties of actin and myosin which produce the cyclic activity of the cross bridges responsible for contraction? What causes rigor mortis?

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Solution:

Myosin, the larger of the two molecules, is shaped like a lollypop (see Figure 1). The myosin molecules are arranged within the thick filaments so that they are oriented tail-to-tail in the two halves of the filament; the globular ends extend to the sides, forming the cross briges which bind to the reactive site on the actin molecule. Actin is a globular-shaped molecule having a reactive site on its surface that is able to combine with myosin. These globular proteins are arranged in two chains which are The globular end of the myosin molecule, in addition to being a binding site for the actin molecule, contains a separate binding site for ATP. This active site has ATPase activity, and the reaction that is catalyzed is the hydrolysis of ATP: H_2O + ATP \textemdash\textemdash> ADP + Pi However, myosin alone has a very low ATPase activity. It appears that an allosteric change occurs in the active site of myosin ATPase when the myosin cross bridge combines with actin in the thin filaments, considerably increasing the ATPase activity. The energy that is released from the splitting of ATP produces cross bridge movement by an as yet unknown mechanism. It is believed that the oscillatory movements of myosin cross bridges produce the relative movement of thick and thin filaments, resulting ultimately in the shortening of a muscle fiber (see Figure 3). Since many cycles of activity are needed to produce the degree of shortening observed during muscle contrac-tion, the myosin bridge must be able to detach from the actin and then rebind again. This is accomplished by the binding of ATP to the myosin in the cross bridge, forming what is known as a low-energy complex. The low-energy complex has only a weak affinity for actin; the actin-myosin bond is broken, allowing the cross bridges to dis-sociate from actin. Shortly after this event, a conformational change occurs in the myosin - ATP complex and a high energy complex is formed. The high energy complex has a very high affinity for actin, and the cross bridges are able to rebind to the actin. In this manner, the cross bridges are able to bind and dissociate from actin in a cycle of coordinated actions. This cycle may be summarized in the following sequence of events: A = actinM= myosin M-ATP\textemdash\textemdash\textemdash\textemdash>M^\textasteriskcentered - ATP (low-energy complex)(high-energy complex capable of binding action) A + M^\textasteriskcentered -ATP\textemdash\textemdash\textemdash\textemdash>A - M^\textasteriskcentered -ATP (with actin bound, myosin is able to split ATP) A - M^\textasteriskcentered -ATP \textemdash\textemdash\textemdash\textemdash>A - M + ADP + Pi (as ATP is split, cross bridge movement occurs) A - M + ATP \textemdash\textemdash\textemdash\textemdash>A + M-ATP (low-energy complex dissociates from actin) Rigor mortis is a phenomenon in which the muscles of the body become very stiff and rigid after death. It results directly from the loss of ATP in the dead muscle cells; the myosin crossbridges are unable to combine with actin and those bonds already formed cannot be broken - thus the rigid condition. At the molecular level, we can identify two specific roles for ATP: 1) to provide energy for movement of cross bridge, and 2) to dissociate actin from the myosin cross bridges during the contraction cycle of the bridges. ATP is also needed to restore Ca^+ in the sarcoplasmic reti-culum following contraction (see next question). Two regulatory proteins, troponin and tropomyosin, are associated with actin. During nervous stimulation of a muscle, there is an increase in free intracellular calcium ions: calcium diffuses in from the terminal cisternae and from the extracellular fluid in the T tubules. Calcium binds to troponin which causes tropomyosin to shift its position along the actin helix. This exposes the binding site on ac-tin for myosin.

In order to solve this problem we must transform the given 2-body problem to an equivalent one-body problem. To do this, we find the equation of motion of each mass shown in the figure. Using Newton's Second Law and his Law of Universal Gravitation, we obtain: F1 2^\ding{217} = m_2r_2^\ding{217} = [Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})]/[\vert r_1^\ding{217} - r_2^\ding{217}\vert^3](1) F2 1^\ding{217} = m_1r_1^\ding{217} = - [{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vert r_1^\ding{217} - r_2^\ding{217}\vert^3}](2) where F_12^\ding{217} is the force exerted on 2 by 1, and similar-ly for F_21^\ding{217}. Rewriting r_2^\ding{217} = (1/m_2)[{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vertr_1^\ding{217} - r_2^\ding{217}\vert^3} ](3) r_1^\ding{217} = -(1/m_1)[{Gm_1m_2(r_1^\ding{217} - r_2^\ding{217})} / {\vertr_1^\ding{217} - r_2^\ding{217}\vert^3}](4) Subtracting (3) from (4), and using the figure to realize that r_1^\ding{217} - r_2^\ding{217} = r^\ding{217}, we obtain r_1^\ding{217} - r_2^\ding{217} = r^\ding{217}= -(1/m_1) [(Gm_1m_2r^\ding{217})/(r^3)] - (1/m_2) [(Gm_1m_2r^\ding{217})/(r^3)] or r^\ding{217} = -[(Gm_1m_2r^\ding{217})/(r^3)] [(1/m_1) + (1/m_2)] Defining the reduced mass \mu as 1/\mu = (1/m_1) + (1/m_2)(5) we find \mur^\ding{217} = -[(Gm_1m_2r^\ding{217})/(r^3)](6) This equation is a one body equation describing the motion of a particle of mass \mu under the influence of a gravitational force. Now, to further reduce the problem, assume that each mass shown in the figure rotates in a circular orbit with the given angular velocities about Q, the center of mass. Using Newton's Second Law for each mass, (m_1v ׳ 1 2 )/(r ׳ 1 ) = (Gm_1m_2)/(r^2)(7) (m_1v_2 ׳ 2 )/(r_2 ׳ ) = (Gm_1m_2)/(r^2)(8) where the primed variables are measured with respect to the point Q. By definition of the center of mass m_1r_1 ׳ \ding{217} + m_2r_2 ׳ \ding{217} = 0 orr_1 ׳ \ding{217} = -(m_2r_2 ׳ \ding{217} )/(m_1)(9) Furthermore,r\ding{217}= r_1 ׳ \ding{217} - r_2 ׳ \ding{217} orr_2 ׳ \ding{217} = r_1 ׳ \ding{217} - r^\ding{217}(10) Inserting (10) in (9), and solving for r_1 ׳ \ding{217} r_1 ׳ \ding{217} = - [(m_2)/(m_1)] r_2 ׳ \ding{217} = -[(m_2)/(m_1)]( r_1 ׳ \ding{217} - r^\ding{217}) r_1 ׳ \ding{217} [1 + (m_2)/(m_1)] = [(m_2)/(m_1)] r^\ding{217} r_1 ׳ \ding{217} = [m_2/m_1r^\ding{217}] / [1 + (m_2)/(m_1)] = [m_2r^\ding{217}] / [m_1 + m_2](11) But using (5) \mu = (m_1m_2)/(m_1 + m_2) Hence, (9) becomes r_1 ׳ \ding{217} = (\mu/m_1) r^\ding{217}(12) Similarly,r_2 ׳ \ding{217} = -(\mu/m_2) r^\ding{217}(13) Hencer_1 ׳ \ding{217} = (\mu/m_1) r r_2 ׳ \ding{217} = (\mu/m_2) r(14) Using (14) in (7) and (8) (m^2_1v_1 ׳ 2 )/\mur = (Gm_1m_2)/(r^2) (m^2_2v_2 ׳ 2 )/\mur = (Gm_1m_2)/(r^2) or(m_1v_1 ׳ 2 )/2= (\muGm_2)/(2r) (m_2v_2 ׳ 2 )/2= (\muGm_1)/(2r) Therefore, the net kinetic energy of the system relative to the center of the mass is T = (1/2) m_iv_1 ׳ 2 + (1/2) m_2v_2 ׳ 2 = [(\muG)/(2r)] (m_1+ m_2)(15) T = (Gm_1m_2)/(2r)(16) by definition of \mu. The total energy is E = T + V = (Gm_1m_2)/(2r) - (Gm_1m_2)/(r) E = - [(Gm_1m_2)/(r)](17) To remove the variable r, we replace it with the angular momentum J^\ding{217} as follows. The total system angular momentum is (relative to Q) J^\ding{217} = r_1 ׳ \ding{217} × m_1v_1 ׳ \ding{217} + r_2 ׳ \ding{217} × m_2v_2 ׳ \ding{217} Since r_1'^\ding{217} and v_i'^\ding{217} are perpendicular, and similarly for r_2 ׳ \ding{217} and v_2'^\ding{217}, we obtain J = m_1r_1 ׳ v_1 ׳ + m_2r_2 ׳ v_2 ׳ (18) From (12) and (13) v_1'^\ding{217} = (\mu/m_1) v^\ding{217}r_1'^\ding{217} = (\mu/m_1) r^\ding{217} v_2'^\ding{217} = -(\mu/m_2) v^\ding{217}r_2'^\ding{217} = -(\mu/m_2) r^\ding{217} orv_1' = (\mu/m_1) v^\ding{217}r_1'= (\mu/m_1) r^\ding{217} v_2' = -(\mu/m_2) v^\ding{217}r_2'= -(\mu/m_2) r^\ding{217}(19) using (19) in (18) J = (m_1)[(\mu/m_1) r] [(\mu/m_1) v] + (m_2)[(\mu/m_2) r] [(\mu/m_2) v] J = (\mu^2/m_1) vr + (\mu^2/m_2) vr = \mu^2 vr[(1/(m_1) + (1/m_2)] By definition of \mu J = \muvr(20) We now eliminate v in (20) so that we may sub-stitute (20) in place of r in (17). We know, from (15) that T = (1/2) m_1v_1^'2 + (1/2) m_2v_2^'2 = (\muG/2r) (m_1 + m_2)(15) Substituting forv_1', v_2'from (19) in (15), Substituting for from (19) in (15), T = (1/2) m_1- [(\mu^2/m_1^2)v^2] + (1/2) m_2 [(\mu^2/m_2^2)v^2] = (\muG/2r) (m_1 + m_2) T = [(\mu^2v^2)/(2m_1)] + [(\mu^2v^2)/(2m_2)] = (\muG/2r) (m_1 + m_2) T = [(\mu^2v^2)/(2)][(1/m_1)+(1/m_2)] = (\muG/2r) (m_1 + m_2) By definition of \mu T = [(\muv^2)/(2)] = (\muG/2r) (m_1 + m_2)(21) T = [(\muv^2)/(2)] = (\muG/2r) (m_1 + m_2)(21) Hencev^2 = (G/r)(m_1 + m_2)(22) Using (22) in (20) J = \mur \surd[{G(m_1 + m_2)}/r] orJ^2 = \mu^2 r G(m_1 + m_2) Thereforer = (J^2) / [\mu^2G(m_1 + m_2)](23) Inserting (23) in (17) E = -[(Gm_1m_2)/2] [ {(\mu^2G(m_1 + m_2)} / J^2] E = -[{G^2 m_1m_2(m_1 + m_2) \mu^2} / 2J^2] Finally, m_1m_2(m_1 + m_2)\mu^2 = m_1m_2(m_1 + m_2) [(m_1m_2) / (m_1 + m_2)] \mu orm_1m_2(m_1 + m_2)\mu^2 = m^2_1m^2_2 \mu ThenE = - [(G^2 \mu m^2_1m^2_2)/(2 J^2)]

The Multiplication Instruction (M or MR): Both types of this instruction have a register for the first operand. But the M-instruction is of register to in-dex (RX) format and has a fullword as its second operand. In contrast, the MR instruction is of register to register (RR) format and has a register as its second operand. Once more let us consider the first operand of the M and MR instructions. In writing the instruction a single register appears as the first operand, but in actuality we are employing two registers. In both cases the register designated as operand 1 must be an even (i.e. 2, 4, 6 etc.) register. Since two registers are employed, by IBM conven-tion, the even register designates the beginning of a "single" field. This 'single' field can be said to have resulted from a conceptual merger of the two 4-byte regis-ters to form a single 8-byte field. Since we perform mul-tiplication from right to left, the number to be multiplied is stored in the low order (right) bytes of the mergered field. Storing is done starting with the rightmost byte of the odd register since this register constitutes the 4 low order bytes of the 8-byte field. In doing multiplication the odd register is first loaded with the number to be mul-tiplied (the multiplicand). The resulting product replaces the multiplicand of the 8-byte field. In many cases the product is small enough to be contained in the 4 low order bytes of the 8-byte field. This is so since 4 bytes can contain 32 binary digits (bits) and each bit represents a power of two. As a result of this, a single register can contain a decimal number as large as four and a third bil-lion. These 4 bytes are of course the odd register. Thus, in storing the product at a convenient location in core, the contents of the odd register is stored. It must be noted that the procedure of storing the single register is done at the discretion of the programmer. Example 1 L7, MULCANSTORING NUMBER TO BE MULTIPLIED IN LOW 8 BYTES M6, MLTPLYRMULTIPLYING REGISTERS 6 AND 7 BY MULTPLYR (FULLWORD) STM6,7,DOUBLESTORING REGISTERS 6 AND 7 IN A DOUBLEWORD OR THE LAST STATEMENT CAN BE REPLACED BY: ST7, FULLSTORES ONLY THE SINGLE REGISTER. Interpretation of: M 6, MLTPLYR This instruction may be seen as: The Division Instruction (D or DR): The operands here are similar in form to those of the multiplication instruction. The D-instruction is of RX-format and takes a register for the first operand and a full-word for the second. The DR instruction is of RR format and has registers as both operands. The aforementioned con-ceptual merger also occurs. The number to be divided (the dividend) is also stored starting with the rightmost byte of the 8 byte field. After the calculation, the quotient is stored as a fullword in the odd register, and the remain-der, which takes the sign of the divider, is stored in the even register. If any of these results do not fit into the registers, a fixed-point exception occurs, and the pro-gram is usually cancelled. It is not difficult to see that the common feature of the IBM 360-370 multiply and divide instructions is that of the conceptual merger. It is this aspect of these instruc-tions that may provide problems for the programmer. If we see these instructions as operations on 8 byte fields de-signated by an even register we should find coding these instructions quite simple. The main idea behind the merger is to have a field large enough to contain the resulting data. It should be noted that the information in the regis-ters are stored in binary form although the examples indi-cate decimal digits.

Photosynthesis in plants is restricted to those cells containing chlorophyll, such as the cells of the leaf. Respiration occurs in all living cells, whether they be the non - chlorophyll - containing root cells or the chlorophyll - containing leaf cells. Photosynthesis requires water and carbon dioxide as raw materials, and sunlight as an energy source. It produces oxygen and simplehexosemolecules, such as glucose. Cellular respiration, on the contrary, consumes oxygen and organic molecules and releases water and carbon dioxide. Photosynthesis is an energy absorbing process, the energy ultimately coming from the sun; respiration releases energy: some is lost as heat and some is stored as chemical bond energy in ATP. Photosynthesis, as the name implies, occurs only when light is present, but respiration takes place continuously, during both night and day. Photosynthesis occurs in the chloroplasts, which contain the necessary enzymes and electron carriers; respiration occurs in the mitochondria, which contain the oxidative enzymes of thetricarboxylicacid cycle (also known as the Krebs cycle) and components of the electron transport system. Finally, photosynthesis results in an increase in the weight of the plant, as more organic substances are manufactured. Respiration results in a decrease in the weight of the plant, as organic materials are broken down for the synthesis and release of energy.

Question:

Suppose a particle of mass m, initially travelling with velocity v^\ding{217}, collides elastically with a particle of mass m initially at rest. Prove that the angle between the velocity vectors of the two particles after the collision is 90

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Solution:

Whenever we are confronted with a collision problem, we may apply the law of conservation of momentum if no external forces act on the system. External forces are forces which act upon the system being considered (in our case, the system consists of the 2 masses) that are due to the environment external to the system, (i.e. friction). Because no external forces are acting in this problem, we may use this conservation law, which is written mathematically as p_10^\ding{217} + p_20^\ding{217} = p_(1)f^\ding{217} + p_(2)f^\ding{217}(1) In this equation, p_10^\ding{217} and p_20^\ding{217} are the initial moment of par-ticles 1 and 2,respectively,and p_(1)f^\ding{217} and p_(2)f^\ding{217} are the final moment of particles 1 and 2, respectively. Note two things: first, momentum is defined as the product of the mass of a particle and its velocity (i.e. p^\ding{217} = mv^\ding{217}) and, secondly, because v^\ding{217} is a vector (i.e. it has direction and magnitude, as in 50 mph EAST) p^\ding{217} is also a vector. Applying equation (1), and noting that v^\ding{217}_20 is 0 because particle 2 is initially at rest, we may write p_10^\ding{217} = p_(1)f^\ding{217} + p_(2)f^\ding{217}(2) Now, we have still not used one last bit of in-formation. The collision is elastic, and, whenever this is the case, kinetic energy is conserved. The law of kinetic energy conservation may be written mathematically as (1/2)m_1 v10^2 + (1/2)m_2 v20^2 = (1/2)m_1 v_(1)f ^2 + (1/2)m_2 v(2)f^2(3) where v_10, v_(1)f, v_20, v_(2)f have the same meaning as previ-ously. Using this equation, we obtain (1/2)m_1 v10^2 = (1/2)m_1 v(1)f^2 + (1/2)m2v(2)f^2(4) But m_1 = m_2 = m, hence (1/2)m v_10 ^2 =(1/2)mv(1)f^2 + (1/2)m v(2)f^2(5) v_10^2 = v(1)f^2 + v(2)f^2(6) Rewriting equation (2), we find mv_10^\ding{217} = mv_(1)f^\ding{217} + mv_(2)f^\ding{217}(7) Dividing both sides of (7) by m, v_10^\ding{217} = v_(1)f^\ding{217}+ v_(2)f^\ding{217}(8) We now want to express equation (8) in terms of magnitudes. To find the magnitude of a vector, we multiply it by itself, using the dot product, and take the square root. The dot product of 2 vectors is defined as A^\ding{217} \textbullet B^\ding{217} = AB cos \texttheta where A is the magnitude of A^\ding{217}, B is the magnitude of B^\ding{217}, and \texttheta is the angle between A^\ding{217} and B^\ding{217}. Note that A^\ding{217} \bullet A^\ding{217} = (A) (A) = A2 because, in this case, \texttheta = 0\textdegree and cos 0\textdegree = 1. Now, dotting each side of equation (8) into itself, we find v_10^\ding{217} \bullet v_10^\ding{217} = (v_(1)f^\ding{217}_ + v_(2)f^\ding{217}) \bullet (v_(1)f^\ding{217}_ + v_(2)f^\ding{217})(9) v_10^\ding{217} \bullet v_10^\ding{217} = v_(1)f^\ding{217} \bullet v^\ding{217}(1)f+ 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} + v_(2)f^\ding{217} \bullet v_(2)f^\ding{217} orv10^2 = v(1)f^2 + v(2)f2+ 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217}(10) Subtracting equation (6) from (10) v10^2\ding{217} = v_(1)f ^2 + v(2)f^2 + 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} -v10^2= v(1)f^2 + v_(2)f ^2 0= 2v_(1)f^\ding{217} \bullet v_(2)f^\ding{217} orv_(1)f^\ding{217} \bullet v_(2)f^\ding{217} = 0(11) v_(1)f v_(2)f cos \texttheta = 0(12) But because v_(1)f, v(2)f\not = 0,this means that cos \texttheta = 0 or \texttheta = 90\textdegree,as was to be shown.

Compound eyes consist of a varying number of units or ommatidia (see Figure). Each ommatidium has its own lens, cornea, and group of neurons. Some arthropods, such as ants, have only a few ommatidia while others, such as dragonflies, have as many as 10,000. Each ommatidium, in gathering light from a narrow sector of the visual field, projects a mean intensity of the light from that sector onto the total retinal field. All the points of light from the various ommatidia form an image, called a mosaic pic-ture. The nature of a mosaic image can be understood by comparing it to the picture of a TV screen or a newspaper photo. The picture is a mosaic of many dots of different intensities. The clarity of the image depends upon the number of dots per unit area. The greater the number, the better the picture. The same is true of the compound eye. Each ommatidial receptor cell with its accompanying nerve-fiber projects a segment of the total picture. The compound eye may contain hundreds of thousands of neurons, clustered in groups of seven or eight per ommatidium. The human eye also forms a retinal picture composed of numerous points, each point corresponding to a rod or cone cell with its accompanying neuron. Considering the fact that the human eye contains approximately 125,000,000 rods and 6,500,000 cones, the concentration of corresponding points is much higher than in the compound eye, and the image formed is thus much finer. In addition, while the compound eye of the arthropod is composed of numerous lenses, the human eye has only one lens\textasteriskcentered for its entire corneal field. Hence, the insect has no structure strictly analogous to the human retina? their critical surface for vision is the outer surface of the compound eye itself, composed of the many closely packed individual lenses. Many arthropods have very wide visual fields? in some crustaceans, the corneal surface covers an arc of 180\textdegree. However, the anthropod cornea and lens are developed from the skeleton, and the eyes are fixed in place. The human eye, has a much smaller visual field, but compensates for this to some extent with movability. Compound eyes are particularly sensitive to motion. This is because the ommatidia recover very rapidly from a light impulse, making them receptive to a new impulse in a very short time. Compound eyes thus can detect flickers at extremely high frequency. Flies detect flickers of a frequency up to 265 per second, as compared to man's limit of about 53 per second. Because flickering light at higher frequency is seen as a continuous light by man, motion pictures are seen as smooth movement, and 60-cycle bulbs give off steady light. The ability of the compound eyes to detect high frequency light changes enables them to rapidly detect motion and facilitates the capture of prey and the avoidance of enemies. Compound eyes are sensitive to a broader range of light wavelengths than human eyes. Insects can see well into the ultraviolet range. Colors that appear identical to man may reflect ultraviolet light to different degrees, and appear strikingly different to insects. Compound eyes are also able to analyze the plane of polarization of light. A sky appearing evenly blue to man,reveals different patterns in different areas to insects, because the plane of polarization varies. Honeybees use this ability as an aid to navigation.

By symmetry, the center of mass lies on the axis and the center of mass of each part is midway be-tween its ends. The volume of the disk is: v_d = \pir^2h = \pi(1 in)^2 (1 in) = \pi in^3. The volume of the rod is: v_r = \pir^2h = \pi[(1/2) in]^2 (6 in) = (3/2) \pi in^3 Since the disk and the rod are both constructed of the same homogeneous material, the ratio of their masses will equal the ratio of their volumes: (mass of disk)/(mass of rod)= (m_d)/(m_r) = (\rhov_d)/(\rhov_r) = (\pi in^3)/[(3/2) \pi in^3] = (2/3) where \rho is their common density. The formula for the distance of the center of mass from a given origin 0 is: x = (m_d x_d + m_r x_r)/(m_d + m_r) where x_d and x_r are the distances of the centers of mass of m_d and m_r respectively from 0. Take the origin 0 at the left face of the disk, on the axis. Then x_d = 1/2 in., and x_r = 4 in. Since m_d = 2/3 m_r x= [(2/3)m_r (1/2 in) + m_r (4in)]/[(2/3) m_r + m_r] = 2.6 in. The center of gravity is on the axis, 2.6 in. to the right of 0.

We wish to analyze the annihilation of a neutron-antineutron pair. We may relate the energies and momenta of the products of the annihilation to the energies and momenta of the neutron-antineutron pair by the laws of conservation of energy and momentun. Initially, the neutron and antineutron are at rest, and therefore, have only rest energy. Furthermore, the neutron and antineutron have equal masses, and 2m_n c^2 = E_\pi(1) + E_\pi(2)(1) where "n", "\pi_1", "\pi_2" refer to the neutron and 2 pions respectively. Applying conservation of momentum, 0 = p^\ding{217}_\pi(1) + p^\ding{217}_\pi(2) ,(2) since the neutron and antineutron are initially at rest. But, relativistically , P^\ding{217} = (m_0v^\ding{217}) / \surd{1 - (v^2 /c^2)} where v^\ding{217} is the velocity of the particle whose momentum we wish to measure, and m_0 is its rest mass. By (2) (m_0\pi(1)v_1^\ding{217}) / \surd{1 - (v^2_1 /c^2)} = (- m-_0\pi(2) v_2^\ding{217}_ ) / \surd{1 - (v^2_2 /c^2) } (3) Butm_0\pi(1) \approx m0\pi(2), and v_1^\ding{217} / \surd{1 - (v^2_1 /c^2) } = - v_2^\ding{217}/ {\surd1 - (v^2_2 /c^2) }(4) Taking the dot product of each side of (4) with itself, we have (since v^\ding{217} \textbullet v^\ding{217} = (v) (v) cos 0\textdegree= v^2) v^2_1 / { 1 - (v^2_1 /c^2) } = v^2_2 / {1 - (v^2_2 /c^2)} v^21= [v^2_2 {1 - (v^2_1 /c^2)}] / {1 - (v^2_2 /c^2)} v^2_1 {1 - (v^2_2 /c^2)} = v^2_2 {1 - (v^2_1 /c^2)} v^2_1 (c^2 - v^22) = v^22(c2- v^2_1) (v^2_1 c^2 ) - (v^21v^2_2)_ = (v^22c^2) - (v^21v^2_2) v^21c^2= v^22c2 v^21= v^2_2(5) Therefore, the magnitudes of v_1^\ding{217} and v_2^\ding{217} have the same value, say v. Using (2) and (3) 0 =[(m_0\pi(1) v^\ding{217}) / {\surd1 - (v^2 / c^2) }] + [ (m-_0\pi(2) v^\ding{217}) / {\surd1 - (v^2 /c^2)}](6) Hence the momenta of the 2 pions are equal and opposite. From (1), we may calculate the kinetic energies of \pi_1(T_\pi(1)) and \pi_2(T_\pi(2)). Noting that the energy of any particle equals the sum of its rest energy and kinetic energy, we may write E = m_0c^2 + T . With (1) 2m_nc^2 = m_\pi(1) c2+ T_\pi(1) + m_\pi(2) c2+ T_\pi(2) Butm\pi(1)= m\pi(2)and 2m_nc^2 \therefore 2m_\pic^2 = T_\pi(1) + T_\pi(2)(7) The mass-energy of the two neutrons is E_nc= 2m_nc^2 = (2) (1.67 × 10^-27kg) (3 × 10^8 m/s)^2 = 3 × 10^-10 J The rest mass-energy of the two neutral pions is E_\pi =2m_nc2= (2) (2.4 × 10^-28kg) (3 ×10^8m/s)^2 = 4.3 × 10^-11 J Therefore , T_\pi(1)^ + T_\pi(2) = 3 × 10^-10 J - 0.43 × 10^-10 J T_\pi(1)^ + T\pi(2)= 2.57 × 10^-10 J(8) To find how this energy is divided among the pions, note that E = [m_0c^2 / \surd{1 - (v^2 / c^2)}] where v is the velocity of the particle whose energy we wish to measure. Hence E _\pi(1) = [m0\pi(1)c^2]/ [\surd{1 - (v^2_1 / c^2)}] E _\pi(2) = [m0\pi(2)c^2] / [\surd{1 - (v^2_2 / c^2)}] Butv_1 = v_2 ,andm_\pi(1) = m\pi(2). whence E _\pi(1) = E _\pi(2) Orm_\pi(1)c^2+ T_\pi(1) = m_\pi(2)c^2+ T\pi(2) AndT_\pi(1) = T_\pi(2) . Using T_\pi(1) = T_\pi(2) = 1.29 × 10^-10 J.

Question:

Describe the chemical reactions that occur in the human retina bringing about visual sensation.

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Solution:

Rhodopsin is a visual pigment in the human eye that is responsible for the perception of an enormous range of light intensities. Located in the rod cell of the retina, rhodopsin has a chromophore (retinal1) bound to a protein (opsin). Retinal is the group) of vitamin A, also called retinol (see Figure 1) Retinal can be synthesized from vitamin A via an oxi-dation reaction, catalyzed by the enzyme alcohol de-hydrogenase. Retinal exists in the cis-configuration before light hits the rhodopsin molecule. When light strikes the rhodopsin molecule, cis-retinal absorbs the light energy and is converted to the trans form (Figure 2). Simultaneously, the sensation of vision, the stimulation of nerve impulses by the rod cells, occurs. The conversion causes a rearrangement of the molecular structure, resulting in an unstable compound called lumirhodopsin. Lumirhodopsin rapidly decomposes into free trans-retinal and opsin. This is due to the inability of trans-retinal to fit onto the opsin molecule as does the cis form. Rhodopsin has to be resynthesized from cis-retinal and opsin to provide for further excitation of the rods. Trans-retinal is first reduced to form trans-retinol, which is subsequently enzymatically converted into cis- retinol. The cis-retinol then undergoes oxidation to regenerate cis-retinal, which then combines with opsin to yield rhodopsin. The series of reactions from the light-induced breakdown of rhodopsin to its resynthesis, collectively termed the visual cycle, is outlined in figure 3. Because vitamin A is the precursor of retinal, synthesis of rhodopsin is dependent on its presence. In vitamin A deficiency, vision is most conspicuously affected. This deficiency is responsible for a condition known as xerophthalmia ("dry eyes") and may cause blindness. An early sign of vitamin A deficiency is night blindness, which is a maladaptation of the eyes to the dark.

According to the de Broglie wave theory of matter, every particle has wave-like characteristics. To every particle there is associated a wave with wavelength given by: \lambda = (h/p) = (h/mV). For an electron (of mass 9.1 × 10^-31 kg.) and velocity 4.43 × 10^6 m/s we then have \lambda = [(6.6 × 10^-34 Js) / {(9.1 × 10^-31 kg)(4.43 × 10^6 m/s)}] = 1.64 × 10^-10 m The wavelength for a "matter wave" is now called the particles's de Broglie wavelength. The Davisson-Germer apparatus is used to ex-hibit the interference effects of the electron "matter wave."

Question:

The mass of an electron is 9.109 × 10^-28 g. What is the atomic weight of electrons?

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Solution:

Atomic mass is defined as the weight of one mole of particles. There are 6.02 × 10^23,or Avogadro's number, of particles in 1 mole. Therefore, the atomic weight of electrons is equal to 6.02 × 10^23 times the mass of one electron. atomic weight of electron = 9.109 × 10^28 g × 6.02 × 10^23 / mole =5.48 × 10^-4 g/mole.

Question:

A single circular loop of wire is rotated in a uniform field of magnetic induction B^\ding{217}. We can suppose that the magnetic field is in the +z-direction, and the axis of rotation of the loop is in the x-direction along a diameter. If the speed of rotation \omega is constant, find the induced EMF in the coil.

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Solution:

The flux through the circular loop is, since B^\ding{217} is constant, \varphi M = B \bullet A = BA cos \texttheta Where \texttheta is the angle between B and A (as shown in the Figure.) The angular velocity \omega, for constant speed of rotation, is \omega = (\texttheta/t)\texttheta = \omegat Therefore, \varphi_M = BA cos \omegat According to Faraday's law of induction, the EMF induced in a coil, due to a changing flux, is given by \epsilon = - [(d\varphi_M )/(dt)] where the negative sign indicates that the EMF is induced in such a direction as to oppose the change in flux which created it. Substituting for \varphi_M and differentiating, we then have \epsilon = \omegaBA sin \omegat = \omegaBA sin \texttheta In this case the induced EMF varies with time in a sinusoidal way.

This problem deals with polygenic inheritance; that is, the situation in which two or more independent pairs of genes have similar and additive effects on the same characteristic. Examples of such inheritance are height and skin color in man, and commercially important characteristics in animals and plants, such as the amount of eggs and milk produced, the size of fruit, and so on. In the cross between a 6-pound squash plant and a 3-pound squash plant : We are told that each dominant gene adds (1/2) pound to the weight. The presence of 3 dominant genes in F_1 (A, B, and C) would increase the weight by (1/2) + (1/2) + (1/2) or 1 (1/2) lbs. Therefore each squash from F_1 weighs 3 + 1 (1/2)or4 (1/2) pounds. In the cross between two F_1 plants, we must first determine the possible gametes from each plant. The number of possible gametes is obtained using the 2^n rule, where n is the number of heterozygous traits and 2n is the total number of different gametes formed. For each parent, n is equal to 3. Therefore the number of gametes is 2^3 or 8. The genotypes of these gametes can be obtained by dichotomous branching, which ensures that all possible combinations are considered. Each possible gene from each allelic pair is matched to every possible gene combination of the other two pairs as follows : Gametes ABC B C ABc A c AbC b C Abc c aBC B C aBc a c abC b C abc c Looking at the cross : Separating the trihybrid cross into three monohybrid crosses, and then using dichotomous branching to determine all possible combinations of the results of these crosses, one obtains: Summarizing:1/64weighs 6 pounds Summarizing:1/64weighs 6 pounds 6/64weighs 51/2 pounds 15/64weighs 5 pounds 20/64weighs 41/2 pounds 15/64weighs 4 pounds 6/64weighs 31/2 pounds 1/64weighs 3 pounds It is important to be able to associate quantitative characteristics with polygenic inheritance. The cross itself is not difficult but may at times be tedious (Note: this problem could also have been done using the Punnett square, but dichotomous branching is more frequently used in crosses involving three or more traits).

Blood flowing to the kidneys first undergoes glomerular filtration. This occurs at the junction of the glomerular capillaries and the wall of Bowman's capsule. The blood plasma is filtered as it passes through the capillaries, which are freely permeable to water and solutes of small molecular dimension yet relatively im-permeable to large molecules, especially the plasma pro-teins. Water, salts, glucose, urea and other small species pass from the blood into the cavity of Bowman's capsule to become the glomerular filtrate. It has been demonstrated that the filtrate in Bowman's capsule contains virtually no protein and that all low weight crystalloids (glucose, protons, chloride ions, etc.) are present in the same concentrations as in plasma. If it were not for the process of tubularreabsorption, the composition of the urine would be identical to that of the glomerular filtrate. This would be extremely wasteful, since a great deal of water, glucose, amino acids and other useful substances present in the filtrate would be lost. Tubular reabsorption is strictly defined as the transfer of material from the tubular lumen back to the blood through the walls of the capillary network in intimate contact with the tubule. The principal portion of the tubule involved in reabsorption is the proximal convoluted tubule. This tubule is lined with epithelial cells having many hair-like processes extending into its lumen. These processes sure the chief sites of tubular reabsorption. As the filtrate passes through the tubule, the epithelial cells reabsorb much of the water and virtually all the glucose, amino acids and other substances useful to the body. The cells then secrete these back into the blood-stream. The secretion of these substances into the blood is accomplished against a concentration gradient, and is thus an energy consuming process - one utilizing ATP. The rates at which substances are reabsorbed, and therefore the rates at which wastes are excreted (because what is not re-absorbed is eliminated), are constantly subjected to physiological control. The ability to vary the excretion of water, sodium, potassium, hydrogen, calcium, and phos-phate ions, and many other substances is the essence of the kidney's ability to regulate the internal environment. Reabsorption also occurs in the distal convoluted tubules, where sodium is actively reabsorbed under the influence of aldosterone, a hormone secreted by the adrenal cortex. When this occurs, chloride passively follows due to an electrical gradient; water is also reabsorbed because of an osmotic gradient established by the reabsorption of sodium and chloride. In addition, reabsorption of water takes place in the distal convoluted tubule and collecting duct, stimulated by the posterior pituitary hormone vaso-pressin, also known as antidiuretic hormone (ADH) . ADH in-creases the permeability of the distal convoluted tubule and collecting duct to water, allowing water to leave the lumen of the nephron and render the urine more concentrated. The kidney also removes wastes by means of tubular secretion. This process involves the movement of addi-tional waste materials directly from the bloodstream into the lumen of the tubules, without passing through Bowman's capsule. Tubular secretion may be either active or passive, that is, it may or may not require energy. Of the large num-ber of different substances transported into the tubules by tubular secretion, only a few are normally found in the body. The most important of these are potassium and hydrogen ions. Most other substances secreted are foreign substances, such as penicillin. In some animals, like the toadfish, whose kidneys lack glomeruli and Bowman's capsules, secre-tion by the tubules is the only method available for excre-tion.

Question:

Write a program which uses a bubble sort procedure to put given strings in lexical order.

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Solution:

The function of BUBBLE sort is based on the function of a simple exchange sort, i.e., when an element, compared to the previous, or next one, is found to be out of order, it is moved to a preceding or following place respectively. The described procedure is continuously performed until that element is properly placed. The solution of this problem is as follows: \textasteriskcenteredBUBBLE SORT PROGRAM &TRIM = 1 DEFINE ('SORT(N)I') DEFINE('SWITCH(I)TEMP') DEFINE('BUBBLE(J)') \textasteriskcentered1:F(ERROR) N = INPUT A = ARRAY(N) \textasteriskcentered2 READI = I +1 A = INPUT:F(GO)S(READ) \textasteriskcentered3 GOSORT(N) \textasteriskcentered4 M = 1 PRINTOUTPUT = A:F(END) M = M + 1:(PRINT) \textasteriskcentered5 SORTI = LT(I,N-1) I +1:F(RETURN) LGT(A,A+ 1>):F(SORT) SWITCH(I) BUBBLE(I):(SORT) SWITCHTEMP =A A = A A =TEMP:(RETUN) BUBBLEJ = GT(J,1) J- 1:F(RETURN) LGT(A,A):F(RETURN) SWITCH(J):(BUBBLE) END Any statement in SNOBOL IV, starting with an asterisk in column 1, is a comment (or remark) statement. These statements are not executable and are usually used to explain the functions of different parts of the program. There are three functions used in this program, which are defined in the beginning of the program. Their procedures are given right after the comment numbered 5. The first card of the input data for this program must contain the number of strings to be sorted, otherwise the program stalls. The part of the program between the comments numbered 1 and 2 reads that number, assigns it to N, and sets up an array of N members. The next program segment (between comments 2 and 3) fills the array with the input items. Notice that counter I is not initialized to zero, because in SNOBOL IV the initial value of the variables is a null string. The statement between the comments 3 and 4 calls and executes the SORT function. Since the entry point label is omitted in the DEFINE statement, the computer looks for the statement labeled SORT, as the initial statement of the function procedure, and finds it right after the comment 5. The function LGT(X, Y), included in the second statement of the SORT function procedure, compares lexically strings X and Y, and succeeds if X follows Y alphabetically. Thus, LGT (SPACE, BLANK) will succeed, while LGT (CAT, DOG) will fail. The next two statements of the SORT function procedure call and execute functions SWITCH and BUBBLE. Function SWITCH makes the two strings switch their places, while the BUBBLE function compares the string with the preceding one and switches them around, if needed, continuously, until the first string is finally properly placed. Thus, the major function SORT works as follows: The first statement (I = LT(I,N-1) I + 1) checks if a currently considered string is numbered 1 through N-1 (there is no need to specially consider the last string N, because, when the string N-1 is considered, it is compared with the string numbered N). The next statement checks the alphabetical order of that string and the one following it in the array. If the order is correct, the next string is considered. If the order is wrong, the fol-lowing string moves up one place, is compared with the next string, moves up again, if needed, and so on, until it is properly placed. When all the input items are considered, the program execution returns to the main part, i.e., to the section between comments 4 and 5 and one by one prints all the input strings in a just arranged alphabetical order. For the input:3 COMPUTER ARRAY BRANCH THE OUTPUT IS:ARRAY BRANCH COMPUTER

Question:

The limits of the visible spectrum are approximately 400 nm to 700 nm. (a) Find the angular breadth of the first-order visible spec-trum produced by a plane grating having 15,000 lines per inch, when light is incident normally on the grating, (b) Show that the violet of the third-order spectrum overlaps the red of the second-order spectrum:

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Solution:

(a) The diffraction grating, shown in Fig. 1 has a grating spacing d given by d = [(2.54 cm)/(in.)] / [(15000 lines) (in.)] = 1.69 × 10^-4 cm. The nth order angular deviation for wavelength \lambda is sin \texttheta = n\lambda/d therefore the angular deviation of the violet is sin \texttheta_v= (\lambda_v)/(d)n = [(4 × 10^-5 cm)]/[(1.69 × 10^-4 cm)] = 0.237, \textthetav= 13\textdegree 40'. The angular deviation of the red is sin \textthetaR= (R)/(d)n = [(7× 10^-5 cm)] / [(1.69 × 10^-4 cm)] = 0.415, \texttheta_R= 24\textdegree30'. The first order deviations (n = 1) for the highest and lowest fre-quencies (violet and red respectively) of the visible spectrum will have an angular difference \texttheta_R - \texttheta_v. Hence, the first order visible spectrum includes an angle of \textthetaR- \textthetav= 24\textdegree30' - 13\textdegree40' = 10\textdegree50'. (b) For n = 3, violet light has the angular deviation sin \texttheta_v= (3\lambda_v)/(d) = [3 × (4 × 10^-5 cm) /[d] The second - order deviation for the red light is sin \texttheta_R = (2\lambda_R)/d = [2 × (7 × 10^-5 cm)] / [d] The first angle \texttheta_v is smaller than the second angle \texttheta_R, therefore whatever the grating spacing, the third order will always overlap the second, as can be seen in Fig. 2.

Question:

Write a program in PL / I to use the Gauss-Seidel method for solving linear equations. Specialize to 3 equations with 3 unknowns.

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Solution:

Consider the system a_11x_1 + a_12x_2 + a_13x_3 = b1 a_21x_1 + a_22x_2 + a_23x_3 = b_2 a_31x_1 + a_32x_2 + a_33x_3 = b_3 The solution algorithm is based on rewriting this system to a form a_11x_1 = b_1 - (a\textasteriskcentered_11x_1 + a_12x_2 + a_13x_3) a_22x-_2 = b_2 - (a_21x_1 + a\textasteriskcentered_22x_2 + a_23x_3) a_33x_3 = b_3 - (a_31x_1 + a_32x_2 + a\textasteriskcentered_33x_3) Let us introduce a P matrix associated with the A matrix, but modified so that the diagonal elements of the matrix A are replaced by 0. Using the new matrix P, one can rewrite the system as a_11x_1 = b_1 - (P_11x_1 + P_12x_2 + P_13x_3) a_22x-_2 = b_2 - (P_21x_1 + P_22x_2 + P_23x_3) a_33x_3 = b_3 - (P_31x_1 + P_32x_2 + P_33x_3) or in contact summation notation a_iix_i= bi- \sum_jP_ijx_jfori= 1, 2, 3. In PL/I notation this last equation becomes X (I) = [B (I) - SUM (P (I, \textasteriskcentered)\textasteriskcenteredX) / A (I, I). Note that we SUM over the subscript \textasteriskcentered corresponding to j. Also note that P_ij =a_ij,i\not = j;P_ii= 0. For this program we read in the dimension N of the square matrix A, and the value of K -the number of iterations. The program looks as follows: START: PROCEDURE; /\textasteriskcenteredGAUSS-SEIDEL\textasteriskcentered/ RPT:GET LIST (N, K) ; BEGIN; DECLARE A (N, N), B (N), X (N) ,P (N, N); GET LIST (A\textasteriskcenteredB); P = A; X = 0; DO I = 1 TO N; P (I, I) = 0; END; DO J = 1 TO K; DO I = 1 TO N; X (I) = [B (I) - SUM (P (I,\textasteriskcentered)\textasteriskcenteredX)) / A (I, I); END; PUT SKIP EDIT (X) (F (10,4)); END; END; PUT SKIP (2); GO TO RPT; END START;

Question:

Two masses of 200 gm and 300 gm are separated by a light rod 50 cm In length. The center of mass of the system serves as the origin of a Cartesian coordinate system. The rod lies in the xy plane and makes an angle of 20\textdegree with the y axis. Find the inertial coefficients I_xx and I_xy with respect to the center of mass.

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Solution:

Since we must calculate the moment of inertia of the rod about an axis through the center of mass, we must first locate the center of mass. Let us find the distance of the center of mass from the 200-gm mass. By definition of center of mass R_c.m. = (M_1x_1 + M_2x_2)/( M_1 + M_2) where x_1, x_2 are the distances of M_1 and M_2 from the origin (in our cases, the 200 gm. mass). Hence, using figure (A), R_c.m. = [(200 gm)(0) + (300 gm)(50cm)]/(500 gm) R_c.m. = 30 cm Looking at figure (B), the Cartesian coordinates of the 200 gm mass (denoted as M_1) are referred to the center of mass as origin, x_1 = (30 cm) sin 20\textdegree = (30 cm)(.342) \approx 10.3 cm y_1 = (30 cm) cos 20\textdegree = (30 cm)(.940) \approx 28.2 cm(1) z_1 = 0 The Cartesian coordinates of the 300 gm mass (denoted as M_2) are x_2 = (-20 cm) sin 20\textdegree \approx -6.8 cm y_1 = (-20 cm) cos 20\textdegree \approx -18.8 cm(2) z_2 = 0 Using these values of the coordinates, we proceed to evaluate the in-ertial coefficients defined by the equations I_xx = \sum_i m_i (y^2_i + z^2_i) I_xy = -\sum_i m_i x_i y_i For our problem, these reduce to I_xx = M_1 (y^2_1 + z^2_1) +M_2 (y^2_2 + z^2_2) I_xx = -M_1 x_1 y_1 - M_2 x_2 y_2 From (1) and (2) I_xx = (200 gm)[(28.2 cm)^2 + 0] + (300 gm) [(-18.8 cm)^2 + 0 ] I_xx = 265080 gm - cm^2 = 2.65 × 10^5gm \bullet cm^2 I_xy = -(200 gm)(10.3 cm)(28.2 cm) - (300 gm)(-6.8 cm)(-18.8 cm) I_xy = -96444 gm - cm^2 = -.96 × 10^5 gm - cm^2 Now suppose that the rod rotates about the x axis with angular velocity \omega. Find the components of J. In general, J_x I_xxI_xyI_xz \omega_x J_y I_yxI_yyI_yz \omega_y J_z I_zxI_zyI_zz \omega_z Where J_x, J_y, J_z, are the vector components of J^\ding{217}, and \omega_x, \omega_y, \omega_z are the componentsof \omega^\ding{217}. The quantities I_xx, I_xy, etc., are the products and moments of inertia of the system we are studying. In our problem, \omega_z = \omega_y = 0 and I_zx = 0. And J_x I_xxI_xyI_xz \omega_x J_y I_yxI_yyI_yz 0 J_z 0I_zyI_zz 0 Therefore, by the definition of matrix multiplication J_x I_xx \omega_x00 J_y I_yx \omega_x00 J_z 000 andJ_x= I_xx \omega_x J_y= I_yx \omega_x J_z= 0. Then(J_y)/(J_x) = (I_yx)/(I_xx) = (-.96 × 10^5)/(2.65 × 10^5) = -.363

Question:

A thin rigid rod of weight W is supported horizontally by two props as shown in Figure A. Find the force F on the remaining support immediately after one of the supports is kicked out.

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Solution:

The moment the support is kicked out, the rod starts rotating about the other support as the free end of the support falls (Figure B). Let x be the displacement of the center of mass of the rod. Immediately after the kicking of the support, x is very small and is vertical. In this case, (d^2x)/(dt^2) becomes the downward acceleration of the center of mass: m[(d^2x)/(dt^2)] = W - F(1) where m is the mass. The torque on the rod about the remaining support is \cyrchar\cyrt = W(L/2) = I[(d^2\texttheta)/(dt^2)] where I is the moment of inertia with respect to the axis of rotation. The moment of inertia of a thin rod with respect to an end is known to be (1/3)mL^2 , hence (1/2)WL = (1/3)mL^2 [(d^2\texttheta)/(dt^2)] or (d^2\texttheta)/(dt^2) = (3/2)(W/mL) For small x, x \approx (L/2)\texttheta or(d^2x)/(dt^2)= (L/2)(d^2\texttheta)/(dt^2). Form (1) and (2), m(L/2)[(d^2\texttheta)/(dt^2)]= m(L/2)[(3W)/(2mL)] = W - F (3/4)W= W - F givingF= W - (3/4)W = (1/4)W.

Question:

A capacitor of capacitance 0.1 \muF is charged until the difference in potential between its plates is 25 V. Then the charge is shared with a second capacitor which has air as the dielectric between its plates; the potential difference falls to 15 V. The experiment is repeated with a dielectric between the plates of the second capacitor, and the final potential difference is 8 V. What is the dielectric coefficient of the dielectric used?(See figures (a), (b) and (c)).

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Solution:

The situation is illustrated in figures (a) through (c). The charge possessed by the first capacitor initially, is, by the definition of capacitance (see figure (a)) Q = C_1 V_1 = 10^-7 F × 25 V = 2.5 \muC. This charge is shared with a second capacitor of capacitance C_2, no charge being lost. Hence Q = Q_1 + Q_2 where Q_1 is the charge on C_1 and Q_2 is the charge on C_2 (see figure (B)). Therefore Q = (C_1 + C_2 ) V_2or C_2 = (Q/V_2 ) - C_1 = [(2.5 x 10^-6 C)/(15 V)] - 10^-7 F = (2/3) × 10^-7 F. If a dielectric of coefficient \kappa is placed between the plates of capacitor C_2 and the experiment is re-peated, the capacitance of C_2 is increased to C_3 = \kappa C_2, where \kappa is the dielectric coefficient. In this case, Q is shared by C_3 and C_1. Hence Q = (C_1 + \kappa C_2) V_3 . \therefore\kappa = (1/C_2 ) [(Q/V_3 ) - C_1 ] = (3/2) × 10^7 F^-1 [{(2.5 × 10^-6 C)/(8 V)} - 10^-7 F] = 3/2 × 10^7 × 17/8 × 10^-7 = 3.2. In this analysis, we have used the fact that \kappa for air is equal to \kappa for vacuum, or \kappa_air = \kappa_vac = 1.

Question:

(a) How much heat energy Is produced by a 5-kg rock that falls a vertical distance of 10 m before it strikes the surface of the earth? Assume that the rock was initially at rest. (b) How much would the temperature of 1 kg of water be raised by the rock striking the earth's surface? (4.19 × 10^3 J of heat energy is required to raise the temperature of 1 kg of water 1\textdegreeK.)

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Solution:

(a) For this motion the kinetic energy is initially zero since the rock was released from rest. Just before colliding with the earth, the kinetic energy is a maximum. The conservation of energy requires that the increase in kinetic energy of the rock must equal the change in its potential energy, U - U_0 =mgh- mgh_0 = mg(h - h_0) Since h - h_0 = 0m - 10m, m = 5 kg, and g = 9.8 m/s^2 , the change in the potential energy is U - U_0 = (5 kg)(9.8m/s^2)(-10 m) = -4.9 × 10^2 J The minus sign means that the potential energy has decreased. The in-creased kinetic energy is then 4.9 × 10^2 J . This kinetic energy is all converted into heat upon collision with the earth. Consequently, the heat energy Q of the rock and earth is Q = 4.9 × 10^2J (b) Since 4.19 × 10^3 J of heat energy will raise the temperature of 1 kg of water 1\textdegreeK, 4.9 × 10^2 J of heat energy will cause a temperature rise of T = (4.9 ×10^2 J)/( 4.19 × 10^2 J/\textdegreeK) = 1.2 × 10^-1 \textdegreeK in 1 kg of water.

Question:

An object of mass 100 g is at rest. A net force of 2000 dynes is applied for 10 sec. What is the final velocity? How far will the object have moved in the 10-sec interval?

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Solution:

Since the force is constant, the acceleration is also. To find a, apply Newton's second law of motion. a = F/m a = F/m = (2000dynes) / (100g) = (2000dynes) / (100g) = 20 cm/sec^2 = 20 cm/sec^2 . The kinematics equations for constant acceleration can be used to find the final velocity. The initial velocity is zero in this problem, since the object is initially at rest. v= v_0 + at v= at = (20 cm/sec^2) × (10 sec) = 200 cm/sec. To find the distance traveled in 10 seconds, use s = v_0t + (1/2)(at^2), and since v_0 = 0 s= 1/2(at^2) = 1/2(20 cm/sec^2)×(10 sec)^2 = 1000 cm = 10 m.

Question:

Let the magnitudes and directions of the emf's, and the Let the magnitudes and directions of the emf's, and the magnitudes of the resistances in the figure be given. Solve for the currents in each branch of the network.

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Solution:

We assign a direction and a letter to each unknown current. The assumed directions are entirely arbitrary. Note that the currents in source 1 and re-sistor 1 are the same, and require only a single letter I_1. The same is true for source 2 and resistor 2; the current in both is represented by I_2. At any branch point we may apply Kirchoff's current law, which states that the current entering a branch point must equal the current leaving a branch point. There are only two branch points, a and b. At point b, \sumI = I_1 + I_2 + I_3 = 0(1) Since there are but two branch points, there is only one independent "point\textquotedblright equation. If the point rule is applied at the other branch point, point a, we get \suml = - I_1 - I_2 - I_3 = 0(2) which is the same equation with signs reversed. We now apply Kirchoff's voltage law to the two loops of the figure. This law states that the sum of the voltage drops around any closed loop of a circuit roust be zero. Hence, for loop defc, v_1 - I_3 r_3 + I_1 r_1 = 0(3) and for loop aefb v_2 - I_3 r_3 + I_2 r_2 = 0(4) We first solve for I_3. Solving (1) for I_1, I_1 = - I_2 - I_3(5) Substituting this in (3) v_1 - I_3 r 3 + (- I 2 - I_3 )r_1 = 0 orv_1 - I_3 r_3-I2r_1-I_3 r_1=0 - 2 r - = v_1 - I_3 (r_1 + r_3) - I_2 r_1 = 0(6) Solving this for I 2 I_2 r_1 = v_1 - I_3 (r_1 + r_3) I_2 = [{v_1 - I_3 (r_1 + r_3)} / (r_1 )](7) Substituting (7) in (4) v_2 - I_3 r_3 + [{v_1 - I_3 (r_1 + r_3)} / (r_1)] r_2 = 0 Solving for I_3 v_2 - I_3 r_3 + [(v_1 r_2)/(r_1)] - I_3 [(r_1 + r_3)/(r_1 )] r_2 = 0 orI_3 [r_3 + {(r_1 + r_3)/(r_1 )} r_2 ] = v_2 + [(v_1 r_2 )/(r_1 )] Hence I_3 = [(v_2 + {(v_1 r_2)/(r_1)}] / [r3 + {(r_1 + r_3)/(r_1)} r_2 ](8) Now, we solve for I_2 by substituting (8) in (4) v_2 - [([(v_2 + {(v_1 r_2)/(r_1)}] r_3 ) / (r_3 + [{(r_1 + r_3) r_2 } / (r_1 }])] + I_2 r_2 = 0 I_2 r_2 = [v_2 + {(v_1 r_2 )/(r_1 )} r_3] / (r_3 + [{(r_1 + r_3) r_2 } / (r_1 )]) - v_2 I_2 = [{[v_2 + {(v_1 r_2)/(r_1)}] (r_3 /r_2)} / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] - (v_2/r_2 )(9) Substituting (9) and (8) in (5), we may solve for I_1 I_1 = (v_2 /r_2 ) - [{[v_2 + {(v_1 r_2)/(r_1)}] (r_3 /r_2)} / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] - [[v_2 + {(v_1 r_2)/(r_1)}] / {r_3 + [{(r_1 + r_3) r_2} / (r_1)]}] Hence I_1 = (v_2 /r_2 ) -[[v_2 + {(v_1 r_2)/(r_1)}] / {r_3 + [{(r_1 + r_3) r_2}/(r_1)]}] [(r_3/r_2 ) + 1]

Question:

A cubic foot of sea water at the surface weighs 64.0 lb. What volume of water weighs 100 lb at the sea bed, where the water pressure is 4000 lb\bulletft^-2? The compressi-bility of sea water is 36 × 10^-7 in^2\bulletlb^-1.

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Solution:

Using the definition of weight density D D = (Weight)/(Volume) thenV = W/D = (100 lb)/(64 lb/ft^3) = 1.5625 ft^3 where V is the volume 100 lb of water occupies at the surface. Where the water pressure P is 4000 lb \textbulletft^-2 = 4000 lb \textbulletft^-2 × 1 ft^2/144 in^2 = 4000/144 lb \textbulletft^-2, the volume will have decreased by ∆V due to this additional pressure. The compressibility k of sea water is defined by 1/K = (Stress)/(Strain) = P/(∆V/V) We are given that k = 36 × 10^-7 in^2\bulletlb^-1 \therefore ∆V/V = kp = (4000/144) lb \textbulletin^-2 × 36 × 10^-7 in^2\bullet lb^-1 = 10^-4. \therefore 1 - (∆V/V) = (V -∆V)/V = 1 - 10^-4or V - ∆V = (100/64) (1 - 10^-4) ft^3. = 1.5623 ft^3 Thus the volume occupied at the lower level is 1.5623 ft^3.

Question:

When a flat plate of glass and a lens are placed in contact, a dis-tinctive interference pattern, known as Newton's Rings, is observed. (See figure A). Derive a formula giving the location of the fringes of the interference pattern relative to the center of the lens. (See figure B).

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Solution:

Destructive interference will result when the waves reflected from the apparatus shown in figure (B) are 180\textdegree out of phase. Let us trace the path of an incident ray. The ray will be partially reflected and partially transmitted at surface (1). When a ray of light is trams- mitted from a region of low refractive index to a medium of higher re-fractive index, it undergoes a phase change of 180\textdegree . Hence, the ray reflected at surface (1) is 180\textdegree out of phase with the incident ray. The transmitted ray next encounters surface (2). At this surface there is no phase change, since the light leaves an area of high refractive index and enters a region of low refractive index. In addition, part of this light is reflected at surface (2). The light transmitted at surface (2) next encounters surface (3) and is reflected with a 180\textdegree or \lambda/2 phase change. (\lambda is the wavelength the light). Hence, the ray reflected at (2), and the ray reflected at (3) are 180\textdegree out of phase. Now, the ray reflected at (3) travels a distance 2t greater than the ray reflected at (2). We will see destructive interference whenever 2t is an integral number of wavelengths, since the additional \lambda/2 re-quired for destruction is provided by the phase change due to reflection. Hence 2t = n\lambdan = 0, 1, 2,...destructive interference(1) 2t = [n + (1/2)]\lambdan = 0, 1, 2,..constructive Interference We must now find the location of the interference fringes in terms of the geometry of figure (B). from figure (B), t = R - \surd(R^2 - d^2)(2) t = R - R\surd(1 - d^2/R^2) t = R {1 - \surd(1 -d^2/R^2)}(3) But d << R and d/R << 1. (This means that the radius of curvature of the lens is large). We may therefore approximate the square root in (3) by the binomial theorem. Therefore, \surd(1 - d^2/R^2) \approx 1 - d^2/R^2(4) Substituting (4) in (3) t = R(1 - 1 + d^2/2R^2) t = d^2/2R d = \surd(2tR) Using (1)d = \surd(n\lambdaR)n = 0, 1, 2,...destructive interference d = \surd[{n+(1/2)}\lambdaRn = 0, 1, 2,...constructive interference The first equation locates the dark rings relative to the center of the lens, and the second equation locates the bright rings.

Question:

A shunt generator has a terminal potential difference of 120 volts when it delivers 1.80 kw to the line. Resistance of the field coils is 240 ohms, and that of the armature is 0.400 ohm. Find the EMF of the generator and the efficiency.

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Solution:

An electric circuit containing a shunt generator (that is, a generator whose field resistance R_F is in parallel with the generator) can be described schematically as shown above. The potential difference at the terminals where a load (or resistance) is connected is the terminal voltage, V. The load, when connected will draw a line current I_L, from the circuit. Therefore, the power deliver-ed to the load is P = I_L V. Solving for I_L = (1800 watts/120 volts) = 15.0 amp Using Ohm's Law,V = I_f R_f,(1) orI_F = (V/R_F) = (120 volts/240 ohms) = 0.500 amp By Kirchoff's current law, the sum of the currents entering a junction equals the sum of the current leaving the junction. Therefore, at point Q I_A = I_F + I_L = 0.50 amp + 15.0 amp = 15.5 amp. By Kirchoff's voltage law, the sum of the voltage drops about a loop is zero. Applying this law to loop (1), we obtain \epsilon = I_fR_f + I_AR_A Using (1) \epsilon = V + I_AR_A \epsilon = 120 volts + (15.5 amp)(.4 ohms) \epsilon = 126 volts The input power P_i = I_A \epsilon = 15.5 amp × 126 volts = 1950 watts and it is delivered by the generator. The output power is the power delivered to the line, and it is given as 1800 watts. Hence, Efficiency = (P_o/P_i) = (1800 watts/1950 watts) = 0.923 = 92.3%.

Question:

For the following voltaic cell, write the half reactions, For the following voltaic cell, write the half reactions, designating, which is oxidation and which reduction. Write the cell reaction and calculate the voltage (E\textdegree) of the cell from the given electrodes. The cell is Cu; Cu^+2││ Ag^+2 ; Ag.

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Solution:

In a voltaic cell, the flow of electrons creates a current. Their flow is regulated by 2 types of reactions occur concurrently, oxidation and reduction. Oxidation is a process where electrons are lost and reduction where electrons are gained. The equation for these are the half-reactions. From the cell diagram, the direction of the reaction is always left to right. Cu \rightarrow Cu^+2 + 2e^-oxidation Ag^+2 + 2e^- \rightarrow Agreduction . Therefore, the combined cell reaction is Cu + Ag^+2 \rightarrow Cu^+2 + Ag. To calculate the total E\textdegree, look up the value for the E\textdegree of both half--reactions as reductions. To obtain E\textdegree for oxidation, reverse of the sign of the reduction E_O . Then, substitute into E\textdegree_cell = E\textdegree_red + E\textdegree_ox.If you do this, you find E\textdegree_cell= -(E\textdegree_redCu) + E\textdegree_red Ag^+2 = -.34 + .80 = .46 volt .

Question:

The atomic weight of iron is 55.847amu. If one has 6.02 g of iron, how many atoms are present?

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Solution:

A mole is defined as the weight in grams of a substance divided by its atomic weight: mole = amount in grams / atomic weight If one calculates the number of moles of iron, then the number of atoms present can be calculated. There are 6.02 × 10^23 atoms per mole. no. of moles of Fe = 6.02 / (55.847 g / mole) = 1.08 × 10-1moles no. of Fe atoms present= (1.08 × 10^-1 moles) × (6.02 × 10^23 atoms / mole) = 6.49 × 10^22 atoms.

Question:

A string of freight cars weighs 200 tons. The coefficient of rolling friction is 0.005. How fast can a freight engine of 400 hp pull the string of cars along?

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Solution:

Power is defined as Work divided by time. P = W/t and W = Fs (where F and s are in the same direction and are force and distance respectively). Then, P = Fs/tand since s/t = v P = Fv The force of friction can be calculated from F =\muN, where \mu is the coefficient of rolling friction and N is the normal force: F = 0.005 × (200 ton × 2000 lb/ton) = 2000 lb Using P = Fv from above and the conversion factor [(550 ft-lb/sec)/(1 hp)] to convert 400 hp to power in units ft-lb/sec: 400 hp (550 ft-lb/sec/hp) = 2000 lb × v v = 110 ft/sec = 75 mi/hr.

Question:

What is the magnetic field intensity at a point in air 30 cm away from and on the extension of the line of a bar magnet 10 cm long whose magnetic moments is 400 poles cm, if the point in question is nearer the north pole of the magnet?

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Solution:

Draw a diagram. Recall that field Intensity H is a vector quantity, which with respect to a pole strength m is given by the expression H = (m/\mur^2 ) which follows directly from Coulomb's Law F = (mm')/(\mur^2) where, by definitionH = f/m' where m' is a test pole with a north polarity. Hence H = (m/\mur^2 ) . In this problem, the pole strength can be evaluated because the nagnetic moment is given by the defining equation M = mL, where M is 400 poles cm, and L is 10 cm. m = (M/L) = (400/10) = 40 poles . It follows that due to the north pole of the magnet, the field intensity H_1 at P is H_1 = (m/\mur^2) = [{40 poles}/{(1) (30cm)^2}] = (40/900) (poles/cm^2 ) = .044 oersteds to the right. Also, the field intensity H_2 due to the south pole is H_2 = [40 poles/{1 (40cm)^2}] = (1/40)(poles/cm^2 ) ThereforeH is the vector sum of H_1 and H_2 or H = .044^\ding{217} - .025^\ding{217} = .019 oersteds to the right. This problem is completely analogous to the problem of finding the electric field intensity at point p, due to two charges, one positive and one negative, at a distance 30 cm and 40 cm from the point P. Coulomb's law for the electrostatic force between charges would be the electro-static analogue for Coulomb's law for the magnetic force between poles.

Question:

Render the following "nonsense"pseudocodeinto FORTRAN, including appropriate comments.Name the law ofpropositional logic you need to render the DO-WHILE and IF-THEN-ELSE constructs: inputL, J, KOUNT, N I \leftarrow 1 dowhile L < J\textasteriskcentered\textasteriskcentered3 and I < KOUNT KOUNT \leftarrow KOUNT + 2 ifJ < 0 or L < 0 thenJ\leftarrow J + 1 elseL \leftarrow L + 1 endif I \leftarrow I + 1 enddo-while outputKOUNT, J, L endprogram

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Solution:

The name of the law requiredIsoften referred to asDeMorgan's Law, which is stated thus: not(P and Q) \equiv not P or not Q Actually,DeMorgan'slawIncludesthe Boolean equation P or Q \equiv not [not P and not Q] Here, the two constructs use the former equivalence. INTEGER L, J, KOUNT,N READ (5,100) L,J,KOUNT,N 100FORMAT (4 I 5) I =1 . CDO WHILE L LESS THAN J\textasteriskcentered\textasteriskcentered3 AND CI IS GREATER THAN KOUNT 10IF (L. GE.J\textasteriskcentered\textasteriskcentered3.OR.I. GE. KOUNT) GO TO 20 KOUNT = KOUNT + 2 CIF J LESS THAN 0 OR L GREATER CTHAN ZERO THEN DO IF (.NOT. (J.GE.0.AND.L.LE.0)) GO TO 30 CELSE DO L = L + 1 GO TO 40 30J = J + 1 40CONTINUE CENDIF I = I + 1 GO TO 10 CEND DO-WHILE 20CONTINUE STOP END

Question:

How does the Watson-Crick model of DNA account for its observed properties?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0612.htm

Solution:

By 1950, several properties of DNA were well established. Chargaff showed that the four nitrogenous bases of DNA did not occur in equal proportions, however, the total amount ofpurinesequaled the total amount ofpyrimi-dines(A+G = T + C). In addition, the amount of adenine equaled the amount of thymine (A = T), and likewise for guanine and cytosine (G = C). Pauling had suggested that, the structure of DNA might be some sort of an \alpha-helix held together by hydrogen bonds. The final observation was made by Franklin and Wilkins. They inferred from x-ray diffraction studies that the nucleotide bases (which are planar molecules) were stacked one on top of the other like a pile of saucers. On the basis of these observed properties of DNA, Watson and Crick in 1953 proposed a model of the DNA molecule. The Watson and Crick model consisted of a double helix of nucleotides in which the two nucleotide helices were wound around each other. Each full turn of the helix measured 34 \AA and contained 10nucteotidesequally spaced from each other. The radius of the helix was 10 \AA. These measurements were in agreement with those obtained from x-ray diffraction patterns of DNA. To account for Pauling's observation, Watson and Crick proposed that the sugar-phosphate chains of DNA should be on the outside, and the purinesandpyrimidineson the inside, held together by hydrogen bonds between bases on opposite chains. When they tried to put the two chains of the helix together, they found the chains fitted best when they ran in opposite directions.Moveover, because x-raydiffragtionstudies specified the diameter of the helix to be 20 \AA, the space could only accommodate onepurineand onepyrimidine. If twopurinebases paired, they would be too large to fit into the helix without destroying the regular shape of the double stranded structure. Similarly, twopyrimidineswould be too small. Moreover, for onepurineto be hydrogen-bonded with onepyrimidine properly, adenine must pair with thymine and guanine with cytosine. An A -T base pair is almost exactly the same in width as a C - G base pair , accounting for the regularity of the helix as seen in x-ray diffraction pictures. This concept of specific base pairing explained Chargaff's observation that the amounts of adenine and thymine in any DNA molecule are always equal, and the amounts of guanine and cytosine are always equal. Two hydrogen bonds can form between adenine and thymine, and three hydrogen bonds between guanine and cytosine (Fig. 2). The specificity of the kind of hydrogen bond that can be formed provides for correct base pairing during replication and transcription. The two chains are thus complementary to each other, that is, the sequence of nucleotides in one chain dictates the sequence of nucleotides in the other. The strands are alsoantiparallel, i.e., they extend in opposite directions and have their terminal phosphate groups at opposite ends of the double helix.

Question:

An air-core toroid of cross-sectional area A and of mean cir-cumferential length l is closely wound with N turns of wire. If N = 100, A = 10 cm^2 , l = 0.50 m, find its self inductance, L.

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Solution:

The magnetic flux density B throughtout the coil will not deviate appreciably from its value at the mean radius r_0 of the torus, if its width is much smaller than l = 2\pir_0 . Applying Ampere's Law around the circle with radius r_0, we get \oint B.dl = \mu_0 NI orBl = \mu_0 NI B = [(\mu_0 NI)/l] and the flux is \varphi_B = BA = [(\mu_0 NAI)/l] . If a current I sets up a magnetic flux, \varphi_B a coil, then \varphi_B and I are related through the self inductance of the coil as follows, L = [(N\varphi_B )/I] where N is the number of turns. Hence L = [(\mu_0 N^2 A)/l] orL = [{4\pi × 10^-7 (w/a-m)} × (100)^2 × {1m^2 /(10^4 m^2 ) × 10 cm^2 }] \textbackslash(0.50 m) \approx 25 × 10^-6 henry .

Question:

Two trains moving along parallel tracks in opposite directions converge on a stationary observer, each sounding its whistle of frequency 350 cycles. s^-1. One train is traveling at 50 mph. What must be the speed of the other if the observer hears 5 beats per second. The speed of sound in air is 750 mph.

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Solution:

When a source is moving toward a stationary observer, the latter hears a frequency for the emitted note which is related to the frequency of the source by the expression f =uf_s/(u -v_s) where u is the speed of sound inair,v_sis the speed of the source andf_sis the frequency of the sound emitted by the source. If one of the trains is moving toward the observer with v_1- and the other with speed v_2, then since both have the same frequency whistle, f_1 = (uf_s)/( u - v_1)and f_2 = (uf_s)/( u - v_2) = (750 mph × 350 s^-1) / [(750 - 50) mph] = 375 cycles \bullet s^-1 But the observer hears 5 beats per second. This corresponds to a frequency difference f_1 = f_2 = \pm 5 cycles \textbullet sec^-1. Hence, f_1 = 370 cycles \textbullet s^-1 or 380 cycles \textbullet s^-1 \therefore f_1 = (uf_s)/( u - v_1) = 370 s^-1or380 s^-1 \therefore u - v_1 = (750 mph × 350 s^-1) / (370 s^-1)or (750 mph × 350 s^-1) / (380 s^-1) = 709.5 mphor690.8 mph \thereforev_1 = 40.5 mphor59.2 mph.

Question:

Complete the equations in Figure A. Name the reactants and the products.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0777.htm

Solution:

Compound I is an acid chloride and compound II is an anhydride. Both of these substances are derivatives of carboxylic acids. Such species undergo nucleophilic substitution reactions. In these reactions, a stronger base displaces a weaker base. Thus, to complete these reactions, investigate whether H_2O is a stronger base than the Cl- ion or the carboxylate anion of the acid halide and the anhydride, respectively. A stronger base will replace a weaker base in a compound. Cl- ion and carboxylate anion are weak bases, so that the following occurs The stronger base OH- ion, replaced the weak CI- ion base. Reacton 2 is shown in Figure B. Again, OH- replaced the weaker carboxylate anion base.

Question:

How much water must be added to 5 gallons of 90% ammonia solution to reduce the solution to a 60% solution?

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Solution:

The volume of the original solution is 5 gallons. If we let x represent the number of gallons of water added to the original solution, then the final solution will have a volume of (5 + x) gallons. No ammonia is added so the only ammonia in the final solution is the ammonia in the original solution. We can now record our given information in tabular form: Volume Percent (Gallons) Percent Concentration Gallons of 100%Ammonia Original solution 5 90 4.5 Final solution 5 + x 60 4.5 Note: The gallons of 100% ammonia for the original solution was obtained in the following way: multiply the total volume of the original solution by the percent concentration of the original solution. Therefore, 5 × 90% =5× (90/100_20) = (90/20) = (9/2) = 4.5 gallons of 100% ammonia. Also, the gallons of 100% ammonia for the final solution is the same as the gallons of 100% ammonia for the original solution; that is, 4.5 gallons, since no ammonia was added. The formula for the amount of substance in solution for the final solution is (total volume) × (percent concentration) = gallons of 100% ammonia. Therefore: (5 + x) × 60%= 4.5 or(5 + x) (60/100)= 4.5 or(5 + x).60= 4.5 or.60 (5 + x)= 4.5 3 + .60x= 4.5 .6x= 4.5 - 3 .6x= 1.5 x= (1.5) / .6 = 2.5 gallons. Therefore, 2.5 gallons of water must be added to reduce the concentration to 60%. Check: If 2.5 gallons of water are added to the original solution, we will have 7.5 gallons of solution, The amount of 100% ammonia remains the same 4.5 gallons. The concentration equals (4.5 / 7.5) × 100 = (450 / 7.5) = 60%.

Question:

To a good approximation, the force required to stretch a spring is proportional to the distance the spring is extended. That is, F = kx where k is the so-called spring constant or force constant and depends, of course, on the dimensions and material of the spring. Many elastic materials, if not stretched too far, obey this simple relationship - called Hooke's law after Robert Hooke (1635-1703), a contemporary of Newton. Suppose that it requires 100 dynes to extend a certain spring 5 cm. What force is required to stretch the spring from its natural length to a length 20 cm greater? How much work is done in stretching the spring to 20 cm?

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Solution:

The force constant is k = F_1/x_1 = (100 dynes)/(5 cm) = 20 dynes/cm Therefore, the force required to extend the spring to 20 cm is F_2 = kx_2 = (20 dynes/cm) × (20 cm) = 400 dynes The average force required to stretch the spring to 20 cm is F= (F_2 + F_0)/2 Where: F_0 = initial force exerted on spring to keep its equilibrium length = kx_0 = k \bullet 0 = 0 F_2 = force required to stretch the spring to 20 cm = kx_2 ThenF= (F_2 + 0)/2 = kx_2/2 and the work expended is the average force multiplied by the distance: W =Fx_2 = (1/2) kx^2_2 = (1/2) × (20 dynes/cm) × (20 cm)^2 = 4000 dynes - cm = 4000 ergs since 1 erg = 1 dyne - cm.

Question:

What are the fourextraembryonicmembranes foundin animals which lay their eggs on land, and what are their functions ?

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Solution:

The development ofextraembryonicmembranes freed early reptiles from an obligatory embryonic de-velopment in water and set the scene for the flourishing of the land-dwelling vertebrates. The mammals have capitalized on their inheritance ofextraembryonicmembranes from the reptiles by utilizing them in intra-uterine development. Thechorion, orserosa, is an outer covering which, in reptiles, rests in contact with the inner surface of the shell, and which in mammals is established next to the cells of the uterine wall. This layer is the site where exchange of substances, largely gases, foods and wastes takes place between embryonic tissue and the maternal environment. To facilitate the exchange , various adaptations have been acquired. In the pig, the chorion elongates tremendously. Its great surface area affords abundant contact with the uterine wall. In most mammals, thechoriondevelops complex outgrowths known as chorionicvilli. These penetrate into the tissue of the uterine wall and greatly increase the area and intimacy of contact between fetal and maternal tissues. Usually there is a specialized region of thechorionwhere exchange between the mother and embryo is most efficient. Such an area is called the placenta. The humanchorion also secretes human chorionicgonadotropin(HCG), a hormone which functions to maintain the corpusluteum. The amnion is a thin fluid-filled, membranous sac which surrounds the embryo and provides it with protection from pressure, abrasion, irritation , and loss of water. The fluid within the amnion is the amniotic fluid . During labor of humans and other mammals, the pressure of the contracting uterus is transmitted via the amniotic fluid and helps to dilate the neck of the uterus. Later, shortly before the fetus is born, the amnion normally ruptures, re-leasing about a liter of amniotic fluid. This is referred to as the breaking of the water bag. Sometimes it fails to burst and the child is born with the amnion still enveloping its head. The amnion is essentially anoutfoldingof the body wall of the embryo consisting of ectodermal andmesod-ermaltissues. The inner part of this fold forms the amnion and the outer part forms thechorion. A thirdextraembryoniclayer found in reptiles, birds, and mammals is the yolk sac, which is an out-growth of the digestive tract of the embryo. This sac is richly endowed with blood vessels and functions to store and supply food. The human egg cell contains minimal yolk, but the sac has other functions. In addition, the yolk sac is the first source of blood cells and blood vessels of the embryo, and also performs biochemical functions which are later taken over by the liver. Theallantois, like the yolk sac, is an outgrowth of the digestive tract . It grows between the amnion andchorionand it fills almost all the space between the two. In reptiles and birds, theallantoisserves as a depot for nitrogenous wastes. The products of nitrogen metabolism are excreted as uric acid by the kidney of the developing embryo. The poorly soluble uric acid is deposited as crystals in the cavity of theallantoisand is discarded along with theallantoiswhen the young hatch out of the egg shells . In birds and reptiles, theallantoisfuses with thechorionto form a compound chorioallantoicmembrane, across which gas exchange takes place . Recall that in mammals both food and gases are exchanged across the chorion. In reptiles and birds, only gas is exchanged across this membrane which is here associated with theallantois. The egg shell is porous and oxygen coming in through it is picked up by blood vessels in the chorioallantois. In the human, wastes are transported through the placenta into the mother. Theallantoisis thus less important as an exchange site. But theallantoicstalk becomes the umbilical cord, linking the chorionto the embryo.

Question:

One electron is removed from O_2 and one from N_2. The bonding in O_2 is strengthened, while the bonding in N_2 is weakened. 1) Explain these findings, and 2) predict what happens if an electron is removed from NO.

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Solution:

There are two types of molecular orbitals: bonding and antibonding. A chemical bond is strengthened by electrons in bonding orbitals and weakened by electrons in antibonding orbitals. For every bonding orbital, there is a corresponding antibonding orbital. Each orbital can hold 2 electrons. Bond order measures bond strength by giving an indication of the number of electrons in bonding versus antibonding orbitals. Bond order is defined as one half the number of electrons in the bonding orbital less one half the number of electrons in the antibonding orbital. Thus, the higher the bond order, the stronger the bond. To find the original bond order, consider the valence electrons, the outermost electrons, since they are the only ones that participate in bonding. For N, Z = 7. Its electron configuration is 1s^2 2s^2 2p^3. The outer-most electrons are in 2p^3. This means that in N^2 there are a total of six valence electrons. There exist 3 bonding p orbitals. They can accommodate the six valence electrons. Since no electrons need be in antibonding orbitals, bond order = (1/2) (6) - (1/2) (0) = 3. When an electron is removed, the bonding orbitals have only 5 electrons. Thus, the bond order becomes (1/2) (5) - (1/2) (0) = 2.5. Since the bond order went from 3 to 2.5, the bond is weakened by removing the electron from N_2. For O, Z = 8. Since its electron configuration is 1s^2 2s^2 2p^4, each O atom has 4 valence electrons. O_2 has a total of 8 valence electrons. The three bonding p orbitals can hold only 6 .of these electrons. Thus, two electrons are in antibonding orbitals. The bond order is (1/2) (6) - (1/2) (2) = 2. When one removes an electron, it is removed from the antibonding orbitals, if electrons exist in such orbitals. Therefore, the O_2, after removal of the electron, has only 1 electron in an antibonding electron. This means the bond order becomes (1/2) (6) - (1/2) (1) = 2.5. The bond increased from 2 to 2.5, which means that bond strength increases when one removes an electron from O_2. To predict what happens to NO bond strength, consider the bond order before and after the electron removal. Recall, an O atom has 4 valence electrons and N has 3 valence electrons. The total in NO is seven. The 3 p bonding orbit-als can hold six of these electrons. This means 1 electron is in on antibonding orbital. Bond order = (1/2) (6) - (1/2) (1) = 2.5. If one removes one electron, the antibonding orbitals contain zero electrons. Thus, the bond order becomes (1/2) (6) - (1/2) (0) = 3. The bond increased from 2.5 to 3, which means the bond strength increases. Therefore, one can pre-dict the chemical bond strength of NO increases when an electron is removed.

Question:

You are handed a stack of IBM cards, each of which is supposed to contain a number between 0 and 50. Write a FORTRAN program to sum the results of the following operations on these numbers: a) Multiply numbers between 0 and 10 inclusive by 3 b) Divide numbers between 11 and 30 inclusive by 2 c) Subtract 5 from all numbers from 31 to 50 inclusive. Also have the program sum the number of errors in the input deck, i.e., how many cards contain numbers outside the specified range. The program should terminate when the value 99.9 is encountered in the deck.

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Solution:

A flowchart outlines the structure of the algorithm to be used: The idea is to check each number to find out the range in which it lies. The specified operation is performed upon every number; each answer (stored in the variable TEMP) is added to the final answer - the sum of all the adjusted numbers. Control passes to the output stage when the number 99.9 is reached. DATA ERRSUM, SUM, TEMP/0.0, 0.0, 0.0/ 10READ (2,100) X 100FORMAT (F3.1) IF (X.EQ. 99.9) GO TO 98 IF (X. LT.O. .OR.X.GT.50.) GO TO 20 20ERRSUM = ERRSUM + 1 GO TO 10 IF (X.GE.0. AND.X.LE.10.) GO TO 30 30TEMP = X{_\ast}3.0 SUM = SUM + TEMP GO TO 10 IF (X.GE.11.AND.X.LE.30.) GO TO 40 40TEMP = X/2.0 SUM = SUM + TEMP GO TO 10 TEMP = X - 5.0 SUM = SUM + TEMP GO TO 10 98WRITE (6,101) SUM, ERRSUM 101FORMAT (1X ,' SUM =', F8.4,' ERRORS IN DECK = ',F4.0) STOP END

Question:

A lead ball, initially at rest, is dropped from a certain height. Assume it takes the ball 5 seconds to reach the ground. Develop a program segment to give position and velocity at one-second intervals.

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Solution:

The initial velocity v_0 zero. The formula for the change of position is y = (1/2)gt^2 , where g = -9.8 meters/sec^2. We include the negative sign since the ball travels in the negative y-direction. Now, with a constant acceleration, the instantaneous velocity is V_n =V_o + at , where n = 1,2,3,4,5. By iteration, we can simulate the ball 's^-motion. WRITE (5,100) 100FORMAT (1X,'POSITION', 5X, 'VELOCITY') DO 20 NT = 1,5 Y = 0.5 {_\ast} -9.8 {_\ast}(NT{_\ast}{_\ast}2) V = -9.8 {_\ast} (NT) WRITE (5,101) Y,V 101FORMAT (1X, F10.4, 5X, F10.4) 20CONTINUE

Question:

A coil of resistance 10 ohm and inductance 0.1 H is in series with a capacitor and a 100\rule{1em}{1pt}V 60\rule{1em}{1pt}cycle \textbullet s^\rule{1em}{1pt}1 source. The capacitor is adjusted to give resonance in the circuit. Calculate the capacitance of the capaci-tor and the voltages across coil and capacitor.

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0777.htm

Solution:

The (complex) impedance Z (seen looking into the terminals a\rule{1em}{1pt}b) of the series R\rule{1em}{1pt}L\rule{1em}{1pt}C circuit (see figure) is Z^\ding{217} = (V^\ding{217}/I^\ding{217}) = R + j (X_L - X_C) where the reactances X_L = 2\pifL = \omegaL and X_C = 1/2\pifC = 1/\omegaC. f is the frequency of variation of voltage. The negative sign is introduced above because the inductive and capacitive reactances tend to cancel one another's effect. In the series combination, the voltages tend to cancel because they are in phase opposition. The (complex) voltage V^\ding{217} = \vertv^\ding{217}\vert e^j\texttheta1 where \vertv^\ding{217}\vert is the magnitude of the voltage and \texttheta_1 is the phase angle of the voltage relative to the current I^\ding{217} (= \vert I^\ding{217} \vert e^jo = \vert I^\ding{217} \vert). The magnitude and phase of the impedance are respectively \vert Z^\ding{217} \vert = \surd{R^2 + (X_L - X_C)^2}\texttheta = arctan {(X_L - X_C)/R} ThenZ^\ding{217} = \vert Z^\ding{217} \vert e^j\texttheta At resonance, X_L = X_C . (1/\omegaC) = \omegaL \therefore C = [1/(\omega^2 L)] = {1/(4\pi^2 × 60^2 s^\rule{1em}{1pt}2 × 0.1 H)} = 70.4 \muF. Further, the circuit becomes purely resistive for \vert Z^\ding{217} \vert = \surd(R^2 + O^2) or \vert Z^\ding{217} \vert = R, and \texttheta = arctan 0 = 0\textdegree. Thus \vert I^\ding{217} \vert = (\vert V^\ding{217} \vert/R) = (100 V/10\Omega) = 10 A. The voltage difference across the inductor is V^\ding{217}_L = I^\ding{217}Z^\ding{217}_L where Z_L is the impedance looking into the terminals a \rule{1em}{1pt} c (see the figure). However, the only impedance present is the inductive reactance X_L . Then \vert Z^\ding{217}_L \vert = X_L = 2\pifL \vert V^\ding{217}_L \vert = \vert I^\ding{217} \vert X_L = 10A × 2\pi × 60 s^\rule{1em}{1pt}1 × 0.1 Henry = 377 V. The voltage difference across the capacitor is, similarly, V^\ding{217}_C = I^\ding{217} Z^\ding{217}_C where Z_L is the impedance looking into the terminals c \rule{1em}{1pt} d (see the figure). But \vert Z^\ding{217}_C \vert = X_C = 1/(2\pifC), hence \vert V^\ding{217}_C \vert = \vert I^\ding{217} \vert X_C = 10 A × {1/(2\pi × 60s^\rule{1em}{1pt}1 × 70.4 × 10^\rule{1em}{1pt}6 F)} = 377 V. Thus, \vertV^\ding{217}_C \vert = \vert V^\ding{217}_L \vert. However V^\ding{217}_C and V^\ding{217}_L are in phase opposition or \vert \texttheta_C - \texttheta_L\vert = \pi, hence, their effects cancel. Therefore, at any time, V_ad = V_ac + V_cd = 0 and the voltage of the source appears across the resistor only. Notice that \vert V^\ding{217}_C \vert and \vert V^\ding{217}_L \vert , individually, may have values greater than source voltage.

Question:

The equilibrium constant and standard Gibbs free energy change for ammonia synthesis at 400\textdegreeC or 673\textdegreeK are 1.64 × 10^-4 and 11, 657 cal/mole, respectively. The equation for this reaction is (1)N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g) Calculate the equilibrium constants and standard free energy changes for (2)(1/2)N_2 (g) + 3/2 H_2 (g) = NH_3 (g)and (3)2NH_3 (g) \rightarrow N_2 (g) + 3H_2 (g).

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Solution:

This problem illustrates the necessity of knowing the balanced chemical equation in order to determine the numerical value of the equilibrium constant. For ideal gases, the equilibrium constant,K_p, is the product of the partial pressure of the product gases, each raised to the power that corresponds with itsstoichiometriccoefficient in the balanced equation, divided by the product of the partial pressure of the reacting gases, each raised to the power that corresponds with itsstoichiometricco-efficient in the balanced equation. The quantity ∆G\textdegree is the standard change in Gibbs free energy. It is the difference between the Gibbs free energy of the products and the Gibbs free energy of the reactants. The Gibbs Free Energy and the equilibrium constant are related in the equation ∆G = - RT InK_p where R is the gas constant, T the absolute temperature andK_pthe equilibrium constant calculated using partial pressures. The equilibrium constant for reaction 1 is The equilibrium constant for reaction 1 is K_1 = [(P^2_NH3)/(P_N2 P^3_H2 )] = 1.64 × 10 K_1 = [(P^2_NH3)/(P_N2 P^3_H2 )] = 1.64 × 10 -4 Thus, for (2) (1/2)N_2 (g) + (3/2)H_2 (g) = NH_3 (g) Thus, for (2) (1/2)N_2 (g) + (3/2)H_2 (g) = NH_3 (g) K_2 = [(P_NH3)/(P^1/2_N2 P^3/2_H2 )] = K^1/2_1 = (1.64 × 10^-4 )^1/2 = 1.28 × 10^-2 By definition, at equilibrium, ∆G\textdegree = - RT In K_2 = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) × log(1.28 × 10^-2) = 5829 cal/mole. And for (3) 2NH_3 (g) = N_2 (g) + 3H_2 (g) K_3 = [(P_N2 P^3_H2 )/(P^2_NH3)] = (1/K_1) = [(1)/(1.64 × 10^-4 )] = 0.6098 × 10^4 ∆G\textdegree_3 = - RT In K_3 = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) × log (0.6098 × 10^4) = - 11, 657 cal/mole. If a reaction is reversed, the equilibrium constant becomes the reciprocal of that for the first reaction; the standard Gibbs free energy change has the same magnitude but the opposite sign.

Question:

Explain the features of the following program. EXAMPLE: PROCOPTIONS(MAIN); DCL BIT_STRINGBIT(4) VARYING; GETLIST(BIT_STRING); IF BIT_STRING \lnot=0 THEN PUT LIST (BIT_STRING, 'NOT ALL BITS ARE ZERO'); ELSE PUTLIST(BIT_STRING,'ALL BITS ARE ZERO'); END EXAMPLE,

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Solution:

The program illustrates the features of bit strings inPL/I. In the declare statement BIT_STRING is the name of a bit string variable. A bit string variable can take bit string data as values, just as arithmeticvariables take on numeri-cal values. A bit string constant consists of a sequence of one or more of the digits0 and 1. The constant is enclosed within quotes. The terminating quotessymbol is immediately followed by the letter B indicating a bit string. e.g.'111'B means that 111 is a bit string. (3)'101'B means that 101101101 is a bit string. ''Bmeans a null length bit string. Any bit string variable must always be explicitly de-clared as in the givenexample. The bit attribute consists of the keyword BIT followed by aninteger enclosed in paren-theses indicating the length of the bit string. Hence (4) in the DCL statement specifies the length of the variable named asBIT_STRING. The attribute VARYING specifies that the variable BIT_STRING is of varying length, and its maximum length is 4. The bit stringvariable can also be initialized using the attribute INITIAL. e.g. DECLARE INTEREST VARYING BIT(5) INITIAL(' 011 'B); initializesthe variable INTEREST to the bit string value 011. The second statement in the given program is a value ac-quisition statement. The rules regarding value acquisition are that all data on the datacard (input stream) be in the form of bit string constants without repetitionfactors. If the value in the input is longer in length than the declaredlength, truncation occurs and the rightmost excess digits are deleted. e.g. If on the data card we have '001100'B then, the value acquired will be BIT_STRING=0011. Thus, the last two bits are truncated. Similarly, if the value specified in the input is shorter in length than thedeclared length, padding, by means of zero bits occurs at the right. e.g. If on the data card the value is '01'B then, the value acquired by the BIT_STRING is equal to 0100. No padding occurs if the VARYING attribute is used. The third statement in the given program is of decision making, usingbbit strings. The variable which is named as BIT_STRING is tested, b inorder to arrive at a certain decision. For decision-making, the PL/I machine accepts a bit string of arbitrarylength. The True path is taken if the bit string contains at least onebit with the value 1. The False path is taken if none of the bits in the stringhas the value one. The null string ' 'B does not contain any bit; the False path is taken if a null string is present. In the same third statement the second part involves value disposition(PUT statement). It prints out the value of BIT_STRING and alsoprints out the remark depending on the decision which is made. Thus, bit-string processing is quite like character-string processing, andhas the facility of using the same built-in functions (SUBSTR, LENGTH, INDEX), concatenation and comparison operators as for character-strings.

Question:

Discuss the chemical composition of DNA.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0611.htm

Solution:

Deoxyribonucleic acid or DNA is one type of nucleic acid present in the nucleus of all cells. Nucleic acids are rich in phosphorus, and contain carbon, oxygen, hydrogen, and nitrogen. DNA is composed of nitrogenous bases, a five-carbon sugar called deoxyribose, and phosphate groups. There are four kinds of nitrogenous bases: adenine, guanine, cytosine, and thymine. Adenine, and guanine belong to the class of organic compounds called purines, and cytosine and thymine belong to the pyrimidines. Each nitrogenous base is attached to deoxyribose molecule via a glycosidic linkage, and the sugar is attached to the phosphate by an ester bond. This combination of base-sugar-phosphate comprises the fundamental unit, termed a nucleotide, of nucleic acid. Four kinds of deoxyribonucleotides are found in DNA, each containing one of the four types of nitrogenous bases. The nucleotides are joined by phosphate ester bonds into a chain. Two complementary chains of nucleotides form a DNA molecule.

Question:

A chemist has an E. coli bacterium cell of cylindrical shape. It is 2\mu (microns) long and 1\mu in diameter and weighs 2 × 10^-12 g. (a) How many lipid molecules are present, assuming their average molecular weight to be 700 and lipid content to be 2%? (b) If the cell contains 15,000 ribosomes, what percent of the volume do they occupy? You may assume them to be spherical with a diameter of 180 \AA; also 10^4 \AA = 1 micron.

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Solution:

(a) Because you are told the molecular weight of an average lipid molecule, you need only compute their total weight in the E. coli cell; and then find the number of moles present. A mole is defined as weight in (grams / mo-lecular weight) (MW) . From this, you use the fact that in one mole of any substance, (6.02 × 10^23) molecules (Avogadro's number) exist. The lipid content is given as 2%. If the total weight of the cell is (2 × 10^-12) g, the lipid molecules must have a total weight of (.02) (2 × 10^-12) or 4 × 10^-14 grams. The average molecular weight of a lipid molecule is given as 700 (g / mole). Thus, you have 4 × 10^-14 / 700 = 5.71 × 10^-17 moles of lipid molecules. If 6.02 × 10^23 mo-lecules are in a mole, then in 5.71 × 10^-17 moles, there are 6.02 × 10^23 5.71 × 10^-17 = 3.44 × 10^7 lipid molecules. (b) Calculate the volume of both the E. coli cell and the total volume occupied by the ribosomes. Volume of the E. coli cell: It is of cylindrical shape. The volume of a cylinder can be found from the product of its length and the area of its base. The area of the base, a circle. is \pir^2 , where r = radius. Because the diamater = 2r,the radius of the cylindrical E. coli cell is 1\mu divided by 2 or 0.5 \mu. This is converted to 5 × 10^3 \AA, by the conversion factor 10^4 \AA = 1 micron. Similarly, 2\mu = (2 × 10^4 \AA) = length of E. coli cell. Its total volume = (area of circle) (length) = [\pi(5 × 10^3)^2 (2 × 10^4] = 1.57 × 10^12. Volume of ribosomes: Each ribosome has a diameter of 180 \AA or a radius of 90 \AA. It is given that it is a sphere. The volume of a sphere is given by the formula (4 / 3 \pi^3), where r = radius. Substituting the volume of one ribosome = (4 / 3)\pi (90)^3 3.05 × 10^6. If there exists 15,000 ribosomes, their total volume must be (15000) (3.05 × 10^6) = 4.58 × 10^10. Therefore, the percentage volume it occupies = (total volume of ribosomes) / (volume of E. coli cell) × 100 = [(4.58 × 10^10) / (1.57 × 10^12)] × 100 = 2.9%.

Question:

Catalase, the enzyme that assists the breakdown of hydrogen peroxide into oxygen and water, has a turnover number of 2 × 10^8. In one hour, what weight of hydrogen peroxide could be decomposed by one molecule of catalase?

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Solution:

Enzymes may be defined as catalysts. A substrate is the compound that the enzyme acts upon to form the product. One aspect of enzymatic reactions is the rate at which the enzyme-substrate complex is formed and shifted to product. One can express this rate as the turnover number. It is defined as the number of molecules of substrate converted to product by one molecule of enzyme in one minute at optimum conditions. In this problem, you are given one molecule of enzyme, and excess substrate. The turnover rate is given as 2 × 10^8. You are asked for the weight of the substrate decomposed in one hour. To find the weight, you need the number of molecules decomposed. According to the definition of turnover number, 2 × 10^8 molecules, decompose in one minute. Thus, in 1 hour, 60 minutes × 2 × 10^8 = 1.2 × 10^10 molecules of hydrogen peroxide (H_2O_2) is decomposed. To calculate the total weight of these molecules, one uses Avogadro's number. It states that there are 6.02 × 10^23 molecules in one mole. Thus, 1.2 × 10^10 molecules represents 1.2 × 10^10 / 6.02 × 10^23 = 1.99 × 10^-14 moles of H_2O_2 decomposed. A mole is defined as weight in grams/mo-lecular weight. The molecular weight of H_2O_2 = 34. Thus, weight of H_2O_2 decomposed = (1.99 × 10^-14 moles) (34 g/mole) = 6.766 × 10^-13 grams.

Question:

Design a FORTRAN function to simulate trials according to the geometric distribution f(x) =pq^x, where p is the probability of success on any trial, and q = 1 - p is the probability of failure.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0476.htm

Solution:

This perfectly binary probability structure is derived from the notion of Bernoulli trials, in which the only two possible outcomes are "success" and "failure". If a sequence of these independent trials is performed, the probability of obtaining X failures before reaching a success can be exhibited by the geometric distribution. There are two approaches to this problem, both depending on the value of p. If p is close to 0 or 1, we can generate a series of inde-pendent trials, and count the number of trials necessary before a suc-cess appears. As q = 1 - p, we can generate a series of random numbers between 0 and 1 and can call the series S. Let Si be thei-thterm of the series S. As we generate a new term Si, we compare its value with q. As soon as Si > q at any point during iteration, procedure stops, andi- 1 turns out to be the value of the geometric variable X. The alternate approach, when p is not close to 0 or 1, uses the formula [i= log r / log q], in whichiis the geometric variable, r is a random number such that 0 \leq r \leq 1, q = 1 - p, and the brackets indicate that only the integer part of the result is desired. This method is unreliable when p is close to 0 or 1 because the ALOG library function is computed with a power series that, on most machines, is imprecise when calculating values near the parameters of the chosen interval. The function procedure is described below. Comments highlight the two different methods of solution. INTEGER FUNCTION BERN(P) CRETURN TO THE MAIN PROGRAM IS CONTINGENT CUPON FINDING THE FIRST SUCCESS IF (P.LE. .\O5 .OR. P. GE. .95) GO TO 1\O CLOGARITHMIC METHOD X = RNDM (NUM) BERN = INT(1.\O + (ALOG 1\O (X) / ALOG 1\O(1.\O - P))) RETURN CTHIS IS THE COUNTING METHOD, WHICH CAVOIDS THE AFOREMENTIONED PRECISION DIFFICULTIES 1\ODO 2\O J = 1, 1\O\O\O X = RNDM (NUM) IF (X.LT.P) GO TO 3\O 2\OCONTINUE CIF NO SUCCESSES, BERN = 1\O\O\O\O BERN = 1\O\O\O\O RETURN CSUCCESS! 3\OBERN = J RETURN END Some FORTRAN compilers contain random number generation function as a library function. For those which do not, the function has to be defined and called. In this program, the random number generation function is called by means of X = RNDM (NUM) statement and has the following (generally used) procedure: FUNCTION RNDM (NUM) IF (NUM. GT. 0) INUM = NUM INU = 65539 \textasteriskcentered INUM IF (INU.LT. 0) INU = INU + 2147483647 + 1 RNDM = .4656613E - 9 \textasteriskcentered FLOAT (INU) INUM = INU RETURN END

Question:

Differentiate between fission, budding, and fragmentation as means of asexual reproduction.

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Solution:

Most protozoans reproduce asexually by fission, which is the simplest form of asexual reproduction. Fission involves the splitting of the body of the parent into two approximately equal parts, each of which becomes an entire, new, independent organism. In this case, the cell division involved is mitotic. Hydras and yeasts reproduce by budding, a process in which a small part of the parent's body separates from the rest and develops into a new individual. It may split away from the parent and take up an independent existence or it may remain attached and maintain an independent yet colonial existence. Lizards, starfish, and crabs can grow a new tail, leg, or arm if the original one is lost. In some cases, this ability to regenerate a missing part occurs to such an extent that it becomes a method of reproduction. The body of the parent may break into several pieces, after which each piece regenerates its respective missing parts and develops into a whole animal. Such reproduction by fragmentation is common among the flatworms, such as the planaria.

Question:

Write a FORTRAN program that calculates the coefficients of a linear Diophantine equation of the form MX + NY = (M,N).

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Solution:

A Diophantine equation of this type is a polynomial with integer coefficients for which only integer solutions are permitted. The set (M,N) is called the solution of the equation. The simplest Diophantine equation is the one-variable, first-degree equation of the form MX = p where M, p are integers, and neither is equal to 0. By definition, this equation has an integer solution if and only if p/M is an integer. Two-variable linear Diophantine equations may be solved by applying Euclid's algorithm for finding the greatest common divisor (GCD). The program used incorporates Euclid's algorithm into the following main algorithm: 1) Input coefficients M and N. 2) Check integers; if M = 0, exit, else, continue. 3) Initialize: MO = N1 = 1, and NO = Ml = 0. 4) Divide M by N; store quotient in Q and remainder in R. 5) Calculate two new coefficients for M and N, which are, respectively M2 = MO - Q\textasteriskcenteredM1 N2 = NO - Q\textasteriskcenteredN1 6) Redefine the variables according to the following schedule: MO = M1 NO = N1 M1 = M2 N1 = N2 7) Redefine the dividend as the previous divisor and the divisor as the previous remainder (this is Euclid's strategy). 8) Loop back to step (4). 9) The final nonzero remainder is (M,N); if the first remainder is zero, (M,N) = │N│. Output M,N,M1,N1 and (M,N) in the form M \textbullet M1 + N \textbullet N1 = (M,N) One more operation must be performed. The absolute values of M and N must be stored in MABS and NABS, respectively, so that the signs of the solutions can be included in the output. Thus, the program is presented below. Comments are added to make it easier to follow the prose description of the algorithm. The program is terminated when 0 is entered as the value of M. Also note, that the program assumes M and N \leq 99999. INTEGER Q,R 5READ (5,100) M,N 100FORMAT (215) CDO WHILE M NOT EQUAL TO ZERO IF (M.EQ.O) GO TO 99 CINITIALIZE VARIABLES MO = 1 M1 = 0 NO = 1 N1 = 0 CFIND ABSOLUTE VALUES TO OBTAIN SIGNS OF M,N MABS = IABS(M) NABS = IABS(N) MSIGN = M/MABS NSIGN = N/NABS CCALCULATE QUOTIENT AND REMAINDER 20Q = MABS/NABS R = MABS - NABS \textasteriskcentered Q CIF R GREATER THAN ZERO, REAPPLY EUCLID'S CALGORITHM IF (R.LE.O) GO TO 30 M2 = MO - Q \textasteriskcentered M1 N2 = NO - Q \textasteriskcentered N1 CREDEFINE VARIABLES MO = M1 NO = N1 M1 = M2 N1 = N2 MABS = NABS NABS = R GO TO 20 CGIVE COEFFICIENTS OF M AND N WITH CTHEIR PROPER SIGNS 30Ml = MSIGN \textasteriskcentered M1 N1 = NSIGN \textasteriskcentered N1 COUTPUT RESULTS; NABS IS THE FINAL NON- CZERO REMAINDER WRITE (6,101) M,M1,N,N1,NABS 101FORMAT (1X,'(' ,I6, ') \textasteriskcentered (', I6,') + (', I6, ') \textasteriskcentered (', 1 I6, ') = ', I5) GO TO 5 99STOP END

Question:

Write a PL/I program to determine whether a given integer be-tween0 and 1000 (exclusive) is prime or composite.

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Solution:

As we know a prime is any positive integer p greater than 1 whose onlypositive divisors are 1 and p. Integers greater than 1 which are not primeare called com-posite. According to the fundamental theorem of arithmetic, every composite number n has at least two prime factors p and qsuchthat 11 has noprime factor less than or equal to \surdn, then n is prime." Thus we have analgorithm to determine whether a given number n is prime or composite: Divide n successively by the primes 2, 3, 5,7, ...p_n, wherep_ndenotes the largestprime that does not exceed \surdn, if none of these primes divides n, thenn is prime, other-wise n is composite. In our example to test integers less than 1000 we will need to divide by the primesless than \surd1000 = 31.6. These are 2, 3, 5, 7, 11, 13, 17, 19, 23,29 & 31.We can now de-tail the steps that our program is to perform. 1. Place the eleven primes into memory. 2. Read an integer n into memory. 3. (a) If 11000 THEN PUT LI ST ('OUT OF RANGE'); ELSE; /\textasteriskcentered CHECK THAT N IS WITHIN RANGE \textasteriskcentered/ RN = N; /\textasteriskcentered CONVERT N TO REAL NUMBER \textasteriskcentered/ NROOT=SQRT(RN); /\textasteriskcentered CALCULATE SQUARE ROOT \textasteriskcentered/ DOI = 1 TO 11; IF(NPR(I) -NROOT)>0 THEN PUT LIST(N,'THE NO IS PRIME'); This statement is followed by comment /\textasteriskcentered IS CURRENT PRIME GREATER THAN SQUARE ROOT \textasteriskcentered/ ELSE IF (N/NPR(I) \textasteriskcenteredNPR(I) -N)=0 THEN PUT LIST (N, 'THE NO IS COMPOSITE'); /\textasteriskcentered WE CHECK IF THE CURRENT PRIME DIVIDES N \textasteriskcentered/ ELSE; /\textasteriskcentered THIS IS A NULL ELSE AND SIGNIFIES THAT IF NEITHER OF THE ABOVE POSSIBILITIES HOLD, THEN THE LOOP IS TO BE REPEATED AND A NEW VALUE OF NPR.IS TO BE OBTAINED \textasteriskcentered/ GOTO LABEL; END PRIME;

Question:

The thymus gland is a two-lobed, glandular-appearing structure located in the upper region of the chest just behind the sternum. What are the two principal functions that have been attributed to this gland?

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Solution:

It is thought that one of the functions of the thymus is to provide the initial supply of lymphocytes for other lymphoid areas, such as the lymph nodes and spleen. These primary cells then give rise to descendent lines of lymphocytes, making further release from the thymus unnecessary. This first function of the thymus is non-endocrine in nature. The second function attributed to the thymus is the release of the hormone thymosin which stimulates the differ-entiation of incipient plasma cells in the lymphoid tissues. The cells then develop into functional plasma cells, capable of producing antibodies when stimulated by the appropriate antigens. To summarize, full development of plasma cells requires two types of inducible stimuli. They are: (1) stimulation by thymosin which initiates differentiation of all types of incipient plasma cells, and (2) stimulation by a specific antigen, which affects the functional maturation of only those cells with a potential for making antibodies against that particular antigen.

Question:

A biochemist performed several experiments on rats. In one, heused two groups of young rats. GroupArats were fed on adiet of purified casein (cheese protein), starch, glucose, lard, minerals, and water. Group B rats were fed the same dietwith the addition of 3 cm^3 of milk daily. It was observed thatGroup A rats stopped growing and lost weight, while Group B rats gained steadily in weight and size. After eighteendays the milk was given to Group A rats and removedfrom the diet of Group B. Group A rats now resumedgrowth and gained in weight, while Group B rats stoppedgrowing and lost weight. The investigator knew that 3 cm^3 of milk has an insigni-ficantfood value in terms of carbohydrates, fats, proteinsand minerals. Consider yourself in his position, confrontedwith only the information given so far. a) What conclusions could you reach about the presencein milk of substances other than these four foods, thepossible functions of the hypothetical substances, and thequantities needed by the rats? b) Why was it necessary to transfer the milkfrom Group B rats to Group A rats half-way through the experiment? c) The rats received only one type of protein (casein) . Why can protein starvationberuled out as a possible cause ofgrowth inhibition and weight loss?

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Solution:

One can conclude that some factors responsible for growth and weightincrease are present in milk. Such factors are called vitamins. Vitamins are complex chemicals which are required in minute amounts in thediet of all heterotrophic organisms. Autotrophic organisms, such as greenplants, also need to procure certain essential vitamins from the environmentfor their growth. Vitamins have no energy value. They play a vital role in many chemicalreactions in metabolism. Most vitamins function as coenzymes. Lack of any vitamin in the diet causes the reaction in which it takes part to slowdown, and since most metabolic reactions are part of a long sequenceof events, an alteration in the rate of any one of them can have widespreadeffects in the body. Since 3 cm^3 of milk is enough for normal growthin rats, it is shown that only small quantities of vitamins are needed. b) By interchanging the diet of Group A and Group B a control experimentis performed, demonstrating that the lack of growth is due to thediet alone and not to any constitutional defects in the rats.I.e, when normallygrowing rats in Group B are fed with the vitamin-deficient diet, theystop growing. c) Casein, a milk protein, is an example of an adequate protein. An adequateprotein is one which contains all the essential amino acids necessaryfor growth. Since the essential amino acids are present, we can ruleout protein starvation, a deficiency in essential amino acids, as a causeof the observed stunted growth.

Question:

Why is it that alveolar air differs in its composition from atmospheric air? Of what significance is this fact?

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Solution:

The respiratory tract is composed of conduct-ing airways and the alveoli. Gas exchange occurs only in the alveoli and not in the conducting airways. The maximum alveolar volume is about 500 ml. Let us consider then what takes place during expiration. Five hundred milliliters of air is forced out of the alveoli, and into the conducting airways of the respiratory tract. Of this air, 150 ml. remains in the respiratory airways following expiration, while 350 ml. of air is exhaled from the body. During the next inspiration 500 ml. of air are taken up by the alveoli, but 150 ml. of this air is not atmospheric, but rather is the air that remained in the tubes following the previous expiration. One can see, then, that only 350 ml. of fresh atmospheric air enters the alveoli during each inspiration. At the end of inspiration, 150 ml. of fresh air also fills the airways but does not reach the alveoli. Hence no gas exchange between this air and the blood can occur. This fresh air will be expelled from the body during the next expiration, and will be replaced again with 150 ml. of alveolar air, thus completing the cycle. From this cycle it can be seen that of the 500 ml. of air entering the body during inspiration, 150 ml. of it never reaches the alveoli, but instead remains in the con-ducting tubes of the respiratory system. The term "anatomical dead space" is given to the space within the conducting tubes because no gas exchange with the blood can take place there. The question arises then, as to the significance of this dead space. Because the tubes are not completely emptied and filled with each breath, "new" air can mix with "old" air. Consequently, alveolar air contains less oxygen and more carbon dioxide than atmospheric air. In addition, the air that remains in the alveoli following expiration (the residual volume) can, to a limited extent, mix with the incoming air and thereby alter its composition.

Question:

One of the first problems in linear programming was the so-called diet problem. The problem was to minimize the cost of eating three meals a day while meeting minimum daily requirements of certain essential nutrients. Suppose we consider only the contribution of vege-tables to the nutritional requirements. How many times should each vegetable be served during the next week in order to minimize cost while satisfying the nutritional and taste requirements? Use the following information: Units per serving Vitamin Vegetable Iron Phosphorus A B Niacin Cost/ serving Green beans Carrots Broccoli Cabbage Beets Potatoes 0.45 0.45 1.05 0.4 0.5 0.5 10 28 50 25 22 75 415 9065 2550 75 15 235 8 3 53 27 5 8 0.3 0.35 0.6 0.15 0.25 0.8 5\textcent 5\textcent 8\textcent 2\textcent 6\textcent 3\textcent Minimum daily requirements from vegetables 6.032517500(USP)245 mg5.0 mg In addition, cabbage cannot be served more than twice during the week, and the other vegetables cannot be served more than four times each during the week. A total of 14 servings of vegetables are required during the week.

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Solution:

A linear programming problem is said to be in canonical form when itis in the form: Min a_1 x1+ a_2 x_2 +.....+a_nx_n= C(1) subjectto b_11x_1, + b_12 x_2 +.........+ b_1n,x_n\geq C_1,(2) b_21x_1 + b_22 x_2 +.........+ b_2nx_n\geq C_2 ::::: : : : : ::: : : : ::::: : : : : ::: : : : b_m1x_1 + b_m2 x_2 +....... .+b_mnx_n\geq C_m andx_i \geq 0i= 1,2,....,n(3) To convert the given problem into canonical form, let x_1= number of times to serve green beans x_2= number of times to serve carrots x_3 = number of times to serve broccoli x_4 = number of times to serve cabbage x_5 = number of times to serve beets x_6 = number of times to serve potatoes The cost of one serving of green beans is $0.05. Thus x_1 servings will cost 5x_1 (in cents) Similarly, x_4 servings of cabbage will cost 2x_4. If we multiply the number of servings for a given vegetable by the cost per serving and sum over all vegetables, we will obtain the cost function (1). Thus, we have Min 5x_1 + 5x_2 + 8x_3 + 2x_4 + 6x_5 + 3x_6 Next, we hunt for the constraints. The minimum daily re-quirements from vegetables of iron, phosphorus, vitamin A, vitamin B and Niacin must be met or exceeded. Thus, for iron, .45x_1 + .45x_2 + 1.05x_3 + .4x_4 + .5x_5 + .5x_6 \geq 42.0 Here the minimum daily requirement of iron, 6.0 is multi-plied by 7 to give the weekly requirement. Proceeding with the other nutrients in a similar manner , we obtain the con-straint inequalities Min 5x_1 + 5x_2 + 8X_3 + 2x_4 + 6x_5 + 3x_6 subject to Iron .45x_1+ .45x_2 + 1.O5x_3 + .4x_4 + .5x_5 +5X_6 \geq 42.0 Phos. 10x_1 + 28x_2 + 50 x_3 + 25x_4 + 22x_5. + 75x_6 \geq 2275 Vitamin A415x_1 + 9065x_2 + 255OX_3 + 75x_4 + 25x_5 + 235x_6 \geq 122500 Vitamin C8x_1, + 3x_2 + 53x_3 + 27x_4 + 5x_5 + 8x_6 \geq 1715 Niacin. 3x1+.35x_2 + .6x_3 + .15x_4 + . 25x_5 + .8X_6 \geq 35 x_1\leq4 x_2\leq4 x_3\leq4 x_4\leq2 x_5\leq4 x_6\leq4 x_1 \geq 0x_2 \geq 0x_3 \geq 0x_4 \geq 0. The next step is to convert the program into a form suitable for computer manipulation . We add slack var-iables S_i to convert all inequalities into equalities . The simplex algorithm is then applied to the resulting system of equations to obtain an optimal solution. The simplex algorithm is available as a packaged pro-gram in most installations .

Question:

It has been found that the following sequence can be used to prepare sodium sulfate, Na_2 SO_4 : S (s) + O_2 (g)\rightarrow SO_2 (g) 2SO_2 (g) + O_2 (g)\rightarrow 2SO_3 (g) SO_3 (g) + H_2 O(l)\rightarrow H_2 SO_4 (l) 2NaOH + H_2 SO_4\rightarrow Na_2 SO_4 + 2H_2 O If you performed this sequence of reactions, how many moles of Na_2 SO_4 could possibly be produced if you start with 1 mole of sulfur? How many moles are possible with 4.5 g of water?

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Solution:

If you had the general equation, aA+bB\rightarrowcC+dD, a moles of A react with b moles of B to produce c moles of C and d moles of D. Thus, if you want to know how many grams of D will be produced, and you know how much A you have, calculate the number of moles of A present. From this, you can determine how many moles of D can be generated, since you know that a moles of A will produce d moles of D. With this in mind, you can proceed to answer these questions. From the first equation, you see that if you start with 1 mole of sulfur (S), 1 mole of SO_2 can be generated (since all the coefficients are one, although 1 is not written). Now that 1 mole of SO_2 is generated, you proceed to the second equation. It states that for every 2molesof SO_2, 2 moles of SO_3 are generated. Thus, in keeping with this relative ratio, 1 mole of SO_3 can be generated from one mole of SO_2. In the third reaction you again have 1 mole of SO_3 yielding 1 mole of H_2 SO_4. At this point, you have 1 mole of N_2 SO_4 (from 1 mole of starting sulfur). The last equation shows 1 mole of H_2S0_4 producing 1 mole of Na_2 SO_4. Therefore, if you where to start with 1 mole of S, you would obtain 1 mole of Na_2 SO_4. Now let us consider H_2 O. Water does not enter the sequence until the third equation. This means you start with SO_3 (g) + H_2 O (l) \rightarrow H_2 SO_4 (l). If you have 4.5 g of H_2 O (MW = 18 g/m), you possess 4.5 g/18 g/mole = .25 moles H_2 O. From the reaction, 1 mole of H_2 O yields 1 mole of H_2 SO_4. Thus, .25molesof H_2 SO_4 are generated from .25 moles of water. You have the same 1 : 1 ratio. The last equation is also a 1 : 1 ratio. Therefore, if you start with 4.5 grams of H_2 O, .25molesof Na_2 SO_4 will be produced.

Question:

Why is the flexor muscle of the crayfish well developed?

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Solution:

In order to answer this question, it is ne-cessary to understand the body plan of the crayfish. (Refer to figure in last solution.) Crustaceans have a head and trunk which is usually composed of a thorax and abdomen. The head and thoracic segments of the decapods are fused together dorsally by a carapace, which hangs over the sides of the animal and encloses the gills. Lobsters and crayfish have a large segmented abdomen, with appendages on each segment. The fourth thoracic appendages are chelipeds, walking legs that function as claws. The next four pairs of thoracic appendages are smaller, thinner walking legs. Lobsters, crayfish, and burrowing shrimps generally crawl with these walking legs. These animals are also capable of swimming backward rapidly. This is accomplished by flexing the abdomen ventrally. The flexor muscle of the crayfish is well-developed and powerful. It enables the animal to flex its abdomen rapidly in swimming. Many non-swimming decapods have a greatly reduced abdomen that is folded up under the thorax. This is often the case in crabs, as most crabs cannot swim. The blue crab, however, is a powerful and agile swimmer. Its last pair of legs are broad flat paddles that act as propellers. Shrimps are generally bottom-dwellers, i.e., they dwell at the bottom of a body of water. They swim intermittently. The abdominal appendages, pleopods, are the principal swimming organs. The pleopods are large and often fringed. Shrimps utilize ventral flexion of the abdomen for quick backward darts and for vertical steering.

Question:

When ammonium nitrate (NH_4NO_3) is heated, laughing gas (N_2O) and water are produced. If the laughing gas is collected at STP (Standard Temperature and Pressure) and one starts with 10 grams of NH_4NO_3, how many liters of laughing gas could be collected?

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Solution:

When any gas is collected at STP (0\textdegreeC and 1atm), one mole of it will occupy 22.4 liters. The balanced equation for this chemical reaction is NH_4NO_3 \ding{217} N_2O + 2H_2O. One mole of NH_4NO_3 yields one mole of N_2O. If the number of moles of NH_4NO3that reacted is known, the number of moles of laughing gas produced is also known. The molecular weight of ammonium nitrate = 80 g / mole. 10 g of ammonium nitrate (10 g / 80 g / mole) = 0.125 moles of NH_4NO_3 produce 0.125 moles of laughing gas. Recall that 1 mole of any gas (at STP) occupies 22.4 liters. To find out what volume 0.125 moles occupy, set up the following proportion: (1 mole / 22.4 liters)= (0.125 / x), where x = the volume. Solving x= 2.8 liters = volume of laughing gas collected.

Question:

What is the equivalent capacitance between the points A and B in diagram (A)? The charge on the 6-\muF capacitor is 90 \muC. What potential difference exists between the points A and R?

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Solution:

In the arrangement shown in diagram (A) there are two sets of capacitors in parallel. Since the net capacitance of n capacitors in parallel is C_net = C_1, + C_2 + . . . + C_n the arrangement is equivalent to that shown in diagram (B). In each branch there are now two capacitors in series. If we have n capacitors in series, their net capacitance is (1/C_net) = (1/C_1) + (1/C_2) + . . . + (1/C_n) For each branch, then, (1/C_net) = (1/C_1) + (1/C_2). For the top branch (1/C_net ) = (1/12 \muF) + (1/4 \muF) = [(4 \muF + 12 \muF)/(48 \muF^2 )] C_net = 3 \muF For the lower branch (1/C_net ) = (1/4 \muF) + (1/2 \muF) = [(2 \muF + 4 \muF)/(8 \muF^2 )] C_net = (4/3) \muF Thus, the arrangement is equivalent to that shown in diagram (C). Here we have two capacitors in parallel. Their capacitances add. Hence, the equivalent capacitance between A and B is C = [3 + (4/3)] \muF = 4 (1/3) \muF. The charge on the 6-\muF capacitor is 90 \muC. Hence, the potential across it is, by definition of capacitance V_DS = [(Q_DS )/(C_DS )] = [(90 × 10^-6 C)/(6 × 10^-6 F)] = 15 V. This potential is common to all the capacitors in the top branch between A and S since they are all in parallel. Hence the total charge on the equivalent 12-\muF capacitor is Q_AS = C_AS V_AS = C_AS V_DS = 12 × 10^-6 F × 15 V = 180 \muC. The conductors connected to S are isolated and initially must have been uncharged. If a charge of - 180 \muC appears on the negative plate of the equivalent 12-\muF capacitor connected to S, a corresponding charge of + 180 \muC must be induced on the positive plate of the 4-\muF capacitor connected to S, leaving the net charge at S zero. A corresponding - 180-\muC charge is induced on the negative terminal of the 4- \muF capacitor, and the voltage between its plates is thus V_SB = (Q_SB /C_SB ) = [(180 × 10^-6 C)/(4 × 10^-6 F)] = 45 V. The total potential difference between A and B is thus V = V_AS + V_SB = (15 + 45) V = 60 V. This is the potential difference between A and B by either branch. Referring to diagram (C), the charge on the equivalent 4/3-\muF capacitor is thus Q_AE = C_AE V_AE = C_AE V = (4/3) × 10^-6 F × 60 V = 80 \muC. This is the charge on the plate attached to A in either the equivalent circuit or the original circuit, since the two produce identical effects. Hence the equivalent 4-\muF capacitor between A and R has charges of \pm 80 \muC on each of its plates. Hence the potential difference across it is V_AR = Q_AR /C_AR = [(80 × 10^-6 C)/(4 × 10^-6 F)] = 20 V.

Question:

A man of mass 80 kg is standing on the rim of a stationary uniform circular platform, of mass 140 kg and diameter 8 m, which is free to rotate about its center. The man throws to a companion on the ground in a direction tangential to the rim a package of mass 1 kg at a speed of 20 m\bullets^-1 rela-tive to the ground. What angular velocity of the man and platform is produced in consequence? The man then walks so as to bring him to a position halfway between the rim and the center of the platform. What is the new angular velocity of the system?

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Solution:

The package is thrown tangentially at a speed of 20 m.s^-1 relative to the ground and consequently has an initial angular momentum about the center of the platform form given by \vertL\ding{217}\vert = \vertr\ding{217}× mv\ding{217}\vert = \vertmvrsin\texttheta\vert = mvr = 1 kg × 20 m\bullets^-1 × 4m \ding{217} \ding{217} \ding{217} = 80 kg\bulletm^2\bullets^-1. where \texttheta(=90\textdegree) is the angle between r\ding{217}and v\ding{217}. \ding{217} \ding{217} Since the platform is free to rotate about its center, there are no external (e.g. in the form of friction) forces acting on the man-package- platform system. There are no external torques acting on this system, but \Gamma\ding{217}= dL\ding{217}/dt \ding{217} \ding{217} where \Gamma\ding{217}is the sum of the external torques acting on the system. Thus \ding{217} 0 = dL\ding{217}/dt \ding{217} L\ding{217}= constant \ding{217} That is, the angular momentum of the system remains the same at all instances of time when no external force acts on the system. This is the principle of conservation of angular momentum. Since the man-package- platform system was initially at rest, L\ding{217}= 0. After the package is thrown, \ding{217} L\ding{217}must still be zero, thus the man-platform must acquire a momentum \ding{217} Equal and opposite to the momentum of the package, to satisfy the condition that the L\ding{217}= 0 for the man-platform-package system at all times. \ding{217} The plat- form is a distributed mass, as opposed to the localized masses of the man and package (which to a very good approximation may be treated as point masses). The fol-lowing expression involving the moment of inertia I of the platform must then be used. (The moment of inertia accounts for this mass distribution). I\cyrchar\cyromega = L(1) where I is the angular velocity of the platform due to the throwing of the package. (This is the rigid body analogue of linear momentum mv = p, where I corresponds to m, v corresponds to \cyrchar\cyromega and L corresponds to P.) I for the uni-form circular platform is I_1 = (1/2) m_1R2 where m_1 is the mass of the platform and R is its radius.For the man, the definition of moment of inertia for adiscrete particle gives I_2 = m_2R^2,where m_2 is the man'smass. Then, since the man also revolves about the center with angular velocity \cyrchar\cyromega, equation (1) becomes (I_1 + I_2)\cyrchar\cyromega = L_1 + L_2 = Lpackage [m_2R^2 + (1/2)m_1R^2)\cyrchar\cyromega = Lpackage (80 × 4^2 + 1/2 × 140 × 4^2) kg\bulletm^2 × \cyrchar\cyromega = 80 kg\bulletm^2\bullets^-1, \cyrchar\cyromega = (80 s^-1)/(80 × 16 + 70 × 16) = 1/30 rad\bullets^-1 . If the man walks toward the center, his moment of inertia about the center decreases. At the halfway posi-tion, his moment of inertia is m_2(R/2)^2 = (80 × 2^2) kg.m^2. But the angular momentum must stay the same, and so the angular velocity must increase to \cyrchar\cyromega', and (m_2R^2)\cyrchar\cyromega = L = [m_2(R/2)^2]\cyrchar\cyromega' (80 × 4^2 + 1/2 × 140 × 4^2)kg\bulletm^2 × \cyrchar\cyromega = (80 × 2^2 + 1/2 × 140 × 4^2) kg\bulletm^2 x \cyrchar\cyromega' . \therefore\cyrchar\cyromega' = (150/90)\cyrchar\cyromega = (1/18) rad\bullets^-1.

Question:

A collimated beam of silver atoms emerges from a furnace at 1500\textdegreeK, passes through a circular hole, and falls on a screen 1m away. Assume that all the atoms travel with the same speed. What size hole is likely to give the smallest spot size on the screen? The mass of a silver atom is 1.8 × 10^-25 kg. (See figure).

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Solution:

If the hole has a radius a, the uncertainty in the z-co-ordinate of an atom passing through the hole is 2a. Thus \Deltaz = 2a . But if p_z is the momentum of the atom in the z-direction, then, at the hole, the Heisenberg Uncertainty Relationship yields \Deltap_z \Deltaz \approx h where \Deltap_z is the uncertainty in the z momentum. But \Deltap_z = m\Deltav_z where \Deltav_z is the uncertainty in the z velocity of the silver atoms. Hence m\Deltav_z \approx (h/2a) or \Deltav_z \approx (h/2am). Thus the atoms have an uncertainty in velocity in the z-direction of the order of h/2am. But the velocity in the z-direction is clas-sically zero. Hence, the atoms, by quantum-mechanical arguments, have velocities around the value h/2am in the z-direction. In the y-direction the atoms have a velocity which is assumed for simplicity to be the same for all atoms. The average kinetic energy possessed by an atom at temperature T is 3/2 kT, where k is Boltzmann's constant and T is given in degrees Kelvin. Thus the silver atoms of mass m have velocities v_y given by v_y = \surd3kT/m. They traverse the distance y in time t. Since no force acts on the silver atoms in the y direction y = v_y t or t = y/v_y In the same time the maximum distance achieved in the z-direction due to the z-velocity of the atoms is, similarly, a = \Deltav_z × t = (y\Deltav_z/v_y) = (hy/2am)\surd(m/3kT) = (hy/2a) \surd[1/(3mkT)] = [(6.6 × 10^-34 J \textbullet s × 1 m)/2a] × \surd[1 / (3 × 1.8 × 10^-25 kg × 1.38 × 10-23 J \textbullet k deg^-1 × 1500 k deg)] = (3.12/a) × 10^-12 m^2. Thus the largest possible distance from the axis at which a silver atom can strike the screen is found by observing the trajectory of the atom at the top of the hole. The furthest it will travel from the center of the hole is r = a + z = a+ (3.12/a) × 10^-12 m^2 . (dr/da) = 1 - (3.12/a^2) ×10^-12 m^2 , and for a minimum of r this quantity must be zero . It can be verified that this is a minimum by a second differentiation. Thus the radius of the hole which gives the smallest spot on the screen is that for which (3.12/a^2) × 10^-12 m^2 = 1 or a = 1.77 × 10^-6 m. In classical physics the spot size would be decreased indefinitely by reducing the size of the hole. Because of the wave nature of the electron, quantum mechanics does not supply the same answer. After a certain point, diminishing the size of the hole increases the size of the spot because of diffraction effects produced by the hole on the silver atoms. A minimum size of spot therefore results for a finite hole size.

Question:

Diffusion is too slow a process to account for the observed rateof transport of materials in active cells. Explain why.

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Solution:

The cytoplasm of a living cell is a watery medium separated by the plasmamembrane from the external environment. The cytoplasm resemblesa loose gel which is somewhat denser than pure water. In order tomove from one part of a cell to another, materials have to move through thisrelatively dense gel. Thus diffusion of materials through a cell is necessarilya much slower process than diffusion of materials through water. Diffusion is a relatively slow method of transport of materials also becausethe molecules in motion are constant-ly colliding with other molecules. This causes them to bounce off in other directions and to take azig-zagpath instead of a straight one. Therefore molecules have to travela longer distance, which requires more time, in order to get to their destinations. Diffusion is only a passive process: it does not re-quire energy. Its ratedepends on the viscosity (thickness) of the medium, the difference in concentrationsof molecules in the two regions, the distance between the tworegions, the surface area of the two regions if they are in contact, and thetemperature. Often the transport of materials in living cells occurs at so rapida rate that it is not possible to be accounted for by simple (passive) diffusionalone. Instead, some active processes involving expen-diture of energyare believed to be responsible.

Question:

In the figure, a fine wire, having a positive charge per unit length \lambda, lies on the y-axis. Find the electric intensity set up by the wire at point P.

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Solution:

Let the wire be subdivided, in imagination, into short elements of length dy. The charge dq on an element is then \lambda dy. Let r represent the perpendicular distance from P to the wire and s^\ding{217} the vector from dq to P. If we view dq as a point charge, then it sets up a field dE^\ding{217} and P given by dE^\ding{217} = k [(\^{s} dq)/s^2] = k [(\^{s}\lambdadq)/s^2], dE^\ding{217} = k [(\^{s} dq)/s^2] = k [(\^{s}\lambdadq)/s^2], where \^{s} is a unit vector in the direction of s^\ding{217}, and k = 9 × 10^9 N \textbullet m^2/c^2, and the resultant intensity E^\ding{217} is the sum of all these infinitesimal fields dE^\ding{217} along the y-axis, or E^\ding{217} = k \int[(\^{s}\lambdadq)/s^2]. The unit vector \^{s} lies in the yz-plane, so its x- component is zero. The magnitude of its y-component is sin \texttheta and that of its z-component is cos \texttheta. The vector equation above is then equivalent to the three scalar equations E_x = 0 E_y = k\lambda ^+\infty\int_-\infty [(sin \texttheta dy)/s^2 ] E_z = k^+\infty\int_-\infty [(cos \texttheta dy)/s^2 ]. The wire is considered to be sufficiently long so that the limits of integration are from - \infty to + \infty. Let us change the integration variable from y to \texttheta, so y = r tan \texttheta, s = r sec \texttheta and \texttheta varies from - (\pi/2) to \pi/2 as y varies between -\inftyand + \infty. Since y = r tan \texttheta, dy = r sec^2 \texttheta d\texttheta. Hence, E_y = k\lambda\pi/2\int_-\pi/2 [{(sin \texttheta)(r sec^2 \texttheta) d\texttheta}/s^2 ] Since sec \texttheta = s/r, E_y = k\lambda\pi/2\int_-\pi/2 [{(sin \texttheta)(s^2/r)d\texttheta}/s^2 ] = k\lambda/r\pi/2\int_-\pi/2 sin \texttheta d\texttheta = 0. This result is not unexpected since in the figure we see that for each dq at y = Y, there is another dq at y = - Y. The y-components of the electric fields of these two charges cancel, whereas the z-components add up. E_z = k\lambda\pi/2\int_-\pi/2 [(dy cos \texttheta)/s^2] = k\lambdar^\pi/2\int_-\pi/2 [{(cos \texttheta)(sec^2 \texttheta) d\texttheta}/s^2 ] = k\lambdar^\pi/2\int_-\pi/2 [{(cos \texttheta)(s^2/r^2)}/s^2 ] = k\lambda/r^\pi/2\int_-\pi/2 cos \texttheta d\texttheta = 2k\lambda/r

Question:

A person seated on a swivel chair is rotated rapidly. The rotation is suddenly stopped and the person is told to stand up. He complains that he feels dizzy. How can you account for his dizziness?

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Solution:

The labyrinth of the inner ear has three semicircular canals, each consisting of a semicircular tube connected at both ends to the utricle. Each canal lies in a plane perpendicular to the other two. At the base of each canal, where it leads into the utriculus, is a bulb-like enlargement (the ampulla) containing tufts of hair cells similar to those in the utriculus and sacculus, but lacking otoliths. These cells are stimulated by movements of the fluid (endolymph) in the canals. When a person's head is rotated, there is a lag in the movement of the endolymph in the canals. Thus, the hair cells in the ampulla attached to the head rotate, in effect, in relation to the fluid. This movement of the hair cells with respect to the endolymph stimulates the former to send impulses to the cerebellum of the brain. There, these impulses are interpreted and a sensation of dizziness is felt.

Question:

The same system which enables starfish to rocky substrates is useless on sandy beaches. Explain.

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Solution:

Starfish (also called sea star) locomotion occurs by means of tube feet. The tube feet are part of a hydraulic system found only in echinoderms called the water vascular system. The under surface of each arm of a sea star has hundreds of pairs of tube feet, which are hollow, thin-walled, sucker-tipped, muscular cylinders. Internally, the tube foot is connected to a muscular sac, an ampulla, at its base. The tube feet are connected via short lateral canals to a radial canal which extends the length of the arm. The radial canals are joined at the central disc of the sea star by a ring canal. The ring canal leads via a short tube called the stone canal to a sieve-like plate, called a madreporite, on the surface of the animal. The entire system is filled with watery fluid. In order to extend the tube foot, the ampulla contracts, forcing fluid into the tube foot. A valve prevents the fluid from flowing into the radial canals. The tube foot attaches to the sea floor (or substratum) by a sucker at its tip, which acts as a vacuum. Longitudinal muscles cause the tube foot to shorten in length; this pulls the sea star forward, and water is forced back in to the ampulla. The tube foot is released from the substratum, and then again. The rapid repetition of this cycle of events enables sea stars to move slowly. One can see that the use of suction in the tube feet would be useless to the sea star in moving on sandy beaches. On a soft surface, the tube feet are employed as legs. Locomotion becomes a stepping process, involving a backward swinging of the middle portion of the podia followed by a contraction, shoving the animal forward. Certain sea stars have tube feet without suckers at their tips? instead the tips are pointed. These sea stars live on soft bottoms, and can feed on buried animals. Starfish are slow-moving carnivorous animals. Their prey are generally sedentary or slow moving animals, such as clams and oysters.

Question:

If the last operation performed on a computer with an 8 bit word was an addition in which the operands were 2 and 3, what would be the values of the following flags? (a) Carry (b)Zero (c)Overflow (d)Sign (e)Even Parity (f)Half-carry What if the operands were -1 (two's complement) and +1?

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Solution:

The addition operation of 2 and 3 is: +2:0000 \vert 0010 +3:0000 \vert 0011 +5:0000 \vert 0101 bit #7654 \vert 3210 (a)The Carry flag is 1 if there is a carry from the most significant bit position (bit 7). Since there is no carry from bit 7 the contents of the Carry flag is 0. (b)The Zero flag is 1 if the result is zero. Since the result is not zero the contents of the Zero flag is 0. (c)The Overflow flag is 1 if the two operands in an addition have the same sign but the result has the oppo-site sign. As an example: +64:01000000 +64:01000000 10000000 This answer is incorrect if the numbers are signed because the answer is negative. Thus, this result is too large for 8-bit two's complement notation and the Overflow flag would be set to 1. Another example: -96:10100000 -64:11000000 01100000 This answer is also incorrect because it is positive and thus the Overflow flag would be set to 1. In the addi-tion of 2 and 3 the sign bit is correct so the contents of the Overflow flag is 0. (d)The Sign flag is 1 if the most significant bit (bit 7) is 1 (corresponding to a negative number) . Thus the contents of the Sign flag is 0 since bit 7 is 0. (e)The Even Parity flag is set to 1 if the result of the last operation contained an even number of 1's. The addition of 2 and 3 produced two l's and hence the con-tents of the Even Parity flag is 1. (f)The Half-carry flag is set to 1 if the last opera-tion produced a carry from the lower half word (bits 0-3). The addition of 2 and 3 produced no carry from bit 3 so the contents of the Half-carry flag is 0. The addition of -1 and +1 is: -1:1111 \vert 1111 +1:0000 \vert 0001 0:10000 \vert 0000 bit #7654 \vert 3210 (a)A carry was generated from bit 7 so the content of the Carry flag is 1. When binary numbers of different signs are being added, the computer will indicate a carry and not an overflow, if the result extends outside of its storage field, as is the case in this example. (b)The addition of -1 and +1 produced zero so the con-tents of the Zero flag is 1. (c)There is no overflow so the contents of the Over-flow flag is 0. (d)The sign bit (bit 7) is 0 so the contents of the Sign flag is 0. (e)There are no 1's in the result so the contents of the Even Parity flag is 1 (zero is an even number). (f)A carry was generated from bit 3 so the contents of the Half-carry flag is 1.

Question:

If we project a body upward with speed 1000 cm/s, how high will it rise?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0228.htm

Solution:

We use the principle of energy conservation to find the height h. Assume that the level of pro-jection is the position of zero potential energy. Then at the point of projection the total energy is purely kinetic E = 0 + (1/2) Mv^2 = (1/2)M ×(10^6 cm^2/s^2) At maximum height v = 0, and the total energy is now purely potential, hence E =Mgh. By equating the two expressions for E, we have Mgh= (1/2)M × (10^6 cm^2/s^2) h = [(1/2) × 10^6 cm^2/s^2] / g = [10^6 cm^2/s^2] / [2(980 cm/s^2] h = .51 × 10^3 cm h = 510 cm.

Question:

What weight of ice could be melted at 0\textdegreeC by the heat liberated by condensing 100 g of steam at 100\textdegreeC to liquid. Heat of vaporization = 540 cal/g, heat of fusion = 80 cal/g.

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Solution:

The quantity of heat necessary to convert 1 g of a liquid into a vapor is termed the heat of vaporization. For water, 540 calories are necessary to change liquid water at 100\textdegreeC into vapor at 100\textdegreeC. In this problem vapor is condensed, thus 540 cal of heat are evolved for each gram of liquid condensed. no. of cal evolved in condensation = 540 cal/g × weight of vapor Here 100 g of vapor is condensed. Thus, no. of cal evolved = 540 cal/g × 100 g = 54000 cal When ice melts, heat is absorbed. About 80 calories of heat are required to melt 1 g of ice, the heat of fusion. Here, 54000 cal are evolved in the condensation. Therefore, this is the amount of heat available to melt the ice. Because 80 cal are needed to melt 1 g of ice, one can find the number of grams of ice that can be mfelted by 54000 cal, by dividing 54000 cal by 80 cal/g. no. of grams of ice melted = (54000 cal) / (80 cal/g) = 675 g.

Question:

When is a condition expression used in COBOL? Give the rules for comparison of numeric or nonnumeric items.

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Solution:

Conditional Expressions: Conditional expres-sions are expressions used in situations where need of test-ing any condition arises. The outcome of a test determines the next logical step to be performed. The conditional expression reduces to a single value, "true" or "false." The truth or falsity is determined by a relational test. This test is based on a comparison of characters. Rules of Comparison: Characters are compared and evaluated on the basis of a computer collating sequence in which the characters have a specified order of magnitude. This collating sequence varies from machine to machine. As an example, the ordering scheme of fig. 1 will be used in the following discussions. Items declared numeric in the DATA DESCRIPTION section cannot be compared with items declared to be nonnumeric. Comparison of Numeric Items: The comparison of numeric items is based on respective values of the item considered purely as algebraic values. The item length, in terms of the number of digits is not itself significant. e.g., comparison of +00003 with +03 results in an equal condition, and +01 is greater than -155. Comparison of Nonnumeric Items: For two nonnumeric items, possibly containing numeric characters, a comparison results with respect to the ordered character set. There are two cases: a) Equal Length Items: If items are of equal length, comparison proceeds by comparing characters in corresponding character positions. Starting from the high-order end, com-parison continues until either a pair of unequal characters is encountered or the lower order end of the item is reached. As an example, consider : 1ABCD and ABBD results in inequality ABBD > ABCD . 2SAM and SAM results in equality. 3NAME 1 and NAME 2 results in inequality, NAME 1 < NAME 2. b) Unequal Length Items: If the items are of unequal length, comparison proceeds as above. If the process exhausts the characters of the shorter item without detection of differ-ence, the shorter item is Less Than the longer item. e.g.,comparingABC & ABCD, every character A, B and C of the first item equals every character of the later item but due to the shorter length of ABC the result is ABC < ABC.

Question:

Describe how J-K, T and D type flip-flops can be derived starting from the S-C type flip-flop, and obtain excita-tion equations for them from the excitation equation of the S-C type flip-flop.

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Solution:

a) J-K type flip-flop can be obtained from an S-C type flip-flop by adding two AND-gates as shown in fig. 1. The input constraint S\bulletC = 0 of the S-C flip-flop is automatically achieved in the J-K flip-flop, as shown in figure 1, because of the AND-gates. And thus, the J and K inputs need not worry about any input constraints. Both J and K can be = 1. The excitation equation of the J-K flip-flop is obtained by substitution in the excitation equation for the S-C flip-flop as follows: y\rightarrow= SY' + c'Y \rightarrow y\rightarrow= (JY')Y' + (KY)'Y (substitution) \rightarrow y\rightarrow= JY'Y' + (K' +Y')Y (by DeMorgans law) \rightarrow y\rightarrow= JY' + K'Y + Y'Y (distributive law) \rightarrow y\rightarrow= JY' + K'Y \rightarrow b) A T type flip-flop is obtained by shorting together the J and K inputs of the J-K flip-flop obtained in (a) as shown, in fig. 2. The excitation equation of the T flip-flop is obtained from the excitation equation of the J-K flip-flop by putting J = K = T \therefore Y\rightarrow= JY' + K'Y(Now put J = K = T) \rightarrow \therefore Y\rightarrow= TY' + T'Y \rightarrow = T + Y c) A D type flip-flop can be obtained from an S-C type flip-flop by giving mutually inverted inputs to the S and C inputs, as shown in Figure3: The inverter automatically ensures that both S and C will never be equal to 1 at the same time. The excitation equation is obtained from the excitation equation of the S-C flip-flop as follows: y\rightarrow= Sy' + C'y = Dy' + (D')'y \rightarrow \therefore y\rightarrow= Dy' + Dy = D (y' + y) = D \textbullet 1 = D. \rightarrow

Question:

Indicate how you would prepare (1) acetone from acetic acid and (2) acetic acid from ethyl alcohol.

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Solution:

This problem is concerned with synthesis of organic compounds. In such problems you want to write out the structures of the initial and final products involved, so that you can identify functional groups. From this, you employ reactions that change the functional groups until the desired product is obtained. If you had calcium acetate, Ca, you could obtain acetone by heating it, since calcium acetate forms calcium carbonate and acetone upon the addition of heat. To obtain calcium acetate, add Ca(OH)_2, which is a base. Because it is a base, the acid reacts with it to produce the calcium acetate, which is the salt, and water. Thus, the sequence of reactions becomes Ethyl alcohol : CH_3CH_2OH. To prepare acetic acid from ethyl alcohol, you must know one important fact; Ethyl alcohol is a primary alcohol, i.e., one where the OH group is located on a carbon atom bonded to only one other carbon atom. Primary alcohols can be directly oxidized to the corresponding carboxylic acid by the addition of KMnO_4, potassium permanganate. Thus, you perform the following reaction:

Question:

Represent the exclusive-OR function, using a) NAND gates only b) NOR gates only Sketch the circuit realizations in each case.

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Solution:

a) Exclusive-OR realization using NAND gates only: The exclusive-OR function can be written as A × B = A \bullet B' + A' \bullet B [Using the Sum of Products form] = [(A\bulletB'+A' \bulletB)' ]' [Complementing Twice] = [(A\bulletB') ' \textbullet (A'\bulletB) ' ]' [Applying deMorgan's Rules on the inner primed bracket] = [(A\uparrowB')\uparrow(A'\uparrowB)] [Using the Shaeffer stroke notation]. We assume that both unprimed variables (A,B) and primed variables (A', B') are available as inputs. Note that the outer bracket represents a NAND function of the two inner brackets. Each inner bracket is a NAND function of two variables each. The circuit realization is shown in figure 1. b) Exclusive-OR realization using NOR gates only. We can write: A × B = (A+B) \textbullet (A'+B')[Using a Product of Sums form] = [ ((A+B) \textbullet (A '+B'))' ]'[Complementing twice ] = [ (A+B)' + (A'+B')']'[Applying DeMorgan's Rules on the inner primed bracket] = [(A\downarrowB')\downarrow(A'\downarrowB')][Using the Pierce Arrow notation]. Observe that in the last two steps the outer square bracket represents a NOR function of the two inner brackets, each of which is a NOR function of two variables. The circuit realization using NOR gates only is shown in figure 2.

Question:

Design a buffer system that will function in the low, near neutral, and high pH levels, using combinations of the three sodium phosphates (Na_3PO_4, Na_2HPO_4, NaH_2PO_4) and phosphoric acid (H_3PO_4). The equilibrium constants for this polyproticacid are k_1 = 7.5 × 10^-3, k_2 = 6.2 × 10^-8, and k_3 = 4.8 × 10^-13. With this information, calculate the pH at both extremes of the buffer, assume 10 : 1 and 1 : 10 ratios, and at the mid-range of the buffer, assume a 1 : 1 ratio. Assume, also, that the acid-salt ratio, in making up the buffer is equal to HA/A^- in solution.

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Solution:

Solutions that contain appreciable amounts of a weak acid, such as phosphoric acid, and its salt or salts, such as sodium phosphates, are called buffers. Their utility is in maintaining a relatively constantpH.This problem requires one to find various pH values for this buffer system. To do this, note that the pH of a buffer is given by pH =pk_a+ log ([A^-] / [HA]) =pk_a+ log ([salt] / [acid]). One knows that to find the pH, one must determine [H_3O^+], since pH = - log [H_3O^+]. To do this consider the ionization of thepolyproticacid, phosphoric acid, which can dissociate to produce 3 protons, k_1, the first ionization constant, corresponds to H_3PO_4 + H_2O \rightleftarrows H_3O^+ + H_2PO_4^- so that k_1 ={[H_3O^+][H_2PO_4^-]} / [H_3PO_4]. k_2, the second ionization constant, corresponds to H_2PO_4^- + H_2O \rightleftarrows H_3O^+ + HPO_4^2- so that k_2 = {[H_3O^+][ HPO_4^2-]} / [H_2PO_4^-]. k_3, the third ionization constant, corresponds to HPO_4^2- + H_2O \rightleftarrows H_3O^+ + HPO_4^3- and k_3 = {[H_3O^+][HPO_4^3-]} / [HPO_4^2-]. In general, the k_a of any acid is given by the expression k_a = {[H_3O^+][A^-]} / [HA]or(H_3O^+) = k_a ([HA] / [A^-]) . It is from this equation that the [H_3O^+] must be calculated, and thus yield the pH value wanted. The three systems: Buffer system with k_1 Consider the ratios of 10 : 1, 1 : 1, and 1 : 10 for the acid : salt ratios, given that the ratio equals [HA]/[A^-]for the low, neutral and high pH values, re-spectively. This means that [H_3O^+] = 10 k_a , k_a and 0.1 k_a for low, neutral, and high, respectively. Here, k_a = k_1 which is = 7.5 × 10^-3.pk_1 = 2.12. Since [H_3O^+] = 10 k_a , k_a , and 0.1 k_a ; pH = - log [H_3O^+] = (pk_a- 1),pk_aand (pk_a+ 1) for low, middle, and high pH values, re-spectively. Thus, with pk_1 = 2.12, pH = 1.12 (low), 2.12 (middle) and 3.12 (high). Buffer system with k_2 One still has pH = (pk_a- 1),pk_a, andpk_a+ 1. But now k_a = k_2 = 6.2 × 10^-8 pk_2 = 7.21, Thus, pH = 6.21 (low), 7.21 (middle), and 8,21 (high). Buffer system with k_3 One still has pH = (pk_a- 1),pk_a, andpk_a+ 1. Now, k_a = k_3 = 4.8 × 10^-13 pk_3 = 12.32. Thus, pH = 11.32 (low), 12.32 (middle), and 13.32 (high).

Question:

From collision theory, it is found that when the rate- determining step involves collision of two A molecules, the rate will be proportional to the square of the A concentra-tion. Explain why.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0453.htm

Solution:

Collision theory makes the assumption that for a chemical reaction to take place, particles must collide. The rate will depend upon the number of collisions per unit time between the particles involved and the fraction of these collisions that are effective. If you increase [A], there are more A molecules which can be hit. Also, the average A molecule gets hit more often. As such, the number of A with A collisions varies as the square of concentra-tion due to this double effect.

Question:

How many amperes will a 720-watt electric iron take from a 120-volt line?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0783.htm

Solution:

P = 720 watts,E = 120 volts P = E × I Then I = (P/E) = (720 watts/120 volts) = 6 amp

Question:

An object is 24 inches from a convex lens whose focal length is 8 inches. Where will the image be?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0872.htm

Solution:

The use of a ray diagram is helpful here. We can see that the image is real (on the far side of the lens and inverted). We then solve mathematically: D_0 is 24 inches and f is 8 inches. Substituting in 1/D_0 + 1/D_I = 1/f 1/24 in. + 1/DI= 1/8 in. Solving D_I = 12 in. Hence D_I = 12 in. which means that the image is 12 inches from the lens on the side away from the object.

Question:

Why is the mitochondria referred to as the "powerhouse of the cell " ?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0036.htm

Solution:

Mitochondria are membrane-bounded organelles concerned principally with the generation of energy to support the various forms of chemical and mechanical work carried out by the cell. Mitochondria are distributed throughout the cell, because all parts of it require energy. Mitochondria tend to be most numerous in regions of the cell that consume large amounts of energy and more abundant in cells that require a great deal of energy (for example, muscle and sperm cells). Mitochondria are enclosed by two membranes. The outer one is a continuous delimiting membrane. The inner membrane is thrown into many folds that extend into the interior of the mitochondrion. These folds are called cristae. Enclosed by the inner membrane is the ground substance termed the matrix (see accompanying diagram). Many enzymes involved in the Krebs cycle are found in the matrix. Enzymes involved in the generation of ATP by the oxidation of NADH_2 , or, the electron transport reactions, are tightly bound to the inner mitochondrial membrane. The enzymes for the specific pathways are arranged in sequential orders so that the products of one reaction do not have to travel far before they are likely to encounter the enzymes catalyzing the next reaction. This promotes a highly efficient energy production. The reactions that occur in the mitochondria are all related in that they result in the production of ATP (adenosine triphosphate), which is the common currency of energy conversion in the cell. Some ATP is produced by reactions that occur in the cytoplasm, but about 95 percent of all ATP produced in the cell is in the mitochondria. For this reason the mitochondria are commonly referred to as the powerhouse of the cell.

Question:

It is estimated that 3 × 10^5 tons of sulfur dioxide, SO_2, enters the atmosphere daily owing to the burning of coal and petroleum products. Assuming an even distribution of the sulfur dioxide throughout the earth's atmosphere (which is not the case) , calculate in parts per million by weight the concentration of SO_2 added daily to the atmosphere. The weight of the atmosphere is 4.5 × 10^15 tons. (On the average, about 40 days are required for the removal of the SO_2 by rain).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0013.htm

Solution:

Here, one is asked to find the number of tons of SO^2 per 10^6 tons, i.e. per million, of atmosphere. This is done by using the following ratio: Let x = no. of tons of SO_2 per 10^6 tons of atmosphere. (3.0 × 10^5 tons SO_2 ) / (4.5 × 10^15 tonsatm) =( x / 10^6 tonsatm) x = (3.0 × 10^5 tons SO_2 ×10^6 tonsatm) / (4.5 × 10^15 tonsatm) x = 6.67 × 10\Elzbar5tons SO_2 or 6.67 × 10\Elzbar5ppmSO_2 .

Question:

What is the significance of hydrophobic and hydrophilic moieties in biological systems?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0026.htm

Solution:

All cells possess a cell membrane, which serves to regulate the materials entering and leaving the cell, and to give the cell shape and protection. The regulation of the movement of materials across the cell membrane (which materials and how much of each) is referred to as selective permeability. Since the cell's internal and external environment are mainly aqueous, the cell membrane must have a unique structure to be able to allow for the passage of water-soluble as well as water- insoluble materials. Water-soluble substances are called hydrophilic ("love of water") and water-insoluble substances are called hydrophobic ("fear of water"). The cell membrane is composed of a double layer of phospholipid with protein molecules dispersed throughout the membrane. Phospholipids and proteins are molecules that have both a hydrophilic and a hydrophobic portion. It is precisely these qualities which bestow the selective permeability property to the cell membrane. The hydrophilic portions (the polar or charged portions) are arranged so that they face the inside or outside of the cell. The hydrophobic portions are arranged so that they lie within the cell membrane. The polar phosphate moiety of the phos-pholipid is hydrophilic and thus faces the inner or outer surface. The fatty acid tails, the hydrophobic regions of the phospholipid, are oriented towards the middle of the bilayer. Membrane proteins are classified as peripheral (extrinsic) or integral (intrinsic) Peripheral proteins are on the inner or outer surface, exposed to the aqueous solution. Integral proteins may span the entire membrane or just the bilayer. Only the polar sections face the surfaces. Note: The above model is not a monomeric unit of the cell membrane. Cell membranes have diversity both externally and internally.

Question:

Suppose a body of mass 0.5 kg slides down a track of radius R^\ding{217} = 1 m, like that in the Fig., but its speed at the bottom is only 3 m/sec. What was the work of the frictional force acting on the body?

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Solution:

Note that we cannot calculate the work done by friction from the definition of work because W_friction = \intF^\ding{217}_friction \bullet ds^\ding{217} and we do not know the functional form of F^\ding{217}_friction. Since work is a form of energy, our next thought may be to try to calculate the work done by the frictional force by using the principle of conservation of energy. At position 1, the mass is at rest, and its energy is equal to its potential energy E = mgR(1) In sliding down the track, some of this energy will be dissipated by friction and be transferred into heat energy Q and some will be transformed into the kinetic energy of the mass. Hence, at position 2, E = Q + (1/2)mv^2(2) Note that the potential energy of the mass at position 2 is zero, since this is our reference position for potential energy. Combining equations (1) and (2) mgR = Q + (1/2)mv^2 andmgR - (1/2)mv^2 = Q or(.5 kg) (9.8 m/s^2) (1 m) - (1/2)(.5 kg) (9 m^2/s^2) = Q 4.9 kg \textbullet (m^2/s^2) - 2.25 kg \textbullet (m^2/s^2) = Q Q = 2.65 Joules This is the heat energy produced by friction, and it is positive because heat energy has been gained by the system. Therefore, the work done by friction must be negative because this energy was lost to heat. Hence W_f = - Q = - 2.65 Joules

Question:

Compute the r.m.s. speed of the molecules of oxygen at 76.0 cm. Hg. pressure and 0\textdegree C , at which temperature and pressure the density of oxygen is 0.00143 gm/cm^3.

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Solution:

The pressure exerted on a wall by an ideal gas can be calculated as follows. In Fig. (a), the elastic collision of a molecule with the wall is shown. If the force exerted on the mole-cule during the collision is perpendicular to the wall, then the x and y components of the molecule's velocity are not affected. The collision is assumed to be elastic, therefore the z component of the velocity is reversed in direction but remains unchanged in magnitude v'_x= vx v'y= v_y v'_z = \rule{1em}{1pt}vz The z-component of momentum of the molecule changes by an amount ∆p_z = p'_z - p_z = m(v'_z - v_z) = - 2mv_z. Now, we need to know the number of molecules that strike the wall with Velocities v^\ding{217}, per unit time per unit area. Let us consider an in-finitesimal area dA on the wall and the molecules incident on it with velocities parallel to v^\ding{217}. These molecules will be included in the tube shown in Fig. (b). Molecules in the tube that strike the wall with a velocityv^\ding{217} in time dt must be at most a distance l = vdt from the wall. Therefore, if we restrict the length of the tube to vdt, all the molecules inside it with velocity v^\ding{217} will strike the wall within the time dt. The molecules lying outside this tube with the same velocity v^\ding{217} will not strike dA during the same period. Since the molecules move in random directions, we expect that the number of molecules per unit volume, moving with velocity v^\ding{217}, does not depend on the direction of v^\ding{217} but rather on its magnitude v. If the density of the molecules with speed v is n(v), the total number of the molecules in the tube with speed v is = n(v) × volume = n(v) × vdt dA cos \texttheta = n(v) v_z dt dA . Because of the randomness of the motion, the average number of mole-cules moving toward the wall with velocity v^\ding{217} is equal to the average number moving away from the wall with velocity -v^\ding{217}. Therefore, the number of the molecules in the tube striking the wall is half the number we found for those with speed v; dN(v^\ding{217}) = (1/2)n(v) v_z dt dA. Therefore, the total number of particles incident on an area dA per unit time is [dN(v^\ding{217})]/[dt] = (1/2)n(v) v_z dA. The total rate of change of momentum is the number of collisions per unit time multiplied by the momentum change in each collision. Using Newton's Second Law. dF_z = [(dp_z)/(dt)] = (1/2) n(v) v_z (\rule{1em}{1pt}2mv) dA = - mv_z^2 n(v) dA where dF_z is the element of force which the area dA exerts on the molecules which strike it. The reaction force exerted on the wall by the molecules, using Newton's third law, is dF = -dFz thus the pressure on the wall becomes p = -(dF_z/dA) = mv_z^2 n(v). Now we have to find the total average pressure, since not all mole-cules have the same v^\ding{217}. For this, we re-express the average of the quantity v^2_z n(v) as < v^2_z n(v) >_av =v^2_zn wherenis the average number of particles per unit volume; n = (total number)/(total volume) = N/V. The average of v^2, for a gas, is the sum of the averages of its com-ponents v^2 =v^2_x+v^2_y+v^2_z. The motion is completely random, therefore all the directions of motion are equally probable. Therefore we may take v^2_x=v^2_y=v^2_z= (1/3)v_2 which gives p = (1/3)mv^2(N/V). In this problem, the pressure of oxygen is the weight of a 0.76 m. mercury column with unit area; p = volume × density × g = h\rho_Hg^g = 0.76 m × 13.6 × 10^3 kg/m^3 × 9.8 m/sec^2 = 1.01 × 10^5 nt/m^2. Pressure can be expressed as p = (1/3)v^2(mN/v) = (1/3)v^2[(M_total)/(v)] = (1/3)v^2\rho_ox givingv^2= 3p/\rho_ox Hence, the r.m.s speed is v^2= (3 × 1.01 × 10^5 nt/m^2)/(1.43 kg/m^3) = 2.12 × 10^5 m^2/sec^2 v_rms = \surd(v^2) = 461 m/sec.

Question:

A current-carrying wire in the form of a semicircle lies in a plane at right angles to the direction of a uniform magnetic induction. Show that the force on the wire is the same as that experienced by a straight wire lying along the diameter between the ends of the semicircle.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0730.htm

Solution:

Let the semicircle carrying a current i have radius r, and let the magnetic induction have magnitude B. Consider Fig. A, in which a vector element d1^\ding{217} of the current-carrying wire is shown in a direction from the center 0 of the semicircle making an angle \texttheta with the radius of symmetry. The force on that element due to the magnetic field is dF^\ding{217} = i(d1^\ding{217} × B^\ding{217}). Here d1^\ding{217} and B^\ding{217} are at right angles, and thus dF^\ding{217} lies perpendicular to both d1^\ding{217} and B^\ding{217}, or along the radius from d1^\ding{217} to 0. Also dF^\ding{217} has components, in the x- and y- directions shown, of - i d1B sin \texttheta and - i d1B cos \texttheta, respectively(see figure B). But there is a corresponding element d1^\ding{217} in a direction from 0 making an angle - \texttheta with the radius of symmetry. This element is subjected to a force which has x- and y-components of + i d1B sin \texttheta and - i d1B cos \texttheta, re-spectively. The x-components of the forces of these two elements thus cancel out, and this is true of all pairs of elements chosen at all possible angles \texttheta on the semicircle. It follows that the total force F^\ding{217} on the semicircle has a y-component only. Now dl/r = d1/r = d\texttheta radians or d1 = rd\texttheta. Also, when 1= 0, \texttheta = \pi/2 andwhen 1 = unity, 0 = + \pi/2. Therefore, F = ^1\int_0dF_y= F = ^1\int_0 - i d1B cos \texttheta = ^(\pi/2)\int_(-\pi/2)- ird\textthetaBcos \texttheta = - iBr[ sin \texttheta ^(\pi/2)]_(-\pi/2) = - 2i Br . Consider the straight wire; the force F^\ding{217}' on it is F^\ding{217}' = i (2r^\ding{217} × B^\ding{217}). Thus F^\ding{217}' has magnitude 2iBr and points in the negative y-direction. This is the same force as that which acts on the semicircle. One may arrive at this result and a more general result by noting that a closed current-carrying loop lying in a plane at right angles to the magnetic field experiences no net force. If the semicircular loop is closed by allowing the current to return along a wire occupying the vacant diameter (or by any loop whatsoever lying in the plane and joining the two ends of the semicircle), the complete circuit so formed experiences no net force. It follows that the forces on the semicircle and on the return wire are equal and opposite. Since reversing the direction of the current reverses the magnetic force on it, it imme-diately follows that the force on the semicircle is equal to the force on any current-carrying conductor lying in the plane and having the same endpoints.

Question:

Outline briefly the events occurring in each stage of mitosis. Illustrate your discussion with diagrams if necessary.

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Solution:

Mitosis refers to the process by which a cell divides to form two daughter cells, each with exactly the same number and kind of chromosomes as the parent cell. In a strict sence, mitosis refers to the division of nuclear material (karyokinesis). Cytokinesis is the term used to refer to the division of the cytoplasm. Although each cell division is a continuous process, in order for it to be studied, can be artificially divided up into a number of stages. We will describe each stage separately, beginning with interphase. 1) Interphase:This phase is called the resting stage. However, the cell is "resting" only with respect to the visible events of division in later phases. During this phase, the nucleus is metabolically very active and chromosomal duplication is occurring. During interphase, the chromosomes appear as vague, dispersed thread-like structures, and are referred to as chromatin material. 2) Prophase:Prophase begins when the chromatin threads begin to condense and appear as a tangled mass of threads within the nucleus. Each prophase chromosome is composed of two .identical members resulting from duplication in interphase. Each member of the pair is called a chromatid. The two chromatids are held together at a dark, constricted area called the centromere. At this point the centromere is a single structure. The above events occur in the nucleus of the cell. In the cytoplasm, the centriole (a cytoplasmic structure involved in division) divides and the two daughter centrioles migrate to opposite sides of the cell. From each centriole there extends a cluster of raylike fila-ments called an aster. Between the separating centrioles, a mitotic spindle forms, composed of protein fibrils with contractile properties. In late prophase the chromosomes are fully contracted and appear as short, rod-like bodies. At this point individual chromosomes can be distingushed by their characteristic shapes and sizes. They then begin to migrate and line up along the equatorial plane of the spindle. Each doubled chro-mosome appears to be attached to the spindle at its centromere. The nucleolus (spherical body within the nucleus while RNA synthesis is believed to occur) has been undergoing dissolution during prophase. In addition, the nuclear envelope breaks down, and its disintegration marks the end of prophase. 3) Metaphase:When the chromosomes have all lined up along the equatorial plane, the dividing cell is in metaphase. At this time, the centromere divides and the chromatids become completelty separate daughter chromo-somes. The division of the centromeres occurs simulta- neously in all the chromosomes. 4) Anaphase:The beginning of anaphase is marked by the movement of the separated chromatids (or daughter chromosomes) to opposite poles of the cell. It is thought that the chromosomes are pulled as a result of contraction of the spindle fibers in the presence of ATP. The chromosomes moving toward the poles usually assume a V shape, with the centromere at the apex pointing toward the pole. 5) Telophase:When the chromosomes reach the poles, telophase begins. The chromosomes relax, elongate, and return to the resting condition in which only chromatin threads are visible. A nuclear membrane forms around each new daughter nucleus. This completes karyokinesis, and cytokinesis follows. The cytoplasmic division of animal cells is accom-plished by the formation of furrow in the equatorial plane. The furrow gradually deepens and separates the cytoplasm into daughter cells, each with a nucleus. In plants, this division occurs by the formation of a cell plate, a partition which forms in the center of the spindle and grows laterally outwards to the cell wall. After the cell plate is completed, a cellulose cell wall is laid down on either side of the plate, and two complete plant cells form.

Question:

A rectangular tank 6.0 by 8.0 ft is filled with gasoline to a depth of 8.0 ft. The pressure at the surface of the gasoline is 14.7 lb/in^2. (The density of gasoline is 1.325 sl/f^3). Find the pressure at the bottom of the tank and the force exerted on the bottom.

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Solution:

The total pressure at the tank's bottom is the sum of the air pressure at the surface of the fluid and the pressure due to the gasoline above the tank bottom: P_air = 14.7 lb/in^2 Since 1 lb/in^2 = 144 lb/f^2 P_air = 14.7 lb/in^2 = (14.7 ) (144 lb/f^2) = 2120 lb/f^2(1) To find the pressure on the bottom of the tank due to the gasoline, we note that the pressure is equal to the force on the bottom of the tank divided by the area of the bottom P_gas = F/A P_gas = F/A But F_gas = \rhogV where \rho is the density of gasoline, g is 9.8 m/s^2, and V is the volume of the gasoline in the tank. Hence P_gas = \rhogV/A But V = hA, where h is the height of thegasoline in the tank. Therefore P_gas = \rhogh = (1.313 sl/f^3)(32 f/s^2)(8 f) P_gas = 336 lb/f^2(2) Hence, using (1) and (2) P_total = (336 + 2120)lb/f^2 P_total/Noting that = 2456 lb/f^2 P_total = F_total/A andF_total = P_total A = (2456 lb/f^2)(48 f^2) = 117888 lb

Question:

When we hold our head upright, there is very little problem keeping our balance. Yet when our head is turned upside down , as in a head stand, we still manage to maintain our equilibrium . How is balance achieved in the latter case? Discuss your answer in relation to the ear.

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Solution:

In the inner ear, there are two small, hollow sacs known as the utriculus andsacculus. Each sac is lined with sensitive hair cells, upon which rest small crystals of calcium carbonate (otoliths). Normally when the head is upright, the pull of gravity causes theotolithsto press against particular hair cells, stimulating them to initiate impulses to the brain via sensory nerve fibers at their bases. Changes in the position of the head cause these crystals to fall on and stimulate some other hair cells, thereby giving the brain a different set of signals. The relative strength of the signals to the brain indicates the position of the head at any given moment . When a head stand is being done, theotolithsin theutriculusand sacculus are reoriented by gravity and come to rest on other hair cells. These hair cells are stim-ulated and send impulses to our brain, informing us that now our head is in a new position. The brain will then transmit the appropriate response to the effectors so that we can maintain our balance even when our head is upside down. It is important to realize that equilibrium in humans depends upon the sense of vision, stimuli from theproprio-ceptors, and stimuli from cells sensitive to pressure in the soles of the feet in addition to the stimuli from the organs in the inner ear. This is why in certain types of deafness, in which the cochlea as well as the equilibrium organs of the inner ear are impaired , the sense of equi-librium is still present.

Question:

Ethyl chloride can be chlorinated to give two isomers, A and B. A can be chlorinated to give two isomers of C_2H_3Cl_3, while B gives only one isomer of C_2H_3Cl_3 upon further chlorination. Indicate the equations for the chlorination of ethyl chloride and identify A and B structurally.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0754.htm

Solution:

You are told about isomerism in the compounds A and B, and their chlorinated products. When two compounds are isomers they have the same molecular formula, but differ in the sequence of linkages of atoms. These are structural isomers. Ethyl chloride has the structural formula The equation for chlorination is its reaction with Cl_2 in the presence of light. Note that there exist two possible carbons on which the chlorines can bond. These are the two isomers A and B. Thus you can write The question is which one is A, as the other must be B. You are told that A gives 2 isomers, while B yields 1, if they are chlorinated again. Let's start with If you chlorinate it, you would obtain These are two different compounds in the arrangement of atoms. They are structural isomers. Thus, isomer A must be If this is correct, then the chlorination of B should yield only one product. Let's see if this is true. and is the same as which means B yields only 1 isomer.

Question:

Coal drops at the rate of 25 slugs per second from a hopper onto a horizontal moving belt which transports it to the screening and washing plant. If the belt travels at the rate of 10 ft per second, what is the horsepower of the motor driving the belt? Assume that 5% of the energy available is used in overcoming friction in the pulleys. (See figure.)

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0312.htm

Solution:

We take the belt and hopper as our system. By using the principle of conservation of momentum, we can relate the net external force on the system (provided by the motor) to the time rate of change of the system momentum P^\ding{217},or (dP^\ding{217})/(dt) = F^\ding{217}_ext(1) We examine a small element of mass, \Deltam, leaving the hopper and falling on the belt. (See figure.) At t = 0, the momentum of \Deltam in the x direction is P_0^\ding{217} = (\Deltam) (0) and, at t = \Deltat, it is P_f^\ding{217} = \Deltam u^\ding{217} Hence\DeltaP^\ding{217} = P_f^\ding{217} - P_0^\ding{217} = \Deltam u^\ding{217} or\DeltaP^\ding{217}/ \Deltat = u^\ding{217}[(\Deltam/\Deltat)] Taking the limit as \Deltat \rightarrow 0, (dP^\ding{217}/dt) = u^\ding{217} [(dm/dt)](2) Comparing (2) with (1) F_ext^\ding{217} = u^\ding{217} [(dm)/(dt)] The power provided by the motor must be the time rate of change of the work, W, done on the belt by the motor. Hence, P = dW/dt = d/dt [\int F_ext^\ding{217} \bulletds^\ding{217}] where ds^\ding{217} is an element of path traversed by the belt. Assuming F_ext^\ding{217} = constant, and noting that ds^\ding{217} /dt = u^\ding{217}, we obtain P = (d/dt) (F_ext^\ding{217} \bullet \int ds^\ding{217}) = (d/dt) (F_ext^\ding{217} \bullet s^\ding{217}) P = F_ext^\ding{217} \bullet (ds^\ding{217}/dt) = F_ext^\ding{217} \bullet u^\ding{217} Since F_ext^\ding{217} and u^\ding{217} are parallel, we find, using(2) P = (F_ext)(u) = u^2 (dm/dt) P = (10 ft/s)2(25 s^1/s) = 2500 1b \bulletft/s(3) This power must be supplied by the motor to keep the belt moving at a uniform rate, assuming that none of it is dissipated in friction. Now, if 5% of the power supplied by the motor is used in overcoming friction, only 95% remains to power the belt. Since the power needed to move the belt is given by (3), we obtain, P = 95% P' orP' = (100/95) P = (100/95) (2500 lb \bullet ft/s) P' = 2631.6 ft \textbullet lb/s Since1 hp = 550 ft \textbullet lb/s P' = 4.8 hp Taking friction into account, the rate of working of the motor is 4 Taking friction into account, the rate of working of the motor is 4 .8 hp.

Question:

Consider the collision of 2 particles of mass M_1 and M_2 , that stick together after colliding. Let M_2be at rest initially, and let v^\ding{217}_1 be the velocity of M_1 before the collision. (1) Describe the motion of M = M_1 + M2after the collision. (2) What is the ratio of the final kineticenergy to the initial kinetic energy? (3) What is the motion of the center of mass of this system before and after collision? (4) Describe the motion before and after the collision in the reference frame in which the center of mass is at rest.

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Solution:

The basic principle used in solving a collision problem is the law of conservation of total momentum. This principle may be applied to any collision, so long as there are no external forces (forces due to the outside environment) acting on the system. From figures (1) and (2), we see that the initial momentum is M_1v^\ding{217}_1, and the final momentum is (M_1 + M_2)v^\ding{217}, and we obtain M_1v_1^\ding{217} = (M1+ M_2) v^\ding{217} v^\ding{217} = [M1/ (M_1 + M_2)] v_1^\ding{217} Hence, (M_1 + M_2) moves with velocity v^\ding{217}, parallel to v_1^\ding{217}. (2) The kinetic energy k_f after the collision is K_f = (1/2) (M1+ M_2)v^2 = (1/2) (M_1 + M_2) [(M1^2v1^2) / (M_1 + M_2)^2] Kf= [(M_1^2 + v_1^2)] / [2(M_1 + M_2)] The initial kinetic energy, k_i is (1/2) M_1v1^2, hence kf/k_i = (M_1^2v_1^2)/[2(M_1 + M_2)] \bullet [1/(1/2 M_1V_1^2)] = [M_1/(M_1 + M_2)] k_f/k_i = [1 - {M_2/(M_1 + M_2)}] The difference, k_i - k_f, is lost to increased internal motion in the (M_1 + M_2) system (i.e. internal excitations and heat). When a meteorite (M_1) strikes and sticks to the earth (M_2),essentially all the kinetic energy of the meteorite will be lost to heat in the earth. This follows from the fact that if M_2 >> M_1, k_f/k_i = [M_1/(M_1 + M_2)] = [1/{1 + (M_2/M_1)}] \approx0 Hence, k_f \approx0 and all the initial kinetic energy is transformed into heat. (3)The center of mass of a system of particles is a fictitious point whose motion is supposed to describe the trajectory of an imaginary bag which contains all the particles in that system. As the particles move around, the shape and the volume of this bag changes but not its momentum. The interactions of particles among themselves cannot result in a net-resultant force or change in momentum of the bag as a result of the action- reaction principle (for each force exerted on one particle by the others, there is an equal and opposite force exerted by this particle on the others). Furthermore, the collisions of particles with each other conserve the momentum of the colliding particles in each collision and cannot change the total momentum of the bag. In this way, we can view the motion of the center of mass as represent-ing the net effect of only the external forces on the system. If there are no external forces, then the center of mass will not change its velocity, irrespective of the final velocities of the particles. The position of the center of mass is given by R_cm^\ding{217} = (Sum of all m_iv_i^\ding{217})/(M_t) where m_i and r_i^\ding{217} are the masses and the positions of individual particles, M_t is the total mass. In our problem, let the collision take place at the origin of our coordi-nate system and at time t = 0. After the collision, the center of mass will just coincide with the mass (M_1 + M_2) and we have R_cm^\ding{217} = vt^\ding{217} = [M_1/(M_1 + M_2)] v_1^\ding{217}_ t The center of the mass velocity is v_cm^\ding{217} = (sum of all p_i^\ding{217})/(M_t) where p_i^\ding{217} are the individual moment. For our problem v_cm^\ding{217} = v^\ding{217}=[M_1/(M_1 + M_2)] v_1^\ding{217} This expression for the velocity of center of mass will be true for all times, i.e. also before the collision (for which t < 0). (4) In this reference frame v_cm^\ding{217} = 0. (This reference frame is attached to the center of mass of the system.) The new velocity of M_1 with respect to this observerwill be the velocity of M_1 in the old frame minus thevelocity of the c.m. frame with respect to the old frame or u_1^\ding{217} = v_1^\ding{217} - v_cm^\ding{217} = [M_2/(M_1+ M_2)] v_1^\ding{217} and, for M2 u_2^\ding{217} = v_2^\ding{217} -v_cm^\ding{217}= - v_cm^\ding{217} = - [M_1/(M_1 + M_2)] v_2^\ding{217} This result for u^\ding{217}_2 could be guessed right away.When the observer was in the old frame,M_2 was station-ary. As the observer moves with v^\ding{217}_cm With respect to this old frame, then with respect to this observer M_2 will appear to move in the opposite direction with equal speed: u_2^\ding{217} = -v_cm^\ding{217} . The total momentum p^\ding{217}_total in this system should add up to zero since v_cm^\ding{217} is zero; indeed we have p\ding{217}_total = M_1u_1^\ding{217} + M_2u_2^\ding{217} = [(M_1M_2)/(M_1 + M_2)] v^\ding{217}_1 - [(M_1M_2)/(M_1 + M_2-)] v_1^\ding{217} = 0 The advantage of the center of mass frame is that the total momentum in it is zero.

Question:

A uniform, spherical bowling ball is projected without initial rotation along a horizontal bowling alley. How far will the ball skid along the alley before it begins to roll without slipping? Assume that the ball does not bounce. (See figure).

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Solution:

The ball is projected along the alley at point A. When it hits the alley at point B, the ball will roll and slip. At point C, the ball has begun to roll without slipping. In order to find the distance at which skidding (or slipping) stops, we must find the acceleration of the ball, and then solve for its position. Applying Newton's Second Law to the horizontal direction of motion of the ball, we obtain ma_x = - f(1) where a_x is the horizontal acceleration of the ball of mass m. Note that we take a_x to be positive in the positive x-direction. Applying the Second Law to the vertical direction ma_y = N - mg But a_y , the y acceleration of the ball, is zero since the ball doesn't bounce. Therefore N = mg(2) In order to calculate torques,\cyrchar\cyrt^\ding{217}, about 0, we use \cyrchar\cyrt^\ding{217} \cyrchar\cyrt^\ding{217}=I\alpha^\ding{217}(3) \cyrchar\cyrt^\ding{217} \alpha^\ding{217} where\alpha^\ding{217}is the angular acceleration of the ball. But, the only torque acting \alpha^\ding{217} on the ball is due to f. Hence \cyrchar\cyrt^\ding{217}= r\ding{217}×f\ding{217} \cyrchar\cyrt^\ding{217} \ding{217} × \ding{217} where r\ding{217}is the location of the point of application of f\ding{217}. Since r\ding{217}and f\ding{217} \ding{217} \ding{217} \ding{217} \ding{217} are perpendicular (see figure), \cyrchar\cyrt = Rf From (3) Rf = I\alpha(4) \alpha where\alphais positive in a direction pointing into the plane of the figure.Now, \alpha Now, if f is constant, (1) tells us that a_x is constant, since a_x = (-f)/m - Hence, we may use the kinematics equations for constant acceleration to find v_x = ^vx_0 [(-f)/m] t(5) (-f)/m] t(5) - x = x_0 + ^vx_0 t [(-f)/(2m)]t^2(6) - where x_0 andvx_0are the initial position and velocity of the ball. Similarly, v x_0 we use (4) to solve for the angular velocity of the ball. \alpha= Rf/I \alpha But\alpha= d\omega/dt where\omegais the angular velocity of the ball. Therefore \alpha \omega \omega d\omega/dt = Rf/I \omega \omega\int_\omega_0 d\omega= Rf/I ^t\int_0 dt \omega \int_\omega \omega \int where\omega_0 is the angular velocity at t = 0. \omega \omega=\omega_0 + [Rf/I]t(7) \omega \omega While the ball is skidding, (7) and (5) are completely independent relations. However, when the ball starts rolling without slipping, they are related by v_x =\omegaR(8) \omega Substituting (7) and (5) in (8) vx_0-(f/m) t' =\omega_0R + [(R^2f)/I] t' v x_0-(f/m) t' =\omega_0R + [(R^2f)/I] t' - \omega But\omega_0 = 0. Solving for t' \omega vx_0 = t'[{(R^2f)/(I)} + (f/m)] v x_0 = t'[{(R^2f)/(I)} + (f/m)] t' = (^vx_0)/[ {(R^2f)/(I)} + (f/m) ](9) At t = t', slipping stops. To find the position of the ball when slipping stops, we substitute (9) in (6) x = x_0 -vx_0 [(^vx_0)/{(R^2f / I) + (f/m)}] - ( f/2m) [(^vx_0^2)/{(R^2f / I) + (f/m)}] v x_0 [(^vx_0)/{(R^2f / I) + (f/m)}] - ( f/2m) [(^vx_0^2)/{(R^2f / I) + (f/m)}] Furthermore, x_0 = 0 (see figure), and x =(^vx_0^2) / [f{(R^2/I) + (I/m)}]-[-f(^vx_0^2)]/[2mr^2{(R^2/I) + (I/m)}^2](10) (^vx_0^2) / [f{(R^2/I) + (I/m)}]-[-f(^vx_0^2)]/[2mr^2{(R^2/I) + (I/m)}^2](10) - - The frictional force law is f = u_kN(11) where N is the normal force of the alley on the ball, and u_k. is the coefficient of kinetic friction. (We use the coefficient of kinetic friction because the ball skids from t = 0 to t = t'.) Substituting (2) in (11) f = u_kmg Substituting this in (10), x = (^vx_0^2) / [u_kmg{(R^2/I) + (1/m)}]-(^vx_0^2) / [2u_km^2g{(R^2/I) + (1/m)}^2] - x = (^vx_0^2) / [u_kmg{(R^2/I) + (1/m)}]-[1 - (1 / [2m{(R^2/I) + (1/m)}])] - For a sphere I = 2/5 mR^2 and R^2/I = R2/ (2/5 mR^2) = 5/2m then x = (^vx_0^2) / [u_kmg{(5/2m) + (1/m)}]-[1-(1 / [2m{(5/2m) + (1/m)}])] - - x = [(^2vx_0^2) / (7u_kg)] [1-(1/7)] =(^12vx_0^2)/(49u_kg) - This is the position at which slipping stops.

Question:

Of the terrestrial arthropods, the insects are among the best adapted for the prevention of water loss. Explain.

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Solution:

Insects are a large group of mostly terrestrial arthropods. Their success of survival .on land is partly due to the adaptation of water conserving mechanisms. Their epicuticle, the outermost exoskeletal layer, is impregnated with waxy compounds. This serves to prevent loss of water by evaporation. Their wastes are excreted as uric acid, a crystalline substance, thus minimizing water loss due to protein metabolism. Uric acid is ex-creted via Malpighian tubules. Malpighian tubules lie more or less free within the hemocoel (blood sinus) , with the proximal end attached to the digestive tract at the junction of the midgut and hindgut (rectum). There may be 2 to 250 tubules. If there are only a few, the tubules are usually long, slender, and convoluted. If there are many, the tubules may be short and grouped in bunches. The cells lining the tubule lumen are cuboidal epithelial cells. The outer layer of the tubule wall is composed of elastic connective tissue and muscle fibers. The tubules are capable of peristalsis. Uric acid, formed by the tissues and released into the blood, is selectively absorbed by the tubule cells from the blood within the hemocoel. Salt, water, and amino acids are also absorbed. Together, the substances are discharged from the Malpighian tubules into the hindgut. Nutritive substances and some of the salts are reabsorbed by the rectal epithelium and are returned to the blood. Much water is also reabsorbed to conserve water inside the body. The uric acid is excreted with the waste products of digestion. There are several other methods of excretion in insects. Some excess salts and other substances are deposited in the cuticle and shed when the insect molts. Some particulate wastes are picked up and degraded intra-cellularly by nephrocytes, cells located on or near the heart.

Question:

What is wrong with each of the following gate representa-tions ?

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Solution:

a) This is an example of the incorrect assumption that a single variable may be controlled simultaneously from two or more elements whose outputs are inconsistent. Even though the definition of a gate structure is fulfilled \rule{1em}{1pt}\rule{1em}{1pt} a set of elements with given behaviors and a set of couplings between elements \rule{1em}{1pt} \rule{1em}{1pt} the contradictions inherent in the structure make its analysis meaningless. For example, if X_1 = X_2 = 0 and X_3 = X_4 = 1, then Y could be either 1or 0. b) Although a circuit may contain dependency loops, this circuit contains a closed loop of inconsistent dependency in which a variable Y depends upon another variable X in such a way that two different values of Y are required si-multaneously. This example fails for Y = 0 and X = 1 be-cause Y must also equal 1 now. c) Let us represent the output in terms of a pair of simul-taneous Boolean equations: Y_1 = X_1 \textbullet Y_2 Y_2 = X_2 + Y_1 When solving these equations for Y_1 and Y_2 by substitution, we obtain either of two sets of equations: Y_1 = Y_2 \textbullet X_1 Y_2 = X_2(a) or Y_1 = X_1 Y_2 = X_2 \textbullet X_1 If we begin with X_2 = 0 and X_1 = 1, set (a) gives Y_1 = Y_2 = 0, but set (b) gives Y_1 = Y_2 = 1. With this state, the response of the circuit is ambiguous.

Question:

An experiment to measure Avogadro's number involves the collection of a beam of alpha particles to which electrons are added to give neutral helium. Suppose that for a particular run it takes a number of electrons equivalent to 6.40 × 10^-5 coulomb/sec to neutralize the beam, and that after 24 hours of collection 1.14 × 10^-4 g of neutral helium are collected. From this data, determine Avogadro's number. The charge on one electron is 1.602 × 10^-19 coulombs.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0114.htm

Solution:

Remembering that it takes two electrons to neutralize one alpha particle, i.e., one positively charged He atom, we will calculate the total charge required during the 24 hour run and, from this, the number of electrons. The number of alpha particles neutralized, or the number of helium atoms produced, is then equal to one-half the number of electrons. Using the mass of helium collected and the atomic mass of helium (4.0026 g/mole), we will calculate the corresponding number of moles. Dividing the number of atoms by the number of moles will give the number of atoms per mole, or Avogadro's number. 24 hours is equivalent to 24 hours × 60 min/hr × 60 sec/min = 86,400 sec. Thus, in 24 hours it took 6.40 × 10^-5 coulomb/sec × 86,400 sec = 5.530 coulomb to neutralize the beam. Since the charge on one electron is 1.602 × 10^-19 coulomb, dividing this number into the total number of coulombs gives the equivalent number of electrons, or number of electrons = coulombs/ coulombs of electron = 5.530 coulombs / (1.602 × 10^-19 coulomb / electron) = 3.452 × 10^19 electrons . It takes two electrons to neutralize one alpha particle, hence the number of alpha particles neutralized by this number of electrons is number of alpha particles = number of electrons / 2 = (3.452 × 10^19) / 2 = 1.726 × 10^19 alpha particles. But one alpha particle produces one helium atom, so that the beam produced 1.726 × 10^19 helium atoms. The mass of helium corresponding to this number of helium atoms produced by the beam is given as 1.14 × 10^-4 g. Dividing by the atomic mass of helium, we convert this mass to moles: mass helium = mass helium / atomic mass helium = (1.14 ×10^-4 g) / ( 4.0026 g/mole) = 2.84 × 10^-5 mole. Hence, 1.726 × 10^19 helium atoms correspond to 2.84 × 10^-5 mole of helium. Dividing these two numbers gives Avogadro's number: Avogadro's number = number of atoms / moles = (1.726 × 10^19 ) / (2.84 × 10^-5 mole) = 6.08 × 10^23 atoms/mole. Comparing this with the currently accepted value of 6.02 × 10^23 atoms/ mole, we see that the estimate calculated is in error by [{(6.08 × 10^23) - (6.02 × 10^23)} / (6.02 × 10^23)] × 100% = [{( 6.08- 6.02 ) / (6.02 )} × 100% ] = (0.06 / 6.02) × 100% = 1% .

Question:

Write a program that finds, for integers m and n, the quotient and remainder when m is divided by n.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0546.htm

Solution:

The Fundamental Theorem of Arithmetic states that if m and n are any integers with n \not = 0, then there exist unique integers q and r, 0 \leq r < │n│ such that m =qn+ r. As examples: 1) m = 43 n = 5; using the division algorithm q = 8 r = 3 and m =nq+ r, or 43 = 8(5) + 3. 2) m = -43 n = -5 q=9 r = 2 m =nq+ r, or -43 = 9(-5) + 2 Here r = 2 < │n│ = │-5│ = 5 . 3) m = 8 n = 32; 8 = 0(32) + 8 . The following computer program accepts any integers m, n (n \not = 0) and computes the quotient and remainder of m/n. CINPUT INTEGERS M AND N. IF N IS CZERO, READ ANOTHER PAIR. 5READ (5,10) M,N IF (N.EQ.0) GO TO 5 10FORMAT (2I6) CFIND THE QUOTIENT IQ = INT (M/N) CFIND THE REMAINDER IR = M - IQ\textasteriskcenteredN COUTPUT RESULTS WRITE (6,60) M,N 60FORMAT (1HO, 10HDIVIDEND =, I6, 10H DIVISOR =, I8) 70WRITE (6,80) IQ, IR 80FORMAT (1H, 10HQU0TIENT =, I6, 12H REMAINDER =, I6) 90STOP END

Question:

What is the magnetic induction B at the center of a Circular cable consisting of 100 turns of wire having a common radius of 20 cm carrying 15 amperes?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0740.htm

Solution:

Consider the circular contour about the circular loop shown in The figure. By the Biot-Savart law the contribution to the magnetic induction at the center of the circular loop, due to an infinitesimal element, dl^\ding{217}, of the loop, is given by dB = (\mu_0 /4\pi) [(NI dl sin \texttheta)/a^2] where \mu_0 = 4\pi × 10^-7 Weber/A \bullet m is the constant of permeability, \texttheta is the angle between dl^\ding{217} and the radius vector a^\ding{217} (\texttheta = 90\textdegree), and NI is the total current through the loop. At the center of the loop, B is directed per-pendicular to the plane of the loop. B = \int dB = (\mu_0 /4\pi) (NI/a^2 ) \int dlfor sin 90\textdegree = 1 B = (\mu_0 /4\pi) (NI/a^2 ) 2\pia = (\mu_0 /2) (NI/a) We are given that N =100 turns, I = 15 amp and a = 20 cm × 1 m/100 cm then B = {2\pi × 10^-7 [W/(A-m)] × 100 × 15A}/(0.2 m) = 1.5 × 10^-4 \pi weber

Question:

Can you distinguish between two gene loci located on the same chromosome that have 50 percent crossing over and two gene loci each located on different chromosomes?

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Solution:

When two linked genes have 50 percent crossing over between them, it means that 50 percent of the progeny will be recombinants. For example, a heterozygous parent genotypically will form four will form four types of gametes if a cross-occurs between A and B: Now if the recombinant types occurred 50% of the time, this means that they are produced in numbers equal to the parental types. This could happen only if crossover occurred between the genes during every meiosis in every individual. In other words, if the genes are so far apart that it is certain that crossover will occur between them, then in any meiotic event, recombinant types will always be formed, and in equal numbers with parental types. Therefore, the four types of gametes occur in a 1:1:1:1 ratio. When testcrossed, the progeny will also occur in a 1:1:1:1 phenotypic ratio. This means that the ratio of the progeny from parents having linked genes showing 50% crossing over is indistinguishable from that of parents with genes that are not linked. Thus when genes are separated by 50 map units, it is impossible to differentiate between linkage and non-linkage. Note that no more than 50% recombination is ever expected between any two loci because only two of the four chromatids are involved in crossing over. In fact, the ratio is usually less than 50% because crossover may not always occur and because double or multiple cross-- overs can reduce the apparent number of crossover events, as shown below.

Question:

Explain briefly the background and applications of BASIC computer language.

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Solution:

BASIC (Beginner's All-purpose Symbolic Instruction Code). In the early 1960's, a project was undertaken at Dartmouth College to make the computer more accessible and easy to use by both students and faculty. One of the objectives was to develop a language that could be quickly learned but powerful enough to be used in solving most small to medium scale problems in any discipline. The result of this effort was BASIC, and it more than met its objective. Since BASIC provides the opportunity for someone with little or no previous computer experience to quickly begin writing programs and using the computer, it is commonly available on time sharing systems. BASIC's statements and operands are, probably, the closest to actual English language and algebra respectively and the easiest to comprehend, which ensures its huge popularity in almost every field of science, business, and engineering.

Question:

Calculate the number of grams of NH_4Cl that must be added to 2 l of 0.1 M NH_3 to prepare a solution of pH = 11.3 at 25\textdegreeC. The K_b for NH_3 = 1.8 × 10^-5. The equilibrium equation is: NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0436.htm

Solution:

To determine how many grams of NH_4Cl must be added to this solution to maintain a pH of 11.3, one must first calculate the [NH_4^+] needed, because [NH_4^+] = [NH_4Cl] in the solution. This is done by using the ex-pression that describes the dissociation, NH_3 + H_2O \rightleftarrows NH_4^+ + OH^-. The dissociation constant (K_b) for this reaction is 1.8 × 10^-5. K_b is equal to the product of the concentrations of the substances formed divided by the concentration of the reactants. These concentrations are raised to the power of their coefficients in the equation for the reaction. [H_2O] is not included because it is assumed to remain constant. K_b = {[NH_4^+] [OH^-]} / [NH_3] = 1.8 × 10-^5 . To solve for [NH_4^+], one must know [OH^-] and [NH_3] first. One is given that NH_3 solution is .1 M, thus [NH_3] = 1. One can find [OH^-] by using the definition ofpOH.pOH= 14.0 - pH = - log [OH^-]. The pH of this solution is 11.3, thereforepOH= 14.0 - 113. = 2.7. One solves for OH^- by using the relationpOH= - log [OH^-]. Solving for [OH^-] 2.7 = - log [OH^-] 10^-2.7 = [OH^-] 2.0 × 10^-3 = [OH^-] One can now use the expression for K_b to solve for [NH_4^+] 1.8 × 10^-5 = {[NH_4^+] [OH^-]} / [NH_3] = {[NH_4^+] [2.0 × 10^-3]} / [.1] [NH_4^+] = {(.1) (1.8 × 10^-5)} / (2.0 × 10^-3) = 9.0 × 10^-4 M One is given that 2 l of this solution has been made up. Because, Molarity= number of moles / volume in liters; number of moles =molarity× volume = [9.0 × 10^-4 (mole/liter)] (2 liters) = 18 × 10^-4 moles. 18 × 10^-4 moles of NH_4Cl must be added to the solution. Because moles = weight in grams/molecular weight weight in grams = moles × M.W. (MW of NH_4Cl = 53.5) = 18 × 10^-4 moles (53.5 g/mole) = 0.096 g of NH_4Cl to be added.

Question:

Determine the radii of the first and second quantum orbits in the hydrogen atom.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0668.htm

Solution:

We will derive the Bohr theoretical expression for the radius of the quantum orbit of an electron. Consider an electron of mass m and charge e moving at a constant distance r from the nucleus with a tangential velocity v, as shown in the accompanying figure. In order for this orbit to be stable, the Coulombic force attracting the electron to the nucleus (e^2/r^2) must be balanced by the centrifugal force on the electron (mv^2/r), which, if not compensated for, would compel the electron to leave its orbit. Hence, the magnitude of these two forces must be equal. (e^2 /r^2) = (mv^2/r) Solving for r, we obtain r = (e^2/mv^2) Bohr introduced quantization of the electron`s orbit by assuming that the electron's angular momentum (mvr) is given by: mvr = n (h/2\pi), where h is Planck's constant and n is an index, called the principal quantum number, which takes on the values n = 1 for the first orbit, n = 2 for the second orbit, and so on. Solving for v, v = [(nh)/(2\pimr)] We can now obtain an expression for r which includes quantization and is in terms of known quantities. Using the above expressions for v and r, r = (e^2/mv^2) = (e^2/m) [(2\pimr)/(nh)] = [(4\pi^2 e^2 mr^2)/(n^2 h^2)], r = [(n^2 h^2)/(4\pi^2 e^2 m)] The radius of the first orbit (n = 1), is called the Bohr radius. Solving for r : e = 4.80 × 10^-10 esu and m = 9.11 × 10^-28 g. r_1 = [(1^2 × h)/(4\pi^2 e^2 m)] = [{1 × (6.626 × 10^-27 erg-sec)^2} / {4\pi^2 × (4.80 × 10^-10 esu)^2 × 9.11 × 10^-28 g)}] = 5.29 × 10^-9 cm = 0.529 \AA. Let the radius of the nth orbit be denoted r_n. Then r_n = [(n^2 h^2)/(4\pi^2 e^2 m)] = n^2 [(h^2)/(4\pi^2 e^2 m)] = n^2 r_1 = n^2 × 0.529 \AA. Hence, the radius of the second quantum orbit is r_2 = n^2 × 0.529 \AA = (2)^2 × 0.529 \AA = 2.12 \AA.

Question:

Describe between an exoskeleton and an endoskeleton.

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Solution:

In most of the largemulticellularanimals, some kind of skeleton is found that serves for protection, support and locomotion. The skeleton of an animal may be located outside or inside the body. If the former is the case , the skeleton is termed an exoskeleton, and is usually composed of a semiflexible substance known as chitin. If the latter is the case, the skeleton is termed an endoskeleton, and is composed of either carti-lage bone , or both. The hard shells of lobsters, crabs, oysters,clanis, and snails are examples of exoskeletons. In fact, themolluscand arthropod phylums are charac-terized by the presence of an exoskeleton. The advantage of an exoskeleton as a protective device is obvious, but a serious disadvantage is the attendant difficulty of growth. Snails and clams solve this difficulty by secret-ing additions to their shells as they grow. Lobsters and crabs have evolved a complicated process, called molting, whereby the outer shell is first softened by the removal of some of its salts . A new similarly soft cuticle is secreted beneath the old shell. The old shell is then split down the back. Following this split, the animal crawls out of the old shell, grows rapidly for a short time, and thenredepositsthe removed mineral salts in the new shell, hardening it. During molting, the animal usually remains well-hidden, because this is a time when it is defenseless and easily killed by its enemies. Human beings and all other chordates have an endo-skeleton. The skeleton of sharks and rays is made of cartilage. The human skeleton consists of approximately 200 bones; the exact number varies at different periods of life as some bones, at first distinct, gradually become fused. The skeleton in man acts as a supporting framework for the other organs of the body. Since the skeleton is internal and grows with the organism, no problem such as that encountered in the growth of lobsters and crabs occurs.

Question:

Trace the movement of the integers 4, 3, 5, 1, 2 as they are sorted in ascending order with the bubble sort. Can you give a general formula for the time needed for sorting N numbers in the best case? In the worst case?

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Solution:

The bubble sort compares each pair of adjacent elements in a list. If in the proper order, the pair is left alone, and the next pair is examined. If out of order, the items are switched, and the next pair is examined. In this fashion, the larger elements are "bubbled\textquotedblright down the list, while the smaller elements move to the head of the list. Of course, if the list is to be sorted in descending order, the opposite process will occur. Let us illustrate the steps necessary to arrange the integers in ascending order: 1st comparison3, 4, 5, 1, 2Compare 3 and 4,no switch 2nd comparison3, 4, 5, 1, 2Compare 4 and 5,no switch 3rd comparison3, 4, 1, 5, 2Compare 5 and 1,switch them 4th comparison3, 4, 1, 2, 5Compare 5 and 2,switch - end of 1st pass 5th comparison3, 1, 4, 2, 5Compare 4 and 1,switch 6th comparison3, 1, 2, 4, 5Compare 4 and 2,switch - end of 2nd pass 7th comparison1, 3, 2, 4, 5Compare 3 and 1,switch 8th comparison1, 2, 3, 4, 5Compare 3 and 2,switch - end of 3rd pass In the best case, i.e., when the data are already order, the bubble sort makes only one pass through the list. No sorting method can do any better than this. However, in the worst case, the bubble sort makes a maximum of N (N - 1) / 2 comparisons, just like other sequential sorting methods. The worst case is realized when the data are in the opposite order from the order you wish to have. For example, if we wish to sort the integers 5, 4, 3, 2, 1 into ascending order, the bubble sort will make 10 switches. 4, 3, 2, 1, 54 switches 3, 2, 1, 4, 53 switches 2, 1, 3, 4, 52 switches 1, 2, 3, 4, 51switches 10 switches total

Question:

Explain the process of gaseous exchange in the leaf.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0255.htm

Solution:

A leaf that is actively carrying on photosynthesis must have a high rate of gas exchange with the environment to remove the excess oxygen and to acquire an adequate supply of carbon dioxide. As carbon dioxide is utilized in photosynthesis, its con-centration within the cell decreases and carbon dioxide diffuses into the cell from the film of water surrounding the cell. Other carbon dioxide molecules pass from the air spaces of the spongy layer of the leaf and dissolve in the water film. They will then diffuse into the cell as the cell rapidly uses up carbon dioxide. Still other carbon dioxide molecules diffuse from the air outside the leaf through the stomata and into the air spaces within the leaf. In short, carbon dioxide moves into the cell along a concentration gradient. Carbon dioxide molecules diffuse from a region of higher concentration such as the atmosphere, to a region of lower concentration, in this case, the cells of the leaf . In a similar manner but opposite direction, oxygen produced within the cells as a result of photo-synthesis passes from cell to water film to air spaces and through the stomata to the exterior by diffusion along a concentration gradient.

Question:

Calculate the molecular weight of a pure isoelectric protein if a 1% solution gives an osmotic pressure of 46 mm of H_2O at 0\textdegreeC. Assume that it yields an ideal solution.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0812.htm

Solution:

The relationship between molecular weight and osmotic pressure is M = (C/ \pi)RT, where M is the molecular weight, c the concentration in grams per liter, R the gas constant (0.082 liter atm deg-1 mole^-2) , T the absolute temperature, and \pi the osmotic pressure in atmospheres. One is given these quantities in the problem, but they are expressed in different units than those needed to solve for molecular weight. Converting these quantities: Concentration: The solution is 1% by weight protein. One liter of water weighs 1000 g; 1% of 1000 is 10 so that c = 10. Osmotic pressure: The osmotic pressure is 46 mm H_2O. To convert to atmospheres, 46 mm H_2O must be converted to mm Hg. mmHg= 46mm × [(.9970 density of H_2O) / (13.534 density of Hg)] =3.39. There are 760 mm Hg in 1 atm. Therefore, the number of atmospheres is equal to (3.39 mm Hg) / (760 mm Hg/atm) = 4.45 × 10^-3 atm Temperature: T = 0\textdegreeC + 273 = 273\textdegreeK. Solving for M: M = [(10 g / litre)] / [4.45 × 10^-3 atm] ×.082 (liter-atm/\textdegreek-mole) × 273\textdegreek = 50,300 g/mole.

Question:

Explain the action of the following procedure which in-cludes asubroutine procedure within itself. VERIFY_TEST_VALUES: PROCEDURE; DCL LENGTH FIXEDDEC(3); CALL GET_AND_TEST_INPUT; \textbullet \textbullet \textbullet \textbullet \textbullet CALL GET_AND_TEST_INPUT' \textbullet \textbullet \textbullet \textbullet \textbullet GET_AND_TEST_INPUT:PROCEDURE; AGAIN:GETLIST(LENGTH); IF LENGTH = 0 THEN GOTO L; IF LENGTH<0 \vert LENGTH>90 THEN DO; PUTLIST('ERROR', LENGTH);GOTOAGAIN; END; /\textasteriskcentered END OF DO GROUP \textasteriskcentered/ END GET_AND_TEST_INPUT; \textbullet \textbullet \textbullet \textbullet \textbullet CALL GET_AND_TEST_INPUT; L:ENDVERIFY_TEST_VALUES;

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Solution:

From three different points in the program, the subroutine procedureis called. This is called a subroutine procedure because no re-sult value is produced. This is what differentiates a sub-routine procedure from a functionprocedure, which initiates an evaluation of a result to be later usedby the main pro-gram. Each time the subroutine is invoked in the above ex-ample, it acquiresa new value length from the data card. Then, the subroutine checksif each length value is between 0 and 90. If the value is <0 or >90 then the PUT LIST statement is executed, andthe control goes to the statementlabelled'AGAIN'. Thus, a new value forlength is obtained. If a value of zero is obtained for length, it is an indicationthat data is over and the program returns to the statement labelledFINISH in the calling program. But, if the value of length lies between0 and 90, then the program does not go to the statementlabelled FINISH, nor does it enter the DO group. Instead, the program control goes tothe next statement after the DO group, which happens to be the last statementof the subroutine procedure. From here, control is transferred backto the calling procedure, which con-tinues with the other steps of the program.

Question:

What is the equilibrium concentration of oxygen gas in fresh water at 25\textdegree C when exposed to air at a pressure of 760 torr? The Henry's law constant for oxygen in water at 25\textdegreeC is 3.30 × 10^7torrs. Express your answer in milligrams of oxygen per liter.

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Solution:

Solving this problem requires the use of Henry's law. It is stated P_2 = X_2 , K_2 , where P_2 is the partial pressure,X_2 is the mole fraction, and K_2 is the Henry's law constant. The subscript, 2, indicates the solute (or component at lower concentration). Since P_2 = 760torrand K_2 = 3.30 × 10^7torrsare known, we can solve for X_2 or the mole fraction of O_2 in water. Thus, X_2 = (P_2 ) / (K_2 ) = (760torr) / (3.3 × 10^7torrs) = (moles of O_2 ) / [(moles of O_2 ) + (moles of H_2O)] The volume is assume to be 1 liter or 1000 grams of H_2O . The number or moles of H_2O is [(1000g) / (18.02g/mole)] 55.49 moles H_2O . In addition, the number of moles of O_2 is negligible in comparison with the number of moles of H_2O, thus the number of moles of H_2O is [(760torr) (55.49 moles H_2O)] / (3.3 × 10^7torrs) , = 1.28 × 10-^3 moles O_2 . The weight in milligrams of O_2 is (1.28 × 10-^3 moles ) (32g/mole) (10^3 mg/g) = 41 mg. Since there is 1 liter of H_2O , then the concentration of O_2 is 41mg/l .

Question:

Baking powder consists of a mixture of cream of tartar (potassium hydrogen tartrate, KHC_4 H_4 O_6, molecular weight = 188 g/mole) and baking soda (sodium bicarbonate, NaHC0_3, molecular weight = 84 g/mole). These two components react according to the equation KHC_4 H_4 O_6 + NaHCO_3 \rightarrow KNaC_4 H_4 O_6 + H_2 O + CO_2. How much baking soda must be added to 8.0 g of cream of tartar for both materials to react completely?

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Solution:

From the equation, we know that one mole of NaHCO_3 reacts with one mole of KHC_4 H_4 O_6. Hence, if we con-vert 8.0 g of KHC_4 H_4 O_6 to moles, we know how manymoles of NaHCO_3 must be added. Finally, all we need to do is to convert moles of NaHCO_3 to grams of NaHCO_3. In order to convert from grams to moles, we use the relationship moles = mass/molecular weight. The number of moles of KHC_4 H_4 O_6 in 8.0 g is moles= [(mass) / (molecular weight)] = [(8.0 g) / (188 g/mole)] = 4.3 × 10^-2 mole. Hence, we must add 4.3 × 10^-2 mole of NaHCO_3. Using the relationship mass = moles × molecular weight, we find that 4.3 × 10^-2 mole of NaHCO_3 corresponds to 4.3 × 10^-2 moles × 84 g/mole = 3.6 g. Hence, 3.6 g of baking soda must be added.

Question:

lf a muscle had to rely on its stored supply of ATP for contraction, it would be completely fatigued within a few twitches. Therefore, if a muscle is to maintainitscontractile activity, molecules of ATP must be made avail-able as rapidly asthey are broken down. What are the sources of ATP in a skeletalmuscle? What is oxygen debt?

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Solution:

There are three sources of ATP in muscle cells: 1)creatine phosphate, 2) substrate levelphos-phorylationduringglycolysisand the TCA cycle, and 3) oxidativephosphorylationin the mitochondria. In an exercised muscle, the increased levels of ADP and inorganic phosphateresulting from the breakdown of ATP ultimately act as positive modulatorsfor oxidativephosphorylation. However, a short time elapses beforethesemultienzymaticpathways begin to deliver newly formed ATP ata high rate. It is the role ofcreatinephosphate to provide the energy for ATP formation during this inter-val.Creatinephosphate can transfer its phosphategroup to a molecule of ADP, converting it to ATP and leaving a creatinemolecule behind. A single enzyme catalyzes this reversible reaction. During moderate exercise, the muscle cell is able to derive ATP fromsources other thancreatinephosphate. The source of ATP during moderateexercise comes from the complete oxidation of carbohydrate (e. g. glycogen) to carbon dioxide and water viaglycolysis, the Krebs (TCA)cycle,and the electron transport chain. During the unavoidable delaybefore adjustments of the respiratory and circulatory systems increasethe oxygen supply to the active muscles, some of the oxygen for aerobicrespi-ration may come fromoxymyoglobin.Myoglobinis a compoundin muscle which is chemically similar to hemoglobin. It forms a loosecombination with oxygen while the oxygen supply is plentiful and storesthe oxygen until the demand for it increases. During heavy muscularactivity, a number of factors begin to limit the cell's ability to replaceATP by oxidativephosphorylation: the delivery of oxygen to the muscle, the availability of substrates such as glucose, and the rates at whichthe enzymes in the metabolic pathways can process the substrates. Any of these may become rate-limiting under various conditions. Since oxidativephosphorylationdepends upon oxygen as the final electron acceptor, the continued formation of ATP depends upon an adequate deliveryof oxygen to the muscle. Should the oxygen deliverybe insufficient, lactic acid (homolactic) fermentation occurs. The muscles obtainthe extra energy they need from this anaerobic process and lactic acid, the end product, accumulates. The muscle thus incurs what physiologistscall oxygen debt. When the violent activity is over, the musclecells consume large quantities of oxygen as they convert the lactic acidintopyruvicacid.Pyruvicacid is oxidized via the TCA and the electrontransport processes. The cells utilize the energy thus obtained to resynthesizeglycogen from the lactic acid that remains. In this manner, theoxygen debt is paid off, and the lactic acid is removed. This is why one continuesto breathe hard for some time after one has stopped high levels ofactivity.

Question:

Determine the free energy change, ∆G, for transforming liquid water at 100\textdegreeC and 1 atm. to vapor at the same conditions. ∆H = 9720 cal (molar enthalpy of vaporization) .

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Solution:

This problem deals with the energies of reactions: enthalpy, entropy and free energy. The entropy of a system is a measure of the state of randomness. As more thermal energy is added to a system, the more random it becomes. Entropy can be measured quantitatively. Entropy or ∆S =q_rev/T, whereq_revis the amount of heat added to the system that can be recovered if the reaction were reversed. For example, the energy needed to melt ice can be recovered by freezing the liquid back to ice. Enthalpy may be defined as the heat content of a system. The concept of energy and entropy may be combined to define the driving force of a reaction. This driving force is referred to as Gibbs free energy (G) where ∆G = ∆H - T∆S, where ∆H is the enthalpy. This energy must be equal to the heat content of the system (∆H) minus the energy necessary to increase the randomness, or entropy, of the same system. Therefore, to solve this problem, you need to sub-stitute into this equation the following values: T = 100\textdegreeC = 373\textdegreeK q = ∆H = 9720 cal Therefore, ∆S = (q/T) = [(9720 cal)/(373\textdegreeK )] Now, ∆G = 9720 cal - 373 \textdegreeK [(9720 cal)/(373\textdegreeK)] = 0. The answer is zero. The system can derive no usable energy to perform work from this change. For any system when ∆G = 0, the system is at equilibrium.

Question:

Briefly discuss some abnormalities of the respiratory tract.

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0403.htm

Solution:

The respiratory tract is subject to various infections and damages which may interfere with normal functioning. Tuberculosis is caused by the bacteriaMycobacterium tuberculosisand results in the destruction of lung tissue. It was once a major cause of death. In-fectious bacteria or viruses can bring about infections in the nasal cavities, the sinuses, the throat, the larynx (laryngitis), the bronchial tubes (bronchitis), and the lungs (pneumonia). Pneumonia, which may result from infection by bacteria or viruses, results in the filling of the alveoli with fluid. As the infection spreads through-out the lungs, the surface area for gas exchange is reduced. Allergic reactions may also occur in the respiratory tract. They occur when a person becomes sensitized to certain substances in the air, such as pollution, smoke or pollen. A person who is allergic to these substances has a condition known as hay fever. In such people, allergic reactions are initiated in the nasal cavities. Should these substances pass into the bronchioles and not be filtered in the nasal cavities, allergic asthma may result. The bronchiole tubes swell and become clogged with mucus, and the person has respiratory difficulty. It has long been known that cigarette smoking can cause cancer in the lungs. It is not as widely known, however, that it can also result in emphysema. Emphysema results from a thickening of the lung tissues, which causes a loss of elasticity and a loss in the ability to absorb oxygen. In extreme cases the vital capacity is so reduced that breathing becomes a major effort. Since exercise requires increased oxygen consumption, physical activity is greatly reduced in people with emphysema. Alveolar cells secrete a detergent-like chemical called surfactant, which intersperses with the water molecules in the alveoli. The air within each alveolus is separated from the alveolar membrane by an extremely thin layer of fluid. Surfactant markedly reduces the water molecules cohesive force, thereby decreasing the surface tension. If the surfactant were absent, this tension would be strong enough to cause the alveoli to collapse. In some newborn infants, predominantly those born prematurely, the lungs do not secrete an adequate amount of surfactant for their inspiratory forces to over-come the surface tension. Hence their alveoli are unable to expand when they inhale. Without treatment, these infants die of suffocation. The disease associated with insufficient surfactant production is referred to as respiratory distress syndrome. The recent discovery that the administration of hormones from the adrenal cortex enhances maturation of the surfactant-synthesizing cells may provide an important means of combating this disease.

Question:

Suppose we have two charges, q_1 = + 4stateCand q_2 = \rule{1em}{1pt}6 stateC, with an initial separation of r_1 = 3 cm. What is the change in potential energy if we increase the separation to 8 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0578.htm

Solution:

The potential energy of two point charges, q_1 and q_2 , separated by a distance r_1, is given V_12 = q_1q_2 / r_1 . Therefore as the separation is increased from r_1 to r_2 the change in potential energy is ∆V_12 = q_1q_2[(1 / r_2) \rule{1em}{1pt} (1 / r_1)] = (4statC)(\rule{1em}{1pt}6statC) × [(1 / 8cm) \rule{1em}{1pt} (1 / 3cm)] = (\rule{1em}{1pt}24 statC^2)(\rule{1em}{1pt}5cm / 24cm^2) = +5[(statC)^2 / cm] = +5 ergs In this case, there is a net increase in the electrostatic potential energy (that is, ∆V_12 > 0) because work was done by an outside agent against the attractive electrostatic force.

Question:

Write a FORTRAN program segment to find a root of the continuous func-tion f(x) via the bisection method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0514.htm

Solution:

The bisection method starts with two values of X, called HI and LO, for which the values of f(x) have opposite signs. The object is to find X when f(x) = 0. We do this by halving the in-terval between HI and LO, alternately from the negative side (LO) and from the positive side (HI). As we continue the process, we can approach X until we reach an approximation with any desired degree of accuracy. [Note that rounding errors can alter the accuracy of the approximation.] The program logic is given below, using the function X3 + 3X - 5 (Figure) as a sample f(x). We have chosen HI = 2.0, so f(HI) = (2)3 + 3(2) - 5 = 9 , and LO = 0.0, so f(LO) = (0)3 + 3(0) - 5 = -5. REAL F, X, HI, LO, ERROR CCHOOSE SOME INITIAL HI AND LO VALUES, CSUCH AS HI = 2.0 AND LO = 0.0. CALSO CHOOSE ERROR TO BE, SAY, = .002. DO 20 J = I, 1000 IF ((HI - LO).LE. ERROR) GO TO 30 X = (LO + HI)/2.0 F = (X \textasteriskcentered\textasteriskcentered 3) + (3.0 \textasteriskcentered X) - 5.0 IF (F.GT.O.O) GO TO 15 LO = X GO TO 20 15HI = X 20CONTINUE 30WRITE (5,\textasteriskcentered) X [Note: This WRITE statement includes an asterisk, a feature provided on some machines. It simply indicates that only variables are to be outputted, without any literals. Its purpose is to save time by eliminating a FORMAT statement.]

Question:

A satellite of mass M_s is placed in a stable circular orbit of A satellite of mass M_s is placed in a stable circular orbit of radius R around the earth. What is its angular momentum about an axis through the earth perpendicular to the plane of its orbit? Assume that R > > radius of satellite.

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Solution:

In the figure, the earth's gravitational pull on the satellite accounts for the satellite's centri-petal acceleration: (M_sv^2)/R = (GM_sM_e)/(R^2) Where M_e is the mass of the earth. Thus:v = \surd[(GM_e)/(R)] Its angular velocity is therefore: \omega = v/R = \surd[(GM_e)/(R^3)] Since R > > radius of satellite, when calculating the satellite's moment of inertia, with respect to the axis A-A', we can take all of its mass to be at a distance R from the axis. This reduces to the case of finding the moment of inertia of a point mass with respect to an axis. This is: I = M_sR^2 The angular momentum L of the satellite is therefore: L = I\omega = M_sR^2\surd[(GM_e)/(R^3)] = M_s \surd(GM_eR)

Question:

A chemist has tert-butane. If he adds Br_2 and light, what are the monobromo substitution products expected? Which one will predominate and why? How would you prepare the predominating compound by an addition reaction?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0753.htm

Solution:

The first question concerns substitution in alkanes. You are performing a halogenation substitution reaction and are asked to predict the monohalogenation products expected for tert-butane. This necessitates the writing of its structure and analyzing its hydrogens, since they will be replaced by the bromine atoms. Tert butane can be written All the hydrogens on the methyl groups are equivalent. This means that if you substitute a bromine atom for one hydrogen atom from a methyl group, it is the same as sub-stituting for any other hydrogen on one of the other methyl groups. The other hydrogen (the one attached to the 3\textdegree carbon) is not equivalent to the hydrogens bound to the 1\textdegree carbons. Thus you expect two monobromo products. They are You could have placed the Br on the other two methyls and the same molecule would be obtained, for Example, When the Br is attached to the 3\textdegree carbon tert-butyl bromide is obtained. This product predominates greatly. To see why, consider the mechanism of this reaction. The mechanism is as follows (let x = a halogen). x_22x \textbullet(radicals denoted by dots) x \textbullet + alkaneHx + alkane \textbullet alkane \textbullet + x_2alkyl halogen + x \textbullet You see that the reacton proceeds by radical formation. To form isobutyl bromide, a 1\textdegree radical. must be formed first. T-butyl bromide goes through a 3\textdegree radical, i.e., A 3\textdegree radical is more stable and, thus, has a longer lifetime than the 1\textdegree radical. Since the life time of the 3\textdegree radicals is longer, there will be more 3\textdegree radicals to react with the bromine radicals, and the amount of t-butyl bromide formed exceeds the amount of iso-butyl bromide formed. You can prepare t-butyl bromide, by an addition re-action. HBr can be added across the double bond of iso-butylene as shown in the following reaction. It has been shown that when an inorganic acid, such as HBr, adds across a double bond in an olefin the hydrogen ion will add to the carbon with the most hydrogen atoms already bound to it. This is called Markovnikov's rule.

Question:

A spaceship from earth enters a circular orbit 22,000 km above the surface of Mars at the equator, which allows it to rotate at the same speed as Mars and thus to stay always above the same point on the planet's surface. The Martians set up a guided missile directly beneath it and fire it vertically upward in an attempt to destroy the intruder. With what minimum velocity must the missile leave the surface in order to suc-ceed in its mission? Mars is a sphere of radius 3400 km and of mean density 4120 kg\bulletm^-3.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0150.htm

Solution:

The missile will just reach the spaceship if its velocity at the latter's position is zero. We must therefore calculate the missile's velocity as a function of position. Using Newton's Second Law, we ob-tain F = ma(1) where m is the missile's mass,ais its acceleration and F is the net force acting on it. Note that F = -GmM_m/r^2(2) where M_m is the mass of Mars and r is the radial distance of the missile from the center of Mars. Assuming that the missile travels in a radial direction, we may write a =dv/dt= (dv/dr)(dr/dt) = v(dv/dr) Substituting (2) and (3) in (1) -GmM_m/r^2 =mv(dv/dr) orv(dv/dr) = -GmM_m/r^2 vdv= -GM_m(dr/r^2) ^v\int_v0 (vdv) = -GM_m^R\int_R0 (dr/r^2) where v = v at r = R, and v = v_0 at r = R_0. (R_0 is the position of the surface of Mars relative to its center). (v^2 - v^2 _0)/2 =GM_m(1/R - 1/R_0) v2_0 = v^2 + 2GM_m(1/R_0 - 1/R) The minimum value of v_0 is found by requiring that themissle'svelocity at the spaceship's position be zero (v = 0). Hence v2_0 _min = 2GM_m(1/R_0 - 1/R)(3) The mass of Mars is its volume times its density (\rho_m). M_m = (4/3)\piR3_m×\rho_m= (4/3)\pi× (3.4 × 10^6m)^3 × 4120\bulletm^-3 = 0.678 × 10^24kg. orv^2 = 2 × 6.67 × 10^-11N\bulletm^2 \bulletkg^-2 × 0.678 × 10^24kg [{1/(3.4 ×10^6m)} - {1/(25.4 × 10^6m)}] = 2.30 × 10^7 m^2\bullets^-2. ThereforeV = 4.8 km\bullets^-1. Note that R in (3) is the position of the spaceship from the center of Mars, orR = R_0 + 2.2 × 10^6m R = 25.4 × 10^6m

Question:

Enzyme behavior may be described by the lock-and-key theory. Explain and describe this theory.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E23-0829.htm

Solution:

Millions of chemical reactions occurring in the body require catalysts to make them go to completion. Enzymes serve as these catalysts. Their behavior and/or function is described by the lock-and-key theory. According to this theory, enzymes have definite three-dimensional structures. They are arranged so that the substrate (the substance upon which the enzyme acts) fits into the enzyme's structure. Only a specific kind of substrate can fit into a given enzyme. When the substrate and enzyme unite, i.e., form an aggregate, the substrate is exposed for a reaction to occur. In Fig. A, the three- dimensional shape of the enzyme is arranged to accommodate the substrate. The active site is that portion of the enzyme that catalyzes the reaction. In Fig. B, the sub-strate is held in position by the enzyme while the re-action occurs. In Fig. C, the product(s) of the reaction leaves the enzyme, thus, freeing the enzyme to catalyze another reaction.

Question:

How can we deduce information about prehistoric man from the fossil teeth, jaws, and craniums found?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0748.htm

Solution:

Of all the fossils discovered relevant to the evolution of man, the craniums, jaws, and teeth have been the most numerous and useful. Teeth have been found in large numbers at archaeological sites because their enamel coating gives them great durability. Teeth are complex structures with many distinctive features. The size and shape of the cusps, and the root size of the fossil teeth can be compared with modern man's dental anatomy and to that of modern apes. In this manner, a fossil man can be placed in a chronological and evolutionary relationship to otherhominiod-like forms, whether they be extinct or extant. Much can also be said about our ancestor's diet by studying the structure of a fossil's teeth. For example, there is an evolutionary trend towards reduction in size of the canine teeth in man's evolution (see Figure). Man and apes are charac-terized by larger canines. Since the canines are used in tearing raw meat, the reduction in their size would seem to parallel the evolution of man's mode of feeding and his diet. A major problem encountered by anthropologists is the absence of teeth from many of the jaws that are found. It becomes difficult to determine the exact number and size of the teeth when the jaw is not complete. As a general rule, the size of the jaw is directly related to the size of the teeth. For instance, man has charac-teristically both smaller teeth and smaller jaws than his predecessors. The shape of the jaw alone is also an important evolutionary indicator. The dental arch, or portion of the jaw where the teeth are attached, is rectangular in the apes and parabolic in man. This indi-cates that in the apes, the back teeth on either side of the jaw are parallel to each other, with the incisors in a straight row. In man and some of his fossil ancestors, parabolic dental arches are present wherein the distance between the canines on either side of the jaw is smaller than that between the molars. The structure and size of the teeth, the ridges on the jaw, and the dimensions and shape of the dental arch indicate how the face may have looked, and the nature of the diet. If the fossil man ate hard, rough objects that required teeth with effective chewing and grinding surfaces, we would most likely observe huge molars that have undergone much wear, and jaws with large ridges for attachment of strong chewing muscles. Enlarged chewing muscles are also accompanied by large ridges on the sides and top of the skull, since thetemporalismuscles used for chewing originate in the skull region in hominoids, although not always in the same place. Huge neck muscles would also be accompanied by large ridges on the skull. This would indicatequadrupedalism. If the animal was quadruped, huge neck muscles would be needed to support the neck, and obvious muscle ridges would be present in the skull. If the animal was bipedal, the neck muscles and ridges would be reduced since the head would now be balanced on the spinal column. The centered position of the head indicatingbipedalismcan be further determined by the position of the foramen magnum.Bidepalorganisms like man have the foramen magnum located underneath the center of the brain, while in apes, the foramen magnum is located toward the rear of the skull. The overall dimensions of the braincase are im-portant in determining an animal's cranial capacity. If the skulls of the various fossil men are put into a chronological sequence, a trend toward increasing brain size is noticed as well as a general thinning of the skull bones. A great deal of the information indicated above is still to a large degree speculative, since skulls, jaws, or teeth are rarely found in their original form without great amounts of weathering and fragmentation.

Question:

A 3000-lb automobile at rest at the top of an incline 30 ft high and 300 ft long is released and rolls down the hill. What is its speed at the bottom of the incline if the average retarding force is 200 lb?

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Solution:

The potential energy at the top of the hill is available to do work against the retarding force F^\ding{217} and to supply kinetic energy. The work done by the retarding force F^\ding{217} is W = \int F^\ding{217} \bullet ds^\ding{217} where the integral is evaluated over the path of motion of the auto. If F^\ding{217} is constant, W = F^\ding{217} \bullet \int ds^\ding{217} = F^\ding{217} \bullet s^\ding{217} s^\ding{217} being the vector displacement of the auto. Hence, by the principle of energy conservation, mgh = F^\ding{217} \textbullet s^\ding{217} + (1/2) mv^2 where v is the velocity at the bottom of the incline. Since F^\ding{217} and s^\ding{217} are parallel and in the same direction, we may write Wh = Fs + (1/2) mv^2 (1) Since the height of the incline is much less than the length of the base, we may use 300 feet as an approxima-tion to the length of the hypotenuse (see figure). Then, substituting into (1) 3000 lb × 30 ft = 200 lb × 300 ft + (1/2) × 94 slugs × v^2 9.0 × 10^4 ft-lb - 6.0 × 10^4 ft-lb = (1/2) × 94 slugs × v^2 v^2 = (3.0 × 10^4 ft-lb)/(47 slugs) = 640 ft^2/sec^2 v = 25 ft/sec.

Question:

Two long, straight wires, each carrying a current of 9 A in the same direction are placed parallel to each other. Find the force that each wire exerts on the other when the separation distance is 1 × 10^-1 m.

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Solution:

The currents carried bv the wires i_a = 9 A, i_b = 9 A, and the distance between them, d = 1 × 10^-1 m, are given. The first current carryingwire (a) produces a field of induction B at all nearby points around it. The magnitude of B is B = (\mu_0 /2\pi) (i_a /d)(1) where i_a is the current through wire a, d is the distance a separating the two wires and the permeability constant \mu_0 = 4\pi × 10^-7 W/(A \bullet m). The other wire will find itself immersed in the field due to the first wire. The magnetic field it creates, that is its own self field, has no influence on its behavior. The magnetic force on this second current carrying wire (of length l) is F^\ding{217} = i_b l^\ding{217} × B^\ding{217} where i_b is the current through the second wire. Since the wire length is perpendicular to the magnetic field vector B, we then have F = i_b lB(2) F is directed inward toward the first wire. It is perpendicular to both B and to the length vector. There-fore, combining equations (1) and (2) we have F = (\mu_0 /2\pi) [(i_a i_b)/d] l The force per unit length on a wire is then (F/l) = (\mu_0 /2\pi) [(i_a i_b )/d] = 2 × 10^-7 (N/A^2) [(i_a i_b)/d] = 2 × 10^-7 (N/A^2) [{(9A) (9 A)} /(1 × 10^-1 m)] = 1.62 × 10^-4(N/m)

Question:

What basic nutritional requirements do all livingorganisms have in common? Comparephototrophsandchemotrophs, autotrophs andheterotrophs.

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Solution:

Organisms, ranging from bacteria to man, share a set of nutritional requirements necessary for normal growth. These requirements must be known in order to cultivate microorganisms in the laboratory, and purecul- tures may be obtained through the preparation of appropriate selective growth media. All organisms require a source of energy. Green plants and some bacteria can utilize radiant energy (from the sun) and are thus called phototrophs . Animals and non-photosyn-thetic bacteria must rely on the oxidation of chemical compounds for energy and are thus called chemotrophs . All organisms require a carbon source. Plants and many bacteria require only carbon dioxide as their carbon source. They are termed autotrophs . Animals and other bacteria require a more reduced form of carbon , such as an organic carbon compound. Sugars and other carbohydrates are examples of organic carbon compounds. Organisms which have this requirement are termed theterotrophs. They depend upon autotrophs for their organic form of carbon, which they use as both a carbon source and an energy source. All organisms require a nitrogen source. Plants utilize nitrogen in the form of inorganic salts such as potassium nitrate (KNO_3), while animals must rely on organic nitrogen- containing compounds such as amino acids. Most bacteria utilize nitrogen in either of the above forms, although some bacteria can use atmospheric nitrogen. All organisms require sulfur and phosphorous. While phosphorous is usually supplied by phosphates, sulfur may be supplied by organic compounds , by inorganic compounds, or by elementary sulfur. All organisms need certain metallic elements, and many require vitamins . The metallic elements include sodium, potassium, calcium, magnesium , manganese, iron, zinc, and copper. While vitamins must be furnished to animals and to some bacteria, there are certain bacteria capable of synthesizing the vitamins from other nutrient compounds. Finally, all organisms require water for growth. For bacteria and plants , all the above nutrients must be in solution in order to enter the organism . Bacteria show considerable variation in the specific nutrients required for growth. For example, all heterotrophic bacteria require an organic form of carbon, but they differ in the kinds of organic compounds they can utilize. Different bacteria utilize nitrogen in its various forms. Some require several kinds of amino acids and vitamins, while others require only inorganic elements.

Question:

A heavy particle of mass M collides elastically with a light particle of mass m (see the figure below) . The light particle is initially at rest. The initial velocity of the heavy particle is v_h^\ding{217} = u_h\^{\i}, the final velocity is w_h^\ding{217}. If the particular collision is such that the light particle goes off in the forward (+\^{\i}) direction, what is its velocity w_1^\ding{217}? What fraction of the energy of the heavy particle is lost in this collision?

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Solution:

This problem can be solved using the principle of conservation of linear momentum. The linear momentum before collision must equal the linear momentum after collision. The initial momentum p_i of the system is: p_i = Mv_h + mv_1 = Mv_h since the smaller mass m is initially at rest. The final momentum p_f of the system is: p_f = Mw_h + mw_1 By the conversation of linear momentum, p_i = p_f : Mv_h = Mw_h + mw1 Thus:mw_1 = Mw_h - Mwh w_1 = (M/m) (v_h - w_h) The energy of the heavy particle before the collision is(1/2) Mvh^2. After the collision the kinetic energy is (1/2) Mwh^2. The fraction of its original kinetic energy that the heavy mass M retains after the collision is: (E_f)/(E_i) = [{(1/2) Mv_h^2} / {(1/2) Mw_h^2}] = (v_h/w_h)^2 where E_i and E_f are the initial and final kinetic energies respectively. The fraction of the energy of the heavy particle that is lost in the collision is Fractional energy loss = (E_i - E_f) / (E_i) = 1 - (Ef/ E_i) = 1 - (vh/ w_h)^2 .

Question:

Estimate the minimum voltage needed to excite K radiation from an x-ray tube with a tungsten target, and calculate the wavelength of the K_\alpha line.

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Solution:

Moseley found experimentally that the x-ray line spectrum varies in a regular way from element to element, unlike the irregular variations of optical spectra. This regular variation occurs because characteristic x-rays are due to transitions involving the innermost electrons of the atom. Because of the shielding of the other elec-trons the inner electron energies do not depend on the com-plex interactions of the outer electrons, which are respon-sible for the complicated optical spectra. Also, the inner electrons are shielded from the interatomic forces respon-sible for the binding of the atoms in solids, According to the Bohr model of the atom, the energy of an electron in the first orbit is proportional to the square of the nuc-lear charge. Mosely therefore, reasoned that the energy, and thus the frequency, of a characteristic x-ray photon should vary as the square of the atomic number of the element. He plotted the square root of the frequency of particular x-ray lines versus the atomic number Z of theelement. He found that his curves could be described by the general equation, equation, f = A^2 (Z - b)^2 f = A^2 (Z - b)^2 Where f is the frequency of the x-ray and A and b are constants for each line. One family of lines, called the K-series, has b = 1. If the bombarding electron in an x-ray tube knocks an electron from the n = 1 inner orbit of a target atom out of the atom, photons are emitted corres-ponding to transitions of other electrons to the vacancy in the n = 1 orbit. The lowest frequency line corresponds to the lowest energy transition, from n = 2 to n = 1. This line is called the K_\alpha line. Bohr's relation for the frequency of a photon emitted from a one electron atom is, f = Z^2 [(2\pi^2m e^4) / {(4\piE_0)^2 h^3}] [{1/(n_2^2)} - {1/(n_1^2)}](2) The energy E of a photon is given by hf. Using Z - 1 in place of Z in eq (2) so as to fit the form of Moseley's equation (1) for the K-series, we find, E = hf = [(2\pi^2m e^4) / {(4\piE_0)^2 h^2}] [{1/(n_2^2)} - {1/(n_1^2)}](z - 1)^2(3) For K radiation to occur, an electron must be knocked from the n_1 = 1 orbit to infinity (n_2 = \infty). This is the minimum energy that must be transmitted to the atom for K radiation and is, E_ionization = [(2\pi^2me^4) / {(4\piE_0)^2 h^2}] (Z - 1)^2 [(1/\infty) - 1] = 2.18 × 10^-18 (Z - 1)^2joules = 13.6 ev (Z - 1)^2 For tungsten, Z = 74, and the energy input required is (13.6 ev) (73)^2 = 72,800 ev. So the anode must be at least 72,800 volts above the cathode potential. The first line in the K series of tungsten has a wavelength given by E = hf. But f\lambda_K\alpha = c and E = hf = hc/(\lambda_K\alpha) = 13.6 ev (Z - 1)^2 [(1/1^2) - (1/2^2)] \lambda_K\alpha = [{6.625 × 10^-34 joule-sec (3.0 × 10^8 m/sec)} / {(13.6 × 1.60 × 10^-19 joule) (73)^2 (1 - 1/4)}] = 0.21 × 10^-10m = 0.21 A

Question:

An iron steam pipe is 200 ft long at 0\textdegreeC. What will its increase in length when heated to 100\textdegreeC? (\alpha = 10 x 10^\rule{1em}{1pt}6 percelsiusdegree).

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Solution:

The change in length, \DeltaL, of a substance due to a temperature change is proportional to the change, \DeltaT, and to the original length,L_0,of the object: \DeltaL = \alpha L_0 \DeltaT where \alpha is the proportionality constant and is called the coefficient of linearexpension. L_0 = 200 ft,\alpha = 10 × 10^\rule{1em}{1pt}6 per C\textdegree, T = 100\textdegreeC,T_0 = 0\textdegreeC. Increase in length= \DeltaL = \alpha L_0\DeltaT = (10 × 10^\rule{1em}{1pt}6) (200) (100) = 0.20 ft.

Question:

A particle of mass m is released from rest at point a in Fig. 1 falling parallel to the (vertical) y-axis. (a) Find the torque acting on m at any time t, with respect to origin O. (b) Find the angular momentum of m at any time t, with respect to this same origin, (c) Show that the relation \tau = d1/dt yields a correctresult when applied to this problem.

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Solution:

(a) The torque is given by \tau = r × F, its magnitude being given by \tau = rF sin\texttheta. r = sin\texttheta = b and F = mg so that \tau = mgb = a constant Note that the torque is simply the product of the force (mg) times the moment arm (b). The right-hand rule shows that \tau is directed perpendicularly into the figure. (b) The angular momentum is given by 1 = r × p, its magnitude being given by 1 =rp sin\texttheta. r sin\texttheta = b and p = mv = m(gt) so that 1 = mgbt. The right- hand rule shows that 1 is directed perpendicularly into the figure, which means that 1 and \tau are parallel vectors. The vector 1 changes with time in magnitude only, its direction always remaining the same in this case. (c) Since d1, the change in 1, and \tau are parallel, we can replace the vector relation \tau = d1 / dt by the scalar relation \tau = d1 / dt . Using the expression for \tau and 1 from (a) and (b) above mgb = (d/dt) (mgbt) = mgb, Which is an identity. Thus the relation \tau = d1 / dt yields correct results in this case. Canceling the constant b out of the first two terms above substituting for gt the equiva-lent quantity v, mg = (d/dt) (mv). Since mg = F and mv = p, this is the familiar result F = dp / dt. Thus, relations such as \tau = d1/ dt, though often vastly useful, are not new basic postulates of classical mechanics but are rather the reformulation of the Newtonian laws for rotational motion. Note that the values of \tau and 1 depend on choice of origin, that is, on b. In particular, if b = 0, then \tau = 0 and 1 = 0.

Question:

Summarize the roles of the various enzymes involved in digestion.

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Solution:

Enzymes in Digestion Enzyme Source Optimum pH Product(s) Salivary amylase Salivary glands Neutral Maltose Pepsin Stomach Acid Peptides Rennin Stomach Acid Coagulated casein (A milk protein) Trypsin Pancreas Alkaline Peptides Chymotrypsin Pancreas Alkaline Peptides Lipase Pancreas Alkaline Glycerol, Fatty acids, Mono- andDiglycerides Amylase Pancreas Alkaline Maltose Ribonuclease Pancreas Alkaline pentoses , nucleotides and nitrogen bases Deoxyribonuclease Pancreas Alkaline pentoses , nucleotides and nitrogen bases Carboxypeptidase pancreas Alkaline Free amino acids Aminopeptidase Intestinal Glands Alkaline Free amino acids Maltase Intestinal Glands Alkaline Glucose Sucrase Intestinal Glands Alkaline Glucose and fructose Lactase Intestinal Glands Alkaline Glucose andgalactose

Question:

Explain briefly the background and applications of Pascal.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0394.htm

Solution:

Pascal is a modern high-level programming langu-age developed in the early 1970's in Switzerland and named in honor of 17th century French mathematician and inventorBlaisePascal. Originally intended as a language for teach-ing structured programming, it is now used for a variety of scientific and commercial applications. An important boost for Pascal has come from its widespread implementation on microcomputers. Being one of the newest general purpose programming languages, Pascal features modern control structures for se-quence, selection, and iteration , which makes it particularly, well-suited for structured programming , makes programs read-able, and allows for implementation of various data-structures.

Question:

[(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B](1) The + operator denotes the inclusive OR function and the \bullet operator denotes the AND function. NOT is represented by \textasciitilde \bullet (b) Change statement (1) into its equivalent switching circuit. (c) Show the gate representation of statement (1).

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Solution:

(a) Set up the truth table by specifying all possible combinations of A,B,C, and D and evaluating the terms of the state-ment for each combination. Since there are 4 variables, the truth table has 2^4 = 16 rows of entries as shown in figure 1. First, evaluate the innermost terms: A B \backsimB C D A + B A\bullet\backsimB A + C A + D 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 Fig. 1 Now, transfer the results in columns 6, 8, and 9 to a second truth table, shown in figure 2 which evaluates the first bracketed term. A + B A + C A + D (A + B) + (A + C) + (A +D) 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Fig. 2 Finally, transfer the result in the last column of this truth table and the result in column 7 of the first truth table to a third truth table, shown in figure 3, which evaluates the entire statement (1). A \bullet \backsim B (A + B) + (A + C) + (A +D) Statement (1) 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 Fig. 3 The entry in each row of column 3 is the result of performing the operation on the entries in columns 1 and 2 of that row. (b) The switching circuit equivalent to statement (1) is obtained by connecting, in parallel, variables which are operands of the + operation and connecting, in series, variables which are operands of the \bullet operation. First, find equivalent switching circuits for the innermost terms. These are shown in figure 4. The switching circuit equivalent to the first bracketed term is a parallel connection of the three parallel circuits A + B, A + C, A + D, as shown in figure 5. The switching circuit equivalent to statement (1) is a series con-nection of the parallel circuit [(A + B) + (A + C) + (A + D)] and the series circuit [A \textbullet \backsim B]. The complete switching circuit is shown in figure 6. [(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B] Note that the circuit conducts if and only if A is closed and B is open, regardless of the settings of C and D. This agrees with the truth table, since statement (1) has value if and only if A = 1 and B = 0. (c) Variables which are operands of the + operation are inputs to an OR gate. Variables which are operands of the \bullet operation are inputs to an AND gate. A variable that is the operand of the \textasciitilde operation is the input to a NOT gate. The gate representation of the terms in parentheses are shown in figure 7. The composition of 2 operations is done by plugging the output of 1 gate into the input of another, so that the gate representation of [A \textbullet \backsim B] is shown in figure 8. Finally, by the same reasoning, the complete gate representation of [(A + B) + (A + C) + (A + D)] \textbullet [A \textbullet \textasciitilde B] is shown in figure 9.

Question:

The barometric pressure on the lunar surface is about 10^-10 torr. At a temperature of 100\textdegreeK, what volume of lunar atmosphere contains (a) 10^6 molecules of gas, and (b) 1 millimole of gas?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0052.htm

Solution:

This problem is an application of the ideal gas equation, PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature. Solving for V, V = nRT/P, in the first part of this problem, we use the definition n = [(number of molecules)/(Avogadro's number)] = (N/A). We must then substitute into the formula V = nRT/P = (N/A) (RT/P) in order to obtain the volume corresponding to N molecules. In the second part of this problem we can use V = nRT/P directly, remembering that 1 millimole = 10^-3 mole. Hence, for the first part of the problem, V = (N/A) (RT/P) = [(10^6 molecules)/(6.02× 10^23 molecules/mole)] × [(0.0821 l -atm/\textdegreeK - mole × 100\textdegreek)/(1.316 × 10^-13 atm)] = 1.04 × 10^-4 l, where we have used P = 10^-10 torr = 10^-10 torr × [(1 atm)/(760 torr)] = 1.316 × 10^-13 atm. For the second part of the problem, V = (nRT/P) = [(10^-3 mole × .0821 l -atm/\textdegreeK - mole × 100\textdegreeK) / (1.316 × 10^-13 atm)] = 6.24 × 10^10 l.

Question:

All the silicon and most of the oxygen in the earth's crust exists in the form of silicate minerals, such as granite. If the basic chemical unit of all silicates is the Sio_4^4- ion, which consists of an Si^4+ ion and four O^2- ions, write the electronic configurations of the Si^4+ and O^2- ions in silicate materials.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E27-0907.htm

Solution:

An electronic configuration of an element (or ion) gives a picture of the arrangement of electrons that surround a nucleus. The atomic number gives the total number of electrons of the atom (notthe ion). With this in mind, write the electron configuration of the oxygen and silicon atoms. The electronic configuration of oxygen with atomic number 8 is 1s^22s^22p^4 and for silicon with atomic number 14 it is 1s^22s^22p^63s^23p^2. To find the electronic configurations of the ions Si^4+ and O^2- note each ion's charge. The charge of 4+ on Si indicates that it lost 4 electrons (electronsare negatively charged). Thus, if the silicon atom had a configuration of1s^22s^22p^63s^23p^2, and it loses 4 electrons, the configuration becomes 1s^22s^22p^6 (ions will form by losing only high energy electrons). The oxygen atom configuration was 1s^22s^22p^4. The fact that the charge was 2- indicates that the atom gained two electrons. Consequently, the configuration becomes 1s^22s^22p^6.

Question:

The electron of a hydrogen atom, revolving in the first Bohr orbit can be thought of as being equivalent to a current loop, as shown in the figure. Find the magnetic moment \mu_B of this loop (which is called one Bohr magneton).

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0710.htm

Solution:

The average charge, per unit time, passing a point of the orbit, or the average current I is e/t, where T is the period of rotation. The angular mo-mentum of the first Bohr radius is given by L = mvr = (h/2\pi) , orv = [h/(2\pimr)] . Then, the period and average current are T = (2\pir)/v = (4\pi^2 r^2 m)/h . I = (eh)/(4\pi^2 r^2 m) . The magnetic moment of the loop is \mu_B = IA, where A is its area, \pir^2 . Therefore, \mu_B = (h/4\pi) \bullet (e/m) = [(6.62 × 10^-34joule - sec × 1.76 × 10^11 coul/kq) / (12.57)] = 9.27 × 10^-24 amp \textbullet m^2 .

Question:

A rocket blasts off from Earth at 12 noon. The original mass of the rocket is M_0 , and the exit velocity of the fuel jet isV_e_. Assume that fuel is burned at a constant rate, dm/dt. How high will the rocket be at 12:15 PM? (assuming a straight-line trajectory). Write a FORTRAN program to find the height at intervals of 0.5 minutes.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0302.htm

Solution:

Force may be expressed as the change of momentum, as given by Newton's second law: F =dp/dt. Now, we establish that positive y is the vertical direction upwards. If M_0 is the total mass of rocket and fuel before the launch, and m is the mass of the exiting fuel jet, then the rate of change in momentum is given by: F = (d/dt)(MV) + (V -V_e) (dm/dt) We must also take into account the force of gravity, which is acting against the motion of the rocket. The formula will not be derived step-by-step here; a text on differential equations will have the thorough explanation. However, all variables will be explained, after presenting the general equation. The initial condition v = 0 at t = 0 yields the equation: V = -gt-V_eln[(m_0 -ut) / u] then, by integrating, the equation of motion is obtained, which expresses the height y as a function of time t: y = -(1/2)gt2+V_e[t + {( m_0 -ut) / u} 1n {(m_0 -ut) / m_0}] The variables are as follows: y = vertical height; -(1/2)gt^2 = the effect of gravity pulling the rocket downward; hence the use of the negative sign. V_e= the exit velocity of the fuel jet M_0 = original mass of the rocket plus theunburaedfuel u = dm/dt= the rate of change of the mass of the unburned fuel Note that as the rocket rises, the mass of the fuel, and ultimately, the total mass of the rocket is decreasing. Also notice that in the program, a decision should be made upon appropriate values for M_0, \mu ,V_e. It is assumed that these have already been read in. Also, since g is defined in meters per sec^2 , it may be best to use the MKS system of units. CSIMULATE IHE VERTICAL ASCENT OF AN CUNCONTROLLED ROCKET FOR 15 MINUTES CAT 30-SECOND INTERVALS Y = 0.0 G = 9.80616 WRITE (5,100) 100FORMAT (1X,'TIME', 5X, 'HEIGHT') DO 20 I = 1,30 T = 0.5 {_\ast} FLOAT(I) Y = -0. 5{_\ast}G{_\ast} (T{_\ast}{_\ast}2) + (VE{_\ast} (T + (RMD - (RMU{_\ast} 1T))RMU){_\ast}L0G((RMO - RMU){_\ast}T/RMO)) WRITE (5,101) T,Y 101FORMAT (2X, F5.2, 4X, F12.4) 20CONTINUE STOP END SAMPLE OUTPUT: TIMEHEIGHT 0.50 1.00 1.50 2.00

Question:

Calculate ∆m, the difference in mass between the final and initial nuclei in grams per mole for the emission of a \gamma ray, ^19_8O\textasteriskcentered (excited state)\rightarrow^19_8O (ground state) + \gamma (∆E = 1.06 × 10^8 Kcal/mole).

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Solution:

One solves for ∆m by using Einstein's Law, which relates ∆m and ∆E. ∆E = ∆mc^2, where ∆E is the energy, ∆m is the change in mass, and c is the speed of light, 3 × 10^10 cm/sec. When using Einstein's Law, the ∆E, in Kcal/mole, must be first conver-ted to ergs/mole. There are 4.18 × 10^10ergsin one Kcal, therefore one converts Kcal to ergs by multiplying the number of Kcal by 4.18 × 10^10 ergs/Kcal. no. of ergs= 4.18 × 10^10 ergs/kcal × 1.06 × 10^8 kcal/mole = 4.43 × 10^18 ergs/mole. One can now solve for ∆m. ∆E = ∆mc^2∆E= 4.43 × 10^18 ergs/mole c= 3 × 10^10 cm/sec ergs= (g cm^2) / (sec^2) 4.43 × 10^18 {(g cm^2) / (sec^2 mole)} = ∆m[3 × 10^10 (cm/sec)]^2 [(4.43 × 10^18) / (3 × 10^10)^2] [g/mole] = ∆m 4.92 × 10^-3 g/mole = ∆m When a \gamma particle is emitted, one knows that the final state weighs less than the initial state, therefore, ∆m = - 4.92 × 10^-3 g/mole.

Question:

If we have a list of tabular values represented by a one-dimensionalar-ray, then the first differences of the list areformed by subtracting each element except the last [sincenothing follows the last element] from the element immediatelyfollowing it. Suppose we have a one--dimensionalarray named X that contains 50 elements. Compute the 49 elements of another array named DX from DX(I) = X(I + 1) - X(I) I = 1,2,...,49. Write a program segment to perform this calculation.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0511.htm

Solution:

Dimension both arrays according to the given dimensions and then writea simple DO loop to define the difference array DX. Thus we get: DIMENSIONX(50), DX(49) DO 50 I = 1, 49 50DX(I) = X(I + 1) - X(I) STOP END

Question:

A laboratory analyst puts Fehling solution in a test tube and then adds the urine of a diabetic which contains some sugar. On heating, what red copper compound precipitates?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0935.htm

Solution:

Copper forms two series of compounds: Copper (I) (cuprous), and copper II (cupric) compounds. Copper (I) oxide (Cu_2O) occurs naturally as the mineralcuprite. This red oxide may also be obtained by careful oxidation of copper in air or by precipitation of Cu+ ions from solutions with sodium or potassium hydroxide. Copper (I) hydroxide is not known. Copper (I) oxide is also obtained as a re-duction product of Fehling's solution, a reagent used in the diagnosis of diabetes to test for the presence of reducing sugars in the urine. Fehling's solution consists of an aqueous solution of copper (II) sulfate, sodium hydroxide, and sodium potassiumtartrate. Thetartrate forms a complex with Cu^2+. As a result, so few free Cu^2+ are left in solution that no precipitate of copper (I) hydroxide is obtained when the sodium hydroxide is added. Reducing sugars reduce Cu^2+ to Cu+, and Cu_2O separates as a reddish precipitate.

Question:

How has the role of internal body secretions in behavior been demonstrated in silk worms?

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0804.htm

Solution:

Although much of the stimuli which trigger specific behavioral responses have an external origin, many responses occur as a result of internal stimulation. Much of this internal stimulation is initiated by hormones. The silkworm larva must spin its cocoon at a certain time in life, just following its larval stage. The stimulus for the spinning behavior comes from the disappearance of a specific substance, juvenile hormone. This is secreted by the corporaallata, a pair of glands located behind the brain of the silkworm. The secretion of juvenile hormone prevents the spinning of the cocoon. As the larva grows, the level of juvenile hormone drops, allowing for the spinning reaction to occur. No external sign stimuli is necessary, only internal hormonal levels. The cocoon-spinning behavior of the larvae of silk worms is an example of a behavior under the control of hormones.

Question:

100 ml of 0.1 M NaOH is mixed with 100 ml of 0.15 M HOAC. K_a = 1.75 × 10^-5 for acetic acid, what is the pH of this mixture?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0421.htm

Solution:

To find the pH of the mixture, first determine the H_3O^+ concentration, since pH = - log [H_3O^+]. To find H_3O^+, consider what happens when acetic acid is added to NaOH, a base. When an acid is added to a base, a neuralization reaction occurs such that water and a salt are produced. Thus, This reaction does not show the presence of any H_3O^+ ions, which may lead one to incorrectly conclude that its con-centration was zero. There are 100 ml of 0.15 M HOAC. M = Molarity = moles/liter. Thus, 0.15 = moles/0.10 liters (1 liter = 1000 ml). Therefore,(0.15) (0.10) = 0.015 moles of HOAC. For NaOH, M = 0.10 = moles/0.10 liters. Thus, the number of moles of NaOH = (0.10) (0.10) = 0.010. From the previously written neutralization equation, one mole of acid reacts for every mole of base. Only 0.01 moles of HOAC could react with NaOH. Thus, 0.015 - 0.010 = 0.005 moles of HOAC was NOT neutralized by the base, NaOH. This excess 0.005 moles of HOAC is, in the presence of its salt NaOAC. Thus, it is a buffer solution. Solutions that con-tain a weak acid, such as HOAC, and its salt are called buffers. Therefore, consider the dissociation of acetic acid; it is in excess and is the source of the hydronium ion concentration. HOAC + HOH \rightleftarrows H_3O^+ + OAC^-. The equilibrium constant for this reaction measures the ratio of the concentration of the products to reactants, each raised to a power based on its coefficient in the chemical equation. Thus, K_a , the equilibrium constant, = {[H_3O^+] [OAC^-]} / [HOAC]. Water is not included since it is a constant. One is solving for [H_3O^+] to find pH, let x = [H_3O^+]. From the dissociation equation x = [OAC^-] also, since they are formed in equimolar amounts. Recall, however, that the neutralization reaction produced OAC^- also. There were 0.010 moles each of NaOH and HOAC which reacted. From the equation, 0.010 moles of OAC^- should have been produced. The volume was 200 ml. Thus, M = concentration = (0.01 moles) / (200 ml / 1000 ml/l) =0.05 M. The total concentration of OAC^-, from the acid dissociation and neutralization, is x + .05. The acetic acid concentration is the initial amount minus the amount that dissociated. Recall, there were 0.005 moles of acetic acid in 200 ml. to give (0.005 moles)/ (200 ml / 1000 ml/l) = 0.025M. x moles/liter of it dissociated. Thus, the final concentration is (0.025 - x). K_a = {[H_3O^+] [OAC^-]} / [HOAC]. 1.75 × 10^-5 = {[H_3O^+] [OAC^-]} / [HOAC]. Substituting final concentrations, 1.75 × 10^-5 = {(x) (x + 0.5)} / (0.025 - x). The excess OAC^-, from the dissociation of HOAC, x, is small, Thus, {(x) (x + 0.05)} / (0.025 - x) \approx (0.050 x) / (0.025) = 1.75 × 10^-5. Solving for x, one finds [H_3O^+] = x = 8.8 × 10^-6 M. Thus, pH = - log [H_3O^+] = - log [8.8 × 10^-6] = 5.06.

Question:

Write a program in PL/I to calculate a table of returns on an investment of original amount A which is compounded yearly at P percent return, for Y years.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0445.htm

Solution:

After Y years the original amount A will grow to T =A(1 + P/100)^Y The program below reads in values of A,P, and Y. Then it calculates the corresponding value of T and prints out A, P, Y, and T. /\textasteriskcenteredCOMPOUND INTEREST\textasteriskcentered/ ON ENDFILE (SYSIN) GO TO L60 L10:GET LIST (A, P, Y); L20:T = A\textasteriskcentered(1 + P/100) \textasteriskcentered\textasteriskcenteredY; L30:PUT EDIT (A, P, Y, T) (F (16,4)) ; L40:PUT SKIP; L50:GO TO L10; L60:END

Question:

If 1.588 g of nitrogen tetroxide gives a total pressure of 1 atm when partially dissociated in a 500-cm^3 glass vessel at 25\textdegree, what is the degree of dissociation, \alpha? What is the value of K_p ? The equation for this reaction is N_2O_4\rightarrow 2NO_2 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0324.htm

Solution:

The determination of the density of a partially dissociated gas provides one method for measuring the extent to which the gas dissociates. When a gas dissociates, more molecules are produced, and at constant temperature and pressure the volume increases. The density at constant pressure then decreases, and the difference between the density of the undissociated gas and that of the partially dissociated gas is directly related to the degree of dis-sociation. If \alpha represents the fraction of gas dissociated, then 1 - \alpha represents the fraction of gas undissociated.Assuming one starts with 1 mole of gas the number of moles of gaseous products in the balanced chemical equation is 1 +\sumv_i . (where \sumv_i is the summation of all the coefficients; positive coefficients for the products and negative co-efficients for the reactants) . Therefore, the number of moles of gas present at equilibrium is (1 - \alpha) + (1 + \sumv_i )\alpha = 1 + \alpha\sumv_i The density of an ideal gas at constant pressure and temperature is inversely proportional to the number of moles for a given weight and the ratio of the density \rho_1 of the undissociated gas to the density \rho_2 of the partially dissocciated gas is given by the expression \rho_1 / \rho_2 = 1 + \alpha \sumv_i \alpha = (\rho_1 - \rho_2 ) / (\rho2\sumv_i) If there is no dissociation, then \alpha = 0 and \rho_1 = \rho_2 , if dissociation is complete, then \alpha = 1, \rho2\sumv_i = \rho_1 - \rho_2 and \rho_1 = (1 + \sumv_i ) \rho_2 . Molecular weights are proportional at constant tem-perature and pressure to the gas densities. Therefore, molecular weights can be substituted for the densities: \alpha = ( M_1 - M_2 ) / (M_2 \sumv_i) where M_1 is the molecular weight of the undissociated gas, and M_2 is the average molecular weight of the gases when the gas is partially dissociated. The dissociation of nitrogen tetroxide proceeds by the following equation: N_2O_4= 2NO_2andK_p = (P^2 _N(O)2) / (P_(N)2(O)4) The degree of dissociation is \alpha; (1 - \alpha) is pro-portional to the number of moles of undissociated N_2O_4; 2\alpha is proportional to the number of moles of NO_2 ; and (1 - \alpha) + 2\alpha or 1 + \alpha is proportional to the total number of moles. If the total pressure of N_2O_4plus NO_2 is P, the partial pressures are: P_N2O4 = { (1 - \alpha) / (1 + \alpha) } P and P_(NO)2 = { 2\alpha / (1 + \alpha) } P Then,K_p = [{2\alpha / (1 + \alpha)} P]^2 / [{ (1 - \alpha) / (1 + \alpha) } P] = (4 \alpha^2 P) / (1 - \alpha)^2 M_1 = 92.02 g/mole. Using the Combined gas Law, PV = n RT and setting n, the number of moles of substance, equal to the mass of substance, m, divided by its molecular weight, M_2 , one obtains PV = (m / M_2 )RT Rearranging and solving for M_2 , M_2 = (RT/P) × (m/V) M_2 = (RT/P)(m/V) = {(0.082 liter atm \textdegreeK-^1 mole-^1 ) (298\textdegreeK) (1.588 g)} / {(1 atm ) (0.500 liter)} = 77.68 g/mole \alpha = ( M_1 - M_2 ) / (M_2 \sumv_i) where \sumv_i for the reaction equals [(coefficient of NO_2 ) - (coefficient of N_2O_4) = (2 - 1) = 1] and M_1 = 92.02 and M_2 = 77.68. \alpha= (92.02 - 77.68) / 77.68 = 0.1846 K_p = (4 \alpha^2 P) / (1 - \alpha^2) = { 4 (0.1846)^2 (1 atm) } / {1 - (0.1846)^2 } = 0.141.

Question:

As part of a certain program, there is a declare statement as follows: DCL (A,C) FIXED(5,2), B CHAR(5) VAR, D FIXED(2,1), E CHAR(4), F BIT(7) VAR, G FIXED (3,2); Also, as part of the same program, there is an EDIT-directed output statement as follows: PUT EDIT(A,B,C,D,E,F,G) (LINE(40), COL(15), F(7,2), PAGE, A(3), F (6,2), LINE(10), F(7,3), X (3), A(5), COL(30), B(6), SKIP(7), X(5) , F(3,1) ) ; Show what will be printed out, if the contents of the memory storage are as follows: Name of Memory location Contents A - 123\textbullet45 B B2GLb b C 123\textbullet45 D - 1\textbullet8 E CARS F 1011bbb G - 7\textbullet94 Note:bis a symbol used to denote a blank space.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0325.htm

Solution:

The PUT EDIT statement states that the values for the variables A, B, C, D, E, F and G are to be printed out. The format in which these are to be printed out is given in the second set of parentheses ( ). The first format instruc-tion is LINE(40). Hence, the printer skips to line (40) on the current page. This assumes that the printer did not pass line 40 already. If the printer was indeed past line 40, e.g., on line 45, then, the printer skips the rest of the page and goes to line 40 on the next page. Now, the next format in-struction says COL(15), which is a short form for COLUMN(15). This is an instruction to the printer to go to column 15 on the current line, which is line 40. However, if in case column 15 has already been passed as a result of carrying out some previous format instruction, the printer then skips to col. 15 on the next line. That is, it would skip to column 15 on line 41 in this case if the present position of the printer was, for example, column 59 on line 40. Thus, after having moved to the correct page, line and column, the computer reads the next format instruction which specifies F(7,2). This specification of F(7,2) corresponds to the first element A in the data list of the PUT EDIT statement. Therefore, the computer retrieves the value for A from its memory. This value is - 123\textbullet45, and it agrees with the specification of the F(7,2) format, which requires a number of the type XXXX\textbulletXX. Hence, starting from column 15, of the current line (say, line 40), the printer prints out - 123\textbullet45. After this printing, the printer is on column 22 (15 + 7 = 22). The minus sign and the decimal point also count as one column position in - 123\textbullet45. Now the computer wants to print out the next data item, which is B. Therefore, it checks the format list and sees that the next format instruction says PAGE. This causes the printer to leave blank the rest of the current page and go to the first line, first column of a new page. Next comes the instruction A(3), which means, that the data item B has a character field of 3 characters to be printed. Thus, the value of B, which is, B2GLbis retrieved from memory and as only 3 characters are to be printed, the printer prints B2G. The L is cut off, and the printer now has come to column 4 of the first line of the new page. The next format instruction F(6,2) is for the third item C from the data list of the PUT EDIT statement. There-fore the value of C is fetched from memory as 123\textbullet45. The format specification requires a form XXX\textbulletXX. Hence, 123\textbullet45 is printed out from column 4 onwards, and the printer comes to column 10. Next comes the turn for item D. For this, the first format instruction specifies LINE(10). Therefore the printer jumps from the current LINE(1) to LINE (10). The next format instruction is F(7,3). The value for D from memory is - 1\textbullet8 now, F(7,3) requires a form XXX\textbulletXXX. Therefore - 1\textbullet8 is con-verted to 0 - 1\textbullet800. Now, the print out is: 0 - 1\textbullet800, starting at column 1. Note that the leading zero is suppressed. This brings the printer to column 8. Now, the next item on the data list, which is, E is to be printed out. The first format instruction for E says X(3). This means that 3 columns are to be skipped. The printer now skips from the present column 8 to column 11. Now, a character field A(5) is specified. The value fetched for E from memory is CARS, which has only 4 characters, therefore a blank is added at the end of CARS to make up the 5 characters specified by the format. The printer now prints out CARSb(note that the 5th character is a blank). b The printer thus comes as a result to column 16 of line 10. Next, the item F is to be printed. For this, the format first specifies COL(30); as a result the printer jumps from column 16 to column 30 of the same line 10. Then, the computer reads the specification B(6), which means that a Binary bit string is to be printed out from memory. The value for F is obtained as 1011bbb. This string has 4 characters, and 3 blanks, but the format specification, B(6), requires 6 bits to be printed out. Hence, zeroes are filled in to make up the required string length, and the printer prints out 101100. Now, the last item on the data list, which is G is to be printed out. The format specification for G specifies first a SKIP(7). This causes the printer to jump 7 lines from the present line 10 to line 17. The next format specification says X(5). Hence, the printer jumps 5 columns from column 1 of line 17 to column 6 on the same line 17. The next format instruction specifies F(3,1) for G. The value for G as obtained from memory is -7.94. But the format specification requires a form X\textbulletX. Hence, the value of G is truncated at both ends, and the printer prints out 7\textbullet9. Note that the minus sign and the digit 4 have been truncated. The truncation of the decimal place (digit 4) is not very serious. At most is affects the accuracy of the number. However, the truncation of the minus sign from the left is very serious. It changes a negative number into a positive number. Trunca-tion from the left can also cut off the higher order digits. Hence, the computer is programmed to regard truncation of a number from the left as an error condition. This error is called a SIZE condition. Also, the printer prints out a warning that a SIZE condition has occurred. To summarize, the print-out for this problem would be as shown in fig. 1.

Question:

A string under a tension of 256newtonshas a frequency of 100 vibrations per second. What is its frequency when the tension is changed to 144newtons?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0816.htm

Solution:

The tension is only (144/256) as much as it was at first. The square root of (144/256) is (12/16), or (3/4) . Since the frequency is directly proportional to the square root of the tension, the new frequency is (3/4) × 100, or 75vps.

Question:

Using FORTRAN rules what would be the value of sum in each case.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0275.htm

Solution:

The rules adopted by the creators of FORTRAN state that evaluation of arithmetic operations is conducted from left to right, except when the succeeding operation has a higher "binding power" than the one currently being considered. The binding power of "+" and "-" is the lowest in strength; operators "{_\ast}" and "/" are of middle strength, and raising to a power ({_\ast}{_\ast}) has the highest priority. The concept of parenthesizingsubexpressionswas added to the rules to achieve a sequence in operations different from the rules of binding power of operators. Therefore, using this knowledge we find the solutions to be : to 3. The result is 7.

Question:

The square ABCD has a side of 10 cm length. Equal positive charges of +50 statcoulombs are placed at A and B and equal negative charges of\rule{1em}{1pt}100 statcoulombs are placed at C and D. Calculate the electric field at P, the center of the square.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0591.htm

Solution:

P is the point where the two diagonals AC and BD cross. Let AP = BP = CP = DP = R. Then, applying Pythagoras' theorem to the right angle triangle APB, AP^2 + BP^2 = AB^2 2R^2 = 10^2 cm^2 R^2 = 50 cm^2 R = 7.07 cm The charge q_A at A produces a contribution E_A to the field at P given by definition as EA= q_A / R^2 = 50 statcoul /50 cm^2 Since 1 (statcoul / cm^2 ) = 1 (dyne / statcoul) E_A = 1 dyne per statcoulomb Since the charge at A is positive, E_A points directly away from A, (see figure) . Similarly, the charge at B produces a contribution E_B = 1 dyne per statcoulomb which points directly away from B. The negative charge q_C at C produces a contribution E_C = q_C /R^2 = (100 statcoul) /(50 cm)^2 = 2 dyne per statcoulomb Since q_C is negative this contribution points directly toward C. Similarly, the charge at D produces a contribu-tion E_D = 2 dyne per statcoulomb which points directly toward D. Adding together E_A and E_C, which point in the same direction, we get a vector of magnitude 3 dyne per stat-coulomb pointing toward C. Adding E_B to E_D we get a vector of magnitude 3 dyne per statcoulomb pointing toward D. This is shown in the figure. This figure also gives the triangle of vectors PQR used to add the two vectors shown. Here we are using the parallogram law of vector addition. The resultant field E =PQ^\ding{217} is obviously vertically downward and its magnitude is obtained from Pythagoras' theorem PQ^2 = PR^2 + RQ^2 E^2 = (3^2 + 3^2)dyne^2 / statcoul^2 = 18 dyne^2/statcoul^2 E = 4.24 dyne per statcoulomb The resultant field at P is therefore 4.24 dyne per statcoulomb pointing in the directionAD^\ding{217}.

Question:

A 0.96-lb ball A and a 1.28-lb ball B are connected by a stretched spring of negligible mass as shown in the diagram. When the two balls are released simultaneously, the initial acceleration of B is 5.0 ft/sec^2 westward. What is the initial acceleration of A?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0124.htm

Solution:

If we take the balls and spring as our system, as shown in the figure, no external forces act. The only forces acting are then internal, and consist of the force of m_A on m_B, and the force of m_B on m_A,(f\ding{217}_ABand f\ding{217}_BA, respectively). But by Newton's Third Law, f\ding{217}_AB = -f\ding{217}_BA(1) at each instant of time. By Newton's Second Law, f\ding{217}_AB = m_B a\ding{217}_B f\ding{217}_BA = m_A a\ding{217}_A(2) since f\ding{217}_AB is the only force on m_B, and similarly for f\ding{217}_BA and m_A. Using (2) in (1) m_B a\ding{217}_B = -m_A a\ding{217}_A orm_B g a\ding{217}_B = -m_A g a\ding{217}_A(3) Butm_B g = W_B m_A g = W_A where W_A and W_B are the weights of A and B. Hence, (3) yields W_B a\ding{217}_B = -W_A a\ding{217}_A Solving for a\ding{217}_A, a\ding{217}_A = -(W_B/W_A) a\ding{217}_B(4) To find a\ding{217}_A at t = 0, we use the value of a\ding{217}_B at t = 0 in (4) giving a\ding{217}_A (t = 0) = -{1.28 lb/.96 lb}(5 ft/s^2 westward) a\ding{217}_A (t = 0) = -6.7 ft/s^2 westward a\ding{217}_A (t = 0) = 6.7 ft/s^2 eastward since eastward is the opposite of westward.

Question:

The process of ovulation continues until a period of the female's life called the menopause. What are the changes that occur during this period?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0564.htm

Solution:

The menopause usually comes between the ages of 45 and 50. It is apparently attributable to two reasons: (1) the cessation of secretion of LH by the pituitary as a result of declining activity of the LH- stimulating center in the hypothalamus, and (2) the declining sensitivity of the ovaries to the stimulatory activity of gonadotropins. In the absence of LH, secretory activity in the ovaries falls off. Neither estrogen nor progesterone can be produced in significant quantities. Ovulation and menstruation become irregular and ultimately cease completely. Some secretion of estrogen generally continues beyond these events but gradually diminishes until it is inadequate to maintain the estrogen-dependent tissues, most notably the breasts and genital organs, which become atrophied. Sexual drive, however, is frequently not diminished and may even be heightened. Severe emotional disturbances are not uncommon during menopause and are generally ascribed not to the direct effect of estrogen deficiency, but to the disturbing nature of the wholeperiod-the awareness that reproductive potential has ended. Woman during menopause often experience hot flashes. These result from dilation of the skin arterioles, causing a feeling of warmth and marked sweating. The reason for this is unknown. Many of the symptoms of the menopause can be reduced by the administration of small amounts of estrogen. Just recently, another aspect of menopause was discovered, that being the relationship between plasma estrogen and plasma cholesterol. Estrogen significantly lowers the level of plasma cholesterol. This explains why women have a significantly lower incidence of arte-riosclerosis (a disease positively correlated to the level of cholesterol in the plasma) than men until the menopause, when the incidence becomes similar in both sexes. Furthermore, decreased levels of estrogens are implicated in osteoporosis since estrogen normally stimulates osteoblasts to form new bone.

Question:

The unaided eye can perceive objects which have a diameterof 0.1 mm. What is the diameter in inches?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0002.htm

Solution:

From a standard table of conversion factors, one can find that 1 inch= 2.54 cm. Thus, cm can be con-verted to inches by multiplying by 1 inch/2.54cm. Here, one is given the diameter in mm, which is .1 cm.Milli-- metersare converted to cm by multiplying the number of mm by .1 cm/1 mm. Solving for cm, you obtain: 0.1 mm × .1 cm/1 mm = .01 cm. Solving for inches: .01 cm × [(1 inch) / (2.54 cm)] = 3.94 ×10\Elzbar^3 inches.

Question:

One group of compounds that may become quite important in birth control is the prostaglandins. What are the advantages of prostaglandins in birth control?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0561.htm

Solution:

The compounds known as prostaglandins have been identified as cyclic, oxygenated, 20-carbon fatty acids. There are now 16 known natural prostaglandins. The prostaglandins, which are secreted by the seminal vesicles and other tissues, appear to mediate hormonal action by influencing the formation of cyclic AMP. The end result of their actions is varied. Though they are expected to be valuable in treating many ailments, from nasal congestion to hypertension, their most significant use may be in birth control. Prostaglandins induce strong contractions in the uterus, and hence they may be powerful agents for abortion. As abortive agents, the prostaglandins efficient in almost all of the cases where they are used. When more is known about the side effects of pros-taglandins, it is believed that their application in abortion would be safer than mechanical means. The pros-taglandins are also being considered as a once-a-month replacement for birth control pills. The powerful uterine contractions induced by prostaglandins prevent the fertilized ovum from being implanted. Since they can be introduced directly into the uterus through the vagina, the prostaglandins will probably be safer contraceptives than the oral chemical agents which must travel through the blood and may affect other parts of the body.

Question:

\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Three forces acting at a point are F_1\ding{217} = 2i˄ - j˄ + 3k˄, F_2\ding{217} \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= - i˄ + 3j˄ + 2k˄, and F_3\ding{217} = - i˄ + 2j˄ - k˄. Find the direction \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2and magnitudes of F_1\ding{217} + F_2\ding{217} + F_3\ding{217}, F_1\ding{217} - F_2\ding{217} + F_3\ding{217}, and \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} + F_2\ding{217} - F_3\ding{217}. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0002.htm

Solution:

\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2When vectors are added (or subtracted) , their components in the \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2directions of the unit vectors add (or subtract) algebraically. Thus since \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} = 2i˄ - j˄ + 3k˄, F_2\ding{217} = - i˄ + 3j˄ + 2k˄, F_3\ding{217} = - i˄ + 2j˄ - k˄, \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2then it follows that \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} + F_2\ding{217} + F_3\ding{217} = (2 \"{\i}\textquestiondown1⁄2 1 \"{\i}\textquestiondown1⁄2 1)i˄ + (-1 + 3 + 2)j˄ + (3 + 2 \"{\i}\textquestiondown1⁄2 1)k \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= 0i˄ + 4j˄ + 4k˄. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Similarly, \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2F_1\ding{217} - F_2\ding{217} + F_3\ding{217} = [2 \"{\i}\textquestiondown1⁄2 (- 1) \"{\i}\textquestiondown1⁄2 1]i˄ + [- 1 \"{\i}\textquestiondown1⁄2 (3) + 2] j˄ + [3 \"{\i}\textquestiondown1⁄2 (2) \"{\i}\textquestiondown1⁄2 1]k˄ \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= 2i˄ - 2j˄ + 0k˄ \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 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It therefore has a magnitude of \surd(4^2 + 4^2) units = 4 \surd2 units \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2= 5.66 units, and lies in the y- z plane, making an angle \texttheta with the \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2y - axis, as shown in figure (a), where tan \texttheta = 4/4 = 1. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Thus \texttheta = 45\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Similarly, F_1\ding{217} - F_2\ding{217} + F_3\ding{217} has a magnitude of 2 \surd2 units = 2.82 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2units, and lies in the x - y plane, making an angle \textphi with the x-axis, as \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2shown in figure (b), where tan \textphi = + 2/-2 = - 1. Thus \textphi = 315\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2Also, F_1\ding{217} + F_2\ding{217} - F_3\ding{217} has a magnitude of \surd(2^2 + 6^2) units = 2 \surd10 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2units = 6.32 units, and lies in the x - z plane at an anglexto the x-axis, as \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 x \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2shown in figure (c), where tanx= 6/2 = 3. Thusx= 71\"{\i}\textquestiondown1⁄234\"{\i}\textquestiondown1⁄2. \"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2\"{\i}\textquestiondown1⁄2 x x

Question:

The transformation of a spermatid into a functioning sperm is a process of differentiation. It varies in detail in different species, but the basic features are the same. Outline this process of differentiation in humans.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0574.htm

Solution:

Within the cytoplasm of the human spermatid, one can find proacrosomic granules, centrioles, and mitochondria. During differentiation of the spermatid, a process called spermiogenesis, the proacrosomic granules, which were formed from the Golgi complex, gather toward one pole of the cell and condense into the form of a bead. This acrosomal bead contains appreciable amounts of carbohydrates and several enzymes of lysosomal nature. This region of the differentiating spermatid eventually becomes the acrosome, and is chiefly concerned with enzymatic penetration of the egg. The two centrioles of the spermatid move toward the opposite side of the nucleus. Here the centriole nearest the nucleus spins a filament that grows outward and becomes the axis of the sperm's tail or flagellum. This flagellum shows the usual 9 + 2 tubule pattern, and is often accompanied by additional fibrous structures that run alongside the tubules. (It is interesting to note that whereas basal bodies usually form flagella, here the centriole forms the flagellum.) The centrioles then move apart. The proximal one remains close to the posterior surface of the nucleus. The distal one forms a ring around the axial filament and recedes a short distance. Some of the mitochondria now gather around the axial filament in the space between the two centrioles and form a spiral in an area called the middle piece. Most of the cytoplasm pinches away and is discarded, but a little remains as a delicate sheath over the head area. Thus the resultant sperm consists of 3 pieces. The head con-tains an anterior acrosome, a posterior nucleus, and little cytoplasmic material, and by virtue of its lytic enzymes, functions in penetration of the egg. The midpiece contains the mitochondria necessary for energy-requiring motility. Finally, the tail with a 9 + 2 microtubule arrangement, contains the contractile machinery for whiplike movements. It was shown that before fertilization, sperm in the vicinity of an ovum in the oviduct undergo a sequence of structural changes called the acrosome reaction. The outer membrane of the acrosome fuses at multiple points with the overlying plasma membrane of the sperm head to create numerous openings, through which the enzyme-rich contents of the acrosome are liberated in a process not unlike the release of secretory products from a glandular cell. The released enzymes help the sperm to digest its way through the protective coating of the egg. The result is the entry of the sperm into the egg cytoplasm.

Question:

An electron initially at rest is accelerated though1cm by an electric field of 3×10^4 V/m. What is the terminal speed?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0621.htm

Solution:

In order to solve for kinematical variables such as position, velocity, and acceleration, we must use Newton's Second Law to relate the net force on our system to its acceleration. We then integrate this equation to obtain v(t) and x(t). In this problem, the only force acting on the electron is that due to the electric field E^\ding{217}. By definition E^\ding{217} = F^\ding{217}/q where F^\ding{217} is the force acting on q. Hence, F^\ding{217} = qE^\ding{217} Since the electron charge is - e, where e = 1.6 x 10^\rule{1em}{1pt}19 coul. we may write F^\ding{217} = -eE^\ding{217} But, by the second law, F^\ding{217} = ma^\ding{217} where m and a^\ding{217} are the electron mass and acceleration, respectively. Therefore, ma^\ding{217} = \rule{1em}{1pt} eE^\ding{217} ora^\ding{217} = (\rule{1em}{1pt} e / m) E^\ding{217} For our problem, E^\ding{217} is constant. Since e and m are also constants, the same must be true of a^\ding{217}. Looking at the figure, we see that E^\ding{217} is in the negative x direction. If \^{\i} is a unit vector in the positive x direction E^\ding{217} = E\^{\i} anda^\ding{217}= (\rule{1em}{1pt} e/m) (\rule{1em}{1pt} E\^{\i}) = (eE / m)\^{\i}(1) Noting that a^\ding{217} is only in the x direction we may drop the vector notation in (1) and resort to scalar notation. a = eE/m(2) where a is positive when in the positive x direction. But we want to know v in terms of x. Using the fact that a= dv/dt = [dv/dx] [dx/dt] =[dv/dx] (v) we obtaina = v(dv / dx) oradx = vdv Using (2), (eE/ m) dx = vdv (eE/ m) ^(x)f\int_(x)0 dx =^(v)f\int(v)0vdv(3) where the subscripts "f\textquotedblright and "\textdegree" indicate final and initial values, respectively. Integrating(3) (eE/m) (xf\rule{1em}{1pt} x_0) = 1/2 v2(v)f│_(v)0 = 1/2 (v2f\rule{1em}{1pt} v2_0) Solving for vf (eE/m) (x_f \rule{1em}{1pt}x_0) = 1/2 (v2f\rule{1em}{1pt} v2_0) (2eE/ m) (x_f \rule{1em}{1pt} x_0) + v2_0 = v2f vf= \surd[(2eE/m) (x_f \rule{1em}{1pt} x_0) + v^2 _0 ] But v_0 = 0 since the electron is initially at rest, and x_f \rule{1em}{1pt} x_0 = 1 cm. Therefore, v_f = \surd[{2(1.6 × 10^\rule{1em}{1pt}19 coul) (3× 102v/cm) (1cm)} / (9.11 × 10^\rule{1em}{1pt}31 kg)] v_f = \surd(1.054 × 10^14 m/s^2 ) v_f = 1.03 × 10^7 m/s

Question:

Find the angular width of the central maximum of a Fraunhofer dif-fraction pattern if the light used has wavelength \lambda = 6000\textdegreeA and the slit width is .1 × 10^-3 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D30-0904.htm

Solution:

The angular location of the minima of a Fraunhofer diffrac-tion pattern is given by sin \psi = m\lambda/a(m = 1,2,...) where a is the slit width. The central maximum is flanked on either side by the first order minima (m = 1). The angular position of this minimum is sin \psi = [{(1) (6000\textdegreeA)} / {(.1×10^-3 m)}] Since1\textdegreeA = 10^-10 m sin \psi = (6 × 10^3 × 10^-10 m) / (1 × 10^-4) m sin \psi = 6 × 10^-3 = .006 \psi = 1/3\textdegree Hence, the angular breadth of the central maximum is (see figure) 2\psi = 2/3\textdegree.

Question:

An elevator is accelerated upward at 2 ft/sec^2. If the elevator weighs 500 lb, what is the tension in the supporting cable?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0110.htm

Solution:

The net force acting on the elevator is \sumF = T - mg where T is the cable tension, and mg is the elevator's weight. (Note that the positive direction is taken as upward). By Newton's Second Law, this must equal the product of the elevator's mass and acceleration, whence T - mg = ma Solving for T(1) T = m(g + a) We don't know the mass, m, of the elevator, but we do know its weight W. Since W = mg (1) becomes T = (W/g) (g + a) Using the data provided T = [(500 lb) / (32 ft/s^2)](32 ft/s^2 + 2 ft/s^2) T = (34/32) 500 lb T = 531.2 lb.

Question:

atom, isotope, ion. Could a single particle of matter be all three si-multaneously?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0005.htm

Solution:

An atom is the smallest particle of an element that can retain the chemical properties of that element. It is composed of a nucleus, which contains positively charged protons and neutral neutrons, around which negatively charged electrons revolve in orbits. For example, a helium atom contains 2 protons, 2 neutrons, and 2 electrons. An ion is a positively or negatively charged atom or group of atoms. An ion which is negatively charged is called an anion, and a positively charged ion is called a cation. Isotopes are alternate forms of the same chemical element. A chemical element is defined in terms of its atomic number, which is the number of protons in its nucleus. Isotopes of an element have the same number of protons as that element, but a different number of neutrons. Since atomic mass is determined by the number of protons plus neutrons, isotopes of the same element have varying atomic masses. For example, deute-rium (^2H) is an isotope of hydrogen, and has one neutron and one proton in its nucleus. Hydrogen has only a proton and no neutrons in its nucleus. A single particle can be an atom, an ion, and an isotope simultaneously. The simplest example is the hydrogen ion H+ . It is an atom which has lost one electron and thus developed a positive charge. Since it is charged, it is therefore an ion. A cation is a positively charged ion (i.e. H+) and an anion is a negatively charged ion (i.e. Cl-). If one compares its atomic number (1) with that of deuterium (1) , it is seen that although they have different atomic masses, since their atomic numbers are the same, they must be isotopes of one another.

Question:

Explain how the metabolism of a fermenting yeast cell differs from that of a fatigued muscle cell.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0077.htm

Solution:

We shall first examine the anaerobic metabolism of yeast. When yeast is placed in an environ-ment devoid of oxygen, their cells are able to extract energy from glucose in the absence of molecular oxygen via glycolysis, which occurs in both aerobic and anaerobic organisms. There are no hydrogen acceptors due to the absence of molecular oxygen, the final acceptor. When O_2 is not available, all intermediates of the respiratory chain are reduced, and the TCA cycle and electron transport system are turned off. Pyruvate, the degradation product of glucose, cannot therefore be oxidized to acetyl CoA and enter the TCA cycle for further energy extraction. Instead, pyruvate itself accepts the electrons (via acetaldehyde) and is reduced to ethyl alcohol (ethanol). This process, known as alcoholic fermentation, converts each glucose molecule to two ethanol molecules and two CO_2. Alcoholic fermentation also serves to regenerate oxidized NAD+, so that glycolysis can continue. Two net ATP's are formed per glucose molecule in alcoholic fer-mentation. In man and other organisms, anaerobic respiration can supplement aerobic respiration for short periods of time. In a fatigued muscle cell, pyruvic acid, produced via glycolysis, again serves as an electron acceptor since oxygen is unavailable. However, unlike yeast, human cells produce lactic acid instead of ethanol and carbon dioxide: This process, sometimes called homolactic fermentation, occurs during strenuous exercise when muscle cells cannot obtain enough oxygen to support aerobic respiration. Instead, the energy expended comes from glycolysis (2 ATP per glucose molecule) with the build-up of lactic acid, the cause of muscle fatigue. Although both processes, alcoholic and homolactic fermentation, enable energy production, they are not as efficient as aerobic respiration. A yeast cell or skeletal muscle cell can theoretically obtain 18 times as much ener-gy by oxidizing a glucose molecule completely to carbon di- oxide and water, instead of stopping at ethanol, if it em-ploys the aerobic type of respiration.

Question:

What is the volume in liters of a rectangular tank which measures2.0 m by 50 cm by 200 mm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0009.htm

Solution:

One liter is equal to 1000cc; therefore, one should find the volume ofthe tank in cubic centimeters first and then convert to liters. This method isbest for this problem because the sides of the tank are given in units whichcan quickly be converted to centimeters. There are 100 cm in 1 m.Thus ,2 m is equal to 2.0 m × 100 cm/1 mor 200 cm. There are 10 mm in 1cm,therefore 200 mm is equal to 200 mm×1 cm/10 mm or 20 cm. Solving for the volume of the tank in cubic centimeters: volume= 200 cm × 50 cm × 20 cm = 2.0 × 10^5 cc. To convert from cubic centimeters to liters multiply by 1 liters/1000cc. volumein liters = 2.0 × 10^5 cc × 1 liters/1000cc = 200 liters.

Question:

Define the operating system, and describe the resource man-agement functions it carries out.

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Solution:

An operating system is a set of programs that: a) Provides an interface between the users and the hardware. This generally means creating a more hospitable environment for the users. Examples are interactive use of the computer, convenient execution of the programs, and receiving meaningful error messages when programs fail to run. b) Manages the sharing of computer resources. This means the management of hardware, software, and information reliably and efficiently, as discussed next. OS Resource Management Functions: The operating system (also called the executive, monitor, master, supervisor, etc.) manages the following resources: 1. CPU or Multiple Processing Elements (PEs) : This is done by multiprogramming and multitasking when there are several PEs. Multitasking refers to loading (assigning tasks to) of process-ing elements in a balanced manner. The CPU scheduling is used to assign and reclaim CPU to processes. 2. Primary Memory and Cache Memory : This is done through vari-ous allocation anddeallocationof memory to user programs. One popular method is demand paging, where fixed-size program sec-tions (pages) are allocated to the equal size physical memory regions (frames) on demand. Cache memories (faster than regular primary memory) are used to keep track of allocation of pages to frames. 3. Secondary Storage and Files: The disk space is managed through contiguous, linked, or indexed allocation methods. Some-times, the amount of free disk space is kept track of using bitmaps (sequence of bits indicating whether a disk block is free or used) . The disk space management is part of the file management functions. File management allows creation, move, copy, and deletion of files. In addition, it supports file op-erations such as open, read, write, position, append, close. 4. I/O and Communication Devices : The Operating System (OS) keeps track of the I/O devices such as tapes, terminals, and printers and communication devices such as channels, multiplex-ers, and modems. The OS assigns these devices to user processes in a fair, reliable, and efficient manner; and reclaims them when they are no longer needed. 5. Data and Information : The operating system also keeps track of and allows efficient sharing of data such as system buffers and stacks contents, the messages stored in mailboxes (shared communication buffers), and the like. The information management functions include the accounting for CPU usage, and system status information such as the time, date, directory contents, available disk space, free memory space, etc.

Question:

Discuss the separate half-reactions and net cell reaction that occur in a flashlight battery. If you have a current of .01 amp, how long can the current flow if the battery contains 25.0g of a paste of MnO_2 , NH_4Cl, and ZnCl_2 of which 20% is MnO_2 ?

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Solution:

A standard flashlight dry cell, or Leclanche cell, is composed of a graphite rod in a moist MnO_2 - ZnCl_2- NH_4Cl paste, all in a zinc wrapper. The zinc wrapper serves as the electrode for the oxidation half-- cell (the anode), while the graphite rod is the electrode for the reduc-tion half-cell (the cathode). Thus, the probable half-reactions are: at the anode: Zn(s) \rightarrow Zn2++ 2e^- and at the cathode: 2MnO_2(s) + 2H_3O^+ + 2e^- \rightarrow 2MnO(OH) + 2H_2O . The net reaction for the overall process is Zn(s) + 2MnO_2 \rightarrow ZnO \bullet Mn_2O_3 . To determine how long the current can flow, calculate the number of moles of MnO_2 present and the number of Faraday's required to reduce it per mole. You are told that 20% of 25 grams of paste is MnO_2.Thus, you have (.20)(25g) = 5.0g of MnO_2 .The molecular weight of MnO_2 = 86.94g . Thus, you have (5g) / (86.94g /mole) = .0575 moles of MnO_2 . From the net equation, you see that 2 electrons are transferred. A Faraday = 96,500 coul and is capable of reducing one equivalent posi-tive charge, which means Avogardro's number of individual unit charges (one mole of electrons). Since 2 electrons were transferred per 2 moles, it takes 2 Faraday's to reduce 2 moles of MnO_2 . According to the net equation 2 moles of MnO_2 are reduced for every mole of ZnO \bullet Mn_2O_3 obtained. Thus, you can write that the current will last for (.0575 moles MnO_2) (2F / 2moles MnO_2) (96,500 coul / F) (amp /coul/sec) (1 / .01 amp) = 154 hours.

Question:

A dc series motor operates at 120 volts and has a resistance of 0.300 ohm. When the motor is running at rated speed, the armature current is 12.0 amp. What is the counter EMF in the armature?

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Solution:

An electric circuit containing a series motor (that is a motor whose coil resistance is in series with the motor) is drawn schematically above. The motor draws current, and uses it to produce the mechanical motion of a shaft. By Kirchoff's voltage law, the sum of the voltage drops around the circuit must be zero. Using this fact along with Ohm's Law (V = IR) we obtain V = \epsilon + IR or\epsilon = V \rule{1em}{1pt} IR = 120 volts \rule{1em}{1pt} 12.0 amp × 0.300 ohm = 116 volts.

Question:

Distinguish between a hormone and a vitamin, and a hormone and an enzyme.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0531.htm

Solution:

A hormone is an organic substance synthesized by a specific organ or tissue and secreted into the blood, which carries it to other sites in the body where the hormonal actions are exerted. Vitamins are chemical compounds, a small quantity of which is essential for the survival of certain organisms. Unlike hormones, most vitamins cannot be synthesized by the body but must be obtained from external sources (i.e. diet). Like hormones, vitamins are necessary in only small quantities to ensure proper body functioning. Vitamins ordinarily function as coenzymes, which are non-protein portions of enzymes that help in binding the enzyme to the substrate. Enzymes are organic compounds that function as bio-chemical catalysts produced by living organisms, and are primarily constituted of protein. Unlike hormones which may travel considerable distances to target sites, enzymes generally are used at, or very near to, their site of production. For example, pancreatic enzymes exert their effect in the duodenum, which is nearby the pancreas. Protein hormones that act through the adenylcyclasesystem generally elicit their responses by activating certain enzymes in their target cells. Whereas enzymes are always made up of protein, hormones may consist of amino acids or steroids. Also, enzymes are not released into the blood as are hormones.

Question:

A wire 0.5 mm in diameter is stretched along the axis of a cylinder 5 cm in diameter and 25 cm in length. The wire is maintained at a temperature of 750\textdegreeK by passing a current through it, the cylinder is kept at 250\textdegreeK, and the gas in it has a thermal conductivity of 6 × 10^-5 cal\bulletcm^-1\bulletC deg^-1\bulletsec^-1. Find the rates at which the heat is dissipated both by conduction through the gas and by radiation, if the wire is perfectly black.

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0549.htm

Solution:

The heat flow due to conductivity through a hollow cylinder is given by H = [2\piKL(t_2 - t_1)] / [1n (R_2/R_1)] where R_1 and R_2 are the inner and outer radii of the cylinder, L is its length, and K is the coefficient of thermal conductivity of the material of which the cylinder is composed. Also, t_1 is the cylinder temper-ature at its inner radius, and similarly for t_2. In this case, therefore H = [2\pi × 6 × 10^-5cal \bullet cm^-1 \bullet C deg^-1 \bullet s × 25 cm × (750-250)C deg] / [1n (2.5 cm/0.025 cm)] = 1.02 cal \bullet s^-1. The rate of emission of energy per unit area by radiation is the net outflow according to Stefan's Law. Thus, in the usual notation, R = e \sigma (T^4 - T^4_0) where T is the absolute temperature of the outer cylinder surface, and similarly for T_0, e is the emissivity of the cylinder's surface, and \sigma is a constant. However, H' = RA = e \sigma A(T^4 - T^4_0) where A is the area of the surface which emits the radiation. Since the wire is emitting the energy, and it is to be considered a black body, e = 1 andA = 2\pi R_1 L H' = [5.67 × 10^-12W \bullet cm^-2 \bullet Kdeg^-4 × 2\pi × 0.25 cm × 25 cm(750^4 - 250^4)(K deg)^4] / [4.186 J\bullet cal^-1] = 1.67 cal \bullet s^-1.

Question:

How many ounces of silver alloy which is 28% silver must be mixed with 24 ounces of silver which is 8% silver to produce a new alloy which is 20% silver?

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Solution:

Let x = number of ounces of 28% silver to be used. The relationship used to set up the equation is Volume of 28% silver + Volume of 8% silver = Volume of silver in mixture. .28x + .08(24) = .20(x + 24) 28x + 8(24)= 20 (x + 24) 8x= 288 x= 36 ounces of silver Check: Volume of 28% silver= (.28) (36) = 10.08 Volume of 8% silver= (.08) (24) = 1.92 Total amount of silver = 12 ounces The total mixture contains 24 + 36 = 60 ounces, and 12 ounces is 20% of 60 ounces.

Question:

The analysis of a sample of soil weighing 0.4210 g yields a mixture ofKClandNaClweighing 0.0699 g. From theKCl, 0.060 gAgClis precipitated out. Calculate the percent of KCl in the soil.

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Solution:

To solve this problem (1)setup the equation for the reaction ofKCltoAgCl (2)determinethe number of moles ofAgClprecipitated (3)fromthestoichiometryof equation determine how many moles of KClwere consumed (4)calculatethe weight ofKClconsumed (5)calculatethe percentage ofKClin the soil by dividing the weight ofKClby the total weight of soil and multiplying by 100. The equation for the reaction is KCl + Ag^+ \ding{217}AgCl+ K+ The number of moles ofAgClis (weightofAgCl) / (MW ofAgCl) = [(0.60 g) / (143.5 g / mole)] = 4,18× 10^-4 moles From thestoichiometryof the equation, 4.18 × 10^-4 moles ofKClis consumed . The weight ofKClconsumed is the number of moles ofKCltimes its molecular weight. Thus, weightofKCl= (4.18 × 10^-4 moles) (74.6 g / mole) = 0.0312 g The percentage ofKClin the soil is [(weight ofKCl) / (weight of soil)] × 100 = (0.0312 gKCl) / (0.4210 g Soil) = 7.41 %.

Question:

Two men, each of whom weighs 150 lb, stand opposite each other on the rim of a small uniform circular platform which weighs 900 lb. Each man simultaneously walks clockwise and at a fixed speed once around the rim. The platform is free to rotate about a vertical axis through its center. Find the angle in space through which each man has turned. (The platform's moment of inertia is I = (1/2)MR^2 , where M is its mass).

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Solution:

The figure (parts (a) through (c)) illustrates various posi-tions of the 2 men, and the platform, relative to an observer on the ground. Our ultimate goal is to find an equation, for each man, de-scribing his angular position, \texttheta, as a function of time, relative to his initial position figure (a)). It is equally acceptable to find an equation for the angular velocity \omega^\ding{217} of each man, since this may be integrated to find \texttheta. Rather than jumping right into a dynamic analysis of this problem, it might be worth our while to see if we can use any conservation relations in solving this problem. Note that the angular momentum of the system consisting of the 2 men and platform is constant in time, since no external torques (i.e., friction) act on the system. Furthermore, using angular momentum as our conserved quantity will give us a relation between the angular kinematical variables (\alpha,\omega,\texttheta) at 2 times during the motion of the system. We take these 2 times to be as illustrated in figures (a) and (b). The initial angular momentum of the system, L_0, is zero, since figure (a) shows the system at rest. The final angular momentum, L_F^\ding{217}, is due to the angular momentum of each man(L_1^\ding{217}.L_2^\ding{217}) plus the platform's angular moment-um, L_p^\ding{217}. Hence L_f^\ding{217} = L_1^\ding{217} + L_2^\ding{217} + L_p^\ding{217}(1) But, by the definition of the angular momentum of a particle about point C, L_1^\ding{217} = r_1^\ding{217} × p_1^\ding{217} L_2^\ding{217} = r_2^\ding{217} × p_2^\ding{217} where p_1^\ding{217} and p_2^\ding{217} are the linear momentum of each men, and r_1^\ding{217} and r_2^\ding{217} are as illustrated in figure (a). Also, form figure (b) r_1^\ding{217} = -r_2^\ding{217} p_1^\ding{217} = -p_2^\ding{217} since each man walks with the same velocity. Then L_1^\ding{217} = r_1^\ding{217} × p_1^\ding{217} andL_2^\ding{217} = r_1^\ding{217} × p_1^\ding{217}(2) Furthermore, the angular momentum of the platform is L_p^\ding{217} = I \omega^\ding{217}(3) where \omega^\ding{217} and I are its angular velocity and moment of inertia relative to C. (See figure (a)). Using (2) and (3) in (1) Lf^\ding{217} = 2r_1^\ding{217} × p_1^\ding{217} + I \omega^\ding{217}(4) But, the momentum of particle l(p_1^\ding{217}) is related to its angular velocity (\omega_1^\ding{217}) relative to an external observer by p_1^\ding{217} = m_1\omega_1^\ding{217} × r_1^\ding{217} Therefore, r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} ×(\omega_1^\ding{217} × r_1^\ding{217})(5) Since, for any vectors A^\ding{217}, B^\ding{217} and C^\ding{217} A^\ding{217} × (B^\ding{217} × C^\ding{217}) = (A^\ding{217} \bullet C^\ding{217})B^\ding{217} - (A^\ding{217} \bullet B^\ding{217})C^\ding{217} we obtain, using (5) r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} ×(\omega_1^\ding{217} × r_1^\ding{217}) = m_1r^2_1 (\omega_1\ding{217}) - m_1(r_1\ding{217}\bullet \omega_1\ding{217})r_1^\ding{217}(6) \ding{217} \ding{217} \ding{217} ^\ding{217} ^\ding{217} Now (\omega_1^\ding{217}). is perpendicular to the plane of the platform, which contains r_1^\ding{217}. Then \omega_1^\ding{217} \bullet r_1^\ding{217} = 0 and (6) becomes r_1^\ding{217} × p_1^\ding{217} = m_1r_1^\ding{217} × (\omega_1^\ding{217} × r_1^\ding{217}) = m_1r^2_1 \omega_1^\ding{217}(7) Using (7) in (4) L_f^\ding{217} = 2m_1r^2_1 \omega_1^\ding{217} + I\omega^\ding{217}(8) By the principle of conservation of angular momentum L_f^\ding{217} = L_0^\ding{217} = 0 and (8) yields 2m_1r^2_1 \omega_1^\ding{217} + I\omega^\ding{217} = 0 or \omega_1^\ding{217} = [(-I)/(2m_1r^2_1)] \omega^\ding{217}(9) Since each man travels with the same speed, \omega_1^\ding{217} = \omega_2^\ding{217} = \omega'^\ding{217}(10) \omega_1^\ding{217} = \omega_2^\ding{217} = \omega'^\ding{217}(10) Also, the masses of the 2 men are equal, and m_1 = m_2 = m.(11) Noting that r_1 equals r_2, and that each equals the platform's radius, R, we obtain r_1 = r_2 = R.(12) Using (10) through (12) in (9) \omega'^\ding{217} = [(-I)/(2mR^2)] \omega^\ding{217}(11) Now \omega = (d\texttheta)/(dt) and \omega' = (d\texttheta')/(dt) by definition, where \texttheta and \texttheta' are the angular positions of the plat-form and either man relative to an outside observer. Then (d\texttheta')/(dt) = [(-I)/(2mR^2)][(d\texttheta)/(dt) or \int^\texttheta'_\texttheta'0 d\texttheta'= [(-I)/(2mR^2)] \int^\texttheta_\texttheta0 d\texttheta where\texttheta' = \texttheta'_0 and \texttheta = \texttheta_0 at t = 0. (figure (a)). Finally \texttheta' - \texttheta'_0 = [(-I)/(2mR^2)](\texttheta - \texttheta_0) This relates the net angle traversed by each man (relative to an observer on the ground) to the net angle which the platform rotates through. Using the given data \texttheta' - \texttheta'_0 = -[{(1/2)(MR^2)}/(2mR^2)](\texttheta - \texttheta_0) = (-M)/(4m)(\texttheta - \texttheta_0) \texttheta' - \texttheta'_0 = [(-Mg)/(4mg)](\texttheta - \texttheta_0) = [(-900 lb)/(600 lb)](\texttheta - \texttheta_0) \texttheta' - \texttheta'_0 = -(3/2)(\texttheta - \texttheta_0)(12) Each man makes 1 revolution (or traverses 2\pi radians) relative to the disc. (See figure (c) ). Then, relative to an outside observer, the man traverses an angle (\texttheta' - \texttheta'_0) equal to 2\pi; plus the angle the disc turns through relative to him (\texttheta - \texttheta_0). Hence, (\texttheta' - \texttheta'_0) = (\texttheta - \texttheta_0) + 2\pi(13) Using (12) in (13) (\texttheta' - \texttheta'_0) = -(2/3)(\texttheta' - \texttheta'_0) + 2\pi (\texttheta' - \texttheta'_0) = -(2/3)(\texttheta' - \texttheta'_0) + 2\pi (5/3)(\texttheta' - \texttheta'_0) = 2\pi \texttheta' - \texttheta'_0 = (6\pi)/(5) = 216\textdegree Thus \texttheta - \texttheta_0 = -(2/3)(\texttheta' - \texttheta'_0) = -144\textdegree The negative sign indicates that the men move in a direction opposite to the direction of motion of the disc.

Question:

An incident wavefront of light makes an angle of 60\textdegree with the surface of a pool of water. The speed of light in water is 2.3 × 10^8 m/s. What angle does the refracted wavefront make with the surface of water?

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Solution:

The angle between \texttheta_i the incident ray and the normal to the surface (as shown in the figure),equals the angle between the incident wavefront and the water surface, and \texttheta_i = 60\textdegree. Snell's Law, relating the angle of incidence, \texttheta_i , to the angle of refraction, \texttheta_r of the light, is n_1 sin \texttheta_i = n_2 sin \texttheta_r where n_1 and n_2 are the refractive indices of air and water, respectively. Hence sin \texttheta_r = (n_1/n_2) sin \texttheta_i But n_1 = [{speed of light (vacuum)}/{ speed of light (air)}]n_2 = [{speed of light (vacuum)}/{ speed of light (water)}] Hence sin \texttheta_r = [(2.3 × 10^8 m/s)/(3 × 10^8 m/s)] sin 60\textdegree = .664 or\texttheta_r = 42\textdegree. \texttheta_r also equals the angle the refracted wavefront makes with the water surface.

Question:

Anucleusof mass M emits a \Upsilon ray of energy E_\Upsilon^\ding{217} . The nucleuswas initially at rest . What is its recoil energy after the emission?

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Solution:

From momentum conservation we have 0 =p_\Upsilon+p_nuc In addition we know that E=h\Elzpscrv=hc/\lambda and by thedeBroglierelation p = h/\lambda \textbullet Therefore we have E = pc and p_\Elzpbgam = E_\Elzpbgam/c. Hence, ▏p_nuc ▏= E_\Upsilon / c . The recoil energy is given by K_nuc = 1/2 Mv^2 = (M^2v^2)/ (2M) = (Mv)2/ 2M= (p^2_nuc) / 2M \textbackslash = E_\Elzpbgam^2/ 2Mc^2 The last equality involves substituting E_\Elzpbgam/c for P_nuc . We have assumed that the recoil velocity may be treated as nonrelativistic.

Question:

Calculate the percent composition by weight of ether (C_2H_5)_2O.

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Solution:

To understand this problem, one must first realize that the subscript 2 outside the brackets means that all elements inside the brackets are multiplied by 2. Carrying this procedure out, one obtains the formula Cu_4H_10O for ether. Percent composition is found by dividing the individual weights of the components by the molecular weight. Therefore, first find molecular weight by using the table below. Elements Number of Atoms Atomic Weight Total AtomicWeight C 4 12 48 H 10 1 10 O 1 16 16 M.W. = 48 + 10 + 16 = 74 g/mole. Percent composition is then calculated from the formula % composition = [(total atomic weight) / (M.W.)] × 100 % C = (48/74) × 100 = 64.86 % % H = (10/74) × 100 = 13.51 % % O = (16/74) × 100 = 21.63 %. To double check, make sure that the total percent equals 100.

Question:

esterification, saponification, elastomer and nitration.

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Solution:

Esterification is the process that occurs during the reaction of an alcohol (R'-OH) with a carboxylic acid (R-COOH) to produce an ester . It can be illustrated as follows in the formation of the ester ethyl acetate Saponification is the term used to describe those reactions of an esterwith a base to produce soft and hard soaps. The process can be illustrated by the formation of sodium stearate (C_17H_35COO)_3C_3H_5 + 3NaOH \rightarrow 3C_17H_35COONa + C_3H_5(OH)_3 Ester (tristearin)BaseSoap (sodiumGlycerol stearate) Elastomer is the general name of all rubber sub-stitutes. They tend to be better than natural rubber. For example, most elastomers are more oil resistant and some are less permeable than rubber. Nitration:This is the name of the reaction in which nitric acid (HNO_3) is mixed with sulfuric acid (H_2SO_4) and added to a benzene ring, such that an NO_2 is substituted for a hydrogen atom attached to the ring to form nitro-benzene (fig. A).

Question:

What is the de Broglie wavelength of an electron with a kineticenergy of 1eV?

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Solution:

When calculating the de Broglie wavelength, it is important to know whetheror not to use relativistic formulas in order to calculate the quantitiesappearing in the de Broglie formula. One way of deciding this is torealize that, when the velocity v is very small compared with c, (and, hence, the problem is non-relativistic) the kinetic energy is small when comparedto the rest mass energy, m_0c^2 . The rest mass energy of an electronis 0.512 × 10^6eVwhich is large compared with 1eV. We need notuse relativistic formulas in the present example. In the formula for the de Broglie wavelength, \lambda = (h/mv)(1) itis permissible to insert the rest mass, m_0, for m, m\approx m_0 = 9.11 × ^10-28 gram. The kinetic energy is (1/2)m_0v^2= 1eV. Because 1eV= 1.6 × 10^-12 erg (1/2)m_0v^2 = 1.6 × 10^-12 erg. Therefore v =\surd[(2 × 1.6 × 10^-12 erg)/m_0] =\surd[(2 × 1.6 × 10^-12 erg)/(9.11 × 10^-28 g)] = 5.92 × 10^7 cm/sec. Notice that this is, in fact, very much less than the velo-city of light (3 × 10^10 cm/sec). Inserting these values of m and v into equation (1), \lambda = [(6.625 × 10^-27 erg sec)/{(9.11 × 10^-28 × 5.92 × 10^7)g \textbullet cm/sec}] = 1.23 × 10^-7 cm. This is about ten times larger than the diameter of an atom.

Question:

Give examples of valid variable names in SNOBOL IV. List the basic statements and briefly explain their operations.

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Solution:

The variable names in SNOBOL IV must begin with a letter which may be followed by any number of letters, digits, periods and underscores. Some examples of valid variable names are listed below: A, B1, VARIABLE, D987654321, Y0U.OR.ME, K6.J2 The basic statements are as follows: variable: variable = INPUT - Assigns the first read in string to a variable name. OUTPUT = string or variable - Prints out a given string, PUNCH = string - Punches the given string, variable = string - Assigns the given string to a variable.

Question:

TheRiverdellEnglewoods, a little League team of the future, is offering a prize to anyone who can write a FORTRAN program to keep track of the box scores of their games. An underfinanced team, theEnglewoodshave an ancient card-punching machine in the dugout. Each time an Englewood comes to bat, the scorekeeper prepares a card with the following items: 1)the player's number (2 digits) 2)the result of his at-bat (0 for out, 1 for a single, etc., up to 4 for a home run) 3)whether the player scored a run (either 0 or 1) 4)the number of runs batted in The manager wants the program to output the player's number, the number of at-bats, hits, runs scored, and runs batted in for each game. After the last player has been listed, compute and output the team totals.

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Solution:

We assume that there is a total of 20 players on the team numbered 1 through 20. Thus, we need a loop to compute the required statistics for each player. After each computation, we store the elements in arrays so that team totals may be computed at the end. Variable names are chosen to match closely the baseball terminology. The program stores each player's totals until a card containing all zeros is encountered. This signifies the end of the deck, whereupon the output procedure begins, completed with appropriate headings. CBOX SCORES FOR ENGLEWOODS INTEGER PLAYNO, EFORT, RUNSCO, RBI INTEGER ATBAT, HITNO, RUNNO, RBINO INTEGER TOTBAT, TOTHIT, TOTRUN, TOTRBI DIMENSION ATBAT(20), HITNO(20), RUNNO(20), RBINO(20) CINITIALIZE ALL VARIABLES TO ZERO DATA PLAYNO, EFORT, RUNSCO, RBI/0,0,0,0/ DATA ATBAT, HITNO, RUNNO, RBINO/20{_\ast}0 , 20{_\ast}0, 20{_\ast}0 , 20{_\ast}0/ DATA TOTBAT, TOTHIT, TOTRUN, TOTRBI/0,0,0,0/ CREAD CARDS 10READ (2,100) PLAYNO, EFORT, RUNSCO, RBI 100FORMAT (12,3I1) CDO WHILE PLAYNO IS NOT EQUAL TO ZERO IF (PLAYNO. EQ. 0) GO TO 20 ATBAT (PLAYNO) = ATBAT (PLAYNO) + 1 IF (EFORT. NE. 0) HITNO (PLAYNO) = HITNO (PLAYNO) + 1 RUNNO (PLAYNO) = RUNNO (PLAYNO) + RUNSCO RBINO (PLAYNO) = RBINO (PLAYNO) + RBI GO TO 10 CPRINT HEADINGS WRITE (6,101) 101FORMAT(1X, 'PLAYER NUMBER',5X, 'AT BATS' ,5X, 1'HITS',5X,'RUNS',5X,'RBIS') 20DO 25 J = 1,20 IF (ATBAT (J) .EQ. 0) GO TO 25 WRITE (6,102) J, ATBAT (J), EFORT, RUNSCO, RBI 102FORMAT (7X,I2,13X,I1,11X,I1,2(8X,I1)/) TOTBAT = TOTBAT + ATBAT (J) TOTHIT = TOTHIT + HITNO (J) TOTRUN = TOTRUN + RUNNO (J) TOTRBI = TOTRBI + RBINO (J) 25CONTINUE WRITE (6,103) TOTBAT, TOTHIT, TOTRUN, TOTRBI 103FORMAT (1X, 'TEAM TOTALS',9X, I2,11X, I2,2 (8X, 12) ) STOP END

Question:

Certain substances are maintained in the blood at a high threshold while others are held at a very low threshold. What factor determines what the kidney threshold for a substance will be? Which substances characterize each level?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0448.htm

Solution:

The amounts of the individual substances dissolved in the blood exhibit wide variation. Some substances are high threshold substances, which means that they are maintained in the blood at a high concentration relative to their concentration in the urine. Substances maintained at such levels include glucose, amino acids, fatty acids, vitamins, and hormones. These are filtered out of the blood at theglomerulusand almost fully reabsorbed by the tubules. Recall that it is through active transport by the cells of the proximal convoluted tubule thatreabsorptiontakes place. The threshold level ofreabsorptionis thus a function of the rate of active transport across the tubule surface. Only if the concentration of a high threshold substance in the blood is very great will the amount filtered surpass the amount reabsorbed and cause the substance to spill into the urine. Substances with high threshold values are essential nutrients or other compounds needed in body metabolism. Low threshold substances are those having a high concentration in the urine relative to their concentration in the blood. Only a small fraction of these substances is reabsorbed in the tubules. Low threshold substances are mostly waste products, such as urea; the concentration of urea in the urine may be up to 70 times that in the blood. Various drugs, such as aspirin and antibiotics, are also included in this group. This is not always advanta-geous, for the kidneys may remove a drug so fast that continued application is needed for maximal effectiveness. This problem is solved in some cases by mixing the medication with a binder so that it is released slowly over an extended period of time. Substances having medium thresholds include inorganic mineral ions and certain organic substances. Such substances are roughly equally reabsorbed and excreted. Important ions in this group are the sodium and chloride ions. These ions are involved in the maintenance of the pH of the blood (7.3) by their association with hydrogen and hydroxyl ions. The cellular metabolism of food particles most often results in an excess of hydrogen ions; these are then excreted along with chloride ions, resulting in a slightly acidic urine.

Question:

Describe the evolution of the reptilian excretory system to account for the transition from an aquatic to a terrestrial habitat.

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Solution:

The reptilian excretory system has evolved in a way that enables the animal to conserve most of its water. The conservation of body fluid is a necessary characteristic of a terrestrial animal. Without this feature, the body tissues would become dessicated in the dry environment. Reptiles conserve water by having a coarse, dry, horny skin. In addition, the glomeruli of the reptilian kidney have diminished in size so that less water is filtered from the blood. Another modification is a greater degree of reabsorption of water from the glomerular filtrate by the kidney tubules. This occurs as the rep-tiles have evolved kidney tubules with two highly coiled regions and a long loop of Henle extending deep into the medulla of the kidney. These long portions of the tubule function in the reabsorption of water. Their ability to produce a concentrated hypertonic excretory product is an important adaption to land life. The conservation of water in the reptile has affected the nature of the excretory product. Nitrogenous wastes are excreted as uric acid in reptiles, as opposed to urea in amphibians (whose ancestors gave rise to primitive reptiles). Uric acid is excreted as a watery paste in reptiles. It is less toxic and less soluble in water than urea. Therefore, it is not necessary for reptiles to use water to dissolve their nitrogenous wastes. In fact, only a comparatively very small amount is required to simply flush the uric acid out of the excretory system. In this way, the reptile conserves most of its body water and prevents dehydration.

Question:

Evaluate 16^\rule{1em}{1pt}3/4.

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Solution:

16^\rule{1em}{1pt}3/4= 1/(16^3/4) = [1/(^4\surd16)^3 Note that2^4= 2\bullet2\bullet2\bullet2 = 16, hence ^4\surd16 = 2. Thus, 16^\rule{1em}{1pt}3/4. = 1/2^3 = 1/(2\bullet2\bullet2) = 1/8.

Question:

An E. coli cell is 1.0\mu in diameter and 2.0\mu long and may be assumed to be cylindrical. It contains 80% water. If the intracellular pH is 6.4, calculate the number of H^+ ions in a single cell.

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Solution:

One can find the number of H^+ ions in one E. coli cell from the dimensions of the cell and the pH, because pH = -log [H^+].[H^+] is found in moles per liter and the volume of the cell can be calculated in liters. Solving for [H^+]: 6.4 = - log [H^+] [H^+] = 10^-6.4 = 3.98 × 10^-7 moles / liter To determine the number of ions in the cell, the number of moles present must be determined first. This can be found from the volume. The volume of a cylinder is equal to l\pir^2, where l is the length and r is radius of the cross-section, 1\mu = 10^-4 cm. Volume of the cell = (2 × 10^-4 cm) (\pi) (0.5 × 10^-4 cm)^2 = 1.57 × 10^-12 cm^3 = 1.57 × 10^-15 l If 80% of the cell is H_2O, 80% of the volume equals the volume of H_2O present. The H^+ ions can only exist in the aqueous phase of the cell. Volume of H_2O = .80 × 1.57 × 10^-15 l = 1.26 × 10^-15 l To solve for the number of moles of H^+, multiply [H^+] by the volume of H_2O present. No. of moles of H^+ = 1.26 × 10^-15 l × 3.98 × 10^-7 moles / l = 5.01 × 10^-22 moles. There are 6.02 × 10^23 ions per mole. No. of H^+ ions = 6.02 × 10^23 (ions / mole) × 5.01 × 10^-22 mole = 3.02 × 10^2 ions = 302 H^+ ions.

Question:

Compare the processes of digestion in the hydra, flatworm and earthworm.

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Solution:

Hydra is a member of the coelenterate phylum. Hydra is a multicellular, sac-like organism whose body wall is two cells thick. Many of the cells are semi-specialized, differing in function from their immediate neighbors. The hydra's digestive system, the gastrovascular cavity, has only one opening-the mouth (refer to Figure 1) . The tenta-cles of the hydra have special cells, called nematocysts, that spear tiny organisms swimming near the hydra. The ten-tacles then draw the prey into the mouth. The cells lining the gastrovascular cavity begin digestion by secreting en-zymes. This mode of digestion, occuring outside cells, is extracellular digestion. Extracellular digestion is followed by intracellular digestion in the hydra. When small food fragments come into contact with the cells lining the cavities, these fragments are phagocytized and digestion is completed inside the cell. The indigestible material is expelled through the mouth. Thus, the hydra utilizes extracellular digestion to break down relatively large organisms into small pieces which are then digested intracellularly by individual cells. Flatworms are small animals that are often parasitic. Parasitic flatworms have lost most of their ability to procure food and are provided with nutrients by their host. However, some flatworms such as planaria are non-parasitic and are free-living organisms. The mouth of the planaria is on the ventral side, in the middle of the body (refer to Figure 2). It opens into a tubular muscular pharynx that can protrude through the mouth directly onto the prey. The pharynx opens into a gastrovascular cavity similar to those other coelenterates but much more branched. The extensive branching brings nutrients to all parts of the body and helps increase the absorptive surface area of the cavity. Some extracellular digestion occurs in the cavity but most food is phagocytized and digested intracellularly in the cells lining the cavity. An earthworm has a complete digestive tract. It has two openings, the mouth and anus, (refer to Figure 3), with wastes expelled through the anus and food ingested through the mouth. Various regions of the tract are specialized: the pharynx to suck in food, the esophagus to transport food to the crop, the crop to store the food, the gizzard to break the food up, and the intestine to digest and absorb the food. Most digestion is extracellular, utilizing the enzymes secreted by cells lining the tract. The digestive tract is not widely branched, but is rather like a simple gastrovascular cavity. However, large numbers of folds are found in the tract lining, which greatly increase the absorptive surface area. Since the earthworm is a land animal, it must also conserve its body fluids by reabsorbing water from the indigestible wastes. This occurs in the posterior region of the intestine, analogous to the human large intestine.

Question:

In an electrolytic cell, a liter of a 1M aqueous solution of MnO_4^- is reduced at the cathode. Determine the number of faradays required for each of the following to be made a) a solution that is .01 M MnO_4^2- ; b) 1 gram of MnO_2 ; c) 1 gram-equivalent of Mn metal.

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Solution:

For each part, you need to write the reaction that occurs. The masses of substances produced during the process are proportional to their equivalent weights. The unit electrical equivalent = 1 Faraday or 1F. One Faraday = 96,500 coul and is capable of reducing one equivalent positive charge, an equivalent positive charge contains Avogadro's number of individual unit charges. You proceed as follows: reaction is MnO_4^- + e^- \rightarrow Mn_4^2-.One electron is being transferred. You want [MnO_4^2-] = .01 M . Since the solution has a volume of 1 liter, .01 M = .01 moles, since M = molarity = moles / liter. Due to the fact that only 1 electron is transferred, only one Faraday is needed to reduce a mole of MnO_4^-, since it would require only a mole of electrons. You need only, however, .01 moles of MhO_4^2- . Since a mole of MnO_4^2- is formed from every mole of MnO_4^- oxidized you need to reduce .01 moles of MnO_4^2- . To do this, requires, therefore, .01 F . Mn0_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H-_2O . Here, 3 electrons are required to reduce MnO_4^- to MnO_2 . Thus, one Faraday can only reduce (1 / 3) of a mole of MnO_4^-,. Recall, one Faraday is Avogadro's number of electrons and this reaction requires three moles, or 3F per mole of MnO_4^-. You want lg of MnO_2 (atomic wt. = 86.94 g / mole), which is (1g) / (86.94g / mole) or (.0115) moles of MnO_2 . Since one mole of MnO_2 will be produced for every mole of MnO_4^- , you must reduce .0115 moles of MnO_4^- . One faraday will reduce (1 / 3) of a mole of MnO_4^- . To reduce all of it, you want 3 Faraday's / mole. Thus, number of Faraday's required = (3F / mole)(.0115 mole) = .0345F . MnO_4^- + 8H^++ 7e^- \rightarrow Mn + 4H_2O to occur. 7 electrons are to be transferred. Thus, 1 Faraday can only reduce (1 / 7) mole of material. However, you want 1 gram-equivalent of material, which is the amount deposited by 1F. Hence, 7 equiv. of Mn = 1 mole Mn. Thus, to deposit this amount, you use (1.00 equiv. Mn)(F / equiv. Mn) = 1.00F.

Question:

A coil of 600 turns is threaded by a flux of 8.0 × 10^-5 weber. If the flux is reduced to 3.0 × 10^-5 weber in 0.015 sec, what is the average induced EMF?

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Solution:

We note that there is a change in magnetic flux. This immediately implies the use of Faraday's Law, which relates a change in magnetic flux to an induced E.M.F. (electromotive force). The flux linking the coil will induce an E.M.F. in the coil. Faraday's Law states: E.M.F. = [(-Nd\varphiB)/dt](1) where N is the number of turns in the coil, and \varphi_B is the magnetic flux linking the coil. We may write this using average values as E.M.F. = - N [(∆\varphiB)/∆t](2) where ∆\varphi_B is the change in magnetic flux over the interval ∆t, andE.M.F. is the average value of the resulting E.M.F. Substituting the values given in the statement of the problem into equation (2) , we obtain E.M.F. = - [(600) (8 × 10^-5 weber - 3 × 10^-5 weber)]/(.015 s) = - 2 [(weber)/(sec)] But1 weber = 1 [(Joule - sec)/(coul)] Therefore, E.M.F . = - 2 [(Joule - sec)/(coul - sec)] = - 2 Volts

Question:

Four moles of PCl_5 are placed in a 2-liter flask. When the following equilibrium is established PCI_5\rightleftarrows PCI_3 + CI_2 , the flask is found to contain 0.8 mole of Cl_2 . What is the equili-brium constant?

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Solution:

The equilibrium constant (Keq) for this reaction is: Keq= {[PCI_3 ] [CI_2]} / [PCI_5] where [ ] indicate concentrations. One is given that, at equilibrium, there are 0.8 mole of Cl_2 present. From the equation for the re-action, it follows that there is 0.8 mole of PCl_3 present also. This means that 0.8 moles of PCl_5 has reacted. Originally 4 moles of PCI_5 were present, thus, at equilibrium, 4.0\Elzbar 0.8 moles or 3.2 moles are present. These components are present in a two-liter flask. Therefore, the concentrations will be expressed in moles/2 liters. One can now solve for the equilibrium constant. Keq= {[PCI_3] [CI_2] } / [PCI_5][PCI_3] = 0.8 moles/2 liters [CI_2]= 0.8 moles/2 liters [PCI_5] = 3.2 moles/2 liters Keq={(0.8/2) (0.8/2)} / {(3.2/2)} = {(0.4M) (0.4M) = {(0.4M) (0.4M) } / {(1.6M)} = (0.16M^2) / (1.6M) = 0.1M.

Question:

A spacecraft, with a mass m = 2000kg, rotates with an angular velocity of \cyrchar\cyromega = (0.05 rad / s)\^{\i} + (0.15 rad / s)\^{\j}. Two small jets, located at points A and B, are turned on in a direction paral-lel to the z-axis. Each jet has a thrust of 25N. The radii of gyration of the spacecraft are k_x = k_z = 1.5m and k_y = 1.75m. Determine the required time of operation for each jet, so that the angular velocity of the spacecraft reduces to zero.

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Solution:

Choose a coordinate system fixed in the body such that the rockets at A and B lie in the x, y plane and the origin is at the center of mass. The positions of A and B are r^\ding{217}_1 and r^\ding{217}_2 respectively. Qualitatively, what happens is both the rockets oppose the original motion about the y axis but B will increase the original motion about the x- axis while A will oppose it. Thus, A will have to be fired longer in order to compensate for the increase about the x- axis due to B. The torque, \tau^\ding{217}, produced by A and B is \tau^\ding{217}_B = r^\ding{217}_1 × F^\ding{217}_1 = (y_1F_1i^\ding{217}) - (x_1F_1y^\ding{217})(1) \tau^\ding{217}_A = r^\ding{217}_2 × F^\ding{217}_2 = -(y_2F_2i^\ding{217}) - (x_2F_2j^\ding{217}). From Newton 's Second Law, \tau^\ding{217} = l\alpha^\ding{217} where I is the moment of inertia and \alpha^\ding{217} is the angular acceleration. In components this becomes \tau^\ding{217} = I_x\alpha_xi^\ding{217} + I_y\alpha_yj^\ding{217} + I_z\alpha_zk^\ding{217}(2) where I_x , I_y and I_z are the moments of inertia about the x, y, z axes. Comparing equations (2) and (3) gives \alphaA_x = [(d\cyrchar\cyromegaA x )/(dt)] = -[(y_2F_2) / I_x] = -c_1; [(d\cyrchar\cyromegaA_y)/(dt)] = -[(x_2F_2 ) / I_y] = -c2;; [(d\cyrchar\cyromegaA_z) / (dt)] = 0(3) [(d\cyrchar\cyromegaB_x)/(dt)] = -[(y_1F_1 ) / I_x] = c_3; [(d\cyrchar\cyromegaB_y)/(dt) = -[(x_1F_1 ) / I_y] = -c_4 ; [(d\cyrchar\cyromegaB_z) / (dt)] = 0 These equations can be integrated directly to give ∆\cyrchar\cyromegaA_x = -c_1t_A ; ∆\cyrchar\cyromegaA_y = -c_2t_A ; \cyrchar\cyromegaA_z = constant(4) ∆\cyrchar\cyromegaB_x = c_3t_B ; ∆\cyrchar\cyromegaB_y = -c_4t_B ; \cyrchar\cyromegaB_z = constant where ∆\cyrchar\cyromega = \cyrchar\cyromega(t) - \cyrchar\cyromega(0) is the change in angular velocity about each axis due to firing the rockets. Since initially \cyrchar\cyromega^\ding{217} = 0.2 (rad / sec)i^\ding{217} + 0.1 (rad /sec)j^\ding{217}; w_z(0) = 0, the constants in equation (4) must also be zero. Since we want the final angular velocity to be zero, the sum of the initial angular momentum plus the additional in-crements about each axis must be zero. So, \cyrchar\cyromega_(x)0 + ∆\cyrchar\cyromegaA_x + ∆\cyrchar\cyromegaB_x = 0 \cyrchar\cyromega_(y)0 + ∆\cyrchar\cyromegaA_y + ∆\cyrchar\cyromegaB_y = 0. Substitution from equation (3) gives (C_1t_A) - (C_3t_B) = \cyrchar\cyromega(x)0 (C_2t_A) + (C_4t_B) = \cyrchar\cyromega_(y)0 These are two simultaneous equations in the unknowns t_A and t_B. From the theory of equations, the solutions are t_A = (1 / ∆)\mid\cyrchar\cyromega_(x)0-c_3\mid= (1 / ∆)[(c_4\cyrchar\cyromega_(x)0) + (c_3\cyrchar\cyromega_(y)0)](5) \mid\cyrchar\cyromega_(y)0c_4\mid t_B = (1 / ∆)\midc_1\cyrchar\cyromega_(x)0\mid= (1 / ∆)[(c_1\cyrchar\cyromega_(y)0) - (c_2\cyrchar\cyromega_(x)0)] \midc_2\cyrchar\cyromega_(y)0\mid where ∆ =\midc_1-c_3\mid= (c_1c_4) + (c_2c_3). \midc_2c_4\mid For this problem x_1 = 2.5 m, y_1 = 4 m, x_2 = 4 m, y_2 = 2.5 m m = 2 × 10^3kg, F_1 = F_2 = 25 n, k_x = k_z = 1.5 m and k_y = 1.75 m. From equation (3) the C values are c1= (y_2F_2) / (I_x) = [(2.5 m 25 n) / {2 × 10^3 kg(1.5 m)^2}] = 0.0139(rad /sec^2) c2= (x_2F_2) / (I_y) = [(4 m 25 n) / {2 × 10^3 kg(1.75 m)^2}] = 0.0163(rad /sec^2) c3= (y_1F_1) / (I_x) = [(4 m 25 n) / {2 × 10^3 kg(1.5 m)^2}] = 0.0222(rad /sec^2) c4= (x_1F_1) / (I_y) = [(2.5 m 25 n) / {(2 × 10^3 kg(1.75 m)^2}] = 0.0102(rad /sec^2) ∆ = (c_1c_4) + (c_2c_3) = 0.000503 (rad / s^4). Substitution in eq. (5) gives the times t_A = [(c_4\cyrchar\cyromega_(x)0) + (c_3\cyrchar\cyromega_(y)0)] / (∆) = [0.0102 × 0.05 + 0.0222 × 0.15] / (0.000503) = 7.634 sec. t_B = [(c_1\cyrchar\cyromega_(y)0) - (c_2\cyrchar\cyromega_(x)0)] / (∆) = [0.0139 × 0.15 - 0.0163 × 0.05] / (0.000503) = 2.525 sec. Thus, it is necessary to fire A approximately three times as long as B in order to stop the capsule's rotation. This agrees with our initial qualitative remarks regarding the motion.

Question:

Which of the following would maintain a constant body temperature in spite of changes in the environmental temperature: frog, robin, fish, dog, or lizard?

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Solution:

Animals that can regulate their internal body temperature so that it remains constant in varyingenvirinmentaltemperatures are known as homeotherms. Of the phylumChordata, only mammals and birds can thermregulate- i.e., they possess temperature-regulating sys-tems. They are considered to be warm-blooded animals. From the list of animals given, two of them arehomeotherms. The robins, since it is a bird, belongs to the class Aves, and the dog is a member of the class Mammalia. Hence these two animals arehomeothermicand will maintain a constant body temperature. The remaining animals arepoikilotherms. They possess no temperature-regulating system and as a result, their internal body temperature varies directly with the environmental temparature. Thispoikilothermiccharacteristic places strict limitations on the environment exploited by the animal, which in turn affects its behavioral patterns. The frog is an amphibian whereas the lizard is a reptile. A fish belongs to thesuperclasspisces. These three animals are poikilothermsand hence do not maintain a constant body temperature independent of changes in environmental temperature.

Question:

Explain how a change of one base on the DNA can result in sickle-cellhemoglobin rather than normal hemoglobin.

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Solution:

If one base within a portion of DNA comprising a gene is altered, andthat gene is transcribed, the codon for one amino acid on the mRNA willbe dif-ferent. The polypeptide chain which is translated from that mRNA maythen have a different amino acid at this position. Although there are hundredsof amino acids in a protein, a change of only one can have far-reaching effects. The human hemoglobin molecule is composed of two halves of protein, each half formed by two kinds of polypeptide chains, namely an alphachain and a beta chain. In persons suffering from sickle-cell anemia,.amutation has occurred in the gene which forms the beta chain. As a result, a single amino acid substitution occurs, whereglutamicacid is replacedbyvaline. Onecodonforglutamicacid is AUG, while one for valineis UUG. (A = adenine, U = uracil, G = guanine). If U is substituted forA in thecodonforglutamicacid, thecodonwill specifyvaline. Thus, it iseasy to see how a change in a single base has resulted in this amino acidsubstitu-tion. This change is so important, however, that the entire hemoglobinmolecule behaves differently. When the oxygen level in the blooddrops, the altered molecules tend to form end to end associations andthe entire red blood cell is forced out of shape, forming a sickle- shaped body. These may aggregate to block the capillaries. More importantly, thesesickledcells cannot carry oxygen properly and a person havingthem suffers from severe hemolytic anemia, which usually leads to deathearly in life. The condition is known as sickle-cell anemia. In a normal hemoglobin molecule, theglutamicacid residue in question , being charged, is located on the outside of the hemoglobin moleculewhere it is in contact with the aqueous medium. When this residueis replaced byvaline, a non-polar residue, the solubility of the hemoglobinmolecule decreases. In fact, the hydrophobic side chain of valinetends to form weak vanderWaals bonds with other hydrophobic sidechains leading to the aggregation of hemoglobin molecules.

Question:

Delivery trucks which operate by making use of the energy stored in a rotating flywheel have been in use for some time in Germany. The trucks are "charged up" before leaving by using an electric motor to get the flywheel up to its top speed of 6000 rev\bulletmin^-1. If one such flywheel is a solid homogeneous cylinder of weight 1120 lb and diameter 6 ft, how long can the truck operate before returning to its base for "recharging\textquotedblright, if its average power requirement is 10 hp?

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Solution:

The angular speed of the flywheel is \omega = 6000 rev\bulletmin^-1 × 2\pi rev\bulletmin^-1 × 1/60 min\bullets^-1 = 200\pi rev\bullets^-1 The kinetic energy stored in the flywheel is given by E_k = (1/2) I\omega^2 where I is the moment of inertia. For a disc of radius r and mass M, I = (1/2) Mr^2. Therefore, E_k = (1/2) x (1/2) Mr^2 x \omega^2 = (1/4) Mr^2\omega^2 = (1/4) × (1120/32) slugs x 9 ft^2 x 4\pi^2 x 10^4 s^-2 = (63\pi^2)/2 x 10^5ft\bulletlb. The truck consumes this energy at a rate of P = 10 hp = 10 x 550 ft\bulletlb-s^-1, Thus, assuming that the frictional losses are negligible, the truck can work for t = (E_k)/p = (63\pi^2 x 10^5ft\bulletlb)/(2 x 5500 ft\bulletlb\bullets^-1) = [(63\pi2x 10^5)/(2 x 5500 x 60)] min = 94.2 min. The flywheel must therefore be "recharged" before this time has elapsed.

Question:

Describe the light reactions of photosynthesis. What are the products of these reactions?

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Solution:

The essential requirements for the energy - requiring dark reactions of photosynthesis are ATP and NADPH _2. These molecules are produced via the light reac-tions of photosynthesis, which occur entirely within the chloroplasts. This was proven by Daniel Arnon in 1954 who showed that intact chloroplasts carefully isolated from spinach leaves were able to carry out the complete photo-synthetic reaction. Scientists now agree that there are two reaction pathways occurring in the chloroplasts which generate ATP molecules from light energy, one of the pathways being designated cyclic photophosphorylation, the other noncyclic photophosphorylation. Both reactions require the presence of light. Each pathway is a photsystem. A photosystem is an energy-trapping center which receives packets of en-ergy from excited chlorophyll molecules and converts them into a stream of high - energy electrons which can then travel down the electron transport chain. Photosystem I is called P700 due to the wavelength at which chlorophyll a absorbs light (700 nm). Photosystem II is similarly designated P680 due to the response by chlorophyll b. In cyclic photophosphorylation, light striking a chlorophyll a molecule excites one of the electrons to an energy level high enough to allow it to leave the molecule. The chlorophyll a+ molecule, having lost an electron, is now ready to serve as an electron acceptor because of its net positive charge. However, the ejected electron does not return ti its ground state and the chlorophyll a^+ molecule directly; instead it is taken up by ferredoxin and passed along an electron transport chain. As the electron passes from ferredoxin to the cytochromes, to plastocyanin and finally back to chlorophyll a+ , two ATP molecules are produced. In this way, light energy is con-verted into chemical bond energy, and not lost as heat as it would have been had the excited electron returned di-rectly to its ground state. In noncyclic photophosphorylation, oxygen is produced by the photolysis of water. NADPH and ATP molecules are formed. The excitation of chlorophyll a by light at Photosystem I ejects high energy electrons. These electrons pass to ferredoxin and then to NADP^+, reducing it to NADPH. To restore chlorophyll a+ to its original state, another set of chlorophyll molecules known as chlorophyll b come into play at Photosystem II. When light energy is absorbed by chlorophyll b, it also ejects high energy electrons. These pass to an electron ac-ceptor called Q and then via plastoquinone, cytochromes and plastocyanin to chlorophyll a+. During these steps, ADP is phosphorylated to ATP. The electrons necessary to restore chlorophyll b+ to its ground level come from the splitting of water due to the high affinity that chlorophyll b+ has for electrons. The water molecule is thus split into component protons, electrons, and oxygen. The oxygen, after combining with another oxygen atom, is released in gaseous form as O_2. The two elec-trons are picked up by cytochrome b and the hydrogen ions are used in the conversion of NADP^+ to NADPH.

Question:

What was the Appalachian Revolution? When did itoccur andwhat were its effects?

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Solution:

The Appalachian Revolution refers to the general folding of the earth'scrust which raised the great Appalachian mountain chain running fromNova Scotia to Alabama. It occurred at the end of the Permian Period, the final period of Paleozoic Era, some 280 million years ago.The Appalachian Revolution also brought into existence some of the mountain rangesof Europe. Because of a greatglaciationwhich affected most of thelands existing at that time, the climate became colder and drier. Many ofthe Paleozoic forms of life were unable to adjust to the climatic changes andbecame extinct. Many marine forms were lost during the Appalachian Re-volution, due to the cooling of the water and the decrease in the amountof available space caused by the drying up of the shallow seas. Other new forms of organisms arose, and the Paleozoic Era gave way to the Mesozoic Era.

Question:

One end of a fingernail file is clamped in a vise and the other end is given a to-and-fro vibration. The motion of the free end is approximately S.H.M. If the frequency is 10 vibrations per second and the amplitude is 4 millimeters, what is the velocity when the dis-placement of the free end is 2 millimeters ?

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0370.htm

Solution:

The problem states that the motion is S.H.M. Therefore, we know that the displacement of the file is x = A sin (\omegat+ \alpha)(1) where A is the amplitude and \alpha is a constant.\omegais the angular frequency of the vibration. If f is the frequency of the motion \omega = 2\pif. The velocity of the end of the file is, differentiating (1), v = A \omegacos(\omegat+ \alpha)(2) We need the velocity when x = 2 mm. At this position, using (1) 2 mm= 4 mm sin (\omegat+ \alpha) sin(\omegat+ \alpha)= 1/2 whence (\omegat+ \alpha)= 30\textdegree Hence, using (2), v= A \omegacos(30\textdegree) orv= A (2\pif)cos(30\textdegree) v= (4 mm) (6.28) (10 per sec) {\surd(3/2)} v= [(40 mm)/(sec)] (6.28) (.866) = 218 mm/sec.

Question:

What is the gravitational force on a person whose mass is 70 kg when he is sitting in an automobile that accelerates at 4 m/s^2?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0101.htm

Solution:

The mass, m = 70 kg, and the acceleration, g = 9.8 m/s^2, are the known observables. The gravitational force on the person is given by Newton's Second Law, F = mg where g is the acceleration due to gravity. F = mg = (70 kg)(9.8 m/s^2) = 6.86 × 10^2N. Although the person will experience the force causing the acceleration, 4 m/s^2, his weight is unaffected by the car's motion. This occurs since the acceleration of the auto-mobile is perpendicular to that caused by gravity and has no effect upon the person in the downward direction.

Question:

Given log_102 = 0.3010, find log_1032.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0960.htm

Solution:

Note that, 32 = 2 \bullet 2 \bullet 2 \bullet 2 \bullet 2 = 2^5. Thus, log_1032 = log_102^5. Recall the logarithmic property, log_bx^y = y log_bx. Hence,log_1032 = log_102^5= 5 log_102 = 5 (0.3010) = 1.5050

Question:

In microscopy, small spherical bodies are often seen attached to the network of endoplasmic reticulum. What are these bodies? What function do they serve in the cell? The role of the ribodome in protein synthesis.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0038.htm

Solution:

The small spherical bodies that we see studding the endoplasmic reticulum - more accurately, the rough endoplasmic reticulum - are the ribosomes. The rough endoplasmic reticulum owes its rough appearance to the presence of ribosomes. The smooth endoplasmic reticulum appears smooth because it lacks ribosomes. Ribosomes consist of two parts, a large subunit and a small subunit. Both the large and small subunits are made of proteins and ribonucleic acid (RNA). However, the two subunits differ both in the number and in the type of proteins and RNA they contain. The large subunit contains large and more varied RNA molecules than the small subunit. It also has more protein molecules than the smaller one. An interesting point to note is that when we put together all the chemical components of a ribosome, under favorable conditions, these parts will rearrange themselves and come together, without direction from pre- existing ribosomes, to form a functional assembly. This ability to self- assemble may provide us with a clue to the origin of living things. Ribosomes are the sites of protein synthesis in the cell. Messenger RNA (mRNA), which carries genetic in-formation from the nucleus, associates with the small ribosomal subunit first and then binds to the large sub-unit as a prelude to protein synthesis. This association of mRNA to ribosomes holds the components of the complex system of protein synthesis together in a specific manner for greater efficiency than if they were dispersed freely in the cytoplasm. The mRNA then pairs with complementary molecules of transfer RNA (tRNA), each carrying a specific amino acid. The linking up of tRNA molecules into a chain complementary to mRNA brings together amino acids which bind with each other to form a highly specific protein molecule. Thus ribosomes are the sites where proteins are synthesized under genetic control.

Question:

How and where is seminal fluid produced in man?

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0584.htm

Solution:

Semen containing seminal fluid and sperm is transferred to the vagina during sexual intercourse. There are three pairs of glands that together contribute to pro-duce the two to five ml of seminal fluid per ejaculation. These are the paired seminal vesicles, the prostate gland, and Cowper's glands. While the contents of semen have been determined, the contributions from each gland are only suggested below. The seminal vesicles secrete mucus and nutrients (such as fructose) into the ejaculatory duct. The single prostate gland secrets a milky, alkaline fluid which increase the pH of the semen. At the base of the erectile tissue of the penis lies a third pair of glands, called Cowper's glands, which contribute more mucus to the seminal fluid. The seminal fluid that results contains glucose and fructose which are metabolized by the sperm for energy, acid-base buffers to protect the sperm from the acidic secretions of the vagina, and mucous materials that lubricate the passages through which the sperm travel.

Question:

Liquid helium boils at 4\textdegreeK. What is the boilingtemperature onthe Fahrenheit scale?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0020.htm

Solution:

The temperature in Kelvin is the temperature in degrees Centigrade addedto 273. In this problem, the boiling point is given in \textdegreeK. Hence, the temperatureshould be converted to \textdegreeC and, then, to Fahrenheit using the relation \textdegreeF = 9/5 \textdegreeC + 32 The boiling point of helium can be converted to \textdegreeC by subtracting 273 from theboiling point in \textdegreeK. \textdegreeC = \textdegreeK\Elzbar 273 \textdegreeC = 4\textdegreeK\Elzbar 273 =\Elzbar 269\textdegreeC. After the temperature is converted to the Centigrade scale, the temperatureon the Fahrenheit scale can be determined. \textdegreeF = 9/5 \textdegreeC + 32 \textdegreeF = 9/5 (\Elzbar 269\textdegreeC) + 32 =\Elzbar 452\textdegreeF The boiling point of helium on the Fahrenheit scale is\Elzbar 452\textdegreeF.

Question:

Assume that 90% of the electron density is representative of the volume of the atom. The following atomic radii have been obtained for 7 elements across the second period of the periodic table and for six elements down through the first family: H 0.37 LiBeBCNOF 1.23.89.80.77.74.74.72 Na 1.57 K 2.03 Rb 2.16 Cs 2.35 Explain the observed trends of these atomic radii.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0655.htm

Solution:

Atomic radii increase going down a family, but decrease going across a period. To explain this, one must consider the change in the elements in going across a period or down a family and see how it could affect atomic radii. In moving across a period, the atomic number increases, which means, the nuclear positive charge increases. But the principal quantum number of the outside electrons remains constant. This means that the outer electrons, assuming all other factors, are constant, will remain at the same distance from the nucleus. Not all factors are constant, though. With increasing atomic number, the nuclear positive charge increases, and, thus, the force pulling the electrons towards the nucleus increases. Therefore, the atomic radii decreases as one moves across a period. Moving down a family presents a similar case of increasing atomic number. But, here, the principal quantum number is also increasing, which means the outermost electrons are farther and farther away from the nucleus. The principal quantum number has a stronger effect than atomic number in determining atomic radii. Thus, the radii increase going down a family.

Question:

Compare the processes by which glucose and fatty acids are activated.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0092.htm

Solution:

Before glucose and fatty acids can be oxidized, they must first be activated in the first steps of their respective pathways. A fatty acid is activated by reaction with ATP and coenzyme A to yield a fatty acyl Co A: R-COOH + ATP + CoA - SH \textemdash\textemdash> R - CO \sim S - CoA + AMP +PPi fatty acylfatty acyl CoA A series of enzymes, called acyl - CoA- synthetases are involved in the catalysis of this reaction. The bond formed between the fatty acid and coenzyme A is a high energy thioester bond. The energy for its formation comes from the hydrolysis of ATP to AMP and pyrophosphate. While a fatty acyl - AMP intermediate is formed, the final activated compound is a thioester. The activated form of glucose, on the other hand, is phosphorylated. Only ATP is involved in the activation, which yields glucose - 6 - phosphate and ADP Unlike fatty acid activation, ATP is hydrolyzed to ADP, not AMP, and only after glucose is converted to pyruvate does coenzyme A serve as an activator, enabling pyruvate to enter the TCA cycle.

Question:

A ring of charge with radius 0.5 m has a 0.02 m gap (see figure (a)). Compute the field at the center if the ring carries a charge of +1 coulomb.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0573.htm

Solution:

The field can be found by superposition of the fields of an imaginary ring of charge, and a negative charge segment located where the gap would be ( see figure (b) ). We must first calculate the linear charge density of the incomplete ring. \sigma = q/(length of ring) = q/(2\piR-0.02) = [1 coulomb] / [{2\pi(0.5) - 0.02}m] = 0.321 coulombs/m where \sigma is the linear charge density and R = 0.5 is the radius. We can approximate the arc length of the gap with the linear distance between the ends since this chordal length is small compared with the radius. The field at the center due to the complete ring is zero. Since all charge elements are diametrically opposite to the center, all field elements must cancel. Thus the field is entirely due to the negative charge segment. Both the ring and the segment must have charge den-sities whose absolute values (since the segment is negative) equal the charge density of the incomplete ring. Thus, the charge on the segment is: q' = (0.02m)(\rule{1em}{1pt}0.321 coulombs/m) = 6.42 × 10^-3 coulombs The field at the center is: E = k(q'/r^2) = 9 × 10^9 (N - m^2/C^2) [(6.42 ×10^-3 C)/(0.5m)^2 ] = 2.31 × 10^8 N/C where we have treated the segment as if it were a point source.

Question:

IfDNAaseis added to a bacterial cell, the DNA is hydrolyzed , the cell cannot make any more proteins and eventually dies. IfDNAaseis added to RNA viruses, they continue to produce new proteins. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0155.htm

Solution:

By means of electron microscopy and x-rays diffraction studies, much has been discovered about the structure and composition of viruses. The infective ability of viruses is due to their nucleic acid composition. An individual virus contains either DNA or RNA but not both as is true for cells . Therefore since the virus proposed in the question contains RNA, it is not affected byDNAase. The RNA replicates, forming a complementary RNA strand which acts as messenger RNA in order to code for the synthesis of new viral proteins. To produce new viral RNA, the viral RNA first synthesizes a complementary strand and thus becomes double- stranded . The double-stranded RNA serves as a template for synthesis of new viral RNA. The virus could have been tobacco mosaic virus, influenza virus or poliomyelitis virus. These are all viruses which contain single- stranded RNA as their genetic material. There are at least two groups of RNA viruses in which the RNA is normally double-stranded, and assumes a double-helical form. DNA viruses, such as the smallpox virus, SV 40 (a tumor- inducing virus), and certain bacterial viruses, such as bacteriophages T_2, T_4, and T_6, contain double-stranded DNA. Yet there are somebacteriophageswhich have asingle-strandedDNA molecule. It does not matter whether the genetic information is contained in DNA or RNA, or if it exists as a single strand or as a double helix. For viruses, the important point is that the genetic message is present as a sequence of nucleotide bases.

Question:

Suppose that all of the electrons in a gram of copper could be moved to a position 30 cm away front the copper nuclei. What would be the force of attraction between these two groups of particles?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0601.htm

Solution:

The atomic mass of copper is 63.5. Therefore, 1 g of copper contains a number of atoms given by Avogadro's number divided by the mass of 1 mole (that is, 63.5 g): No. atoms = [(6.02 × 10^23 atoms / mole) / 63.5 g / mole ] = 0.92 × 10^22 atoms / g The atomic number of copper is 29; in other words, each neutral cop-per atom contains 29 electrons. Therefore, the number of electrons in 1 g of copper is No. electrons in 1 g = 29 × 0.92 × 10^22 = 2.7 × 10^23 electrons Thus, the total change on the group of electrons is q_e = 2.7 × 10^23 ×(\rule{1em}{1pt}e) = 2.7 × 10^23 × (\rule{1em}{1pt}4.8 × 10^\rule{1em}{1pt}10 stateC) = \rule{1em}{1pt}1.3 × 10^14 stateC A similar positive charge resides on the group of copper nuclei. Hence, the attractive electrostatic force (in the OGS system) F_E = (q_1q_2) / (r^2) where r is the distance between charges q_1 and q_2 . Therefore, since the 2 groups of charges with which we are concerned both have magnitude q_e. F_E = (q2_e) / (r^2) = (\rule{1em}{1pt}1.3 × 10^14 stateC)^2 / (30cm)^2 = 1.9 × 10^25 dyne Because the nuclei and electrons have opposite charges, this force is attractive. It is as great as the gravitational force between the Earth and the moon.

Question:

Two species form a predator-prey pair. Their relation is expressed by the rule: The rate of predator population Increase is proportional to an excess in prey population and the rate of decrease of the prey pop-ulation is proportional to the increase of its enemy's population. Let N_1(t), N_2(t) be the prey and predator populations, respectively and A_1, A_2, are positive constants for natural (i.e., in absence of other species) rates of birth (per unit of population per unit of time). Write a FORTRAN program which uses the modified Euler method to simulate this system from time t = 0 to t =t_fif N_1(0) = N_10 and N_2(0) = N_20.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0587.htm

Solution:

Proceeding directly to the mathematical model, one can express the predator-prey relation as: Ṅ̇_1 = -A_2N_2(1) Ṅ̇_2 = A_1N_1(2) The program applies the modified Euler method first to equation (1) and then to equation (2). N1 and N2 are kept in a COMMON block so that the new computed value of N1 could be used in determining N2. A logical flag, TWO, is used to determine which equation is being integrated: REALT/0.0/,TFIN,DT,REALN,ACCUR,N1,N2,N10,N20,A1,A2 INTEGER I,N LOGICAL TWO/.FALSE./ COMMON TW0,N1,N2,A1,A2 READ,N,TFIN,ACCUR,NIO,N20,A1,A2 PRINT,T,NIO,N20 REALN = N DT = TFIN/REALN N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1, ACC UR, DT) TWO = .TRUE. CALL MEULER (T,N2,ACCUR,DT) TWO = .FALSE. PRINT, T,N1,N2 10CONTINUE STOP END FUNCTION G(W) REAL G,W,N1,N2,A1,A2 LOGICAL TWO COMMON TWO,N1 ,N2 ,A1 ,A2 G = (-A2)\textasteriskcenteredN2 IF (TWO) G = A1\textasteriskcenteredN1 RETURN END Note, that if equation (1) is differentiated, the result is: Ṅ̇= -A_2Ṅ_2(3) Substituting equation (2) into equation (3) one gets: Ṅ̇_1 = -A_2A_1N_1(4) Differentiating equation (2) and substituting equation (1) into the result yields: Ṅ̇_2 = -A_2A_1N_2(5) Equations (4) and (5) are of the form ӱ = -Ay where A = -A_2A_1 is constant. The conclusion can be made that this is a free harmonic oscillating system.

Question:

Explain the properties of NAND gates, and use NAND gates to generate AND, OR, and NOT operations. Draw the symbol diagrams, and give truth tables for each operation.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0051.htm

Solution:

The NAND gate is a combination of an AND gate followed by a NOT gate. In fig. 1 the structure of a NAND gate is shown. The key to understanding a NAND gate is to notice that the output is 0 if and only if its inputs are simultaneously 1. The Truth table of the NAND gate is given in fig. 2. ab ab f(a,b) 00 01 10 11 0 0 0 1 1 1 1 0 Fig. 2 F(a,b) =ab NAND gates are usually represented as in fig. 3 when used in circuit diagrams; Several other interesting properties of NAND gates are; f(a , b) =ab=a+b(DeMorgan's Law) f(a , a) =a+a=a A NAND gate with both of its inputs the same acts like a NOT gate. This idea is illustrated in fig. 4. Truth table for the gate illustrated in fig. 4 is shown in fig. 5. a f(a) 0 1 1 0 Fig. 5 Use of NAND gates to generate AND operations; Two NAND gates connected as shown in fig. 6 generate an AND operation at the output. Since AND gates have the truth table shown in fig.7. a b f(a, b) 0 0 1 1 0 1 0 1 0 0 0 1 Fig. 7 It can easily be verified that the output in fig. 4 has the sane operation. When inputs a and b are both 0, output of the first NAND gate is 1. This output is the input of the second HAND, gate, therefore both inputs to the second NAND gate are 1. The output of this HAND gate then is found to be 0, from the definition of NAND gates. The OR gate can be obtained from three NAND gates, connected as shown in fig. 8. When inputs a and b are both 0, the input to the third NAND gate is 11 and its output is 0. In all the other input combinations the output will be 1. The truth table of the OR gate is given in fig. 9. a b f(a, b) 0 0 1 1 0 1 0 1 0 1 1 1 Fig, 9

Question:

In writing a basic assembler language program, our first and foremost concern is that of housekeeping. What is housekeep-ing and why is it necessary?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0188.htm

Solution:

The first six or so instructions of the IBM/360- 370 assembly language are the housekeeping instructions. In almost all programs, there is some housekeeping to be done. These instructions provide standard linkage between the program and the operating system. In general, these instructions are: LABELSTART\varphi \varphi \varphi SAVE(14, 12) BALR3, 0 USING\textasteriskcentered, 3 ST13, SAVE+4 LA13, SAVE These housekeeping instructions have three basic func-tions: 1. They provide standard operating system (OS) program linkage. This is done by means of the SAVE, ST and LA in-structions. 2. They load and identify the base register for the program. The BALR and USING statements perform this task. The third function is not of fundamental concern to the BAL programmer. This step involves preparation of the units and materials needed by the main processing part of the program. In most instances, the task is to open input and output files, a task which the system itself usually fails to perform. Let us now consider the basic functions of each instruc-tion: LABEL START\varphi: The label is usually the name of the program. \varphi START\varphigives the system a location in memory where the program \varphi should be stored. The address coded is almost never obtained so a zero is substituted, later to be re-placed by the assembler. SAVE (14, 12): Here, the contents of 15 (Register 13 excluded) of the system's 16 general purpose registers are saved. It is important to save the contents of these re-gisters since they may hold information pertinent to the system's operation. The save area generally has a length of 18fullwords, and saving of the registers starts in the fourth word. Let us now discuss the 'ST 13, SAVE+4' and 'LA 13, SAVE' instructions. Because of some system operations which the programmer doesn't need to concern himself with present-ly, it was found to be convenient to store the contents of Register 13 in the secondfullword, (SAVE + 4), of the SAVE area. The 'LA 13, SAVE' simply places the address of the SAVE area in Register 13. This is done so that the area can be referred to, and the contents of the register re-stored at the end of the program. Now, going back, we can look at the 'BALR 3,0' and the 'USING \textasteriskcentered,3' instructions. The BALR sets up the base regis-ter for the program in Register 3. It then stores the base address (or just the address) of the program. The 'USING \textasteriskcentered, 3', in effect confirms that Register 3 is the base regis-ter of the program. It tells the computer that starting at that point (indicated by \textasteriskcentered-asterick) the base register of this program is Register 3. The way these two instructions are coded indicates the natural order of these two instruc-tions in most assembly language housekeeping procedures. The choice of the base register, of course, is left up to the programmer. From the above discussion we can clearly see why house-keeping is of foremost concern to the programmer. It allows the computer to preserve, for later use, the present con-tents of its registers, and it enables the assembler to properly place the users program in core.

Question:

What is the binding energy of the lithium nucleus?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1042.htm

Solution:

The binding energy is the energy released when a nucleus is formed; it is what holds the nucleus together. We would have to add this energy again in order to break up the nucleus. The source of this energy, released when the nucleus was destroyed, is in the form of a mass loss in the nucleons.Therefore, by comparing the "standard" mass of all the nucleus, (protons and neutrons) and the actual mass of the nucleus, we can calculate the mass defect - the amount of mass converted into energy in the formation of the atom. Einstein's theory tells us that mass and energy are equivalent. Since there are 931 Mev (energy) per 1 amu (mass) we can claulate the energy released in the destruction of the nucleus. In the isotope L_1^7_13 there are 3 protons and 4 neutrons (7 - 3). So z = 3,A = 7, M = 7.01822 where z is the atomic number, A the mass number, and M the atomic mass of Lithium. Therefore, mass defect = [3(1.00814) + 4(1.00898) - 7.01822] = [(3.02442 + 4.03592) - 7.01822] amu = [7.06034 - 7.01822] = 0.04212 amu Converting mass to energy : binding energy in Mev = 931 × mass defect = 931 × 0.04212 in amu = 39.2Mev

Question:

Determine the entropy change that takes place when 1 mole of ammonia (a) Passes from the liquid state to the gaseous state at its boiling point, - 33\textdegreeC; ∆H_VAP = 5570 cal/mole (b) as a gas at - 33\textdegreeC comes to room temperature, 25\textdegreeC. Assume heat capacity is constant at 8.9 cal/deg - mole for this range.

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Solution:

Entropy, S, may be defined as an indication of the randomness of a system. The change in entropy for any change in the state of a system may be written as ∆S =S_final-S_initial_. .For a reversible process, ∆S = [(q_rev)/(T)], whereq_rev= amount of heat involved in a reversible process and T = the temperature in \textdegreeK at which the amount of heat is reversibly absorbed or evolved. ∆S can also be defined in terms of heat capacity (C_P) and temperature ranges, say T_1 \rightarrow T_2, by the following equation: ∆S = 2.303 C_P log T_2/T_1. With this in mind, proceed as follows: (a)1 mole of NH_3 (ammonia) goes from liquid to gas at - 33\textdegreeC. The ∆H_VAP is the heat necessary to vaporize 1 mole of liquid material at its boiling point at constant temperature. A gas can also be liquefied, thus, this change of state is reversible. Recalling that ∆S =q_rev/T, You can say, therefore, that ∆S = ∆H_VAP /T. Since ∆H_VAP is the only heat involved,q_rev= ∆H_VAP. By substitution into ∆S = ∆H_VAP /T, you obtain ∆S = [(5570 cal/mole)/(240\textdegreeK)]or∆S = 23.2 cal/deg - mole. (b)This question uses heat capacity, which demands use of the expression ∆S = 2.303 C_P log T_2/T_1. Since all the values are known, you can substitute them into the above equation: ∆S = 2.303 C_P log (T_2 /T_1 ) = (2.303) (8.9) log [(298\textdegreeK)/(240\textdegreeK)] = 20.5 log 1.24 = (20.5) (.0934) = 1.91 cal/deg - mole.

Question:

In the figure below, C_1 = 6 \muf, C_2 = 3 \muf and V_ab = 18 v. In the figure below, C_1 = 6 \muf, C_2 = 3 \muf and V_ab = 18 v. Find the charge on each capacitor. What is the value of the Find the charge on each capacitor. What is the value of the equivalent capacitance (see figure)? equivalent capacitance (see figure)?

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Solution:

The charge Q on a capacitor having capacitance C is Q = VC, where V is the potential difference across the capacitor. Then, Q_1 = V_ab C_1 = (18v) × (6 × 10^-6 f) = 108 × 10^-6 C Q_2 = V_ab C_2 = (18v) × (3 × 10^-6 f) = 54 × 10^-6 C The equivalent capacitance must carry the same total charge as the original system, since charge is conserved (none leaks out of the system). Hence Q_net = Q_1 + Q_2 = 162 × 10^-6 C Then, the equivalent capacitance is C_e = (Q_net /V_ab) = [(162 × 10^-6 C)/(18 V)] = 9 × 10^-6 f

Question:

What is differential reproduction? What factors bring about differential reproduction?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0715.htm

Solution:

When the percentage of viable offspring carrying a certain genotype cannot be accounted for by purely random mating, differential reproduction is said to have occurred. Differential reproduction can result from nonrandom mating, differential fecundity, or dif-ferences in either zygote viability or offspring fertility. Nonrandom mating may occur due to genotypic variaton, genetic mutations, or both. There are well established behavior patterns of courtship and mating in many species, which are primarily genetically controlled. For example, in many species of birds and fish, a brightly colored spot on the male serves to stimulate the female for copu-lation. Males having this characteristic will succeed in finding a mate easily and more frequently than males with-out this trait, hence producing in their lifetime more offspring carrying their genes. Males lacking this trait will find it difficult to pass their genes on to succeeding generations. Mutations that lead to, formation of bigger or brighter spots will benefit the males with the mutated gene by rendering them more attractive to females. Conversely, mutations that lead to disappearance of these spots or formation of smaller, duller ones, will have a negative selective value. These mutations are inherited by the progeny, perpetuating differential reproduction through nonrandom mating. Success in mating, however, can neither guarantee successful fertilization nor assure large numbers of off-spring. Such factors can also be important in differen-tial reproduction. Differences in the number of gametes produced by different individuals and the proportion of their gametes that will successfully unite with others to form zygotes can be collectively termed differential fecundity. Differential fecundity may be determined ge-netically. Within a species, the individuals producing large numbers of gametes or having a large percentage of successful matings (resulting in fertilization) will nec-essarily contribute the greatest percentage of genes to the next generation. This depends, of course, on the via-bility of the zygotes formed. In environments which result in low zygote viability, individuals having high fecundity will tend to reproduce in larger numbers. In fish, for example, zygotes must be fertilized and also develop in an external environment; those species producing large numbers of both gametes and zygotes will tend to be preserved. Mammals, on the other hand, whose zygotes develop internally, cannot and need not produce such large numbers of gametes or zygotes. In fact, large numbers of zygotes may actually be a disadvantage, since this limits the amount of care and feeding available to each offspring. Another factor affecting the reproducing ability of a given individual or individuals is the fertility of the offspring. The offspring produced as a result of a given mating may be viable, but unless they are themselves fertile, the result is the same as if they were not viable. Thus, it is the interaction of a variety of factors which is responsible for the reproductive capacity of a given individual or species. Any one of these factors could be responsible for differential reproduction.

Question:

Two equal conducting spheres of negligible size are charged with 16.0 × 10^\rule{1em}{1pt}14C and \rule{1em}{1pt}6.4 × 10^\rule{1em}{1pt}14 C, re-spectively, and are placed 20 cm apart. They are then moved to a distance of 50 cm apart. Compare the forces between them in the two positions. The spheres are connected by a thin wire. What force does each now exert on the other?

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Solution:

The equation giving the force between the spheres, which may be considered as point charges, is by Coulomb's law, F = (1 / 4\pi\epsilon_0)(q_1q_2) / r^2 where q_1, q_2 are the charges on the spheres, and r is their separation. Thus F_1 = (1 / 4\pi\epsilon_0)[(q_1q_2) / (0.2)^2m^2 ] and F_2 = (1 / 4\pi\epsilon_0)[(q_1q_2) / (0.5)^2m^2 ] \therefore F_1 / F_2 = (0.5)^2 / (0.2)^2 = 6(1/4) If the spheres are joined by a wire, the charges, which are attracted to one another, can flow in the wire under the influence of the forces acting on them. The charges will neutralize as far as possible and [16.0 × 10^\rule{1em}{1pt}14 + (\rule{1em}{1pt} 6.4 × 10^\rule{1em}{1pt}14)] = 9.6 × 10^\rule{1em}{1pt}14 C will be left distributed over the system. Neglecting the effect of the wire, by symmetry 4.8 × 10^\rule{1em}{1pt}14 C will reside on each sphere. The force between the two spheres is now F = (1 / 4\pi\epsilon_0)(q^2 / r^2) = 9 × 10^9 N.m^2.C^\rule{1em}{1pt}2 × [(4.8 × 10^\rule{1em}{1pt}14)^2C^2 ] / (0.5)^2m^2 = 8.29 × 10^\rule{1em}{1pt}17N.

Question:

How much work is required to raise a 100-g block to a height of 200 cm and simultaneously give it a velocity of 300 cm/sec?

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Solution:

The work done is the sum of the potential energy, PE = mgh, and the kinetic energy, KE = (1/2) mv^2: PE= mgh = (100 g) × (980 cm/sec^2) × (200 cm) = 1.96 × 10^7 g-cm^2/sec^2 = 1.96 × 10^7 ergs KE= (1/2) mv^2 = (1/2) × (100 g) × (300 cm/sec)^2 = 4.5 × 10^6 g-cm^2/sec^2 W= PE + KE = 1.96 × 10^7 ergs + 0.45 × 10^7 ergs = 2.41 × 10^7 ergs = 2.41 J

Question:

What activation energy should a reaction have so that raising the temperature by 10\textdegreeC at 0\textdegreeC would triple the reaction rate?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0468.htm

Solution:

The activation energy is related to the tem-perature by the Arrhenius equation which is stated k = Ae ^-E/RT where A is a constant characteristic of the reaction; e is the base of natural loragithms, E is the activation energy, R is the gas constant (8.314 J mol^-1 deg^-1) and T is the absolute temperature. Taking the natural log of each side: ln k = ln A - E/RT For a reaction that is 3 times as fast, the Arrhenius equation becomes 3 k = Ae-E/R(T + 10\textdegree) Taking the natural log: ln 3 + ln k = ln A - E/R(T + 10\textdegree) Subtracting the equation for the final state from the equation for the initial state: lnk =lnA - E/RT - (ln3 + ln k = ln A - E/R(T + 10)) -ln3 = - E/RT + E/R(T + 10) Solving for E: -ln3 = - E/RT + E/R(T + 10)R = 8.314 J/mole \textdegreeK T = 0 + 273 = 273 -ln3 = - E/(8.314 J/mole-K) (273K) + E/(8.314 J/mole-K)(283K) -1.10 = - E/2269.72 J/mole + E/2352.86 J/mole (2269.72 J/mole)(2352.86 J/mole) × - 1.10 = (- E/2269.72 J/mole + E/2352.86 J/mole)(2269.72 J/mole) (2352.86 J/mole) - 5.874 × 10^6 J2/ mole^2 = (- E) (2352.86 J/mole) + E (2269.72 J/mole) - 5.874 × 10^6 J2/ mole^2 = - 8.314 × 10^1 J/mole × E 7.06 × 10^4 J/mole = E

Question:

If a wire carrying 2 amperes lies perpendicularly across a uni-form magnetic field of flux density 5 × 10^-2 weber/meter^2 in such a manner that 15 cm of the wire are subjected to the field (i.e., if the pole pieces of the magnet are 15 cm across), how much side thrust is experienced by the wire, and which way does it act?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0738.htm

Solution:

By the right-hand rule, it is seen that the thrust is down-ward. The magnitude of the thrust is given by F^\ding{217} = IL^\ding{217} × B^\ding{217} where B^\ding{217} is the magnetic induction, and I is the current in a wire of length L. Since the angle between L^\ding{217} and B^\ding{217} is 90 o , then F = BLI thereforeF = 5 × 10^-2 w/m^2 × 15cm × (1m/100cm) × 2 amp = 15 × 10^-3 n = .015 nt where we have used the fact that 1 cm = (1/100) n .

Question:

What is the critical angle between carbon disulfide and air? (The index of refraction for Carbon Disulfide is 1.643.) Critical Angle

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0860.htm

Solution:

Carbon disulfide is a more optically dense mater-ial than air. Therefore, as a beam of light passes from carbon disulfide to air, the angle of refraction is larger than the angle of incidence. There is an angle of incidence smaller than 90\textdegree for which the angle of refraction is equal to 90\textdegree, meaning that the beam of light emerges parallel to the boundary between the two mediums. This angle of incidence is called the critical angle. If the angle of incidence is greater than this value, the light will not escape from the carbon disulfide. It will be reflected back into the carbon disulfide following the regular law of reflection. Solving for the critical angle \texttheta_1, let \texttheta_2 be 90\textdegree. The index of re-fraction for carbon disulfide is 1.643 and for air it is 1.00 Using Snell's law n_1 sin \texttheta_1 = n_2 sin \texttheta_2 1.643 sin \texttheta_1 = 1.00 sin 90\textdegree sin \texttheta_1 = (1.00/1.643) = 0.608 \texttheta_1 = 37.4\textdegree.

Question:

Explain why penicillin is effective only against actively growing bacteria. Describe the mode of action of some other antimicrobial agents.

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Solution:

An antimicrobial agent is one that interferes with the growth and activity of microorganisms. Knowledge of the mode of action of a particular agent makes it possible to determine the conditions under which it will act most effectively. Among the known sites of action of antimicrobial agents are the cell wall, cell membrane, protein structure and synthesis and enzyme activity. The cell walls of some gram-positive bacteria are attacked by the enzyme lysozyme, which is normally found in tears, mucous secretions, and leucocytes (white blood cells). Lysozyme breaks down the peptide linkages in the cell wall complex. Some bacteria secrete enzymes which, degrade cell walls of other bacteria or prevent cell wall formation. Without a cell wall to provide support for the bacterium, it will soon lyse and die. The antimicrobial effect of penicillin is attributed to its inhibition of cell wall synthesis. Penicillin prevents the incorporation of the amino sugar, N-acetylmuramic acid into the mucopeptide structure that comprises the cell wall. This is why penicillin only works on actively growing bacteria. If cell wall formation is complete, penicillin has no effect. The cell membrane helps contain the cellular con-stituents and provides for selective transport of nutrients into the cell. Damage to this membrane thus inhibits growth or causes death. The bactericidal (kills bacteria) action of phenolic compounds, such as hexachlorophene, is attributed to their effect on cell permeability. This results in leakage of cellular constituents and eventual death. Proteins are essential to the cell for both structure and enzymatic activity. Protein denaturation (alteration of their natural configuration) causes irreparable damage to the cell. High temperatures, acidity, and alcohol denature proteins. Streptomycin combines with the ribosomes of sensitive bacteria and disturbs protein synthesis. Many agents inhibit enzymes involved in the energy- supplying reactions of the cell. For example, cyanide inhibits cytochrome oxidase in the electron transport chain, fluoride inhibits glycolysis, and dinitrophenol uncouples oxidative phosphorylation. All of these inhibit ATP synthesis. A familiar antibiotic, sulfanilamide, works by blocking the synthesis of folic acid, a necessary substrate for certain reactions in the cell. A precursor of folic acid is para-aminobenzoic acid (PABA) whose structure is very similar to that of sulfanilamide: Sulfanilamide, by effectively competing with this precursor for the binding site on an enzyme involved in the pathway, inhibits folic acid synthesis. Any compound, such as sulfanilamide, that interrupts synthetic processes by substituting itself for a natural metabolite is called an antimetabolite or metabolic analogue.

Question:

A force of 2000 dynes produced an acceleration of 500 cen-timeters per second^2. What was the mass of the object accelerated?

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Solution:

Here, we can apply Newton's Second Law, F = ma. In this case F = 2000 dynes and a = 500 cm/sec^2. Then 2000 dynes= M × 500 cm/sec2 whence M= 4 gm

Question:

Starch (amylose), (C_6H_10O_5)x, is a polymer of the sugar glucose, C_6H_12O_6, linked by - O - bonds; each starch mo-lecule may contain 200 to 1000 glucose units. A mixture of amino acids added to a solution of amylose produces no detectable reaction. One molecule of the enzyme amylase, containing the same amino acids, however breaks (hydrolyzes) 4000 -O- bonds per second, (a) Estimate the mass in picograms of glucose produced by an amylase molecule in 1 day. (b) The heat in calories liberated to a cell by the oxidation of the glucose: C_6H_12O_6 (aq) + 6O_2 (g) \rightarrow 6O_2 (g) + 6H_2O (l). ∆H = - 686 Kcal. How many calories does the cell obtain from the mass of glucose calculated in part (a)?

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Solution:

(a) Five steps can be used to solve for the number of picograms of glucose liberated from the amylose by 1 amylase in a day. (1) find the number of seconds in one day (2) calculate the number of glucose molecules formed per day (3) determine the weight of one molecule of glucose (4) solve for the number of grams of glucose formed per day (5) convert from grams to picograms. Solving: (1) There are 60 seconds in a minute,60 minutes in an hour and 24 hours in a day. One can solve for the number of seconds in a day by multiplying these factors together The units will cancel out. 60 (sec/min) × 60 (min/hour) × 24 (hours/day) = (8.64 × 10^4) (sec/day) (2) The problem states that one amylase molecule liberates 4000 molecules per second. Therefore the number of molecules liberated per day can be found by multiplying the number of seconds in a day by the number of molecules formed per second. no. of glucose molecules formed per day = = (8.64 × 10^4)(sec/day) × 4000 (molecules/sec) = (3.456 × 10^8) (molecules/day) (3) The weight of one molecule of glucose is found by dividing the molecular weight of glucose by the number of molecules per mole. (MW of C_6H_12O_6 = 180 g/mole, Avogadro's Number = 6.02 × 10^23 molecules/mole) weight of one molecule = (180 g/mole) / (6.02 × 10^23 molecule/mole) =2.99 × 10^-22 g/molecule (4) After one knows the number of molecules formed per day and the weight of one molecule, one can find the weight in grams of the glucose formed in one day by one amylase. no. of grams of glucose formed per day = = 2.99 × 10^-22 g/molecule × 3.456 × 10^8 molecules/day = 1.03 × 10^-13 g/day. (5) There are 10^12 picograms in one gram. Grams can be converted to picograms by multiplying the number of grams by the conversion factor 10^12 pg/g no. of pg produced = 1.03 × 10^-13 g/day × 10^12 pg/g = 1.03 × 10^-1 pg/day. (6) From the equation given one knows that 686 Kcal are liberated when 1 mole of glucose is oxidized. One can find the number of calories liberated by 1) determining the number of moles of glucose present 2) solving for the number of Kcal produced 3) converting Kcal to calories. Solving: 1) One has solved for the number of molecules produced per day previously (part (a)(2)). Therefore one can find the number of moles produced by dividing the number of mo-lecules produced by the number of molecules in one mole. (3.456 × 10^8 molecules/day) / (6.022 × 10^23 molecules/mole) = 5.74 × 10^-16 moles/day. 2) There are 686 Kcal liberated per mole of glucose oxidized, therefore the number of Kcal formed when 5.74 × 10^-16 moles are oxidized is equal to 5.74 × 10^-16 moles/day × 686 Kcal/mole = 3.94 × 10^-13 Kcal/day. 3) There are 1000 cal in 1 Kcal, thus Kcal can be converted to cal by multiplying by the conversion factor 1000 cal/Kcal no. of cal produced = 3.94 × 10^-13 Kcal/day × 1000 cal/Kcal = 3.94 × 10^-10 cal/day.

Question:

A 500-kg ore bucket is raised and lowered in a vertical mine shaft using a cable. Determine the upward force exerted by the cable when (a) the upward acceleration is 4 m/s^2 and (b) the bucket is moving upward with a constant velocity.

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Solution:

(a) The known quantities are the mass of the ore bucket, m = 5 × 10^2 kg, and the acceleration, a = 4 m/s^2. The two forces acting on the bucket are the weight and the tension of the cable, as illustrated in the figure. To accelerate the bucket upward, the tension, f, must be larger than the weight of the bucket. This is a consequence of Newton's first law. Newton's first law states that an ob-ject will maintain a constant velocity (or remain at rest) when no net force acts on the object. If the object accelerates, a net force must be acting on it. By vector addi-tion, the resultant upward force, F, is equal to f + w, where w = mg. Since f and w are oppositely directed, the magnitude of their vector sum is equal to the difference of their magnitudes. F = f - w. Using Newton's second law, f - w= ma(1) f= ma + w The weight of the bucket is w= mg andf= ma + mg. Substituting the known observables, f= (5 × 10^2 kg)(4 m/s^2) + (5 × 10^2 kg)(9.8 m/s^2) = 6.9 × 10^3 N. (b) If the bucket moves upward with constant velocity (uniform motion), the acceleration is zero. (This follows from Newton's first law). Equation (1) becomes f = w. The upward force (tension) exerted by the cable equals the weight of the bucket, 4.9 × 10^3 N.

Question:

It is known that both metallic and ionic crystals are good conductors of electricity, yet many of their other proper-ties differ significantly. Explain why.

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Solution:

The similarities and differences of the two will be seen after an investigation of their space lattices, i.e., those patterns of points describing the atomic or molecular arrangements of points in a unit of a crystal. For something to conduct electricity, it must have charges. In an ionic crystal, the lattice points are occupied by positive or negatively charged ions, each of which occupies a position exactly equivalent to every other species of like charge. The electric conductivity of ionic crystals is low but increases with Temp, because more electrons are excited into the conduction band of the crystal. Ionic crystals are held together by electro-static forces between the ions. In a metallic crystal, you have discrete atoms, not ions, at the lattice points. It seems that the species should not conduct electricity. It is important to realize, however, the valence or outer electrons of the metallic crystal are distributed over the crystal as a whole rather than being localized on each atom. The mobile electrons, therefore, which are able to move in an applied electric field, account for the electrical conductivity. This mobility of electrons also explains why a metallic crystal is generally strong, lustrous and malleable. In the ionic crystal, you have strong electro-static forces between the ions, which accounts for the high melting points. However, these crystals also tend to be hard and brittle because the crystal consists of parallel sheets of positively and negatively charged ions, lateral displacement may bring ions of like charge into the vicinity of each other, resulting in electro-static repulsion of like charges. This, then, helps to explain why they possess a facile crystal fracture.

Question:

Explain the Header Card method of stopping a computer job, i.e., a program.

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Solution:

In this, the first executable statement in the program is a statement whichreads in the value of some var-iable, say N, the value of which denotesthe number of data items which will follow. If the program is punchedout on cards, the value of the variable N is punched on the first cardof the data section of the deck of cards. This method is useful when duringdifferent re-runs of the same program, the set of data that is to be usedin the program changes. We have no advance knowledge, while writing the program, how muchdata will be in each set of data, and what will be their range of values. Hence, while writing the program, we just use the variable N. Later, just before running the pro-gram, we punch out a card with the value ofN for the parti-cular set of data we have. We call this card the 'Header Card', and place it as the first card of the data cards. Below is a program to illustrate the use of the Header Card Method. SUM: PROCOPTIONS(MAIN); DCL (N,X,Y)FIXED(3)INIT(0),SUM FIXED(4)INIT (0); GETLIST(N); /\textasteriskcentered WE READ THE VALUE OF N FROM THE HEADER CARD \textasteriskcentered/ LOOP: GET LIST(X) /\textasteriskcenteredWEREADTHE VALUE OF X FROM A DATA CARD \textasteriskcentered/ SUM = SUM+X; Y = Y + 1; /\textasteriskcenteredY IS INCREMENTED BY 1 \textasteriskcentered/ IF Y = N THEN GOTO FINISH; GOTO LOOP; /\textasteriskcentered IF Y IS NOT EQUAL TO N THEN WE GO BACK TO LOOP AND GET ONE MORE VALUE OF X \textasteriskcentered/ FINISH: PUTLIST(SUM); END SUM;

Question:

Army engineers have thrown a pontoon bridge 10 ft in width over a river 50 yd wide. When twelve identical trucks cross the river simultaneously the bridge sinks 1 ft. What is the weight of one truck? The density of water is 1.94 slugs\bulletft^-3.

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Solution:

When the trucks are on the bridge, the extra volume of the bridge immersed is 1 ft deep, 10 ft wide, and 50 yd = 150 ft long, i.e., 1500 ft^3. The up thrust on this extra volume immersed in water is the weight of an equal volume of water according to Archimedes' principle. Thus U =v\rho_wg = 1500 ft^3 × 1.94 slugs\bulletft^-3 × 32 ft\bullets^-2 = 9.312 x 10^4 lb. where \rho_w is the density of water, and g is the gravitation-al acceleration. But this up thrust just balances the weight of the twelve trucks. Hence one truck has a weight W = U/12 = (9.312 × 10^4 lb)/12 = 7760 lb.

Question:

Find the weight density of water at a pressure of 4000 lb/in.^2, taking the weight density at normal atmospheric pressure as 62.4 lb/ft^3. (The bulk modulus of water is 2.97 × 10^5 lb/ft^2).

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Solution:

The change ∆V in the volume of water as a result of a change ∆P in the pressure acting on it is given by -∆V/V = ∆P/B where B is the bulk modulus of water. It is defined as stress/strain = ∆P/(∆V/V) . Initially, the pressure is that of air and it is increased to 4000 lb/in.^2 Therefore, ∆P = 4000 lb/in2- P_air- = (4000 - 15) lb/in.^2 and-(∆V/V) = [(4000 - 15) lb/in^2]/[(144 in^2/ft^2) × (2.97 × 10^5 lb/ft^2)] = 9.35 × 10^-5 where we have converted the bulk modulus to lb/in^2. The weight density of water is mg/V, hence the fractional change in the density of water is ∆D = [(Mg)/(V + ∆V)] - [(Mg)/(V)] = -[(Mg)/(V)] [1 - {1/(1 + ∆V/V)}] where the weight Mg is unchanged. We find that ∆D = - D [(∆V/V)/(1 + ∆V/V)] and, since ∆V/V is much less than 1, we get ∆D/D \approx - (∆V/V) = 9.35 × 10^-5 The final density of water is D_f = D + ∆D = [62.4 + 9.35 × 10^-5 × 62.4] lb/ft^3 = 62.406 lb/ft^3. Since the compressibility of water is very small, there is not an appreciable increase in density.

Question:

When a 2\rule{1em}{1pt}V cell is connected in series with two electrical elements, the current in the circuit is 200 mA. If a 50 cycle. s^\rule{1em}{1pt}1 , 2-V ac source replaces the call, the current becomes 100 mA. What are the values of the circuit ele-ments? Suppose that the frequency is increased to 1000 cycles \textbullet s^\rule{1em}{1pt}1. What is the new value of the current?

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Solution:

Since a dc current can flow through the elements, neither of them can be a capacitor, which offers an infinite resistance to direct current. The resistance of the elements is R = (V/I) = {(2 V)/(200 × 10^\rule{1em}{1pt}3 A)} = 10 \Omega Since the current changes when alternating current is supplied, the circuit must also contain an inductor of inductance L. The reactance (the ac analogue of resist-ance) of an inductor is X = \omegaL where \omega = 2\pin. n is the frequency of variation of voltage. The impedance Z (i.e, the ratio of V to I at a set of terminals of a circuit as shown in Figure A) of the series combination of the resistor and inductor (Fig. B) is Z^2 = R^2 + X^2 Z = \surd(R^2 + \omega^2L^2) = (V'/I') = {2V/(100 × 10^\rule{1em}{1pt}3 A)} = 20 \Omega. \therefore100 \Omega^2 + \omega^2L^2 = 400 \Omega^2orL^2 = (300 \Omega^2/\omega^2). \thereforeL = {(17.32 \Omega)/(2\pi × 50 s^\rule{1em}{1pt}1)} = 0.055 H. When the frequency is increased to 10^3 cycles. s^\rule{1em}{1pt}1 , I' = (V'/Z') = (2V)/{\surd[100 \Omega^2 + (2\pi × 10^3 × 0.055)^2 \Omega^2]} = (2/346) A = 5.77 mA. We see that when the frequency n = 0, I = 200 mA. When n = 50 sec^\rule{1em}{1pt}1, I = 100mA, and when n = 1000 sec^\rule{1em}{1pt}1 , I = 5.77 mA. The inductor does then act as a resistor. Its opposition to current (the reactance) is directly proportional to frequency as is seen by the relation X = (L)\omega = (2\piL)n = Kn where K is a constant.

Question:

A 100-keV X ray is Compton scattered through an angle of 900 . What is the energy of the X ray after scattering?

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Solution:

When X-rays pass through a crystal, some of the X-rays knock electrons out of the crystal. The scattering of electromagnetic waves (light, X-rays, etc.) by the electrons in solids can be considered to be a collision between a photon of energy hѵ and an electron (Fig. 1). The electron absorbs some of the energy of the incident photon and the scattered photon therefore has less energy and thus a lower frequency ѵ'. The collision is elastic, hence total energy as well as total momentum is conserved during the collision (Fig. 2). Hence, E^ph _i + Ee1_i = Eph_f + Ee1_f P^\ding{217}ph _i = P\ding{217}ph_f + P\ding{217}e1_f . The expressions for energies and momenta are P_i = hѵ/c, P_f = hѵ'/c ; P_e = m_ev / [\surd(1 - \beta^2)] E^Ph _i = P_ic, Eph_f = p_fc ; Ee1_i = m_ec^2 ,Ee1_f = m_ec^2 / [\surd(1 - \beta^2)] where m_e is the rest mass of the electron, \beta = v/c . Conservation of momentum is shown schematically in Fig. 2. Using the law of cosines for the triangle in the figure, p2_e = p2_i + p2_f - 2p_ip_f cos \texttheta ; adding and subtracting the quantity 2p_ip_f on the right hand side, we have P2_e= (p_i - pf)^2 + 2p_ip_f(1 - cos \texttheta). The expression for the electron energy is modified to eliminate \beta from the expression for convenience; (E^e1_f)^2 =m^2_ec^4 / (1 - \beta^2) or(E^e1_f)^2 - m^2_ec^4 = [m^2_ec^4 \beta^2]/ (1 - \beta^2)] = [m_ev / \surd(1 - \beta^2)]^2 c^2 therefore(E^e1_f)^2 - m^2_ec4+ p^2_ec^2. From conservation of energy p_ic + m_ec^2 = p_fc + \surd(m^2_ec^4 + p^2_ec^2) or\surd(m^2_ec4+ p^2_ec^2) = (P_i - P_f) c + m_ec^2. Squaring both sides, we obtain another expression for p^2_e ; P^2_e = (P_i - P_f)^2 + 2m_ec(P_i - P_f). Equating the two expressions we obtained for P^2_e ; (P_i - P_f)^2 + 2P_iP_f(1-cos \texttheta) = (P_i - P_f)^2 + 2m_ec(P_i - P_f) orP_iP_f(1 - cos \texttheta) = m_ec(P_i - P_f). Now, we substitute hѵ/c and hѵ'/c for p_i and p_f respectively; ѵѵ'(1 - cos \texttheta) = [m_ec^2 / h] (ѵ - ѵ') , and divide both sides by ѵѵ' to obtain Compton's equation 1/hѵ' = [1/hѵ] + [1/ m_ec^2] (1 - cos \texttheta) Using the values m_ec^2 = 511 keV and cos 900 = 0, we find 1/hѵ' = [1/(100 keV)] + [1/ 511 keV] or hѵ' = [{(100 keV) × (511 keV)} / {(100 keV)+(511 keV)}] = [51100/611] keV = 84 keV The electron, of course, carries off the remainder of the incident energy: KE = 100 keV - 84 keV = 16 keV

Question:

In a certain species of flies, the male presents the female witha silken balloon before mating. Explain the evolutionary derivationof this behavior.

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Solution:

Animals with beneficial adaptive behavior patterns usually survive longerand successfully raise more offspring than those without. These behaviorpatterns will often be continued through successive generations. As they are passed on, they evolve. Thus, new behavior patterns can be seento arise from ancestral behavior patterns. The presentation of a silkenballoon to the female fly (familyEmpididae) before mating is the resultof the evolution of a behavior pattern. Different species of the fly family,Empididae, ex-hibit slightly differentcourtship behaviors. These dif-ferent behaviors may represent the differentstages of the same behavior as it evolved from the ancestral behaviorpattern. The male of the most primitiveEmpididaespecies mates withthe female without giving her anything. This often results in the femalecapturing and eating the male. In more advanced species, the male will first capture an insect and presentit to the female. While the female is occupied with eating the insect, the male mates with her. In even further advanced species, the malecaptures an insect, wraps it in silk and presents this to the female. The female attempts to open the silken wrapping while the male mates withher. In the most advanced species ofEmpididaethe male presents anempty balloon of silk to the female. The female pro-ceeds to unravel the silkenballoon while the male mates with her. At this stage in the evolution ofthis behavior, the silken balloon (not the insect) is the important ele-ment. By studying different species, a clue to the deriva-tion of behavior patternscan be obtained. It is impor-tant to realize that many seemingly oddbehaviors per-formed by animals have an evolutionary origin and have helpedthat animal adapt to its environment.

Question:

Suppose that a satellite is placed in a circular orbit 100 miles above the earth's surface. Determine the orbital speed v and the time t required for one complete revolution of the satellite.

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Solution:

The radius R of the circular path is determined as follows (see the figure), R= R_e + 100 miles where R_e , the earth's radius, is 6.378 × 10^6m; a distance of 1 mile is equal to 1,609m; therefore, 100 miles = 1.61 × 10^5m R = 6.378 × 10^6m + 1.61 × 10^5m = 6.539 × 10^6m. For our purposes, it is sufficient to retain only the first two digits: R = 6.5 × 10^6m The gravitational pull on the satellite is F_g = mg. This pull provides the centripetal force for the circular motion, therefore F_center = mv^2/R = mg orv^2 = gR = (9.8m/s^2)(6.5 × 10^6m) = 63.7 × 10^6m^2/s^2 \approx 64 × 10^6m^2/s^2 v = 8 × 10^3m/s This orbital speed is only approximately correct because it has been assumed that the effect of gravity 100 miles above the earth is the same as at the earth's surface. The gravitational "pull" weakens as one recedes from the earth's surface, but at 100 miles above the earth it is only slightly different from the value g, so the calculation above Is reason-ably accurate. To determine the time interval required for one revolu-tion of the satellite, the distance the satellite travels in one revolu-tion must be calculated. This is just the circumference C of a circle of radius R: C= 2\piR = (2)(3.14)(6.5 × 10^6m) = 4.1 × 10^7m The period of the motion is t= C/v = (4.1 × 10^7m)/(8 × 10^3m/s) = 5.1 × 10^3s t= (5.1 × 10^3s)/(60s/min) = 85 min The time required for one complete revolution of the satellite is about 85 min.

Question:

A G.T.O. has a 22 gal. cooling system. Suppose you fill it with a 50-50 solution by volume of (CH_2OH)_2, ethylene glycol, and water. At what temperature would freezing become a problem? Assume the specific gravity of ethylene glycol is 1.115 and the freezing point depression constant of water is (1.86 C\textdegree / mole).You might have placed in methanol (CH_3OH) instead of (CH_2OH)_2. If the current cost of ethylene glycol is 12 cents / lb and the cost of methanol is 8 cents / lb, how much money would you save by using CH_3OH? And yet, ethylene glycol is the more desirable antifreeze. Why? density = (.79 g / ml) for CH_3OH and 3.785 liter = 1 gallon.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0251.htm

Solution:

To determine the freezing point of the water- ethylene glycol solution, you employ the following equation ∆T_f= [(1000gK_f(N_2)] / [M_1(N_1 + N_2)] where ∆T_fis the change in the freezing point of the solvent, water,K_fthe freezing point depression constant of the solvent, M_1 is the molecular weight of the solvent, N_1 the number of moles of solvent and N_2 the number of moles of solute. You are givenK_ffor water and the molecular weight is 18; hence, you can solve for ∆T_fafter determining N_1 and N_2. You can calculate the number of moles of water and of ethylene glycol present by using their densities and volumes. Use the following steps: 1) Convert 22 gal (the volume of the tank) to liters, volume = (22 gal × 3.785 l / gal) = 83.3 l 2) You are told that the mixture is 50-50 by volume or one half water and one half ethylene glycol. Hence there are (83.3 / 2)l or 41.68l of each. 3) Use the density of each compound to find the number of grams of each present. (density of H_2O = (1.0 and 1000 cm^3 = 1l) mass of H_2O = (1.0 g / cm^3) × (1000 cm^3 / l) × (41.68 l) = 41.68 × 10^3 g H_20 mass of (CH_2OH)_2 = (1.115 g / cm^3) × (1000 cm^3 / l) × (41.68l) = 46.47 × 10^3 g of (CH_2OH)_2 4) The number of moles of each is found by dividing its mass by its molecular weight. (MW of H_2O = 18, MW of (CH_2OH)_2 = 62). moles of H_2O = (41.68 × 10^3 g) / ( moles 18 g / mole) = 2.32 × 10^3 moles of (CH_2OH)_2 = (46.47 × 10^3g) / (62g / mole) = 7.50 × 10^2 moles You can now solve for ∆T_fby substituting into the above equation ∆T_f= [1000 g (1.86 \textdegreeC / mole) (7.50 × 10^2mole)] / [(18 g / mole) (2.32 × 10^3 + 7.50 × 10^2 moles)] The normal freezing point of water is 0\textdegreeC, after adding the ethylene glycol, the freezing point becomes (0\textdegree-25.2\textdegreeC) or (-25.2\textdegreeC). Determine the amount of money it is possible to save by using a methanol-water mixture by calculating how much 41.68l (half of the volume of the tank) of methanol costs and subtracting this value from the cost of 41.68l of ethylene glycol. To calculate the cost of each 1) 1lb = 454 g. Since the costs are given in (cents / 1b) weight of determine the weight in pounds of 41.68 l of each. Earlier the weight of 41. 68 l of ethylene glycol was found to be (46.47 × 10^3g). Converting this to pounds, weight of (CH_2OH)_2 = 46.47 × 103g × (1 lb / 454 g) = 102.41b weight of (CH_3OH)= density × volume = (.79 g / ml) × (10^3 ml / l ) × 41.68l = 33.07 × 10^3 g Converting to pounds weight of CH_3OH = 33.07 × 10^3 g ×(1lb / 454 g) = 72.8 lb. The cost of each is found by multiplying the price per pound by the weight of each. cost of (CH_2OH)_2 = 102.4 lb × $.12 / lb = $12.29 cost of (CH_3OH) = 72.8 lb × $.08 / lb = $5.82 The difference in cost is $12.29 - 5.82 = $6.47. Hence, you save $6.47 by using methanol. You have found that a 50-50 mixture of water and ethylene glycol protects the car to a temperature of -25.2\textdegreeC. The most desirable antifreeze will protect the car at the lowest temperature, hence, to determine whetherehtyleneglycol is more desirable than methanol, calculate the lowest temperature at which the car will be protected by methanol. This is accomplished by using the above equation for freezing point depression. ∆T_f= [(1000 g (K_f) (N_2)] / [(M_1)(N2+ N_1)] Previously, one has determinedK_f, N_1and M_1 to be (1.86\textdegreeC / mole), (2.32 × 10^3) moles and (18 g / mole), respectively. Thus, to solve for ∆T_f determine N_2 first. As shown, the weight of the methanol is 33.07 × 10^3 g. Find N_2 by dividing this weight by the molecular weight of methanol, 32. No. of moles of methanol (N_2) = (33.07 × 10^3 g) / (32 g / mole) = 1.03 × 10^3 moles. Solving for ∆T_f: ∆Tf= [(1000 g)(1.86 \textdegreeC / mole)(1.03 × 10^3 moles)] / [(18 g / mole)(1.03 × 10^3 + 2.32 × 10^3 moles)] = 31.8\textdegreeC. The methanol will protect the car at temperatures as low as (0\textdegreeC-31.8\textdegreeC) or (-31.8\textdegreeC). This is fine while the car is not in motion and is cold, but when the car is driven the engine becomes very warm. Methanol has a boiling point of 60\textdegreeC and will evaporate while the engine is operating, whereas the ethylene glycol has a much higher boiling point and will not evaporate. Thus, if methanol is used as an anti-freeze it will have to be replaced after each use of the car while the ethylene glycol will last all winter. In the end it is far more expensive and troublesome to use methanol.

Question:

Calculate the magnetic induction B along the axis of a very long 2.5 × 10^-1 meters solenoid of 1000 turns of wire if the radius of the coil is 2 × 10^-2meters andthe current in the wire is 15 amperes.

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Solution:

Consider the rectangular contour (C) about the solenoid as shown in the figure. It encloses a total current NI where I is the current through the wire and N is the number of turns. By Ampere's Law \oint_c B^\ding{217} \textbullet dl^\ding{217} = \mu_0 (NI) where the constant of permeability \mu_0 = 4\pi × 10^-7 Weber/amp-m and dl^\ding{217} is the vector infinitesimal ele-ment of contour length. Assume there is no magnetic field outside the solenoid, and neglect fringing effects. Then, \oint B^\ding{217} \textbullet dl^\ding{217} = ^b\int_a B^\ding{217} \bullet dl^\ding{217} + ^c\int_b B^\ding{217} \bullet dl^\ding{217} + ^d\int_c B^\ding{217} \bullet dl^\ding{217} + ^a\int_d B^\ding{217} \bullet dl^\ding{217} = BL cos 0 o + Brcos 90 o + (0)L + Br cos 90 o = BL Hence, Ampere's law for the solenoid becomes BL = \mu_0 NI B = [(\mu_0 NI)/(L)] If N = 1000 turns, I = 15 amp, and L = 2.5 × 10^-1 m then B = {4\pi × 10^-7 [W/(a-m)] × 1000 × 15a}/(2.5 × 10^-1 m) = 7.54 × 10^-2 (weber/m^2 ) = .0754 (W/m^2 ) The radius of the solenoid does not appear in any of the calculations and is extraneous information.

Question:

How would you define ecology? Differentiate between autecologyandsynecology.

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Solution:

Ecology can be defined as the study of the interactions between groupsof organisms and their environment. The termautecologyrefers to studiesof individual organisms or populations, of single species and their interactionswith the environ-ment.Synecologyrefers to studies of various groupsof organisms which are associated to form a functional unit of the environment. Groups of organisms are characterized by three levels of organization- populations,communititesandecocystems. A population is agroup of organisms belonging to the same species which occupy a given area. A community is a unit composed of a group of populations living in a givenarea. The community and the physical environment considered togetheris an ecosystem. Each of these designations may be applied to a smalllocal entity or to a large widespread one. Thus a small group of sycamoretrees in a park may be regarded as a population, as could be thesycamore trees in the eastern United States. Similarly, a small pond andits inhabitants or the forest in which the pond is located may be treatedas an ecosystem. From these examples, we see that the limit of an ecosystemdepends on how we define our ecosystem. However each ecosystemmust consist of at least some living organisms inhabiting a physicalenvironment. Various ecosystems are linked to one another by biological, chemicaland physical processes. The entire earth is itself a true ecosystembecause no part is fully isolated from the rest. This global ecosystemis usually referred to as the biosphere.

Question:

Use the principle of conservation of mechanical energy to find the velocity with which a body must be projected vertically upward, in the absence of air resistance, to rise to a height above the earth's surface equal to the earth's radius, R.

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Solution:

Let the center of the earth be the origin. Then the initial distance of the body is r_1 = R and its final position is r_2 = 2R. Let v_1 be the initial velocity. v_2, the final velocity, of the body of mass m is zero since 2R is the maximum height the body rises. Using conservation of energy, we have KE_1 + PE_1= KE_2 + PE_2 (1/2)mv^2_1 - G(mm_E/r_1) = (1/2)mv^2_2 - G(mm_E/r_2) wherem_E, is the earth's mass. Note that potential energy is negative. Substitution yields (1/2)mv^2_1 - G(mm_E/R) = 0 - G (mm_E/2R), v^2_1 =Gm_E/R.

Question:

What is meant by regeneration? Explain regeneration by usingplanariaas an example.

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Solution:

Regeneration is defined as the repair of lost or injured parts. Planaria , like hydra, has great powers of regeneration. Because of its larger size, theplanariahas been studied the most. If a planarian is cut in two pieces, each part will regenerate into a whole worm; that is, a new head will form on the tail-piece and a new tail will form on the headpiece. If a slice is taken out from the animal immediately posterior to the head, this slice will be seen to regenerate two heads. A larger section cut out from the abdomen will regenerate a head from one end and a tail from the other. It is also possible to cut the head region down the middle, keeping the cut surfaces apart, and have two heads regenerated. The capacity for this regeneration is related to cells calledneoblasts, which are part of the mesoderm, that retain the capability to proliferate and differentiate into the different tissues found inplanaria. The striking degree of regeneration, such as seen in hydra and planaria , becomes restricted in more complex animals due to greater cellular differentiation. A star-fish may regenerate its arms and an earthworm can re-generate a new head, but in man and other warm blooded vertebrates, the adult's capacity for regeneration is restricted to the healing of wounds.

Question:

For the reaction CO_2 (g) + H_2 (g) \rightleftarrows CO(g) + H_2O (g), the value of the equilibrium constant at 825\textdegreeK is 0.137. If 5.0 moles of CO_2 , 5.0molesof H_2 , 1.0 mole of CO , and 1.0 mole of H_2O are initially present, what is the composition of the equilibrium mixture ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0315.htm

Solution:

This problem is an application of the expression for the equilibrium constant . From thestoichiometryof the reaction, one mole of CO and one mole of H_2 O are produced for one mole of CO_2 and one mole of H_2 O that are reacted. Hence, if x moles of CO are produced at equili-brium, then x moles of H_2 O are produced, x moles of CO_2 are consumed, and x moles of H_2 are consumed. Therefore, at equili-brium, there are 1+x moles of CO, 1 + x moles of H_2 O, 5x moles of CO, and 5 \rule{1em}{1pt} x moles of H_2 . If we let v denote the volume of the reaction vessel, the equilibrium concentrations are [CO] = [H_2 O] = (1 + x)/v, and [CO_2] = [H_2] = (5 \rule{1em}{1pt} x)/v . Substituting these values into the expression for the equilibrium constant gives K= 0 137 = { [CO] [H_2 O] } / { [CO_2 ] [H_2 ] } = {[(1 + x)/v] [(1 + x)/v] } / {[(5 \rule{1em}{1pt} x)/v] [(5 \rule{1em}{1pt} x)/v] } = { (1+x)^2 v^2} / { (5\Elzbarx)^2 / v^2 } = { 1+2x+x^2 } / { 25\Elzbar10x + x^2 } . or 0.137(25) \rule{1em}{1pt} 0.137(10)x + 0.137x^2 = 1 + 2x + x^2 0.863x^2 + 3.370x \rule{1em}{1pt} 2.425 = 0 . Using the quadratic equation, x =[\Elzbar3.370 \pm \surd{(3.370)^2\Elzbar 4(0.863) (-2.425)}] / 2(0.863) or x = 0.62 , x =\Elzbar 4.52 . The second of these is nonphysical and is therefore discarded. The equilibrium concentrations are then [CO] = 1 + x = 1 + 0.62 = 1.62 moles [H_2 0] = 1 + x = 1 + 0.62 =1.62 moles [C0_2] = 5 \rule{1em}{1pt} x = 5 \rule{1em}{1pt} 0.62 = 4.38 moles [H_2] = 5 \rule{1em}{1pt} x = 5 \rule{1em}{1pt} 0.62 = 4.38 moles .

Question:

Which of the following organic compounds would you predict to be associated liquids: (a) CH_3OH; (b) CH_3OCH_3; (c) CH_3CI; (d) CH_3NH_2.

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Solution:

To answer this question, one should first understand the term, associated liquids. Liquids whose molecules are held together by hydrogen bonding are associated liquids. Hydrogen bonding is an especially strong kind of attraction between a hydrogen atom (that is bound to a highly electronegative atom) and another highly electronegative atom. This type of bond is formed by purely electrostatic forces between the positive end of one polar molecule and the negative end of another polar molecule. For hydrogen bonding to be important, the electronegative atoms must be F, O, or N. Thus, once the structures of the species are known, one can tell if they are associated liquids. (a) CH_3OH, (methanol); It is an associated liquid because oxygen is bonded to the hydrogen atom. And the hydrogen atom thus forms a bond with another oxygen molecule of methane. Liquid (b), CH_3OCH_3 (dimethyl ether); is not an associated liquid; the oxygen is not bonded directly to any hydrogen atom and therefore cannot form any hydrogen bonds. Liquid (c), CH_3CI, (methyl chloride); is not an associated liquid because Cl is not electro-negative enough to form a stable hydrogen bond. Liquid (d), CH_3NH_2, (methyl amine); is an associated liquid, N is highly electronegative and the hydrogen can form hydrogen bonds with nitrogen atoms of other CH_3NH_2 molecules.

Question:

What are the functions of roots? What are the two typesof rootsystems?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0241.htm

Solution:

Roots serve two important functions: one is to anchor the plant in thesoil and hold it in an upright position; the second and biologically more im-portantfunction is to absorb water and minerals from the soil and conductthem to the stem. To perform these two functions, roots branch andrebranchextensively through the soil resulting in an enormous total surfacearea which usually exceeds that of the stem's. Roots can be classifiedas a taproot system (i.e. carrots, beets) in which the primary (first) root increases in diameter and length and functions as a storage placefor large quantities of food. A fibrous root system is com-posed of manythin main roots of equal size with smaller branches. Additional roots that grow from the stem or leaf, or any structure otherthan the primary root or one of its branches are termed adventitious roots. Adventitious roots of climbing plants such as the ivy and other vines attachthe plant body to a wall or a tree. Adventitious roots will arise from thestems of many plants when the main root system is removed. This accountsfor the ease of vegetative propagation of plants that are able to produceadventitious roots.

Question:

What factors affect the rate of transpiration?

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0257.htm

Solution:

A number of different factors affect the rate of transpiration. The most important of these is temperature. Because temperature is a measurement of the average molecular kinetic energy, a high temperature heating up the plant body will provide the water molecules within the leaves with increased internal energy, and many of them may soon gainadequate kinetic energy to enter the gaseous phase and escape into the atmosphere . The higher the temperature rises, the greater will be the rate of transpiration, all other factors remaining constant. Plants of the desert, thus , must adapt some mechanisms to slow down water loss due to rapid evapo-ration by the hot desert air. Humidity is a second important factor. Water is lost much more slowly into air already laden with water vapor. This is why plants growing in shady forests, where the humidity is generally high, typically spread large luxuriant leaf surfaces since their chief problem is getting enough sunlight , not losing water. In contrast, plants of grasslands or other exposed areas often have narrow leaves and relatively little total leaf surface , since they are able to get sufficient light but are constantly in danger of excessive water loss. Air currents also affect the rate of transpiration. Wind blows away the water vapor from leaf surfaces, allowing more vapor molecules to escape from the leaf surface. If the air is humid, wind may actually decrease the transpiration rate by cooling the leaf, but if the breeze is dry, evaporation is increased and the trans-piration rate is greatly enhanced. Leaves of plants growing in windy, exposed areas are often hairy. These hairs are believed to protect the leaf surface from wind action by trapping a layer of air saturated with water vapor over the surface of the leaf. As before , humid air reduces the rate of evaporation, thus preventing excessive loss of water.

Question:

Write a FORTRAN function subprogram that determines whether a particular item is contained in an ordered binary tree. All items in the tree are assumed to be integers.

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Solution:

A binary tree is said to be ordered if the follow-ing properties hold: For each node in the tree, all numbers contained in the left subtree are less than the number at the node being considered. All numbers contained in the right subtree of a node are greater than the number at that node. When a search is begun, we start at the root, turning left if the number being searched is smaller, or turning right if the number is greater. Consider the tree below: We must set up a linked list, complete with two pointers at each node, one for each branch. We also require a pointer for the root node. To implement this in FORTRAN, we use a two-dimensional array called TREE (100,3). We are searching the tree for an integer between 0 and 100, hence the size of the first parameter. The second parameter refers to the data for each node and the two pointers which check, respectively, the left and right subtrees. The following table shows the initial status of the array: I TREE(I, 1) TREE(I, 2) TREE(I, 3) 1 3 0 0 2 15 1 4 3 19 0 0 4 20 3 5 5 22 0 0 6 25 2 8 7 29 0 0 8 39 7 9 9 40 0 10 10 73 0 0 Array Component Node Data Left Subtree Right Subtree The values of the pointers represent row numbers in the array. We take the root of the tree to be 6. The strategy used involves logical variables. We de-clare the function to return a value of .TRUE, if the element is in the tree, and to return a value of .FALSE, if the ele-ment is not found. Nodes are checked in succession. If the value is not found at a particular node, the left branch is investigated first. Then the right branch is checked. If the node is terminal (i.e., having no branches), we exit from the program. The value of K is used to indicate the node under current consideration. The steps are given as follows. Comments will illumi-nate the modular structure of the program. LOGICAL FUNCTION SEARCH (TREE, ROOT, ITEM) INTEGER TREE (100,3), ROOT, ITEM, K K = ROOT CLOOP BEGINS HERE 10CONTINUE CIF NODE IS FOUND, SET SEARCH = .TRUE. AND EXIT IF (TREE (K,1).NE. ITEM) GO TO 20 SEARCH = .TRUE. GO TO 99 20CONTINUE CTRY SEARCHING THE BRANCHES, STARTING WITH CTHE LEFT IF (TREE(K,1).LT. ITEM) GO TO 30 CIF THIS NODE IS TERMINAL,SET SEARCH = .FALSE. AND EXIT IF (TREE (K,2) . NE. 0) GO TO 40) SEARCH = .FALSE. GO TO 99 40CONTINUE K = TREE (K,2) GO TO 98 CNOW TRY THE RIGHT BRANCH; IF NODE IS CTERMINAL, SET SEARCH\textasteriskcentered.FALSE. AND EXIT IF (TREE(K,3).NE.0) GO TO 50 SEARCH = .FALSE. GO TO 99 50CONTINUE K = TREE (K,3) 98CONTINUE GO TO 10 CEND LOOP 99CONTINUE RETURN END

Question:

Suppose the change HC_2 O_4 ̄ + Cl_2 CO_3 ^2 ̄ +Cl ̄ is to be carried out in basic solution. Starting with 0.10 mole of OH ̄, 0.10 mole of HC_2 O_4 ̄, and 0.05 mole of Cl_2, how many moles ofCl ̄ would be expected to be in the final solution?

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Solution:

The equation for this reaction is HC_2 O_4 ̄ + Cl_2 + 5OH ̄ \rightarrow 2Cl ̄ + 2CO_3 ^2 ̄ + 3H_2 O From this equation one knows that 1 mole of HC2O_4 ̄, 1 mole of Cl_2 / and 5 moles of OH ̄ are needed to form 2 moles ofCl ̄. One needs 5 times as much OH ̄ as H_C O ̄ or Cl_2. This means that if one wishes to react 0.10 moles of HC_2 O_4, or Cl_2 there must be 0.50 moles of OH ̄ present. There are 0.10 moles of OH ̄ present, thus one needs only .02molesof each HC_2 O_4 ̄ and Cl_2. Some of these reactants will re-main unchanged. From the equation, one notes that there are 2 moles ofCl ̄ formed for each Cl_2 reacted. If .O_2 moles of Cl_2 are used then .04 moles ofCl ̄ will be formed. OH ̄ is called the limiting reagent.

Question:

Define the terms \textquotedblleftrange\textquotedblright and \textquotedblleftcenter of origin.\textquotedblright

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0728.htm

Solution:

The range of a given species is defined as that portion of earth in whichit is found. The range may be small or large. Thekangeroos, for example, are found only in Australia, while man is found almost all over theworld. The range of a species does not neces-sarily include its center oforigin. The center of origin of a species is that particular place where it originated, since each species of plant or animal originated only once. It is nowknown that a new species arises in a particular geographic area. If thespecies is favored for survival, it spreads out until it is stopped by a barrierof some kind - physical, such as an ocean; environmental, such as anadverse climate; or biological, such as the absence of food, danger frompredators, the failure to compete with native members, or the inability to produce fertile young.

Question:

What are the chief components and structures of a eucaryotic cell?

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Solution:

Membranes are a crucial component of both animal and plant cells. The plasma membrane surrounds the cell and serves to separate the internal living matter from the external environment. Plant cells have in addition a cell wall external to the plasma membrane. It is composed mainly of cellulose and is fairly rigid but permeable. Selectivity of materials entering the plant cells is not a function of the cell wall but of the plasma membrane. Membranes are also present inside the cell, dividing the cell into compart-ments distinctive in both form and function. All membranes are composed basically of lipids and proteins. For this reason the plasma membrane and the membranes inside the cell are termed unit membranes. Subcellular structures in the cytoplasm are known as organelles. When these are surrounded by membranes, they are termed membrane-bounded organelles. Each organelle performs some specific functions. The nucleus is the controlling center of the cell. The nucleus is surrounded by two layers of unit membrane which form the nuclear envelope. Within the nucleus are found the chromosomes, the substances of inheritance. Chromosomes are composed chiefly of deoxyribonucleic acid (DNA) and protein. Genes located within the chromosomes, direct cellular function and are capable of being replicated in nuclear division. The nucleolus is a specialized region in the nucleus involved in the synthesis of ribosomal RNA , the material making up the ribosomes. Mitochondria and chloroplasts are membrane-bounded organelles involved in energy production in the cell. Since energy can neither be created nor destroyed, these organelles actually convert one form of energy to an-other. Chloroplasts found in plant cells, convert solar energy to chemical energy contained in organic substances; the latter are oxidized in the mitochondria to yield energy useful to the cell in the form of ATP. Cells which utilize food for energy must take food into the cell and degrade it. Lysosomes are membrane- bound organelles whose digestive enzymes break down organic substances to simpler forms, which can be used by the cell to yield energy for its life-sustaining process. Most cells contain membrane-delimited bodies called vacuoles. Small vacuoles may be termed vesicles. These structures may contain the ingested materials taken from the cell exterior or materials to be released by the cell. Mature plant cells usually contain a single large fluid- filled vacuole. This vacuole aids the cell in maintaining an internal pressure and thus rigidity. Organelles involved in the synthesis and transport of cellular components are the ribosomes, endoplasmic reticulum, and Golgi apparatus. Ribosomes are involved in protein synthesis. The endoplasmic reticulum is a system of membranes providing transport channels within the cell. It transports the synthesized protein to other parts of the cell. The Golgi apparatus is involved in the packaging of cellular products before they can be released by the cell to the outside. Organelles involved in the maintenance of cellular shape and in movement are the microfilaments and micro-tubules, also known as the skeleton of the living cell. Microfilaments are involved in the connection of adjacent cells for intercellular communication. They also function in the transport of products within the cell. The micro-tubules are the basic substance in the cilia and flagella of motile cells. By the contracting action of micro-tubules, flagella and cilia beat and propel the cell. The microtubules have also been identified as the fundamental substance in the spindle apparatus during cell division. Both microfilaments and microtubules are protein structures.

Question:

(a) What is the magnitude of the electric field at a distance of 1A\textdegree (= 10^-8 cm) from a proton? (b) What is the electrostatic potential at this distance? (c) What is the potential difference in volts between positions 1A\textdegree and 0.2A\textdegree from a proton? (d) A proton is released from rest at a distance of 1A\textdegree from another proton. What is the kinetic energy when the protons have moved infinitely apart? If one proton is kept at rest while the other moves, what is the terminal velocity of the moving proton? (e) A proton is accelerated from rest by a uniform electric field. The proton moves through a potential drop of 100 volts. What is its final kinetic energy? (Note that 100 volts \cong 0.33statvolts.)

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Solution:

(a) From Coulomb's law E = e/r^2 \cong (5 × 10^-10esu)/(1 × 10^-8cm)^2 = 5 × 10^6statvolts/cm StatVolt/cm is the dimension of electric field. The field is directedradially outward from the proton. (b) Electrostatic potential is given by ^\infty\int_r E^\ding{217} \bulletds^\ding{217}. By convention, we shall assume that the potential is 0 at infinity, and we shall set our limits of integration accordingly. \varphi(r) = ^\infty\int_r(e/r^2)dr= \rule{1em}{1pt}(e/r) ^\infty\vert_r = e/r \cong (5 × 10^-10esu)/(1 × 10^-8 cm) \cong 5 × 10^-2statvolts. (c) The potential difference between two points is ^p2\int_p1 E^\ding{217} \bulletds^\ding{217} where p_1 and p_2 are the two points. Therefore, ^p2\int_p1 E^\ding{217} \bulletds^\ding{217} = ^p2\int_p1 (e/r^2)dr= ^(1×10)-8cm\int_(0.2×10)-8cm (e/r^2)dr = \rule{1em}{1pt}(e/r) ^(1×10)-8cm\vert_(0.2×10)-8 cm \cong \rule{1em}{1pt} [(5 × 10^-10esu) / (1 × 10^-8 cm)] + (5 × 10^-8esu) / (0.2 × 10^-8 cm) \cong 0.2statvolts (d) By conservation of energy we know that the kinetic energy must equal the original potential energy. Potential energy is given by \intF^\ding{217} \textbulletds^\ding{217}, where potential energy is zero at infinity. The force is given byqE. Therefore, ^\infty\int_(1×10)-8 cmqE\bulletds= ^\infty\int_(1×10)-8 cm (e^2/r^2) \bulletdr = \rule{1em}{1pt} (e^2/r)^\infty\vert(1x10)-8 cm = (4.8 × 10^-10esu)^2 / (1 × 10^-8 cm) \cong 23 × 10^-12 erg If one proton is kept at rest while the other moves, the terminal velocity of the moving proton is given by (using conservation of energy) (1/2) Mv^2 \cong 23 × 10^-12 erg v^2 \cong (2 × 23 × 10^-12 erg) / (1.67 × 10^-24 gm) \cong [2 × 23 × 10^-12 (gm - cm^2/sec^2)] / [1.67 × 10^-24 gm] \cong 27 × 10^12(cm^2/sec^2) v\cong 5 × 10^6 cm/sec (e) The kinetic energy will be equal to the change in potential energy. This is equal to ^p2\int_p1 F \textbulletds. Here F =qEand energy is^p2\int_p1qE\bulletds = q^p2\int_p1 E \bulletds= q\varphi_12, or (4.8 × 10^-10statcoul) (0.33statvolt) \cong 1.6 × 10^-10 erg.

Question:

Why would it be disadvantageous for a host-specificparasite tokill its host?

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Solution:

In the course of their evolution, parasites usually develop special featuresof behavior and physiology that make them better adjusted to the particularcharac-teristics of their host. This means that they often tend to becomemore and more specific for their host. Where an ancestral organismmight have parasitized all species in a particular family, its variousdescendants may parasitize only one species of host at each stagein its development. This means that many parasites are capable of livingonly in one specific host species and that if they should cause the extinctionof their host, then they themselves would also become extinct. A dynamicbalance exists where the host usually survives without being seriouslydamaged and at the same time allowing the parasites to moderatelyprosper. Probably most long-established host-parasite relationshipsare balanced ones. The ideally adapted parasite is one that canflourish without reducing its host's ability to grow and reproduce. For everyone of the parasitic species that cause serious disease in man and otherorganisms, there are many others that give their hosts little or no trouble. It is generally true that the deadliest of the parasites are the ones thatare the most poorly adapted to the species affected. Relationships thatresult in serious disease in the host are usually relatively new ones, or onesin which a new and more virulent form of the parasite has recently arisen, or where the host showing the serious disease symptoms is not theprimary host of the parasite. Many examples are known where man is onlyan occasional host for a particular parasite and suffers severe diseasesymptoms, although the wild animal that is the primary host showsfew ill effects from its relationship with the same parasite.

Question:

Compute the equivalent resistance of the network in Figure 1, and find the current in each resistor.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0684.htm

Solution:

Successive stages in the reduction to a single equivalent resistance are shown in Figs. 1 and 2. The 6\Omega and 3\Omega resistors in Fig. 1 are in parallel, hence they are equivalent to the [(6\Omega × 3\Omega)/(6\Omega + 3\Omega)] = 2\Omega resistor in Fig. 2. The series combination of this 2\Omega resistor with the 4\Omega resistor gives the 6\Omega resistor in Fig. 3. The total current I_1, in the circuit is I_1 = [(18v)/(6\Omega)] = 3 amp. The current I_2 through the 60 resistor in Fig. 1 is I_2 = [(V_cb )/(6\Omega)] where V_cb is the potential across the 6\Omega and 3\Omega resistors. Similarly, the current I_3 through the 3\Omega resistor is I_3 = [(V_cb )/(3\Omega)] V_cb is obtained by considering the loop equation for Fig. 2; V_cb = I_1 × 2\Omega = 3 amp × 2\Omega = 6v. Therefore I_2 = [(6V)/(6\Omega)] = 1 amp, I_3 = [(6V)/(3\Omega)] = 2 amp.

Question:

A building society lends money to house purchasers subject to a monthly payment of one per cent of the amount borrowed . This payment covers both capital repayment and interest due, interest being charged at a rate of eight percent per annum, calculated monthly. Write a program which reads in the amount of a loan, tabulates the series of payments required, as a four column table showing the payment number, the interest.duethat month, the capital repayment that month, and the outstanding capital balance. The table should also show the final non-standard payment required to complete the payment.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0417.htm

Solution:

PROGRAM LOANPAYMENT (INPUT, OUTPUT); (\textasteriskcentered This program reads in the amount of a loan \textasteriskcentered) (\textasteriskcentered and prints the repayment table showing \textasteriskcentered) (\textasteriskcentered the payment number, the monthly interest, \textasteriskcentered) (\textasteriskcentered the interest repayment, and the residual \textasteriskcentered) (\textasteriskcentered balance. \textasteriskcentered) CONST ANNUALRATE =8;(\textasteriskcentered per cent \textasteriskcentered) VARNUMBER :1 . . MAXINT; INTEREST, REPAYMENT, LOAN, PAYMENT, RESIDUE, MONTHRATE:REAL; BEGIN READ (LOAN); WRITE ('AMOUNT BORROWED =' ,LOAN: 12:2); PAYMENT: = LOAN/100; WRITELN (' ':10, 'MONTHLY REPAYMENT =', PAYMENT:10:2); WRITELN; WRITELN; WRITELN ('NUMBER INTEREST REPAYMENT RESIDUE'); WRITELN; MONTHRATE: =ANNUALRATE/100/12; NUMBER: = 1; RESIDUE: = LOAN; REPEAT INTEREST: = MONTHRATE\textasteriskcenteredRESI DUE; REPAYMENT: = PAYMENT - INTEREST; RESIDUE: = RESIDUE - REPAYMENT; WRITELN (NUMBER: 7, INTEREST: 10:2, REPAYMENT; 10:2, RESIDUE:10:2 ); NUMBER: = NUMBER +1 UNTIL RESIDUE+RESI DUE \textasteriskcenteredMONTHR ATE <= PAYMENT; WRITELN; WRITELN; WRITELN ('LAST PAYMENT=', RESIDUE+RESIDUE \textasteriskcenteredMONTHRATE: 10:2) END.

Question:

Why does a physician advise reduced salt intake in cases of hypertension?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0389.htm

Solution:

Hypertension or high blood pressure, is caused by abnormalities in arterial pressure regulation. The mean arterial pressure (M.A.P.) is the average pressure throughout the pressure pulse cycle. It is not equal to the average of the systolic and diastolic pressure, since the heart remains in diastole twice as long as in systole. M.A.P. = [(SBP + 2DBP) / 3] = [{120 + 2(80)} / 3] = 93.3 Thus mean arterial pressure is closer to the diastolic than to the systolic blood pressure. Arterial pressure is determined by two factors: cardiac output and total peripheral resistance. The relationship between them is: Pressure = Cardiac Output × Total Peripheral Resistance Cardiac output is the amount of blood pumped by each ventricle per unit time, and is determined by both the heart rate and the stroke volume. Total peripheral resistance is the resistance of the entire systemic circulation. Any factor that causes an increase in cardiac output or total peripheral resistance will cause an increase in mean arterial pressure. One such factor is the secretion of the hormonealdosteroneby the adrenal cortex.Aldosteroneacts on the tubules of the kidney to increase thereabsorptionof sodium. This increased retention of sodium causes more water to be retained, thereby increasing the volume of the blood. This increased blood volume increases the stroke volume and hence the cardiac output. Arterial blood pressure increases. When excessive aldosteroneis secreted (such as by a tumor in the adrenal cortex), hypertension results. Salt, or sodium chloride, is a source of sodium. A decreased salt intake causes less water to be retained, decreasing blood volume, and in turn lowering blood pressure. Blood pressure can also be increased by vasoconstriction (constriction of the blood vessels). Constriction causes an increased resistance to flow (since a decrease in vessel radius increases vascular resistance). A protein calledangiotensinII causes both vasoconstriction and stimulation ofaldosteroneproduction and secretion.AngiotensinII is producedenzymaticallyfromangiotensinI, which is converted from angiotensinogen, a glycoprotein made in the liver. The enzyme rennin, formed in the kidney, catalyzes this conversion. Kidney damage can cause anoversecretionof rennin, which causes an increase of angiotensinII, and thus hypertension. Slight hypertension usually accompanies arteriosclerosis, the deposition of fatty material (such as cholesterol) in the arterial walls. The arteries become fibrous and calcified. This buildup of deposits inhibits the flow of blood, increasing resistance to flow and thus increasing the blood pressure needed to overcome the resistance. Oversecretionof rennin andaldosteroneand arteriosclerosis can thus cause hypertension, but in most cases (90%), the cause is unknown. These cases, known as \textquotedblleftessential\textquotedblright hypertension (because of unknown etiology) appear to be hereditary. Hypertension is harmful because it increases the work load of the heart and damages the arteries by subjecting them to excessive pressure. Limiting the amount of salt in foods may help to alleviate hypertension.

Question:

For underground electrical transmission cables using insu-lating materials, the power loss in the insulator is propor-tional to V^2, where V is the potential applied to the cable. Compare the power loss per meter for two standard transmission voltages, 69,000 V and 345,000 V.

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0793.htm

Solution:

The available power is given by the relation-ship, P = VI(1) where I is the current through the cable. I can be re-placed by I = (V/R), by ohm's Law, where R is the cable re-sistance. Then, from (1), P = (V^2/R) If R is constant, P \propto V^2(2) For the transmission cable, power loss is proportional to the available power and, by (2), is therefore also proportional to V^2 . If the same cable is used for both poten-tials, the following ratio provides a comparison: (power loss of 345,000-V line/ power loss of 69,000-V line) = {(3.45 × 10^5V)^2/(6.9 × 10^4 V)^2} = 25

Question:

Assume X and Y are two arrays, each already sorted in ascending order, with M and N elements respectively. Merge X and Y into a single sorted array Z.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0145.htm

Solution:

One method is to compare the first elements of each array. The smaller of the two will become the first member of the new array, while the greater of the two will come next. This picture will illustrate the idea: The flow chart is as follows: INTERIOR L,M,N,K,NX,NY,X(M) ,Y(N) L = M + N K = 1 NX = 1 NY = 1 20IF (X(NX).LT.Y(NY)) GO TO 50 CTHIS SECTI0N EXECUTES ALL STATEMENTS C\ON THE RIGHT SIDE 0F THE FL0W CHART. Z (K) = Y (NY) K = K + 1 NY = NY + 1 IF (NY.LE.N) GO TO 20 D\O 30 IY = K, L Z (I) = X (NX + IY - K) 30C\ONTINUE G\O T\O 9\O CTHIS SECTION EXECUTES ALL STATEMENTS ON THE CLEFT SIDE OF THE FLOWCHART. 50Z (K) = X (NX) K = K + 1 NX = NX + 1 IF (NX.LE.M) G\O T\O 2\O D\O 40 IX = K, L Z (I) = Y (NY + IX - K) 40C\ONTINUE 90 ST\OP END [Note: Merge sorts of this type are used frequently in business applications, in which large files must be updated with new material.]

Question:

Let us assume that the conduction electrons in a metal are completely free except for a retarding force which is pro-- portional to the speed, a) Find the variation of the electron velocity with time when a constant electric field is set up in a metal, b) How long does it take for the electrons to reach maximum velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0632.htm

Solution:

a) The acceleration of an electron due to an electric field is given by m_e (dv/ d_t)1=Ee where v is the velocity of the electron, and e is the electronic charge. The retarding force gives rise to a deceleration given by m_e (d_v / d_t)_2 = \rule{1em}{1pt}bv where b is some constant. The equation of motion is then m_e (d_v / d_t) = Ee \rule{1em}{1pt}bv The solution of this equation is (see the figure) v(t) = eE / b(1\rule{1em}{1pt} e\rule{1em}{1pt}t / T) where T = m_e / bis called the relaxation time. We see that as t becomes large, a terminal velocity v_t = eE / b is established b)If we write v_t as v_t = (eE / m_e ) (me/ b) = (eE / m_e )T it turns out to be just that velocity which an electron would attain in a time T with acceleration eE / me, v_t = aT = (eE / M_e )T.

Question:

Define stack, queue, anddeque. What characteristics of these three list structures make each of them useful for different applications?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0137.htm

Solution:

A queue is a type of list structure with the lim-itation that items may be added or deleted at either end, but not in the middle. The term queue is also used in con-nection with computer memory to mean a group of items chained together sequentially by addresses. But here, we will dis-cuss the queue as a way of entering data at the macro level. Queue is a general term which describes several types of lists. One type of queue is known as a FIFO queue, which stands for "first-in-first out." Items that are added to a FIFO queue at one end are removed at the other. The other type of queue is a LIFO queue, which stands for "last-in- first-out." This type of queue is known as a stack. Stacks have the property that the items are removed in the reverse of the order in which they are entered. In addition, a stack allows insertions and deletions to take place only at one end. The termdequeis an acronym for "double-ended queue", in which items may be added to, or deleted from, either end. This type of list may be constructed with a double set of pointers, one to keep track of items moving up the list, and one to keep track of items moving down the list. One sample application of queues is in the simulation of customer arrival at a ticket window. New customers coming up to the window can be placed at one end of a queue which represents the line of prospective ticket buyers. When the tickets are bought, the customer is removed from the list at the opposite end. Using this concept, the owner of the ticket booth can simulate the flow of customers and plan his business activities accordingly. Stacks are especially useful in processing data struc-tures such as trees. Stacks can also be used when converting expressions in algebraic notation to Polish expressions. Recursive procedures, sometimes found in PL/1 programs, use the notion of stacks to allocate additional memory space for themselves. Deques, since they can be constructed with two sets of pointers, are ideal for two-way, or undirected, lists. This allows the user to search a list back and forth from either end.

Question:

In man, the kidney performs the bulk of the excretion of wastes from the body. Outline the structure of the human kidney and urinary system.

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0441.htm

Solution:

Located on each posterior side of the human body just below the level of the stomach are the bean-shaped kidneys. Each kidney is about 10 cm long, and consists of three parts: an outer layer called the cortex, an inner layer called the medulla, and a sac-like chamber called the pelvis (see figure 1) . The functional unit of a kidney is the nephron; there are about a million nephrons per kidney. A nephron consists of two components: a tubule for conduct-ing cell-free fluid and a capillary network for carrying blood cells and plasma. The mechanisms by which the kidneys perform their functions depend on both the physical and physiological relationships between these two components of the nephron. Throughout its course, the kidney tubule is composed of a single layer of epithelial cells which differs in structure and function from one portion of the tubule to another. The blind end of the tubule is Bowman's capsule, a sac embedded in the cortex and lined with thin epithelial cells. The curved side of Bowman's capsule is in intimate contact with the glomerulus, a compacttuft of branching blood capillaries, while the other opens into the first portion of the tubular system called the proximal con-voluted tubule. The proximal convoluted tubule leads to a portion of the tubule known as the loop of Henle. This hairpin loop consists of a descending and an ascending limb, both of which extend into the medulla. Following the loop, the tubule once more becomes coiled as the distal convoluted tubule. Finally, the tubule runs a straight course as the collecting duct. From the glomerulus to the beginning of the collecting duct, each of the million or so nephrons is completely separate from its neighbors. However, the collecting ducts from separate nephrons join to form common ducts, which in turn join to form even longer ducts, which finally empty into a large central cavity, the renal pelvis, at the base of each kidney. The renal pelvis is con-tinuous with the ureter, which empties into the urinary bladder where urine is temporarily stored. The urine remains unchanged in the bladder, and when eventually excreted, has the same composition as when it left the collecting ducts. Blood enters the kidney through the renal artery, which upon reaching the kidney divides into smaller and smaller branches. Each small artery gives off a series of arterioles, each of which leads to a glomerulus. The arterioles leading to the glomerulus are called afferent arterioles. The glomerulus protrudes into the cup of Bowman's capsule and is completely surrounded by the epithelial lining of the capsule. The functional signi-ficance of this anatomical arrangement is that blood in the capillaries of the glomerulus is separated from the space within Bowman's capsule only by two extremely thin layers: (1) the single-celled capillary wall, and (2) the one- celled lining of Bowman's capsule. This thin barrier permits the filtration of plasma (the non-cellular blood fraction) from the capillaries into Bowman's capsule. Ordinarily, capillaries recombine to form the begin-nings of the venous system. However, glomerular capil-laries instead recombine to form another set of arterioles, called the efferent arterioles. Soon after leaving the region of the capsule, these arterioles branch again forming a capillary network surrounding the tubule. Each excretory tubule is thus well supplied with circulatory vessels. The capillaries eventually rejoin to form venous channels, through which the blood ultimately leaves the kidney.

Question:

What is meant by the term muscle fatigue?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0471.htm

Solution:

A muscle that has contracted several times, exhausting its stored supplyof organic phosphates and glycogen, will accumulate lactic acid. (Review chapter 3, cellular metabolism.) It is unable to contract any longer andis said to be "fatigued". Fatigue is primarily induced by this accumulationof lactic acid, which correlates closely with the depletion of themuscle stores of glycogen. Fatigue, however, may actually be felt by theindividual before the muscle reaches the exhausted condition. The spot most susceptible to fatigue can be demonst-rated experimentally. A muscle and its attached nerve can be dissected out and thenerve stimulated repeatedly by electric shock until the muscle no longercontracts. If the muscle is then stimulated directly by placing electro-deson the muscle tissue, it will contract. With the proper device for detectingthe passage of nerve impulses, it can be shown that upon fatigue, the nerve leading to the muscle is not fatigued, but remains capableof con-duction. Thus, since the nerve is still conducting impulses andthe muscle is still capable of contracting, the point of fatigue must be atthe junction between the nerve and the muscle, where nerve impulses initiatemuscle contrac-tion. Fatigue is then due in part to an accumulation oflactic acid, in part to depletion of stored energy reser-ves, and in part to breakdownin neuromuscular junction transmission. In contrast to true muscle fatigue, psychological fatigue may cause anindividual to stop exercising even though his muscles are not depleted ofATP and are still able to contract. An athlete's performance depends not onlyupon the physical state of his muscles but also upon his will to perform.

Question:

Determine themolarityof an H_2SO_4, solution, 15.0 ml of which is completely neutralized by 26.5 ml of 0.100 M NaOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0366.htm

Solution:

Acids and bases react with each other to produce salts and water. Such reactions are called neutralizations. To find themolarityof the neutralized acid solution, first write out the balanced reaction. H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2HOH. When neutralization is complete, there is no longer any acid or base left. It is given that 26.5 ml of 0.100 MNaOHis used. From this, one wants to compute how many moles ofNaOHwere present. It is known from the neutralization reaction that the number of moles of acid must be 1/2 the number of moles of base for a complete neutralization. From the number of moles of base present, the number of moles of acid needed are determined. Sincemolarity= moles/liter and one is given the volume, themolaritycan be calculated. Proceed as follows: There exists 26.5 ml of 0.100 MNaOH. There are 1,000 ml in one liter. This means that 26.5 ml is equal to 0.0265 liters. Recalling the definition ofmolarity. 0.1 = [(no. of moles ofNaOH)/(0.0265 l)] . Solving moles = 0.00265 ofNaOH. This, then, must be twice the number of moles of acid. For the acid, therefore,molarity= [(.00265)(1/2)]/liters. One is told that the volume of the H_2SO_4 solution is 15 ml, or 0.015 liters (1000 ml = 1 liter).Molarityof the acid becomes (0.001325/0.015) = 0.088 M.

Question:

What kind of marine life is found in the different regions of the ocean and what are these organisms' adaptations to their environment?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0788.htm

Solution:

The neritic, or coastal, waters support a greater population of marine life forms than do those of the open ocean. This abundance results from the rich supply of nitrates, phosphates, and other nutrients dumped into coastal waters by rivers and streams or brought to the surface by upwellings and turbulence. These substances are required by the producers who are the photosynthetic organisms that form the base of the food chain for animal life. Also, the ocean floor receives much more energy from sunlight here and as a result there is a great luxury of both plants and animals in the neritic waters. More is known about the communities of the intertidal and subtidal regions than about those of the lower neritic region. On a typical sandy shore (Fig. 1) the dominant animals in the intertidal region might include ghost shrimps, burrowing isopods, burrowing amphipods, mole crabs, polychaete worms, and beach clams, and the dominant forms in the subtidal region might include hermit crabs, sand dollars, borrowing shrimp, ascidians, and copepods. In contrast, on a rocky shore (Fig. 2) the dominant animal forms in the intertidal region might include barnacles, oysters, and mussels, with such forms as sea anemones, sea urchins, and corals in the subtidal region. On mud bottoms, various kinds of burrowing clams and snails are found. Many of the bottom dwellers are adapted to feed on plankton and dead organism. The organisms of the intertidal region are subjected to alternating periods of exposure to the air and to the strong action of waves. They are adapted to these en-vironmental factors in a number of ways. For example, the seaweeds have tough outer coverings and pliable bodies. Such animals as crabs, barnacles, mollusks and starfish have hard calcerous shells, whereas others like the sea anemones have a leathery outer covering. Also, many animals are able to burrow in the sand to escape exposure to air. Neritic pelagic communities include a great variety of fish species, larger crustaceans, turtles, seals, whales, and others. Most of the fish are plankton feeders - the adult fish feed mostly on zooplankton and are thus secondary consumers. Some fish, such as sharks, are dangerous predators. The euphotic zone of the oceanic region is inhabited by planktonic organisms and a variety of nektonic forms, including many kinds of fish. Mackeral and herring are among the commercial fish found in this oceanic region. Whales may also be present, as well as many kinds of birds. The bathyal and abyssal zones are incomplete ecosystems since there can be no producers where light cannot reach. The food of the inhabitants consists partly of the dead bodies of plants and animals that rain down from above, and partly of each other. A variety of species of fish, crustaceans, echinoderms, worms and mollusks are found in these zones and on the ocean floor. Many of the inhabitants posses light-producing structures which are significant adaptation in this darkened habitat. The deep sea forms are adapted to withstand great pressure by keeping the pressure uniform inside their bodies and on the outside. In the hadal zone at the great depths of the Pacific, organisms are subjected to pressures greater than 1000 atmospheres. At a depth of 10,500 meters, sea cucumbers, sea anemones, bivalve mollusks and crustaceans have been found. This is close to the deepest part of the ocean yet discovered, which is 10,860 meters.

Question:

In humans, the liver is one of the largest and most active organsin the body. Describe its functions.

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Solution:

One of the most important functions of the liver is to regulate the levelof glucose in the blood stream. The maintenance of a constant level ofblood glucose is essential because specialized cells, such as the brain cells, are easily damaged by slight fluctuations in glucose levels. Excess glucoseis stored in the liver as glycogen while declining blood glucose canbe restored by conversion of liver glycogen into glucose. In addition to theformation of glycogen, the liver can convert glucose into fat, which is theneither stored in the liver itself or in special fat-deposit tissues known asadipose tissues. The production of bile occurs in the liver. Bile is of major importancein aiding the digestion of fats. It is secreted by the liver cells intoa number of small ducts which drain into a common bile duct. Bile is storedin the gallbladder, which releases its contents upon stimulation by thepresence of fats in the duodenum. When proteins are metabolized, ammonia, a toxic sub-stance, is formed. The cells of the liver are able to detoxify ammonia rapidly by transformingit into the relatively inert substance urea. Urea can then be excretedfrom the body in the urine. The liver also produces a lipid which has been frequently associatedwith heart diseases - cholesterol. In fact, a diet with reduced cholesterolcontent may not be effective in lowering cholesterol level in the bodybecause the liver responds by producing more. The liver plays some important roles in blood clotting. It is the site ofproduction ofprothrombinand fibrinogen, substances essential to the formationof a blood clot. In addition, bile secreted by the liver aids in the absorptionof vitamin K by the intestine. Vitamin K is an essential cofactor inthe synthesis ofprothrombinand the plasma clotting cofactors. Vitamin D, produced by the skin, is relatively inactive and requires biochemicaltransformation, first by the liver and then by the kidneys, beforeit is fully able to stimulate absorption of calcium by the gut. The mechanismof activating vitamin D is not yet fully understood. However, thepresence of liver and kidney enzymes is essential to the process.

Question:

Simulate a game of draw poker between a player and the computer. Use the Basic language.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0554.htm

Solution:

In the game, both players (i.e. the computervsthe operator) start with a fixed stake (say $200 each). The com-puter accepts IOUs for certain personal items. 10PRINT \textquotedblleftPOKER 20DIM A (50) , B (5) 30DEF FNA (X) = INT (10\textasteriskcenteredRND (1)) 40DEF FNB (X) = X - 100 \textasteriskcentered INT (X/100) 50PRINT \textquotedblleftWELCOME TO THE TABLE. I WELL OPEN THE\textquotedblright ; 60PRINT \textquotedblleftBETTING BEFORE THE DRAW; YOU OPEN AFTER. WHEN\textquotedblright ; 70PRINT \textquotedblleftYOU FOLD, BET 0; TO CHECK, BET .5\textquotedblright 80PRINT : LET O = 1 90LET C = 200: LET S = 200 95LET P = 0 100PRINT: IF C < = 5 THEN 3340 110PRINT \textquotedblleftTHE ANTE IS $5. I WILL DEAL\textquotedblright 120PRINT: IF S > 5 THEN 140 130GOSUB 3490 140LET P = P + 10: LET S = S - 5: LET C = C - 5 150FOR Z = 1 TO 10: GOSUB 1460: NEXT Z 160PRINT "YOUR HAND :" 170N = 1 180GOSUB 1580 190N = 6: I = 2 200GOSUB 1830 210PRINT: IF I <> 6 THEN 330 220IF FNA (0) < = 7 THEN 250 230LET X = 11100 240GOTO 300 250IF FNA (0) < = 7 THEN 280 260LET X = 11110 270GOTO 300 280IF FNA (0) > = 1 THEN 320 290X = 11111 300I = 7: Z = 23 310GOTO 430 320Z = 1: GOTO 365 330IF U > = 13 THEN 390 340IF FNA (0) > = 2 THEN 360 350GOTO 300 360Z = 0 365K = 0 370PRINT \textquotedblleftI CHECK\textquotedblright 380GOTO 465 390IF U < = 16 THEN 420 400Z = 2 410IF FNA (0) > = I THEN 430 420Z = 35 430V = Z + FNA (0) 440GOSUB 3130 450PRINT \textquotedblleftI OPEN WITH\textquotedblright V 460K = V 465GOSUB 2700 470GOSUB 490 480GOTO 630 490IF I <> 3 THEN 580 500PRINT: PRINT \textquotedblleftI WIN\textquotedblright 510C = C + P 520PRINT "NOW I HAVE $ "C" AND YOU HAVE $" S 530PRINT "DO YOU WISH TO CONTINUE"; 540INPUT H$ 550IF H$ = "YES" THEN 95 560IF H$ = "NO" THEN 3750 570GOTO 530 580IF I <> 4 THEN 620 590PRINT: PRINT "YOU WIN" 600S = S + P 610GOTO 520 620RETURN 630PRINT: PRINT "NOW WE DRAW. HOW MANY CARDS DO YOU WANT ?\textquotedblright; 640INPUT T 650IF T = 0 THEN 780 660Z = 10 670IF T < 4 THEN 700 680PRINT "YOU CAN'T DRAW MORE THAN THREE CARDS" 690GOTO 640 700PRINT "WHAT ARE THEIR NUMBERS?" 710FOR Q = 1 TO T 720INPUT U 730GOSUB 1450 740NEXT Q 750PRINT "YOUR NEW HAND:" 760N = 1 770GOSUB 1580 780Z = 10 + T 790FOR U = 6 TO 10 800IF INT (X/10 \uparrow (U - 6)) <> 10\textasteriskcenteredINT(X/10\uparrow (U - 5)) THEN 820 810GOSUB 1450 820NEXT U 830PRINT: PRINT "I AM TAKING" Z - 10 - T "CARD"; 840IF Z = 11 + T THEN 860 850PRINT "S": PRINT: GOTO 860 860N = 6: V = I: I = 1 870GOSUB 1830 880B = U: M = D 890IF V <> 7 THEN 920 900Z = 28 910GOTO 1060 920IF I <> 6 THEN 950 930Z = 1 940GOTO 1060 950IF U > = 13 THEN 1000 960Z = 2 970IF FNA (0) <> 6 THEN 990 980Z = 19 990GOTO 1060 1000IF U > = 16 THEN 1050 1010Z = 19 1020IF FNA (0) <> 8 THEN 1040 1030Z = 11 1040GOTO 1060 1050Z = 2 1060K = 0 1070GOSUB 2700 1080IF T <> .5 THEN 1170 1090IF V = 7 THEN 1130 1100IF I <> 6 THEN 1130 1110PRINT "I'LL CHECK" 1120GOTO 1180 1130V = 2 + FNA (0) 1140GOSUB 3130 1150PRINT "I BET" V 1160K = V: GOSUB 2710 1170GOSUB 490 1180PRINT: PRINT "NOW WE COMPARE HANDS" 1190J$ = H$ 1200K$ = 1$ 1210PRINT "MY HAND:" 1220N = 6 1230GOSUB 1580 1240N = 1 1250GOSUB 1830 1260PRINT: PRINT "YOU HAVE"; 1270K = D 1280GOSUB 3360 1290H$ = J$ 1300I$ = K$ 1310K = M 1320PRINT "AND I HAVE"; 1330GOSUB 3360 1340IF B > U THEN 500 1350IF U > B THEN 590 1360IF H$ = "A FLUSH" THEN 1420 1370IF FNB (M) < FNB(D) THEN 590 1380IF FNB(M) > FNB(D) THEN 500 1390PRINT "THE HAND IS DRAWN" 1400PRINT "ALL $" P "REMAINS IN THE POT" 1410GOTO 100 1420IF FNB (M) > FNB (D) THEN 500 1430IF FNB (D) > FNB (M) THEN 590 1440GOTO 1390 1450Z = Z + 1 1460A (Z) = 100 \textasteriskcentered INT (4\textasteriskcenteredRND (1)) + INT (100\textasteriskcenteredRND (1)) 1470IF INT (A(Z)/100) > 3 THEN 1460 1480IF A(Z) - 100 \textasteriskcentered INT(A(Z)/100) > 12 THEN 1460 1490IF Z = 1 THEN 1570 1500FOR K = 1 TO Z - 1 1510IF A(Z) = A(K) THEN 1460 1520NEXT K 1530IF Z < = 10 THEN 1570 1540N = A (U) 1550A (U) = A (Z) 1560A(Z) = N 1570RETURN 1580FOR Z = N TO N + 4 1590PRINT Z "- - \textquotedblright; 1600GOSUB 1660 1610PRINT "OF"; 1620GOSUB 1750 1630IF Z/2 <> INT (Z/2) THEN 1640 1640PRINT: NEXT Z: PRINT 1650RETURN 1660K = FNB(A(Z)) 1670IF K <> 9 THEN 1690 1680PRINT "JACK"; 1690IF K <> 10 THEN 1700: PRINT "QUEEN"; 1700IF K <> 11 THEN 1710: PRINT "KING"; 1710IF K <> 12 THEN 1720: PRINT "ACE"; 1720IF K > = 9 THEN 1740 1730PRINT K + 2; 1740RETURN 1750K = INT (A(Z)/100) 1760IF K <> 0 THEN 1780 1770PRINT "CLUBS", 1780IF K <> 1 THEN 1800 1790PRINT "DIAMONDS", 1800IF K <> 2 THEN 1810: PRINT "HEARTS", 1810IF K <> 3 THEN 1820: PRINT "SPADES", 1820RETURN 1830U = 0 1840FOR Z = TO N + 4 1850B(Z) = FNB (A(Z)) 1860IF Z = N + 4 THEN 1890 1870IF INT (A (Z)/100) <> INT (A(Z + 1)/100) THEN 1890 1880U = U + 1 1890NEXT Z 1900IF U <> 4 THEN 1970 1910X = 11111 1920D = A (N) 1930H$ = "A FLUS" 19401$ = "H IN" 1950U = 15 1960RETURN 1970FOR Z =N TO N + 3 1980FOR K = Z + 1 TO N + 4 1990IF B(Z) < = B(K) THEN 2050 2000X = A (Z) 2010A (Z) = A (K) 2020B (Z) = B(K) 2030A (K) = X 2040B(K) = A(K) - 100 \textasteriskcentered INT(A(K)/100) 2050NEXT K 2060NEXT Z 2070X = 0 2080FOR Z = N TO N + 3 2090IF B (Z) <> B (Z + 1) THEN 2130 2100X = X + 11 \textasteriskcentered 10 \uparrow (Z - N) 2110D = A (Z) 2120GOSUB 2420 2130NEXT Z 2140IF X <> 0 THEN 2280 2150IF B (N) + 3 <> B (N + 3) THEN 2180 2160X = 1111 2170U = 10 2180IF B (N + 1) + 3 <> B (N + 4) THEN 2280 2190IF U <> 10 THEN 2260 2200U = 14 2210H$ = "STRAIG" 2220I$ = "HT" 2230X = 11111 2240D = A (N + 4) 2250RETURN 2260U = 10 2270X = 11110 2280IF U > = 10 THEN 2350 2290D = A (N + 4) 2300H$ = "SCHMAL" 2310I$ = "TZ," 2320U = 9 2330X = 11000 2340GOTO 2400 2350IF U <> 10 THEN 2380 2360IF I = 1 THEN 2400 2370GOTO 2410 2380IF U > 12 THEN 2410 2390IF FNB (D) > 6 THEN 2410 2400I = 6 2410RETURN 2420IF U > = 11 THEN 2470 2430U = 11 2440H$ = \textquotedblleftA PAIR\textquotedblright 2450I$ \textquotedblleftOF\textquotedblright 2460RETURN 2470IF U <> 11 THEN 2570 2480IF B (Z) <> B (Z - 1) THEN 2530 2490H$ = \textquotedblleftTHREE\textquotedblright 2500I$ = \textquotedblleft\textquotedblright 2510U = 13 2520RETURN 2530H$ = \textquotedblleftTWO P\textquotedblright 2540I$ = \textquotedblleftAIR,\textquotedblright 2550V = 12 2560RETURN 2570IF U > 12 THEN 2620 2580U = 16 2590H$ = \textquotedblleftFULL H\textquotedblright 2600I$ = \textquotedblleftOUSE,\textquotedblright 2610RETURN 2620IF B (Z) <> B (Z - 1) THEN 2670 2630U = 17 2640H$ = "POUR" 2650I$ = "" 2660RETURN 2670U = 16 2680H$ = "FULL H": I$ = "OUSE," 2690RETURN 2700G = 0 2710PRINT "WHAT IS YOUR BET?"; 2720INPUT T 2730IF T - INT(T) = 0 THEN 2790 2740IF K <> 0 THEN 2770 2750IF G <> 0 THEN 2770 2760IF T = .5 THEN 3060 2770PRINT "NO SMALL CHANGE, PLEASE" 2780GOTO 2710 2790IF S - G - T > = 0 THEN 2820 2800GOSUB 3490 2810GOTO 2710 2820IF T <> 0 THEN 2850 2830I = 3 2840GOTO 3030 2850IF G + T > = K THEN 2880 2860PRINT "IF YOU CAN'T SEE MY BET, THEN FOLD\textquotedblright 2870GOTO 2710 2880G = G + T 2890IF G = K THEN 3030 2900IF Z <> 1 THEN 3070 2910IF G > 5 THEN 2950 2920IF Z > = 2 THEN 3000 2930V = 5 2940GOTO 3070 2950IF Z = 1 THEN 2970 2960IF T < = 25 THEN 3000 2970I = 4 2980PRINT "I FOLD" 2990RETURN 3000IF Z = 2 THEN 3080 3010PRINT "I'LL SEE YOU" 3020K = G 3030S = S - G 3040C = C - K 3050P = P + G + K 3060RETURN 3070IF G > 3\textasteriskcenteredZ THEN 3000 3080V = G - K + FNA (0) 3090GOSUB 3130 3100PRINT "I'LL SEE YOU, AND RAISE YOU" V 3110K = G + V 3120GOTO 2710 3130IF C - G - V > = 0 THEN 3330 3140IF G <> 0 THEN 3170 3150V = C 3160RETURN 3170IF C - G > = 0 THEN 3010 3180IF (O/2) <> INT(O/2) THEN 3260 3190PRINT "WOULD YOU LIKE TO BUY BACK YOUR SHIRT?"; 3200PRINT "AND COAT FOR $50?"; 3210INPUT J$ 3220IF J$ = "NO" THEN 3260 3230C = C + 50 3240O = O/2 3250RETURN 3260IF O/3 <> INT (0/3) THEN 3340 3270PRINT "WOULD YOU LIKE TO BUY BACK YOUR GOLD"; 3280PRINT "CHAIN FOR $50?"; 3290INPUT J$ 3300IF J$ = "NO" THEN 3340 3310C = C + 50 3320O = O/3 3330RETURN 3340PRINT "I'M BUSTED. CONGRATULATIONS!" 3350STOP 3360PRINT H$; I$; 3370IF H$ <> "A FLUS\textquotedblright THEN 3410 3380K = INT (K/100) 3390GOSUB 1760 3400PRINT: RETURN 3410K = FNB (K) 3420GOSUB 1670 3430IF H$ = "SCHMAL" THEN 3450 3440IF H$ <> "STRAIG" THEN 3470 3450PRINT "HIGH" 3460RETURN 3470PRINT "S" 3480RETURN 3490PRINT: PRINT "YOU CAN'T BET WHAT YOU DON'T HAVE" 3500IF O/2 = IN (O/2) THEN 3620 3510PRINT "WOULD YOU LIKE TO SELL YOUR SHIRT AND COAT?"; 3520INPUT J$ 3530IF J$ = "NO" THEN 3620 3540IF FNA(0) > = 7 THEN 3580 3550PRINT "I WILL GIVE YOU $75 FOR IT" 3560S = S + 75 3570GOTO 3600 3580PRINT "THIS COAT LOOKS MOTHEATEN - I'LL GIVE YOU $25" 3590S = S + 25 3600O = O\textasteriskcentered2 3610RETURN 3620IF O/3 <> INT (O/3) THEN 3740 3630PRINT "WILL YOU PART WITH YOUR GOLD CHAIN?"; 3640INPUT J$ 3650IF J$ = "NO" THEN 3730 3660IF FNA (0) > = 6 THEN 3700 3670PRINT "YOU ARE NOW $100 RICHER" 3680S = S + 100 3690GOTO 3720 3700PRINT "THIS IS ROLLED GOLD. $25" 3710S = S + 25 3720O = O\textasteriskcentered3 3730RETURN 3740PRINT: PRINT "YOU' RE BUSTED. HAVE A REVOLVER" 3750END

Question:

A small object lies 4 in. to the left of the vertex of a concave mirror of radius of curvature 12 in. Find the position and magnifica-tion of the image. Concave Mirror

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0852.htm

Solution:

The mirror equation for a concave mirror is (1/p) + (1/q) = (2/R) with a positive R. The object distance is p = +4 in. and R is +12 in. Then, (1/4 in.) + (1/q) = (2/12 in.) q = - 12 in., Magnification m is given by m = - (q/p) = - (-12 in./4 in.) = 3 The image is therefore 12 in. to the right of the vertex (q is negative), is virtual (q is negative), erect (m is positive), and 3 times the height of the object. See the figure.

Question:

Write a program, incorporating the function TAN, to compute tangents of angles 0,10,20,....,360 degrees.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0414.htm

Solution:

PROGRAM TABLEOFTANS (OUTPUT); CONST PI = 3.1415926536; VAR DEGREES: INTEGER; FUNCTION TAN (x:REAL):REAL; BEGIN TAN:=sin(x) /cos( x) END; (\textasteriskcenteredTAN\textasteriskcentered) BEGIN( \textasteriskcenteredTABLEOFTANS\textasteriskcentered) DEGREES: = o; WHILE DEGREES < = 360 DO BEGIN WRITE (DEGREES); IF DEGREES MOD 180 = 90 THEN WRITELN ('INFINITY') ELSE WRITELN (TAN (DEGREES\textasteriskcenteredPI/180)); DEGREES: DEGREES + 10 END END.(\textasteriskcenteredTABLEOFTANS\textasteriskcentered).

Question:

A person whose diet is deficient in B vitamins is likely to be listlessand lacking in energy for normal activities Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0098.htm

Solution:

The vitamin B complex includes (among others) thiamine (B1), riboflavin(B2) pyridoxine (B6), biotin, niacin or nicotinic acid,cobalamin (B12) andpantothenicacid.All these vitamins function as parts of coenzymesused in the oxidation of glucose to form ATP. Thiamine is an essential constituent of thiamine pyrophosphate, the coenzymeused in the oxidativedecarboxylationofpyruvicand \alpha- ketoglutaricacids in the TCA cycle. Characteristics of mild deficiencies are: fatigue, loss of appetite and weakness. Severe deficien-cies enhance thesesymptoms and cause degeneration of nerves and muscle atrophy, a conditionknown as beriberi. Niacin is a component of two coenzymes, NAD+ and NADP+, which serveas hydrogen acceptors in reactions ofglycolysisand the TCA cycle. Deficiency of niacin causes stunted growth and the disease pellagra. Riboflavin forms part offlavinadeninedinucleotide(FAD), a coenzymeofsuccinicdehydrogenasein the TCA cycle. Deficiency results instunted growth, cracks in the corner of the mouth, and cornea damage. Pantothenicacid is especially important since it forms part of coenzymeA, and thus used in a number of reac-tions in the metabolism of carbohydrates, fats and proteins. An expected symptom ofpantothenic aciddeficiency is growth failure. A person with any of these vitamin deficiencies would show signs of slowATP production since these vitamins are necessary providers of coenzymesinvolved in metabolic pathways leading to ATP formation. Since ATP is needed for energy-requiring reactions, a lack of ATP would causethe vitamin-deficient person to be sluggish and to exhibit slower growth.

Question:

Fe(s) + Co^2+(.5M) \rightarrow Fe^2+(1.0M) + Co(s) . Fe^2+ + 2e^- \leftrightarrows Fe(s) with E\textdegree = -.44e and Co^2+ + 2e^- \leftrightarrows Co(s) with E\textdegree = -.28, find the standard cell potential ∆E , the cell potential ∆E and the concentration ratio at which the potential generated by the cell is exactly zero.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0594.htm

Solution:

Assume that the reaction proceeds spontaneously. This means, therefore, that the reaction must have a positive value for ∆E\textdegree. With this in mind, you proceed as follows: You are given two half-reactions: Fe^2+ + 2e^- \leftrightarrows Fe(s)E\textdegree = -.44 Co^2++ 2e^- \leftrightarrows Co(s)E\textdegree = -.28ev Both reactions represent reduction (gain of electrons). But, the over-all reaction in a cell is a combination of both a reduction and an oxidation reaction. Thus, you must reverse one, keeping in mind that the ∆E\textdegree must be a positive value. Recall, also, that ∆E\textdegree = E\textdegree_oxid + E\textdegree_red . You can write Fe^2+ + 2e^- \leftrightarrows Fe^2+;E\textdegree = (-.44) = -44ev Co^2+ + 2e^- \leftrightarrows Co(s) : E\textdegree = -.28ev Fe(s) + Co^2+ \leftrightarrows Fe^2+ + Co(s) (overall reaction) with ∆E\textdegree = .44 - .28 = .16 ev. Notice: By reversing the Fe reaction and combining it with the other, you obtained the overall reaction with a ∆E\textdegree = .16 ev, a positive value, which indicates that the reaction proceeds spontaneously. To find ∆E, use the Nernst equa-tion, which states ∆E = ∆E\textdegree -(.059 / n) log K, where n = number of electrons transferred and K = equilibrium constant of reaction. In this problem, the number of electrons trans-ferred is 2, so that n = 2. K = [Fe^2+] / [Co^2+], i.e., the ratio of the concentrations of products to reactants, each raised to the power of its respective coefficient in the chemical equation. Note, Co(s) and Fe(s) are omitted, because they are solids and, thus, considered constants them-selves. You are given [Fe^2+] and [Co^2+] and you have calculated ∆E\textdegree . Therefore, ∆E = ∆E\textdegree -(.059 / n)log [(Fe^2+) / (Co^2+) ] = .16 -(.059 / n)log[(1.0) / (.5)] =.16 - .0295 log 2 = .16 - .01 = .15 ev. To find the concentration ratio, K, when ∆E = 0, use the Nernst equation; ∆E = ∆E\textdegree -(.059 / n) log K. ∆E = 0, ∆E\textdegree = .16, n = 2 and K = [Fe^2+] / [Co^2+] , so that 0 = .16 - (.059 / n) log [Fe^2+] / [Co^2+] or 0 = .16 - .0295 log [Fe^2+] / [Co^2+]. [Fe^2+] / [Co^2+] = antilog(.16 / .0295) = antilog 5.4 = 2.5 × 10^5 thus, [Fe^2+] / [Co^2+] = 2.5 × 10^5, when ∆E = 0.

Question:

Determine the current in each of the resistors in figure A.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0678.htm

Solution:

We find the resistance between points B and C (R_BC) by using the relation for resistors R_1, R_2 , . . . in parallel (see figure (A)). (1/R_Total ) = (1/R_1 ) + (1/R_2 ) + . . . (1/R_BC ) = (1/6.0 ohms) + (1/9.0 ohms) + (1/18.0 ohms) = [(6.0 )/(18.0 ohms)] R_BC = [(18.0)/(6.0)] ohms = 3.0ohms Using the formula for resistors in series, R_Total = R_1 + R_2 + . . . , we find the resistance between points A and C in the circuit, (see figure (B)) R_AC = R_AB + R_BC = 4.0 ohms + 3.0 ohms = 7.0 ohms The current I_t in the circuit is obtained from Ohm's Law I_t = [(V_AC )/(R_AC )] = [(35 volts)/(7.0 ohms)] = 5.0 amp . The current through each individual resistor is, by Ohm's Law, equal to the voltage across the resistor divided by its resistance. Hence I_FG = [(V_FG )/(R_FG )] = [(15 volts)/(6.0 ohms)] = 2.5 amp I_HK = [(V_HK )/(R_HK )] = [(15 volts)/(9.0 ohms)] = 1.7 amp I_LM = [(V_LM )/(R_LM )] = [(15 volts)/(18.0 ohms)] = 0.83 amp Note that I_t = I_FG + I_HK + I_LM by Kirchoff's Current Law. I_t = (2.5 + 1.7 + 0.8) amp = 5.0 amp.

Question:

Write two FORTRAN programs which will compute n!intwo differentways: (a) First according to the standard formula for a positive integern \geq 1, n! =n(n-1) \textbullet\textbullet\textbullet (3)(2)(1) (b)thenuseStirling'sapproximation formula n! \cong (2\pi)1^/2n^n^+(1/2) e^-n

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0520.htm

Solution:

We will formulate the answers as two function routines which will be usedin later work. Denote these two functions byFACT(N) and FACTST(N). Then by applying the above formulas directly we get the programsbelow: INTEGER FUNCTIONFACT(N) IPROD = 1 IF (N.LE.1) GO TO 200 DO 100 I = 2,N IPROD = IPR0D\textasteriskcenteredI 100CONTINUE FACT = IPROD GO TO 300 200FACT = 1 300RETURN END FUNCTIONFACTST(N) Z1 =EXP(-FLOAT(N)) Z2 = (FLCAT(N))\textasteriskcentered\textasteriskcentered(FLOAT(N) + 0.5) CFLOAT CONVERTS INTEGERS TO REALS Z3 = SQRT (2\textasteriskcentered3.14159) FACTST = Z1\textasteriskcenteredZ2\textasteriskcenteredZ3 RETURN END

Question:

Illustrate using flowcharts the need for debugging when dealing with improper loop sequences.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0186.htm

Solution:

Most beginners have some difficulty understanding and applying the rules dealing with loops. Consequently, many of their debugging concentrates on the proper construction of a loop. Consider the flowchart of figure 1 that develops an algorithm for finding the factors of a number. Now, suppose the instruction NUMBER \leftarrow K is omitted from the inner loop, leaving the (partial) flowchart of figure 2. Suppose NUMBER is 12. The loop begins with J = 2. Since 2 divides 12, K will be an Integer. The computer prints 2 IS A. FACTOR and returns to the start of the loop. Since the value of NUMBER has not been changed, the computer repeats exactly the same computations on the second pass around the loop. 2 IS A FACTOR is printed and the computations repeated. The computer is in a loop for which the normal exit condition will never be satisfied. The program must be aborted. The error here can be easily recognized since 2 or 3 pages of '2 IS A FACTOR' indicate that the value of NUMBER is not changing. Now, suppose that the print statement has also been omitted. In this case there is no output to tell the user that he has constructed an infinite loop. He will not receive any output and so may assume that the system is 'down'. In fact, the computer is slaving away trying to exit out of an infinite loop. In this case, the system will also abort the program before it is complete. The lack of output tells the programmer that he should include some PRINT instructions in his loops to find what caused the program to abort.

Question:

For HF,K_diss= 6.7 × 10^-4 what is the H_3O^+ and OH^- concentrations of a 0.10M HF solution? Assume K_W = 1 × 10^-14 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0419.htm

Solution:

To answer this question, you must set up the equations for the equilibriainvolved. HF is an acid, which means it donates pro-tons (H+). The only base present is H_2O. They react as shown in the following equation: HF + H_2O \rightleftarrows H_3O^+ + F^- . This is not the only source of H_3O^+ . Recall that water also can dissociate into ions. H_2O + H_2O \rightleftarrowsH_3O^+ + OH^- . This means that in the consideration of the H_3O^+ concentration you must look at both sources. Because you are asked the concentrations, you must employ equilibrium constants,K_dissandK_w. Let x = the number of moles/liter of H_3O^+ from the reaction of HF with H_2O, then x is also equal to the number of moles/liter of F^- produced. They have the same mole ratio, as indicated by theequilibria. Since HF is the stronger acid, some of its protons ionize and then join with H_2O to form H_3O^+ . Thus, the concentration of HF is (0.1 - x). If you let y = moles/liter of OH^-, then y = moles/liter of H_3O^+ formed by the dissociation of water. These variables that represent the concentrations at equilibrium must satisfy the two equilibrium conditions: K_diss= {[H_3O^+] [F^-]} / [HF] = 6.7 × 10^-4 = {(x+y)(x)} / (0.1 - x) K_w= [H_3O^+] [OH^-] = 1 × 10^-14 = (x+y)(x) . For both conditions, the H_3O^+ concentration is represented by (x+y) since, at equilibrium, the concentration must be the sum from both sources of H_3O^+ production. Solve these equations simultaneously for x and y, you can avoid cumbersome calculations by noting which contribution is the dominant one. In this way, the other can be neglected to give a fairly good approximation. A general rule for determining which is dominant is to compare the dissociation constants. The larger the dissociation constant, the more it dom-inates the final equilibrium state. From inspection of these con-stants, you find that the contribution of H_3O^+ from HF dominates over that from water. You need consider only HF + H_2O \rightleftarrows H_3O^+ + F^-, therefore, K_diss= {[H_3O^+] [F^-]} / [HF] = 6.7 × 10^-4 = {(x+y)(x)} / (0.1 - x) . Solving, you obtain x = 7.7 × 10^-3M = [H_3O^+]. To find y, go back to K_W = [H_3O^+] [0H^-] = 1 × 10^-14 = (x+y) (y) and substitute in your value for x. Assuming that y is negligible in comparison to x, the answer becomes (1.0 × 10^-14) / (7.7 × 10^-3) = 1.3 × 10^-12M = [OH^-].

Question:

A two-person barbershop has five chairs to accommodate waiting customers. Potential customers arrive at the average rate of 3.7634/hr. and spend an average of 15 min. in the barber's chair. Assume M = 7 (total number of persons in the system including the two barbers.) \lambda = 3.7634/hr. \mu = 4/hr. S = 2 servers. a) What is the probability of a customer getting directly into the barber's chair upon arrival? b) What is the expected number of customers waiting for a haircut? c) What is the effective arrival rate assuming Pn = 1? d) How much time can a customer expect to spend in the barbershop? e) What fraction of potential customers are turned away?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0457.htm

Solution:

a) The probability of a customer getting directly into the barber's chair upon arrival is the same as in the case of no customers in the system. This is computed as follows: b) The expected number of customers waiting for a haircut: c) The effective arrival rate S = number of servers M = maximum number of persons in the system Therefore, since n \leq S, Pn = (1/ n!) (\lambda/\mu)^n Po = (1/1) [(3.7643)/4]^1 Po = (1/1) [(3.7643)/4] (.36133) = .2457 \lambdaeff= \mu [S - ^S-1\sumn = 0(S-n) Pn] = 4[ 2- 2 (0.36133) - 0.339957] = 3.7495 d) The time a customer is expected to spend in the barbershop, W = (L/ \lambdaeff) = {L_q + (2 - 2po - p^1)} / \lambdaeff = (1.183083) / (3.7495) = 0.03155 hr \cong 19 min e) Fraction of potential customers that will be turned away P (7 in system) = P_7 = [1/ (S!S^n-S)] (\lambda/\mu)nPo Since S < n \leq M = 2 < 7 = 7 = [1 / (2!2^7-2)] (3.7634/4)7(0.36133) = 0.00368 = 0.3 percent Thus, only 0.3 percent of the potential customers will be turned away. The Fortran program for the problem follows: \textasteriskcentered\textasteriskcentered INFINITE SOURCE, FINITE QUEUE, MULTIPLE SERVERS \textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered QUEUING PROGRAM \textasteriskcentered\textasteriskcentered WE ASSUME ARRIVALS TO SYSTEM ARE COMPLETELY AT RANDOM - POISSON INPUT ARRIVALS FORM A SINGLE QUEUE INFINITE SOURCE AND FINITE QUEUE FIFO QUEUE DISCIPLINE THERE ARE MULTIPLE SERVERS WITH EXPONENTIAL SERVICE TIME DEPARTURES FROM SYSTEM OCCUR COMPLETELY AT RANDOM TO CALCULATE AND PRINT LAVERAGE # OF CUSTOMERS IN THE SYSTEM LQAVERAGE # OF CUSTOMERS IN QUEUE PNPROBABILITY OF N CUSTOMERS IN SYSTEM AT ANY POINT IN TIME PZEROPROBABILITY OF NO CUSTOMERS IN THE SYSTEM WAVERAGE TIME CUSTOMER SPENDS IN SYSTEM WQAVERAGE TIME CUSTOMER SPENDS IN QUEUE LAMEFFOVERALL EFFECTIVE ARRIVAL RATE FOR FINITE QUEUE WITH INFINITE SOURCE REAL\textasteriskcentered4 TITLE (20) , LAMBDA, MU, N, M, L, LQ, LAMEFF 5READ (5, 10, END = 2000) TITLE 10FORMAT (20A4) WRITE (6, 11) TITLE 11FORMAT ('1', 20A4, //) READ (5, 20) LAMBDA, MU 20FORMAT (3F10.0) READ (5, 20) N, M, S C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PZERO\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IS = S+1 SUM = 0.0 DO 405 NK = 1, IS SK = NK-1 CALL FACT (SK, TFACT) SUM = SUM+ (1/TFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredSK 405CONTINUE SUM2 = 0.0 NM = M DO 410 NK = IS, NM SUM2 = SUM2+ (LAMBDA/ (MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (NK-S) 410CONTINUE CALL FACT(S, SFACT) PZERO = 1/ (SUM+ (1/SFACT) \textasteriskcentered ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredSUM2) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PN AND LQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IF(N.LE.S) GO TO 415 IF (N.GT.M) GO TO 420 PN= (1/ (SFACT\textasteriskcenteredS\textasteriskcentered\textasteriskcentered (N-S))) \textasteriskcentered ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 425 415CALL FACT(N, TFACT) PN= (1/TFACT) \textasteriskcentered ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 425 420PN=0.0 425LQ= (((LAMBDA/MU) \textasteriskcentered\textasteriskcentered (S+1) \textasteriskcenteredPZERO) \textasteriskcentered (1- (LAMBDA/(MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (M-S) - (M-S) \textasteriskcentered ( (LAMBDA/(MU\textasteriskcenteredS)) \textasteriskcentered\textasteriskcentered (M-S) ) \textasteriskcentered (1-LAMBDA/ (MU\textasteriskcenteredS ) ) ) )/(S\textasteriskcenteredSFACT\textasteriskcentered(1-LAMBDA/(MU\textasteriskcenteredS))\textasteriskcentered\textasteriskcentered2) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATELAMEFF, L, W, WQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered NS=S SUM=0 DO 560 NK=1,NS KK=NK-1 SKK=KK CALL FACT(SKK,AFACT) SUM=SUM+ (S-KK) \textasteriskcenteredPZERO\textasteriskcentered (1/AFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredKK 560CONTINUE LAMEFF=MU\textasteriskcentered(S-SUM) L=LQ+LAMEFF/MU W=L/LAMEFF WQ=LQ/LAMEFF C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredPRINT RESULTS\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered WRITE (6,810) LAMBDA, MU, L, LQ, N, PN, PZERO, W, WQ, LAMEFF 810FORMAT(' LAMBDA =',F10.5/' MU = F10.S/' L =',F10.5/\textasteriskcentered LQ =',F10.5/' N =',F10.5/' PN =',F10.5/\textasteriskcentered PZERO =',F10.5/\textasteriskcentered WQ =',F10.5/' LAMEFF =',F10.5) GO TO 5 2000STOP END SUBROUTINE FACT(P, PROD) NUM=P PROD=1. IF (NUM.EQ.0) GO TO 20 DO 10 K=1, NUM 10PROD=PROD\textasteriskcenteredK 20RETURN END /DATA WITH N=1 3.76344. 1 .7.2. 1 .7.2. WITH N=1 LAMBDA= 3.76340 MU= 4.00000 L= 1.18309 LQ= 0.24571 N= 1.00000 PN= 0.33996 PZERO= 0.36133 W= 0.31553 WQ= 0.06553 LAMEFF= 3.74953

Question:

The dorsal root of a nerve leading to the leg of a rat is severed. How will this affect the rat? How would the effect vary if the ventral root were severed instead?

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0508.htm

Solution:

A cross-section of the spinal cord (see problem 26-16) reveals that the paired dorsal roots are the junctions where sensory fibers from the peripheral receptor areas enter, while the paired ventral roots are where axons of the motor neurons leave the spinal cord to effectors, such as muscles. When the dorsal root of a spinal nerve is cut, the sensory fibers are severed and afferent impulses (that is impulses to the nervous system) can no longer reach the spinal cord and thence the brain. If a hot iron is brought to touch the leg of a rat, in which the dorsal root of the nerve leading to the leg is severed, the rat will not feel any burn sensation, as evidenced by its lack of attempt to avoid the stimulus. Since the receptor cells on the skin of the leg are not impaired, they do respond to the stimulus by firing impulses along the sensory fibers to the spinal cord. However, the impulses cannot reach the central nervous system and there-fore cannot be translated and interpreted in the brain. Also, simple reflex action such as the rapid withdrawal of the leg cannot occur in the rat, as impulses from the sensory neurons cannot be transmitted to the motor neurons in the reflex arc. As a consequence, the rat will not feel the stimulation of the hot iron, nor would it perform reflexive withdrawal of the leg from the iron. In higher animals such as man in which memory and experience play an important role in behavior, severing the dorsal root may cause loss of sensation and loss of reflex action, but the somatic response to the same stimulus may be present. For instance, by past experience one has learned that the hot iron causes an unpleasant sensation. Although by lesion of the dorsal root he does not sense the burning sensation, he may remove his hand or leg from the hot iron based on this information stored in the memory center of his brain. If instead of the dorsal root, the ventral root were severed, the injury would be inflicted on the motor neurons. In this instance, the rat would sense the stimu-lation of the hot Iron, but would find itself unable to withdraw the leg since the impulses generated by the motor neurons cannot reach theeffectormuscles of the leg. However, the rat may demonstrate other responses, for example, it may squeak in pain.

Question:

How much power is required to service a city of 100,000 homes, each of which draws an average current of 20 A at a potential of 120 V? If the transmission lines that carry power were capable of carrying a current of 1,000 A, what potential would be required to transmit the power from the source?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0791.htm

Solution:

The total power which must be supplied is equal to the total power dissipated in all the homes. Since power equals the product of the current used (or supplied) by a device and the voltage across the device, we calculate, P = IV for each home. Here I is current and V is voltage. Each home uses power, P = (20A) (120V) \rule{1em}{1pt} 2400 watts But there are 10^5 homes and P_net = (2.4 × 10^3 watts) (10^5) P_net = 2.4 x 10^8 watts If the current is 1,000 A, the required voltage is V = (Pnet/I)=[(2.4 × 10^8Watts)/(1.3 × 10^3 A)] = 2.4 × 10^5V

Question:

An aluminum calorimeter of mass 50 g contains 95 g of a mixture of water and ice at 0\textdegreeC. When 100 g of aluminum which has been heated in a steam jacket is dropped into the mixture, the temperature rises to 5\textdegreeC. Find the mass of ice originally present if the specific heat capacity of aluminum is 0.22 cal/g\bulletCdeg.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0482.htm

Solution:

The heat lost by the cooling aluminum must equal the heat gained by the calorimeter and contents. If a mass y of ice were originally present, the total heat gained would have to include the heat acquired by the ice in melting, the heat gained by the 95 g of water in rising in temperature, and the heat gained by the calorimeter in doing likewise. The heat Q absorbed by a mass m of specific heat c as its temper-ature rises an amount ∆t is: Q = m c ∆T Also one gram of ice absorbs 80 calories of heat in changing into water. Thus, heat gained by the aluminum calorimeter = (50 g)(0.22 cal/g\bulletCdeg) (5 - 0)\textdegreeC heat gained by ice as it melts = (y) (80 cal/g) heat gained by water= (95 g) (1 cal/g\bulletCdeg) (5 - 0)\textdegreeC heat lost by chunk of aluminum = (100 g) (0.22 cal/g\bulletCdeg) (100 - 5)\textdegreeC since steam has a temperature of 100\textdegreeC. Thus 100 g × 0.22 cal/g\bulletCdeg × (100 - 5)\textdegreeC = y × 80 cal/g + 95 g × 1 cal/g\bulletCdeg× (5 - 0)\textdegreeC + 50 g × 0.22 cal/g\bulletCdeg × (5 - 0)\textdegreeC \therefore80 y cal/g = [0.22(9500 - 250) - 95 × 5] cal. \thereforey = 1560/80 = 19.50 g .

Question:

What are the basic structural and functional differences between white blood cells and red blood cells? Describe two different ways that white blood cells act to protect the body from foreign agents such as microorganisms.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0358.htm

Solution:

In addition to red blood cells, human blood contains five types of white blood cells, or leukocytes. Unlike the red blood cells, all types of leukocytes contain nuclei and do not contain hemoglobin. Leukocytes are generally larger than erythrocytes and are far less numerous. There are approximately 5,400,000 red cells per cubic millimeter of blood in an adult male (4.6 million in females), while only about 7,000 leukocytes per cubic millimeter. There are five different types of leukocytes (see Figure). Leukocytes are classified as granular (eosino-phils, neutrophils and basophils) or angular (monocytes and lymphocytes). The granulocytes have granules in their cytoplasm, have lobed nuclei, and are produced in red bone marrow. The agranulocytes are produced in organs such as the lymph nodes, spleen and thymus. Neutrophils make up over 60% of the leukocytes present in the body. Like other leukocytes, they can squeeze through the pores of blood vessels and enter the tissue spaces. They then move by amoeboid movement to sites of infected tissue. For example, when bacteria enter a certain tissue of the body, they can either attack cells or produce damaging toxins. The blood vessels in the in-fected region dilate and allow more blood flow to the site, causing the heat and redness characteristic of inflammation. Blood vessel permeability also increases, causing fluid to enter the tissue and swelling to result. Neutrophils and monocytes pass through the blood vessels and engulf, ingest and destroy the bacteria. Foreign particles and dead tissue can also be engulfed by this process of phagocytosis. These leukocytes are chemically attracted to the inflamed site by products released from the damaged tissues. The monocytes (5.3%) usually appear after the neutro-phils and are more important in fighting chronic infections. Monocytes can phagocytize bacteria but more often, they en-large and become wandering macrophages, which can move more quickly and engulf more bacteria. Eosinophils (2.3%) are weakly phagocytic, and are involved in aller-gic reactions (by releasing antihistamines) and in fight-ing trichinosis (an infection caused by the parasitic worm, trichinella). Basophils ( .5%) liberate the anti-coagulant heparin which combats the coagulative pro cesses that sometimes occurs in prolonged inflammation. The second means by which leukocytes protect the body from invasion is carried out by the lymphocytes. Lymphocytes (30%) function in the process of immunity. Lymphocytes are involved in the production of anti-bodies, proteins made by the body in response to speci-fic foreign substances. These foreign substances are subsequently attacked and destroyed by the anitbodies. Immunity and its agents will be discussed in further questions. The main functions of the leukocytes are thus protective: phagocytosis and immunity. The major function of the erythrocytes is to carry hemoglobin for gaseous transport and exchange, which is essential to the continuation of cellular metabolic processes.

Question:

Distinguish between messenger RNA (mRNA), ribosomal RNA (rRNA), and transfer RNA (tRNA). What is the role of each in protein synthesis?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0624.htm

Solution:

These three types of RNA are all single- stranded and are transcribed from a DNA template by RNA polymerase in the nucleus. However, the function of each type after they leave the nucleus is what determines their differences. Messenger RNA carries the genetic information coded for in the DNA and is responsible for the translation of that information into a polypeptide chain. Each set of 3 bases of the mRNA comprises a codon which directs the incorporation of a specific amino acid into the polypeptide chain. Messenger RNA binds reversibly to the smaller subunit of the ribosome, where protein synthesis is initiated. It can dissociate from the ribosome without jeopardizing the integrity of the ribosome. Degradation of mRNA by ribo-nucleases allows the cell to control its metabolism. Messenger RNA in E. coli has a lifetime of about 2 minutes. Degradation of mRNA is in the 5' \textemdash> 3' direction and may proceed even as ribosome's are translating the mRNA. Ribosomes initiate protein synthesis by binding to the 5' end of the mRNA and reading the codons in the 5' \textemdash>3' direction. If degradation were 3' \textemdash>5', there would be synthesis of incomplete protein chains. Messenger RNA is heterogeneous in size due to the different lengths of polypeptide chains which a cell needs to synthesize, there-fore, it is the most variable type of RNA. While mRNA can dissociate from the ribosomes, rRNA is an integral part of the ribosome and its removal results in the destruction of the ribosome. The 30 S subunit of a bacterial ribosome contains 21 proteins and a 16 S fragment of rRNA. The 50 S subunit is composed of 35 proteins and 2 fragments of rRNA with Svedberg constants of 23 S and 5 S, when placed together, the ribosomal proteins and rRNA undergo spontaneous self-assembly into a fully functional ribosome. This suggests that rRNA interacts with the ribo-somal proteins and helps maintain the characteristic 3-D shape of the ribosome. While the function of rRNA is not completely understood, regions of unpaired bases may be involved in the binding of tRNA and mRNA to the ribosome. Ribosomal RNA has an average molecular weight of 5 × 10^5 to 5 × 10^6 daltons and is the least variable type of RNA, having only the three types mentioned above (5S, 16S, 23S). The single strand of both mRNA and rRNA is able to fold back on itself with subsequent formation of hydrogen bonds between complementary regions, giving rise to apparent "double -stranded" helical regions. (see figure 1) Transfer RNA is the smallest type of RNA. With approximately 80 nucleotides, its molecular weight is 2.5 × 10^4 daltons. While there are at least 20 different types of tRNA's (one specific for each amino acid), the sequence of bases in the different tRNA molecules is highly conserved. This leads to the sharing of the same basic shape among the different types (see figure 2 a). The 3' end always ends in the sequence CCA-3' and the 5' end is a guanine. The amino acid residue is bound to the 3' adenine. Like mRNA and rRNA, tRNA folds back upon itself. Unlike the other two RNA's, tRNA, forms loops and double -stranded sections (see figure 2b). In one of these loops is located the anticodon, which is what distinguishes the different types of tRNA. The function of tRNA is to insert the amino acid specified by the codon on mRNA into the polypeptide chain, and it is through the complementation of anticodon and codon that the appropriate amino acid is incorporated. Hydrogen bonds form between the complementary codons and anticodons, thereby orienting the amino acid for peptide bond formation. A distinguishing feature of tRNA is the presence of unusual bases such as methyl inosine, dimethylguanosine, and ribothymidine. It is the presence of these alkylated bases which protects tRNA from digestion by ribonucleases.

Question:

Using the Periodic Table of Elements, find the following for sodiumdihydrogenphosphate, NaH_2PO4: (a) formula weight, (b) percent composition of oxygen, (c) weight in grams of 2.7 moles, and (d) percentage composition of oxygen in 2.7 moles.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0149.htm

Solution:

This problem encompasses work in chemical stoichiometry. With this in mind, you proceed as follows: (a)The formula weight = molecular weight, which is the sum of atomic masses of all the atoms in the substance. Na = 22.98, H = 1.008, O =15.9999, and P = 30.97. Thus, formula weight = Na + 2H + P + 4O = 22.98+ 2(1.0080) + 30.97 + 4(15.9999) = 119.9.(b) Percentage composition of oxygen is (total weight of oxygen in compound) / (total weight of compound) = [(4 oxygen atoms) × (15.9994 mass/oxygen atom)] /(119.9) = .5334 g of oxygen or 53.34 % by weight in NaH_2PO4. (C) Mole = (mass of substance) / (molecular weight) . You are given that there are 2.7 moles and you calculated the molecular weight. Thus, the weight in grams of 2.7 moles = (119.9 g/mole) (2.7moles) = 324 g of NaH_2PO_4 . (d) Following a procedure similar to the one in part (b), Percent composition = [{(no. of moles of O) × (MW of O)} / ( weight of comp/no. of moles)] × 100 = {(4 moles × 15.9 g/mole ) / (324 g / 2.7moles)} × 100 = 53.34% The percent composition of any element in any compound does not change when the amount of the compound present is changed.

Question:

For which of the following two reactions should ∆S be more positive? Why?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E31-0929.htm

Solution:

If the heat gained by a system is equal to that lost by the surroundings, then the entropy change for the surroundings is the negative of the entropy change for the system; for both the system and surroundings taken together, ∆S is zero if the transfer of heat is carried out reversibly. The entropy of a system increases with the increasing randomness of the molecules. Thus, for reaction (a), ∆S is positive owing to the loss of rigidity or increasing randomness of reactants. For reaction (b), ∆S is negative owing to the loss of independent translational motion of the reactants. In other words, the product is more stable or more ordered than the reactants and, thus, the entropy decreases for the reaction.

Question:

Find log_10 100.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0956.htm

Solution:

The following solution presents 2 methods for solving the given problem. Method I. The statement log_10 x = y is equivalent to 10^y = x, hence log_10 100 = x is equivalent to 10^x = 100. Since 10^2 = 100, log_10 100 = 2. Method II. Note that 100 = 10 × 10; thus log_10 100 = log_10 (10 × 10). Recall: log_x (a × b) = log_x a + log_x b, therefore log_10 (10 × 10)= log_10 10 + log_10 10 = 1 + 1 = 2.

Question:

An elevator and its load weigh a total of 1600 lb. Find the tension T in the supporting cable when the elevator, origin-ally moving downward at 20 ft/sec, is brought to rest with constant acceleration in a distance of 50 ft. (See fig.)

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Solution:

The mass of the elevator is m = w/g = (1600 lb) / (32 ft/sec^2) = 50 slugs where w is the weight of the elevator and its load. From the equations of motion with constant acceleration, v^2 = v2_0 + 2ay,a = (v_2 - v^2 _0)/2y. Let the upward direction be positive and the origin (y = 0) be at the point where the deceleration begins. Then the initial velocity v_0 is -20 ft/sec, the final velocity v is zero, and its displacement during this Interval is y = -50 ft. Therefore a = {0 - (- 20 ft/sec)}^2 / (-2 × 50 ft) = 4 (ft/sec^2). The acceleration is therefore positive (upward). From the free-body diagram (Fig. ) the resultant force is \sumF = T - w = T - 1600 lb. Hence, from Newton's second law, \sumF = ma, T - 1600 lb = 50 slugs × 4 (ft/sec^2) = 200 lb, T = 1800 lb.

Question:

Find the minimum energy of a simple harmonic oscilla-tor using the uncertainty principle\Deltax\Deltap\geqh.

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Solution:

Consider a particle of mass m on a spring of force-constant k. The kinetic and potential energies, respectively, are, K = (1/2) mv^2 = (p^2/2m) U = (1/2) kx^2 = (1/2) m\omega^2 x^2 where p =mvand \omega^2 = (k/m). The total energy is constant (conservation of energy) and is therefore equal to its average value. E = = (< p^2 >/2m) + (1/2) m\omega^2 (1) where < > indicates the time average. As a result of symmetry with respect to the equilibrium point, and are zero. The uncertainty in x and p are defined as, (\Deltax)^2 = < (x - )^2 > = - 2< x> + ^2 = - ^2 = - 0 = and (\Deltap)^2 = < (p -
^2 > = - 2
+
^2 = -
^2 = - 0 =
^2 Therefore (1) can be written as, E = (1/2m) (\Deltap)^2 + (1/2) m\omega^2 (\Deltax)^2(2) Using the uncertainty relation \Deltap\Deltax\geqh or (\Deltap)^2 \geq [h^2/(\Deltax)^2] the total energy (2), becomes, E - (1/2) m\omega^2 (\DeltaX)^2 = (1/2m) (\Deltap)^2 \geq [h^2/{2m(\DeltaX)^2}] or E \geq (h^2/2m) [1/(\DeltaX)^2] + (1/2) m\omega^2 (\Deltax)^2 The value of\Deltaxfor which \DeltaE is a minimum is given by the condition, [dE/{d(\Deltax)}] = 0 Hence, - 2(h^2/2m) (\Deltax)^-3 + 2 (1/2) m\omega^2\Deltax= 0 (\Deltax)^4 = (h^2/m^2\omega^2) (\Deltax)^2 = (h^2/m\omega) E_min= (h^2/2m) (m\omega/h) + (m\omega^2/2) (h/m\omega) =h\omega

Question:

A chemist dissolves 10 g of an unknown protein in a liter of water at 25\textdegreeC. The osmotic pressure is found to be 9.25 mmHg. What is the protein's molecular weight. Based upon the number of moles in 10 g of protein, what would the freezing point depression and boiling point elevation be? Assume R = Universal Gas Constant = .0821 (liter- atm / mole\textdegreeK), k_f = (1.86\textdegreeC / m), and k_b = (.52\textdegreeC / m).

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Solution:

The osmotic pressure (\pi) can be related to the molar concentration (C), Universal Gas Constant (R) and temperature (T) in degrees Kelvin of a solution via the formula \pi = CRT. After solving for C one can determine the molecular weight of the protein .Recall, C = (moles / liter) = (N / V), where N = moles and V = volume in liters. Hence, \pi = (N / V) RT. But N = moles = No. of (grams / molecular) wt = (g / M.W.) Rewriting, \pi = [(q / M.W.) /(V)]RTorM.W. = (gRT / \pi V). Substituting for the known values M.W = [(10g)(.821)(298\textdegreeK)] / (9.25mmHG) × [(760mm) / (atm)] = 20,100 (g / mole). Note, you assumed a volume of 1 liter and multiplied by the conversion factor (760 mm / atm), since R is in atm. and \pi in mmHg. Now that you know that the molecular weight is 20,100, the number of moles in the 10 g of protein is (10 / 20),100 = 4.98 × 10^-4 moles. To find the freezing point depression, ∆T_f, and boiling point elevation, ∆T_b, you must find the molality, since ∆T_b = k_bm and ∆T_f = k_fm, where k_f = molal freezing point depression constant, k_b = molal boiling point elevation constant and m = molality. Molality = moles of solute per 1 kg solvent. Thus, molality = .000498 (mole / 1 kg), since in 1 liter of water you have 1000 ml, which weighs 1kg. Therefore, ∆T_f = (1.86\textdegreeC) (.000498) = 9.3 × 10^-4 \textdegreeCand ∆T_b = (.52\textdegreeC) (.000498) = 2.59 × 10^-4\textdegreeC.

Question:

Given that the average speed of an O_2 molecule is 1,700 km/h at 0\textdegreeC, what would you expect the average speed of a CO_2 molecule to be at the same temperature?

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Solution:

The speed of a molecule is related to the mass and temperature by the equation (1/2) mv^2 = 3/2 kT where m is the mass, v is the speed, k is Boltzmann's constant, and T is the absolute temperature. Since k and T are constant 3/2 kT is constant for this system. Thus 1/2 m_(O)2v^2_(O)2 = 1/2 m_C(O)2v^2_C (O)2 where m_(O)2 is the mass of O_2, v_(O)2 is the speed of O_2, m_C(O)2 is the mass of CO_2, and v_C(O)2 is the speed of CO_2. The masses are equal to the molecular weights here. Solving for v_C(O)2 : v_C(O)2 = \surd[{(1/2) m_(O)2v^2_(O)2} / {(1/2) m_C(O)2}]m_(O)2 = 32 m_C(O)2----_--- = 44 v_(O)2 = 1700 km/h. v_C(O)2= \surd[{(1/2) (32) (1700)^2} / {(1/2) (44)}] = 1700 \surd[32/44] = 1700 × (.85) = 1450 km/h.

Question:

What is the molecular weight of a substance, each molecule of which contains 9 carbon atoms and 13 hydrogen atoms and 2.33 × 10^-23 g of other components?

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Solution:

The molecular weight of a compound is the sum of the weights of the components of the compound. It is the weight of one mole of the substance, thus this compound weighs the sum of the weight of 9 moles of C, 13 moles H and (6.02 × 10^23) × (2.33 × 10^-23 g) . Because each molecule of the third substance weighs 2.33 × 10^-23 g, one mole of it weighs (6.02× 10^23) × (2.33 × 10^-23 g) . There are 6.02 × 10^23 molecules of this other substance in one mole of the compound (Avogadro's number). molecular weight=(9 × MW of C) + (13 × MW of H) + (2.33 × 10^-23g) (6.02 × 10^23 /mole) = (9 × 12.01 g/mole) + (13 × 1.00 g/mole) + (14.03g/mole) = 108.09 g/mole \div 13.0 g/mole + 14.03 g/mole = 135.12 g/mole.

Question:

How many calories of heat are required to raise 1,000 grams of water from 10\textdegreeC to 100\textdegreeC?

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Solution:

The temperature rise is 90 degrees centigrade. The number of calories needed is 1000 × 900 = 90,000 calories, since one calorie is required to raise the temperature of one gram of water one degree centigrade.

Question:

The density of KF is 2.48 g/cm^3. The solid is made up of a cubic array of alternate K^+ and F^- ions at a spacing of 2.665 × 10^-8 cm. between centers. From these data, calculate the apparent value of the Avogadro number.

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Solution:

Avogadro's number is the number of particles in one mole of a substance. Here, one is given the dimensions of a cubic lattice made of (1/2)K^+ ions and (1/2)F^- ions. Due to the fact that KF crystallizes likeNaCl, it can be seen that there are eight ions or 4 formula units of KF in the cube. To calculate Avogadro's number from the data, one should find: (1) the volume of one cube 2) the weight of one cube using the density 3) the number of cubes in one mole of KF 4) Avogadro's Number by multiplying the number of cubes by 4, the number of formula units (KF) per cube. Solving for Avogadro's Number: 1) The volume of the cube is found by cubing the length of the edge. From the crystallization of KF, it is seen that the length of the edge of cube is twice the spacing between ion centers. This is due to the fact that three ions make up an edge, and, as such, 2spacingsbe-tween the ions exist. Therefore, length of edge = 2.665 × 10^-8 cm × 2 = 5.33 × 10^-8 cm volume of cube = (5.33 × 10^-8)^3 = 1.51 × 10^-22cm^3. 2) Using the density, one knows that 1 cm^3 weighs 2.48 g. Therefore, the weight of one cube is equal to the density times the volume of a cube. weight of one cube = 2.48 g/cm^3 × 1.51 × 10^-22 cm^3 = 3.76 × 10^-22 g/cube 3) One mole of KF weight the sum of the weights of one mole of K^+ and one mole of F^-. (MW of K = 39.10, MW of F = 19.00.) MW of KF = 39.10 + 19.00 = 58.10 g/mole The number of cubes in one mole is then found by dividing 58.10 g/mole by the weight of one cube. no. of cubes in one mole = (58.10 g/mole) / (3.76 × 10^-22 g/cube) = 1.55 × 10^23 cubes/mole. 4) One finds the number of KF particles in one mole (Avogadro's Number) by multiplying 1.55 × 10^23 cubes/mole by 4 formula units/cube, because there are 4 KF formula units in each cube. Avogadro's Number = 4 [formula units/cube] × 1.55 × 10^23 [cubes/mole] = 6.2 × 10^23 formula units/cube.

Question:

A body is whirled in a vertical circle by means of a cord attached at the center of the circle. At what minimum speed must it travel at the top of the path in order to keep the cord barely taut, i.e., just on the verge of collapse? Assume radius of circle to be 3 ft.

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Solution:

At the top of the circle, the net force on the body of mass m is \sumF = mg + T wherethe positive direction is taken downward, and T is the cord tension. Since the motion is circular, the net force is centripetal and \sumF = (mv^2)/R where v is the body's velocity and R is the circle's radius. Hence mg +T = (mv^2)/R andT = (mv^2)/R - mg If the cord is just on the verge of collapse, T = 0 and (mv^2)/R - mg = 0 Whencev = \surdgR Using the given datav = \surd[(32 ft/s^2)(3 ft)] = \surd[96 ft^2/s^2] v = 9.8 ft/s

Question:

What weight of sulfur must combine with aluminum to form 600 lbs of aluminum sulfide?

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Solution:

In this problem one wants to find out how much reactant (sulfur) was needed to produce a given amount of product (aluminum sulfide). The balanced equation for this reaction is: 2Al + 3S \rightarrow AL_2 S_3 The first method for solving this problem is the pro-portion method. This procedure involves the ratios of weights and molecular weights. Set up the balanced equation showing the weights and molecular weights. Xlbs600 lbs 2 Al + 3S\rightarrowAL_2 S_3 3(32) lb150 lb M.W. of S is 32. M.W. of Al_2 S_3 is calculated below. Atom Number of Atoms Atomic Weight Total Atomic Weight Al 2 27 54 S 3 32 96 M.W. of Al_2 S_3 = 150. Use the proportion equation, [(weight _reactant ) / (moles × M.W. _reactant ) = [(weight _Product ) / (moles × M.W. _product )] Solve for weight_reaotant. weight _reactant = [(3) (32) (600)] / [150] = 384 lbs. sulfur. The second method for solving the problem is by the mole method. We see that 1 mole of Al_2 S_3 requires moles of S. Therefore, if one can calculate the actual number of moles present of 1 substance one can obtain the number of moles of the other substance by setting up a ratio. Then, knowing the number of moles, one can calculate the weight by using the equation moles = weight/M.W. One is given that 600 lbs. of Al_2 S_3 are present. Therefore, the number of moles is 600/150 = 4 pound moles of Al_2 S_3. Setting up the ratio, [(S) / (Al_2 S_3 )] = (3/1) = (x/4) it is seen that x = 12 lb. moles of S. Next, solve the mole equation for weight of S weight = moles × M.W. = 12 × 32 = 384 lbs.

Question:

A circus clown whose mass is 1 × 10^2 kg steps onto the outer edge of a large disk with a radius of 2.0 × 10^1 m and a mass of 2 × 10^3kg. As-sume that the disk is mounted on a frictionless bearing with a vertical axis of rotation and is initially at rest. If the clown now runs clock-wise around the edge of the disk at a speed of 2m/s, how fast does the disk turn and what is the angular momentum?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0210.htm

Solution:

Since no external torques are acting on the clown-disk system, we may apply the principle of conservation of angular momentum to this system. The angular momentum of the system before the clown stepped on the disk was zero. Therefore, after the clown steps on the disk, the angular momentum of the disk-down system must remain zero, or J\ding{217}_system = 0 \ding{217} But J\ding{217}_system = J\ding{217}_disk + J\ding{217}_clown= 0 \ding{217} \ding{217} \ding{217} and J\ding{217}_disk =-J\ding{217}_clown \ding{217} - \ding{217} This last equation states that the magnitude of J\ding{217}_disk and J\ding{217}_clown are the \ding{217} \ding{217} same, but that J\ding{217}_disk and J\ding{217}_clown are opposite in direction. Hence, because \ding{217} \ding{217} the clown is running clockwise, the disk travels counter-clock- wise. If it is assumed that the clown can be represented as a particle, the orbital angular momentum of the clown, is J\ding{217}= r\ding{217}×p\ding{217} \ding{217} \ding{217} × \ding{217} where p\ding{217}is the linear momentum of the clown, and r\ding{217}is a vector from the \ding{217} \ding{217} axis of rotation to the clown. Because the clown is running in a circle r\ding{217}is \ding{217} perpendicular to p\ding{217}and \ding{217} \vertJ\ding{217}_clown\vert = rp_clown = mrv_clown \ding{217} = (1×10^2kg)(2 m/s) (2×10^1m) × × = 4×10^3J\bullets × \bullet which is also the magnitude of the disk's angular momentum. Now, \vertJ\ding{217}_dlsk\vert = I_disk\omega_disk \ding{217} \omega where I_disk is the moment of inertia of the disk about its axis of rotation. Therefore, since \vertJ\ding{217}_dlsk\vert = \vertJ\ding{217}_clown\vert, \ding{217} \ding{217} \omega_disk = (\vertJ\ding{217}_clown\vert)/(I_disk)(1) \omega \ding{217} The only variable we don't know in (1) is I_disk. This we now calculate. By definition, I_disk =\intr^2 dm \int where r is the distance of a mass element dm from the axis about which we calculate I_disk (see figure), and the integral is taken over the mass of the disk. Now \rho= dm/dv \rho where\rhois the density of the disk and dv is a volume element. \rho Whence I_disk =\intr^2\rhodm I_disk =\intr^2\rhodm \int \rho If we consider the disk to be very thin, dv = ds, an element of area of the disk, and ds = r dr d\texttheta \texttheta I_disk =2\pi\int_0^R\int_0\rhor^3 dr d\texttheta I_disk =2\pi\int_0^R\int_0\rhor^3 dr d\texttheta \pi \int \int \rho \texttheta I_disk =2\pi\int_0\rho(R^4/4) d\texttheta I_disk =2\pi\int_0\rho(R^4/4) d\texttheta \pi \int \rho \texttheta I_disk = (\pi\rhoR^4)/(2) I_disk = (\pi\rhoR^4)/(2) \pi\rho where R is the disk radius. But\rho= M/\piR^2, M being the disk's mass. \rho \pi I_disk = (\piR^4)/(2) (M/\piR^2) = MR^2/2 \pi \pi Using (1), \omega_disk= (\vertJ\ding{217}_clown\vert)/(MR^2/2) \omega \ding{217} = [4×10^3J\bullets]/[(2×10^3kg/2)(4×10^2m^2)] × \bullet × × \omega_disk= [4×10^3J\bullets]/[4×10^5m^2\bulletkg] \omega × \bullet × \bullet = 10^-2 N\bulletm\bullets/kg\bulletm^2 \bullet \bullet \bullet \omega_disk= 10^-2 [(kg\bulletm^2/s^2\bullets)/(kg\bulletm^2)] \omega \bullet \bullet \bullet = 10^-2 s^-1 Furthermore, v_disk =\omega_disk R is the velocity of a point on the rim of the disk. \omega v_disk = (10^-2 s^-1)(2×10m) = 0.2m/s × The disk rotates much slower than the clown runs around the edge.

Question:

What will be the final concentration of a solution of 50 ml of .5M H_2SO_4 added to 75 ml of .25M H_2SO_4 (the final volume is 125 ml)?

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Solution:

Two facts must be known to answer this question: (1) M =Molarity = the number of moles of solute per liter of solution and (2) that the number of moles of H_2SO_4 in the final solution is the sum of the number of moles of H_2SO_4 in each solution. molarity= [(no. of moles)/(liters of solution)] One can rearrange this equation to determine the number of moles present in each solution. no. of moles =molarity× liters of solution The total number of moles in the final solution is equal to the sum of the number of moles of each of the two initial solutions. total number of moles = (0.05l)(.5 moles/l) + (.075l) (.25 moles/l) = 0.025 moles + .019 moles = .044 moles. To obtain the final concentration, you return to the definition of molarity, supplying the known values. Thus,molarity(final concentration)= [(0.044moles) / (.125 liter)] = 35 M H_2SO_4 (It was given that the volume of the final solution equaled 125 ml = .125 liters.)

Question:

How many full strokes of a bicycle pump (chamber 4.0 cm diameter and 40.0 cm long) would you need to make in order to pump up an automobile tire from a gauge pressure of zero to 24 pounds per square inch (psi)? Assume tem-perature stays constant at 25\textdegreeC and atmospheric pressure is one atmosphere. Note, that gauge pressure measures only the excess over atmospheric pressure. A typical tire volume is about 25 liters.

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Solution:

One atmosphere equals 14.7 psi, therefore, the amount of pressure needed to fill the tire is 24 + 14.7 psi or 38.7 psi. Converting back toatm: atm. contained in inflated tire = 38.7 psi × 1atm/14.7 psi = 2.63 atm. When the tire is deflated the pressure is 1atmand the volume is 25 l, Eachatmof pressure occupies 25 l. Therefore, the volume of the tire when inflated is 25l × 2.63 atm. Volume of inflated tire = 25l, × 2.63 = 65.75l Since there is 25l present in the tire before in-flation, the volume that the pump must contribute is 65.75 - 25l or 40.75l. The volume of air forced into the tire at each stroke of the pump is equal to the volume of the pump. Volume of pump= \pir^2 h = \pi × (2 cm)^2 × 40 cm = 503 cm^3 /stroke There are 1000 cm^3 in 1 liter, therefore 40750 cm^3 of air must be pumped into the tire. If 503 cm^3 is pumped in per stroke the number of strokes necessary to fill the [(40750 cm^3 )/(503 cm^3 /stroke)] No. of strokes = [(40750 cm^3 )/(503 cm^3 /stroke)] = 81 strokes.

Question:

TheRydberg- Ritz equation governing the spectral lines of hydrogen is (1/\lambda) = R [(1/n_1 ^2) - (1/n_2 ^2)], where R is the Rydberg constant, n_1 indexes the series under consideration (n_1 = 1 for the Lyman series, n_1 = 2 for theBalmerseries, n_1 = 3 for thePaschenseries), n_2 = n_1 + 1, n_1 + 2, n_1 + 3, . . . indexes the successive lines in a series, and \lambda is the wave- length of the line corresponding to index n_2. Thus, for the Lyman series, n_1 = 1 and the first two lines are 1215.56 \AA (n_2 = n_1 + 1 = 2) and 1025.83 \AA (n_2 = n_1 + 2 = 3). Using these two lines, calculate two separate values of the Rydberg constant. The actual value of this constant is R = 109678 cm^-1.

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Solution:

The first thing to do is to convert the wave - lengths from \AA to more manageable units, i.e. centimeters.Using the relationship 1 \AA = 10^-8 cm, the first two Lyman lines are 1215.56 \AA = 1215.56 × 10^-8 cm for n_2 = 2, and 1025.83 \AA = 1025.83 × 10^-8 cm for n_2 = 3. Solving theRydberg- Ritz equation for R, one obtains : R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1, For the first line, R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1 = [1215.56 × 10^-8 cm {(1/1^2) - (1/2^2)}]^-1 = 109689 cm^-1 and for the second line, R = [\lambda {(1/n_1 ^2) - (1/n_2 ^2)}]^-1 = [1025.83 × 10^-8 cm {(1/1^2) - (1/3^2)}]^-1 = 109667 cm^-1 The first of these is 0.0100% greater than the true value, and the second is 0.0100% less than the true value.

Question:

Find Antilog_10 1.4850.

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Solution:

By definition, Antilog_10 a = N is equivalent to log_10N = a. Let Antilog_101.4850 = N. Hence, Antilog_10 1.4850 = N is equivalent to log_10N = 1.4850. The characteristic is 1. The mantissa is 0.4850. Therefore, the number that corresponds to this mantissa will be multi-plied by 10^1 or 10. The mantissas which appear in a table of common logarithms and are closest to the mantissa 0.4850 are 0.4843 and 0.4857. The number that corresponds to the mantissa 0.4850 will be found by interpolation. Set up the following proportion. (d/.01) = (.0007/.0014) cross-multiplying, .0014d= (.01) (.0007), or d =.01 (.0007/.0014) = (1 ×10^\rule{1em}{1pt}2) [(7 × 10^\rule{1em}{1pt}4) / (1.4 × 10^\rule{1em}{1pt}3)] = [(7 × 10^\rule{1em}{1pt}6) / (1.4 × 10^\rule{1em}{1pt}3)] = (7/1.4) × (10^\rule{1em}{1pt}6/10^\rule{1em}{1pt}3) = 5 × 10^\rule{1em}{1pt}6\rule{1em}{1pt}(\rule{1em}{1pt}3) = 5 × 10^\rule{1em}{1pt}3 = 5 ×.001 = .005 Hence,d= 0.005 x= d + 3.05 = 0.005 + 3.050 = 3.055 Hence, N = Antilog_10 1.4850 = 3.056 × 10 = 30.550 = 30.55 Therefore Antilog_10 1.4850 = 30.55.

Question:

A Rowland ring, made of iron, of mean circumferential length 30 cm and cross section 1 cm^2 , is wound uniformly with 300 turns of wire. Ballistic galvanometer measurements made with a search coil around the ring show that when the current in the windings is 0.032 amp, the flux in the ring is 2 × 10^-6 weber(see figure). Compute (a) the flux density in the ring, (b) the magnetic intensity, (c) the perme-ability,(d) the relative permeability.

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Solution:

a)The flux, \varphi, through the ring is defined as \varphi = \int B^\ding{217} \textbullet dA^\ding{217} where B is the flux density and dA^\ding{217} is an infinitesimal cross sectional element through which the flux passes. B is constant in the iron, and the angle between B^\ding{217} and the area vector is zero. Therefore \varphi = B \int dA = BA where A is the cross section of the iron. Hence B = (\varphi/A) = [(2 × 10^-6 w)/(1 cm^2 × 1 m^2 /10^4 cm^2 )] = 2 × 10^-2 w/m^2 . b)Magnetic intensityH = Ni/l = [(300 × .032)/(.30)] = 32 amp-turns/m. c)Permeability \mu = (B/H)(by definition) = [(2 × 10^-2 w/m^2 )/(32 amp/m)] = [6250 × 10^-7 w / amp-m.] d)Relative permeability K_m = (\mu/\mu_0 ) = [(6250 × 10^-7 )/(12.57 × 10^-7 )] = 498(no units).

Question:

Write condensed structural formulas for all the alkynes, i.e. unsaturated compounds with triple bonds, with a molecular formula of C_5H_8.

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Solution:

A condensed structural formula provides all the information represented by other structural formulas (i.e. Lewis diagrams, bond diagrams), but it is not as cumbersome. To solve this problem, first write all the structural isomers with five carbons in a continuous chain: Second, write all the structural isomers with a four carbon parent molecule and one carbon in a branch: No other structural formulas are possible. Thus, there are only three structural formulas poss-ible for C_5H_8 . Next, one writes these formulas in the condensed form. Namely,

Question:

Explain the types and effects of competition.

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0762.htm

Solution:

Competition is the active demand of two or more organisms for a commonvital resource. There are many ways for organisms to compete butbasically competition can be categorized in two ways. Contest competitionis the active physical confrontation between two organisms whichallows one to win the resource. Extreme contest competition involvesdirect aggression. Scramble competition is the exploitation of a commonvital resource by both species. In scramble competition, one organismis able to find and utilize vital resources more efficiently than another. The aggressive behavior of ant colonies is an example of contest competition. Actual colony warfare can establish territorial limits within and betweenspecies. Often, contest competition takes the form of threatening gestureswhich involves safer means to achieve the same goal. Scramble competition can best be demonstrated with observations ofthe fruit fly Drosophila. Combat and aggressivebehavior havenothing todo with the type of competition these flies demonstrate. It has been observedthat when the food supply was limited, there were four conditionsfor survival for the fly larvae. Those that survived were the quickestfeeders, best adapted for the particular medium, heaviest at the beginningof competition, and most resistant to changes due to population increase. Every species occupies a particular niche that to a large extent helpsavoid competition. But niches do overlap in many areas and competition can never be clearly eliminated.

Question:

An E. coli culture will grow to a limiting concentration of about 10^9 cell per cm^3. At this concentration what percentage of the total volume of the culture medium is occupied by the cells? Assume that an E. coli cell is cylindrical and that it is 2.0\mu long and has a 1.0\mu diameter.

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Solution:

This percentage can be found by first determining the volume that one E. coli cell occupies and then the volume that 10^9 cells occupy. This volume divided by 1 cm^3 multiplied by 100 equals the percent of the total volume that the cells occupy. The volume of a cylinder is equal to its length multiplied by the area of its cross-section. Area of a circle = \pir^2, where r = radius of the circle. 1\mu = 10^-4 cm Volume of 1 E. coli cell= (2.0 x 10^-4 cm) (\pi) × (0.5 × 10^-4cm)2 = 1.57 × 10^-12 cm^3 There are 10^9 cells per cm^3 of solution, therefore the volume that the cells occupy in 1 cm^3 of solution is 10^9 × 1.57 × 10^-12 cm^3 or 1.57 × 10^-3 cm^3. Percent of total volume that the E. coli cells occupy = (1.57 × 10^-3 cm^3)/ (1.0cm^3 × 100) = .157%.

Question:

Compute ther.m.s. speed of 0_2 at room temperature.

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Solution:

Ther.m.s. speed of a gas molecule is v_r.m.s . = \surd[(3RT)/(M)] where M is the molar mass of the gas, T is its temper-ature in degrees Kelvin, and R is the gas constant. Hence, v_r.m.s = \surd{[(3)(8.31 joule/mole\textdegreeK)(273\textdegreeK)]/(32 × 10^-3 kg/mole)} Here, we have used the fact that 0\textdegreeC = (0 + 273)\textdegreeK. v_r.m.s = \surd(2.13 × 105m^2/g^2) v_r.m.s = 4.61 × 10^2 m/s.

Question:

(a) Explain what is meant by error detection through redundancy. For each of the following words, determine if it is a "legal" code word for the (b) binary-coded decimal (BCD) system. (c) excess-3(X-3) code system. (i)0 1 0 1 (ii) 0 1 0 0 (iii) 1 1 0 0

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Solution:

(a) All information within a computer is represented as bits of 1's and 0's. The flow of this information within, to and from the computer occurs along some form of transmission line. A basic physical characteristic of any transmission line is the presence of disturbances which introduce the possibility of errors occurring in the information transmitted on this line. Let us represent a trans-mission system by letting the transmitted information be the binary n-tuple (x1 ,x2 ,...,x_n ) which moves along n parallel data lines, forming n independent networks. Thus, one faulty line can influence only a single bit, and therefore, at worst, introduce the error of complementing one of the n-tuple components in the output: This independent error representation of transmission lines character-izes data buses (bidirectional communication lines between the CPU, memory, and I/O devices, which carry instructions and transfer data). A simplifying assumption that will be useful is that, in a data word consisting of n bits, only one bit will be incorrect. Probability theory can be used to show that for an error rate of one in 1000 trans-mitted bits, the odds of having two erroneous bits in a 5-bit word (n = 5) are less than one in 100,000. So the single-error approxima-tion is a reasonable, though simplistic, assumption. Computers are supplied with automatic error detectors. An error can be detected by redundancy, which is the inclusion of extra information with each transmission. This extra information helps the receiver to decide if the received message has been altered in transmission. The simplest error detecting system, conceptually, would be duplicate transmission of each data unit. Thus, if the first transmission doesn't match the second transmission, the receiver can signal that an error has occurred. If we are willing to surrender a degree of certainty in exchange for efficiency, better methods are available. One such method is the binary- coded decimal (BCD) code system. In the BCD system, four bits are used to code the decimal digits 0-9 in their binary representation. Decimal BCD 0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 From these 10 digits any positive decimal Integer can be constructed. In BCD the binary representations for the numbers 10-15 are not used and, if they show up at a receiver, they can be detected as "illegal" by checking the received datum with a "legal word" list: (b) Since the legal BCD word list consists of the binary representa-tions of the decimal digits 0-9, any received 4 bit word whose decimal equivalent is greater than 9 is illegal. (i) Since binary 0101 = decimal 5, 0101 is a legal BCD code word, (ii) Since binary 0100 = decimal 4, 0100 is a legal BCD code word, (iii) Since binary 1100 = decimal 12 is greater than 9, 1100 is not a legal BCD code word. (c) The excess-3 (X-3) code is obtained by adding binary 0011 to the BCD codes: Decimal BCD Excess-3 0 0000 0011 1 0001 0100 2 0010 0101 3 0011 0110 4 0100 0111 5 0101 1000 6 0110 1001 7 0111 1010 8 1000 1011 9 1001 1100 Note that X-3 does not allow 0000 or 1111. Thus, every legal X-3 word contains at least one 0 and at least one 1, which provides the receiver with the information that the data channel is active and transmitting. Again, the error-detection algorithm involves table look-up and comparison. Since 0101, 0100, and 1100 are all on the legal word list, they are all legal X-3 code words.

Question:

With the aid of a diagram, describe the structures found in the human ear.

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Solution:

Three parts of the human ear can be distinguished: the outer ear, the middle ear, and the inner ear. The outer ear consists of the skin- covered cartilaginous flap, or pinna, and a channel known as the external auditory meatus. At the inner end of this canal is a membrane called the tympanic membrane (eardrum). On the other side of the tympanic membrane is a small chamber, the middle ear, which contains three tiny bones. These three bones are the malleus (hammer), incus (anvil) and stapes (stirrup) and are arranged in sequence across the middle ear from the tympanic membrane to another membrane, the oval window, which separates the middle ear from the inner ear. The middle ear is connected to the pharynx via the Eustachian tube, which serves to equalize the air pressure between the outer and middle ear. The inner ear consists of a complicated labyrinth of interconnected fluid-filled chambers and canals. One group of chambers and canals is involved with the sense of equilibrium. There are two small chambers - the sacculus and utriculus - and three semicircular canals arranged perpendicularly to one another. The utricle and saccule are small, hollow sacs lined with sensitive hair cells and containing small ear stones, or otoliths, made of calcium carbonate. The semicircular canals contain hair cells and are filled with a fluid known as the endolymph. Movement of the endolymph over the hair cells stimulates the latter to send impulses to the brain. Besides the structures involved in maintaining equilibrium, the inner ear also houses the organ for hearing, the cochlea. The cochlea is a coiled tube made up of three canals - the vestibular canal, the tympanic canal and the cochlear canal, each separated from each other by thin membranes. The oval window is linked to the vestibular canal while the tympanic canal is sealed by a membrane, the round window, which leads to the middle ear. These two canals are connected with each other at the apex of the cochlea and are filled with a fluid known as the perilymph. The cochlear canal, filled with endolymph, contains the actual organ of hearing, the organ of Corti. The organ of Corti consists of a layer of epithelium, the basilar membrane, on which lie five rows of specialized receptor cells extending the entire length of the coiled cochlea. Each receptor cell is equipped with hairlike projections extending into the cochlear canal. Overhanging the hair cells is a gelatinous structure, the tectorial membrane, into which the hairs project. The hair cells of the organ of Corti initiate impulses via the fibers of the auditory nerve to the brain.

Question:

What will be the relative concentrations of Zn^+2 and Cu^+2 when the cell Zn; Zn+2II Cu^+2; Cu has completely run down? (E\textdegree = 1.10 volts).

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Solution:

The key phrase to understand before solving this problem is "has completely run down". That means there is no apparent current, leading us to assume a value of zero for the electrode potential, E. Next, one should realize that whenever concentrations are involved one should apply the Nernst Equation, which states E = E\textdegree - (RT / nF)In Q , where Q is the equilibrium constant expression, R is the gas constant (8.314 joules per degree), F is 96,500 coulombs, n is the number of moles of electrons transferred and T is 25\textdegreec (or 298\textdegree K). Since the overall cell reaction is Zn + Cu^+2 - Zn^+2 + Cu , 2 moles of electrons are transferred and log Q equals log {[Zn^+2] / [Cu ]}. Remember that solids are omitted from all concentration equations. This yields (via substitution): E = E\textdegree - (RT / nF) In Q and 0 = 1.1 volts - (.059 / 2) log {[Zn^+2] / [Cu^+2]} , which means log {[Zn^+2] / [Cu^+2]} = 37.3 . Solving, [Zn^+2] / [Cu^+2] = 2 × 10^37 . This means that the concentration of Zn^+2 is far greater than Cu^+2 .

Question:

How is an ARRAY declared inPL/I? Explain using examples.

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Solution:

Every identifier to be used as an array name requires explicit declaration. The declaration must also de-fine the subscripting to be used whenthe array elements are referenced. For declaration of an array variable, the declar-ation has the following format: DCL ARRAY NAME (DIMENSION ATTRIBUTES) DATA ATTRIBUTES; Array names can be any variable names and every element within thearray is referenced by the array name declared. The use of the dimension attribute in an array declara-tion is governedby the following rules. 1. The dimension attribute has the form of one or more boundary specificationsenclosed within a pair of parentheses. If two or more boundaryspecifications are used in the list, they are separated by commas. The number of boundary speci-fications in the dimension attributeestablishes dimensional-ity of the array variable. 2. A boundary specification is composed of two, possib-ly signed, decimalinteger constants, separated by a colon. The first integer value specifiesthe lower bound and the second the upper bound of that particulardimension. For example DECLAREA(-1:2, 8:10); In the above, A is the name of the array. It could have been any other nametoo, e.g., NANCY, PETE, POOR, etc.. But the same name cannot beused to refer to more than one data struc-ture or data item. Also note thatin the above, A is a two dimensional array as there are two boundary, specifications. The first subscript may assume any one of the integer values-1, 0, 1, or 2. The second subscript may assume any one of the integervalues in the range of 8 through 10. Thus, in the first subscript thereare 4 elements and the second sub-script has 3 elements so the arrayA contains 4 × 3=12 elements. 3. In a boundary specification, the lower bound must have its value lessthan or equal to the value of the upper bound. The dimension attributes, e.g.,(3:5), (-3:0, 0:3), (9:9, 1:1), arewell formed. But, (5:3) is not well formed because the lower bound is greater thanthe higher bound. If the lower bound is not specified it is assumed to havea numeric value of one. For example, the dimension attributes (17,26,5) and (1:17, 1:26, 1:5) areequivalent. 4. The dimension attribute must always appear as the first attribute followingthe variable name declared. Using the above 4 rules a well-formed declare statement can be written. E.g., consider the following declare state-ment: DECLAREB(5,12) FIXED DEC(8,3); The above statement is a well-formeddeclarestatement. It establishes a twodimensional array B, with upper bounds 5 and 12. The elements are fixeddecimal numbers with preci-sion of the type XXXXX.XXX. However, thefollowing declare statement is wrong. DECLARE BFIXED(5,12) DEC(8,3); The machine would be unable to distinguish between dimension and precisionattributes when written as above. The declara-tions for more than onearray can sometimes be condensed. For example, consider the following: DECLARE (A, B, MATRIX) (10, 1:9, -2:5) FLOAT; DECLARE (X FIXED, Y FLOAT) (17, 20); This is how the array structure is declared. In the above, A, B and MATRIX are3-dimensional arrays with 10×9×8=720 elements in each. All theelements have a Float-ing Point number representation. X and Y are both 2-dimensional arrays, each of 17 × 20 = 340 elements. But, the elements of X have a Fixed Point repre-sentation, and the elements of Y have a Floating Point re-presentation.

Question:

What is the acceleration of a block on a frictionless plane What is the acceleration of a block on a frictionless plane in-clined at an angle \texttheta with the horizontal?

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Solution:

In order to find the acceleration, a, of the block, we must calculate the net force, F, on the block, and relate this to its acceler-ation via Newton's Second Law, F = ma. (Here m is the mass of the block). The only forces acting on the block are its weight mg and the normal force N exerted by the plane (see figure). Take axes parallel and perpendicular to the surface of the plane and resolve the weight into x- and y-components. Then \sumF_y = N - mg cos \texttheta \sumF_x = mg sin \texttheta. But we know that the acceleration in the y direction, a_y = 0, since the block doesn't accelerate off the surface of the inclined plane. From the equation \sumF_y = ma_y we find that N = mg cos \texttheta. From the equation \sumF_x = ma_x, where a_x is the acceleration of the block in the x direction, we have mg sin \texttheta = ma_x, a_x = g sin \texttheta. The mass does not appear in the final result, which means that any block, regardless of its mass, will slide on a frictionless inclined plane with an acceleration down the plane of g sin \texttheta. (Note that the velocity is not necessarily down the plane).

Question:

Using positive logic, determine the logic function represented by the diode circuit of figure 1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0063.htm

Solution:

The ideal voltage-current characteristic of a diode is shown in fig. 2. The diode conducts when V_D is positive and acts as an insulator when V_D is negative. Hence, the diode is similar to a valve which lets current flow in only one direction. As can be seen in fig. 1, when all inputs are high (greater than \rule{1em}{1pt}5V) the diodes conduct, there is a large positive current in R, and the output voltage is greater than \rule{1em}{1pt}5V. When only one or two inputs are high; those diodes corresponding to the high inputs conduct and those diodes corresponding to the low inputs (less than or equal to \rule{1em}{1pt}5V) act as insulators. The conducting diode(s) pass current to the resistor R, forming a voltage drop. Hence, the output voltage is greater than \rule{1em}{1pt}5V. When all inputs are low (less than or equal to \rule{1em}{1pt}5V) all the diodes act as insulators and there is no current in resistor R, hence, the output voltage is \rule{1em}{1pt}5V. From this information it is seen that the output is low only when all of the inputs are low. Tabulating this result, we get: V IN 1 V IN 2 V IN 3 V IN OUT L L L L L L H H L H L H L H H H H L L H H L H H H H L H H H H H It is evident that this table represents an OR gate.

Question:

Calculate the pressure required to compress 2 liters of a gas at 700 mm pressure and 20\textdegreeC into a container of 0.1 liter capacity at a temperature of - 150\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0039.htm

Solution:

One is dealing with changing volumes, press-ures and temperatures of a gas. Therefore, this problem can be solved using the combined gas law. It states that as the pressure increases, the volume decreases and that as the temperature increases, the volume in-creases. These factors are related by the equation (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1, V_1 and T_1 are the initial pressure, volume and temperature and P_2, V_2, and T_2 are the final values. For any problem dealing with gases, the first step always involves converting all of the temperatures to the degree Kelvin scale by the equation \textdegreeK = \textdegreeC + 273 For this question T_1 = 20\textdegreeC = 20 + 273 = 293\textdegreeK T_2 = - 150\textdegreeC = - 150 + 273 = 123\textdegreeK. This seems to indicate that the pressure would decrease. But one is also told that the volume de-creases, which would have the effect of increasing the pressure. Therefore, one cannot predict the final change in volume. For the sake of clarity, set up a table as given below. P_1 = 700 mmP_2 = ? V_1 = 2 litersV_2 = 0.1 liter T_1 = 293\textdegreeKT_2 = 123\textdegreeK Since one is given 5 of the 6 values, it is poss-ible to use the combined gas law equation to determine P_2 (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 P_2 = [(T_2 V_1 P_1 )/(T_1 V_2 ) = [{123\textdegreeK (2 liters) (700 mm)}/{293\textdegreeK (0.1 liter)}] = 5877 mm.

Question:

The skin is much more than merely an outer wrapping for an animal; it is an important organ system and performs many diverse functions. What are some of the primary functions of the skin in man?

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Solution:

Perhaps the most vital function of the skin is to protect the body against a variety of external agents and to maintain a constant internal environment. The layers of the skin form a protective shield against blows, friction, and many injurious chemicals. These layers are essentially germproof, and as long as they are not broken, keep bacteria and other microorganisms from entering the body. The skin is water-repellent and therefore protects the body from excessive loss of moisture. In addition, the pigment in the outer layers protects the underlying layers from the ultraviolet rays of the sun. In addition to its role in protection, the skin is involved in thermoregulation. Heat is constantly being produced by the metabolic processes of the body cells and distributed by the bloodstream. Heat may be lost from the body in expired breath, feces, and urine, but approximately 90 per cent of the total heat loss occurs through the skin. This is accomplished by changes in the blood supply to the capillaries in the skin. When the air 'temperature is high, the skin capillaries dilate, and the increased flow of blood results in increased heat loss. Due to the increased blood supply, the skin appears flushed. When the temperature is low, the arterioles of the skin are constricted, thereby decreasing the flow of blood through the skin and decreasing the rate of heat loss. Temperature-sensitive nerve endings in the skin reflexively control the arteriole diameters. At high temperatures, the sweat glands are stimulated to secrete sweat. The evaporation of sweat from the surface of the skin lowers the body temperature by removing from the body the heat necessary to convert the liquid sweat into water vapor. In addition to their function in heat loss, the sweat glands also serve an excretory function. Five to ten per cent of all metabolic wastes are excreted by the sweat glands. Sweat contains similar substances as urine but is much more dilute.

Question:

The solar constant, or the quantity of radiation received by the earth from the sun is 0.14 W\bulletcm^-2. Assuming that the sun may be regarded as an ideal radiator, calculate the surface temperature of the sun. The ratio of the radius of the earth's orbit to the radius of the sun is 216.

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Solution:

To calculate the temperature of the sun, T, we use Stefan's Law R = e \sigma T4 Here, e is the emissivity of the radiator, \sigma is a constant, T is the temperature of the radiator in Kelvin degrees, and R is the rate of emission of radiant energy per unit area of the radiator. Hence, T^4 = R/e\sigma(1) Regarding the sun as an ideal radiator, e = 1. Furthermore, R = P/A where A is the surface area of the sun, and P is the power provided by the sun as a result of radiation. Using these facts in (1) T^4 = P/\sigmaA(2) Now, the power per unit area intercepted by the earth is P/A' = .14 W/cm^2 where A' is the surface area of a sphere having a radius equal to that of the earth's orbit. Hence, P = (.14 W/cm^2) A'(3) Using (3) in (2) T^4 = {(.14 W/cm^2)A'}/\sigmaA Now A'/A = (4\pir^2)/(4\piR^2), r where r and R are the radius of the earth's orbit and the radius of the sun, respect-ively. Then, T^4 = [(.14 W/cm^2)/\sigma] (r/R)^2or T^4 = [(0.14 W\bulletcm^-2)/(5.6 × 10^-12 W\bulletcm^-2\bulletK deg^-4)] × (216)^2 \therefore T = 5.84 × 10^3 \textdegreeK.

Question:

Using the pedigree below, determine the method of inheritance of the trait, and as far as possible, fill in the genotypes of each individual.

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Solution:

One should immediately note that the trait is only exhibited by male members of the family. Although the number of individuals is small enough that this could have been a a chance occurrence, it strongly suggests that the trait is sex-linked. Proceeding on this assumption, we can deduce the possible genotype of each member. Letting r represent the recessive, sex-linked allele, we can assign the genotype X^rY to P_1, F_1,7, F_2,1 ,F_2,3 ,F_2,5 ,and F_3,1, since all are males expressing the trait. Since F_2,7 ,a male, does not express the trait, and there-fore does not carry the r allele, his genotype is X^RY, where R represents the normal allele. We know that F_3,1 must have received the r allele from his mother, F_2,6 since to a male offspring, the mother contributes the X while the father contributes the Y chromosome. The genotype of F_2,6 is then X^RX^r, since she does not express the trait. F_1,5 must also be a carrier (X^RX^r) since F_2,6 is a carrier, F _2,5 expresses the trait, and F1,4,who must be X^RY, could not have donated the recessive allele to his offspring. Since we have no information concerning the offspring of F_2,4 ,her genotype could be either X^RX^R or X^Rx^r, depending on whether she received a dominant or recessive allele from F_1,5 . Since F_1,5 is a carrier, and her father P_3 ,who does not express the trait, is X^RY, we can assign P_4, the geno-type X^RX^r , which shows that she is also a carrier. F_1,6 can be either X^RX^R or X^Rx^r, one dominant allele in either case being necessarily donated by P_3. F_1,3 a male not expressing the trait, is X^RY. Since no male offspring of P_1 and P_2 express the trait, it is likely though not necessary, that P_2 .is not a carrier. Her possible genotypes are X^RX^R and X^Rx^r. F_1,2 must have received the recessive allele from P_1 and be a carrier (X^Rx^r). F_1,1 must be X^RY since he does not express the trait. F_2,2 could be either X^RX^R or X^RX^r. Summarizing, we can see that our assumption of sex linkage is consistent with the inheritance pattern of the trait :

Question:

Our eyes are the principal organs involved in seeing. Yet we are able to perform manual acts such as dressing or tying knots when our eyes are closed. Why?

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Solution:

There are many sense organs (receptors) in the body. Our pair of eyes is just one example. Although our eyes are extremely important to our perception of this world, we will not be totally and helplessly lost when we can no longer use our eyes. For instance, we are still aware of the relation of our body to the environment even if our eyes are closed. We know whether we are standing or sitting, we know where our limbs are, and we know where one part of our body is in relationship to another. Such perception without the use of our eyes in achieved with a different set of sense organs known asproprioceptors. Proprioceptorsare receptors found in muscles, tendons and joints, and are sensitive to muscle tension and stretch. They pick up impulses from the movements and positions of muscles and limbs relative to each other and relay them to the cerebellum for coordination. Impulses from the proprioceptors are extremely important in ensuring the coordinated and harmonious contraction of different muscles involved in a single movement . Without them, complicated skillful acts would not be possible. Proprioceptorshave other important functions. They help maintain the sense of balance and give the body general awareness of its environment .

Question:

Calculate the mass of the electron by combining the results of Millikan's determination of the electron charge q and J.J. Thomson's measurement of q/m.

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Solution:

The charge of the electron, q = 1.6 × 10^-19 C, and the ratio of charge to mass, q/m = 1.76 ×10^11 C/kg, are known. q/m = 1.76 × 10^11 (C/kg)but q = 1.6 × 10^-19 C Therefore, m = [{1.6 × 10^-19C} / {1. 76 × 10^11 (C/kg)}] = 9.1 × 10^-31 kg The mass of the electron is about 9.1 × 10^-31 kg.

Question:

Bone always develops by replacement of a preexisting connective tissue. When bone formation takes place in preexisting cartilage it is called endochondral ossification. Describe this method of bone formation.

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Solution:

Bones at the base of the skull, in the vertebral column, the pelvis and the limbs are called cartilage bones because they originate from cartilage. This cartilage, present in the infant is replaced with bone in later years by means of a process called endo-chondral ossification. This can best be studied in one of the long bones of an extremity. We first start with a cartilaginous shaft. This shaft begins to ossify, or harden into bone, around its midportion, due to the deposition of calcium by the cartilage cells (chondrocytes) . At the same time, blood vessels from the surrounding layer of connective tissue grow into the diaphysis. The calcified cartilage cells then die and are replaced by cells called osteoblasts, which form the bone matrix. The thinwalled blood vessels branch and grow toward either end of the cartilage model, forming capillary loops that extend into the blind ends of the cavities in the calcified cartilage. Cells are brought into the interior of the cartilage by these vessels. These cells later form bone marrow or bone matrix. In the continuing growth in length, the cartilage cells in the epiphyses become arranged in longitudinal columns. The epiphyses are, like the diaphyses before them, invaded by blood vessels and begin undergoing ossification. The expansion of these centers of ossifi-cation gradually replaces all of the epiphyseal cartilage except that which persists as the articular cartilage, and a transverse disk of longitudinal columns of cartilage between the original area of ossification and the epiphy-seal area of ossification, called the epiphyseal plate. The epiphyseal plate contains the cartilage columns whose zone of proliferation is responsible for all subsequent growth in length in long bones. Under normal conditions, the rate of multiplication of cartilage cells in this zone is in balance with their rate of replacement by bone. The epiphyseal plate, therefore, retains approximately the same thickness. Growth in length is the result of the cartilage cells continually growing away from the shaft and being replaced by bone as they recede. The net result is an increase in the length of the shaft. At the end of the growing period, proliferation of cartilage cells slows and finally ceases. The remaining cartilage becomes converted to bone and it is at this point that no further growth in length can occur. The growth in diameter of bone does not depend upon the calcification of cartilage but rather is the result of deposition of new bone by the periosteum.

Question:

Of the following pairs, which member should exhibit the largest dipole moment. Use the data from the accompanying table, (a) H-O and H-N; (b) H-F and H-Br; (c) C-O and C-S.

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Solution:

A dipole consists of a positive and a negative charge separated by some distance. Quantitatively, a dipole is described by giving its dipole moment, which is equal to the charge times the distance between the positive and negative centers. The polarity of a bond is measured as the magnitude of the moment of the dipole. Thus, to find which member has the higher dipole moment, determine which bond has the greatest polarity. The polarity of the bond is indicated by the difference in electronegativities of the atoms (i.e., the difference in their tendency to attract shared electrons in a chemical covalent bond). The greater the difference in electronegativities, the greater the bond polarity, giving it the greater dipole moment. From the table, note that in part (a) the difference in electronegativities of H-O is 3.5 - 2.1 or 1.4. For H-N, the difference is 3.0 - 2.1 or 0.9. Therefore, H-O has the largest dipole moment. Using similar calculations, one can determine the bonds with the larger dipole moments in parts b and c. (c) H-F and H-Br For H-F: 4.0 - 2.1 = 1.9 For H-Br: 2.8 - 2.1 = 0.7. Thus, H-F has the larger dipole moment. (c) C-O and C-S For C-O: 3.5 - 2.5 = 1.0 For C-S; 2.5 - 2.5 = 0. Thus, C-O has the larger dipole moment here.

Question:

Occasionally a boy is born withundescendedtestes (cryptorchidism), and he will be sterile unless they are surgically caused to descend into the scrotum. Explain the reason for this.

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Solution:

Human sperm cannot develop at the high temperature found within the body cavity. They can only develop in the testes which are at a slightly lower temperature since they are suspended from the body in the scrotum. When the weather is warm, and the body is radiating considerable heat, the skin of the scrotum loosens so that the testes hang away from the body. When the weather is cold, the skin of the testes contracts and draws the testes closer to the body, thus preventing excess loss of heat. During its de-velopment in the mother, the testes remain inside the body cavity of the male fetus. At birth, the testes descend into the scrotum. If there is a deficiency of male hormone or if the canal is too narrow, the testes will not descend. The result of this would be sterility, the sperm being unable to develop at the higher tempe-rature in the body cavity.

Question:

The volume of oil contained in a certain hydraulic press is 5 ft^3. Find the decrease in volume of the oil when sub-jected to a pressure of 2000 lb/in^2. The compressibility of the oil is 20 × 10^-6 per atm.

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Solution:

The volume decreases by 20 parts per million for a pressure increase of one atm. Since 2000 lb/in^2 = 136 atm., the volume decrease is 136 × 20 = 2720 parts per million. Since the original volume is 5 ft^3, the actual decrease is [(2720)/(1,000,000)] × 5 ft^3 = 0.0136 ft^3 = 23.5 in^3. or, the change in volume of the oil is proportional to the original volume and the pressure exerted on the oil. ∆V = - k V_0p The constant of proportionality, k, equals the compressibility of the oil. Therefore, the change in volume can be found using the above equation. ∆V = - kV_0p = - 20 × 10^-6 atm^-1 × 5 ft^3 × 136atm = - 0.0136 ft^3.

Question:

What advantages domulticellularorganisms haveover single-celledorganisms?

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Solution:

Single-celled organisms represent one of the great success stories ofevolution. They probably comprise more than half the total mass of livingorganisms, and have successfully colonized even the harshest environ-ments. Biochemically, many unicellular organisms, such as bacteria, are far more versatile than man being able to synthesize virtually everythingthey need from a few simple nutrients. With the development of eukaryotic cells, certain evolutionary breakthroughsoccurred. Eukaryotic cells are not only capable of being muchlarger than prokaryotes, they have the ability to aggregate into multicellularfunctional units. The cells ofmulticellularorganisms are specializedfor a variety of functions, and interact in a way that make the organismmore than the sum of its parts. Multicellularorganisms employ a high order of complexity. The constituentcells are specialized for the division of labor. The cells that makeup the body of a man are not all alike; each is specialized to carry outcertain functions. For example, red blood cells carry oxygen from the lungsto the various parts of the body while nerve cells are involved in the transmissionof impulses. This specialization allows each cell to function moreefficiently at its own task, and also allows the organism as a whole to functionmore efficiently. Also, injury or death to a portion of the organism doesnot nec-essarily inhibit the functioning and survival of the individual as awhole. Multicellularorganisms are better adapted to sur-vive in environmentsthat are totally inaccessible to unicellular forms. This is most strikingin the adapta-tion ofmulticellularorganisms to land. Whereasuni- cellularanimals and plants, such as theprotozoansand the one-celled algae, survive primarily in a watery en-vironment, the higher,multicellular organsisms, such as the mammals and angiosperms, are predominantly land-dwellers.Mutlicellularityalso carries the potential for diversity. Millions of different shapes, specialized organ systems, and patterns of behaviorare found inmulti-cellularorgansisms. This diversity greatly increasesthe kinds of environments that organisms are able to exploit.

Question:

How many kilograms of O_2 are contained in a tank whose volume is 2 ft^3 when the gauge pressure is 2000 lb/in^2 and the temperature is 27\textdegreeC?

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Solution:

Assume the ideal gas laws to hold. The molecular weight of O_2 is 32 gm/mole. The volume V is V = 2 ft^3 = 2 ft^3 × 28.3litres/ft^3 = 56.6 liters. Gauge pressure is the pressure above air pressure. Air pressure is 14.7 lb/in^2. Therefore P_abs = 2000 lb/in^2 + 14.7 lb/in^2 = 2015 lb/in^2 = [2015 lb/in^2]/[14.7atm/(lb/in^2)] = 137atm The temperature must be expressed in Kelvin degrees in order for us to be able to use the ideal gas laws. T = t + 273 = 300\textdegreeK. Hence the number of moles of gas in the tank is then n =P_absV/RT = (137atm× 56.6 liters)/(0.082liter\bulletatm/mole\bullet\textdegreeK × 300\textdegreeK) = 315 moles. The mass of this amount of gas is m = (315 moles)(32 gm/mole) = 10,100gm = 10.1kgm Note that we had to convert P and V to the appropriate units in order for them to be consistent with the units of the ideal gas law.

Question:

What is a sigma bond? What is a pi bond? what are their basic differences?

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Solution:

A molecular orbital that is symmetrical around the line passing through two nuclei is called a sigma (\sigma) orbital. When the electron density \sigma in this orbital is con-centrated in the bonding region between the two nuclei, the bond is called a sigma bond. The covalent bonds in H_2 and HF are sigma bonds. In the formation of the bonding orbital between two fluorine atoms, the 2p orbitals overlap in a head-to-head fashion to form a sigma bond. However, there is a second way in which half-filled p orbitals of two different atoms may overlap to form a bonding orbital. If the two p orbitals are situated perpendicular to the line passing through the two nuclei, then the lobes of p orbitals will overlap intensively sideways to form an electron cloud that lies above and below the two nuclei. The bond resulting from this sideways or lateral overlap is called a pi (\pi) bond; the bonding orbital is called a pi orbital. It differs from a sigma orbital in that it is not symmetrical about a line joining the two nuclei. Pi bonds are present in molecules having two atoms connected by a double or triple bond. The sigma bond has greater orbital overlap and is usually the stronger bond? a pi bond, with less overlap, is generally weaker.

Question:

Explain what you understand by the DECLARE statement andlist how it is used for Numeric and Non numeric Variables.

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Solution:

The DECLARE statement, abbreviated as DCL, in PL/I allows for properstorage allocation of programmers' variable. It lets the compiler knowin what form the variable will be stored by declaring its name followedby a list of its attri-butes . Numerical variables: PL/I has a wide variety of DCL state-ment forms for declaringnumeric variables. a) Fixed Decimal Point variables: -Thegeneral form is as follows: DECLARE name scale attribute base attribute (precision attribute); For example, DECLARE Y FIXED DECIMAL (6, 3); The above could be followed by a statement as follows: Y = 12 \textbullet 4; Thus, the stored result will be equivalent to 012\textbullet400. Zeros are added on both sides of 12\bullet4 because Y was de-clared to have6 digits in all, but with 3 digits after the decimal point. Similarly, if the variablehad a lesser to-tal number of digits declared, compared to the precisionand the value of the variable, a truncation of the number takes place. This truncation must be regarded as an error. As an example DECLARE A FIXED DECIMAL (5, 2); GET DATA (A); PUT DATA (A). If A = 5218\textbullet79, in view of the precision of (5, 2) declared above, A in storage = 218\textbullet79. Hence, the PUT DATA statement will lead to a print out of A = 218\textbullet79. A truncation of the left-most bit exceeding the precision attribute, has taken place. Similarly if A = 6179, then,Ain storage will be 179\textbullet00. In this case, thePUT DATA (A); statement will lead to a print-out of A=179\textbullet00; (Totally thereshould be five digits, with two digits after the decimal point.). b) Float decimal point variables: -Thebasic declara-tion reads as follows: DECLARE name FLOAT DECIMAL (W) ; The precision is specified by W, i.e., the total number of digits to be accomodatedin the floating point number. For example, DECLARE YFLOATDECIMAL (8); The above could be followed up with a statement as follows: Y = 112\textbullet4357; Hence the stored result will be as follows: 0\textbullet11243570E + 3. If the value of the variable is specified using more digits than the declarationprovides PL/I will truncate, just as for fixed point variables, as illustratedbelow: For example DCL A FLOAT DECIMAL (5); If A = 317\textbullet04 at input, then A = 3\textbullet1704E + 02 at output. If A = 0\textbullet0278 at input, then A = 2\textbullet7800E - 02 at output. If A = 8241\textbullet356 at input, then A = 8\textbullet2413E + 03 at output. c) Similar to declarations for fixed and floating point decimal variables, there are declarations for Binary varia-bles. For example, DECLARE name FIXED BINARY (w, d); e.g., DCL C BINARY FIXED (8, 3); PL/I requires a 'B' immediately following a binary con-stant, to distinguishit from decimal constants, because otherwise, 101\textbullet10 can be a decimalconstant as well as a bi-nary constant. Now consider the following: DCL C BINARY FIXED (8, 3); The above willaccomodatea value of C as follows, for ex-ample, C = 10011\textbullet000B;( =19 decimal). But, if B is not present, the PL/I compiler takes it to be 10011\textbullet000 decimal andwill then convert it to binary accord-ing to normal practice, producing a value10011100011011. (As can be verified by continuously dividing 10011 by 2, thus converting 10011 to a binary form).For the floating point representationof BINARY numbers we can have a declara-tion of the type illustratedbelow: DECLARE name FLOAT BINARY (W); d) Non-Numeric Variables: -Theallocation of stor-age for non- numericvariables must always be explicit. i) CHARACTER STRINGS: -Forcharacter strings, the declaration isas follows: DECLARE name CHARACTER (length); 'length' is a decimal integer specifying the number of char-acters to be accomodated. Character strings have to be shown between single quotationmarks for instance DECLARE X CHARACTER (5) X = 'PARIS' As a result of the above declaration, the compiler saves5 storage locationsfor X, & the five character 'PARIS' will be placed in X. When a stringlength differs from the stor-age area made available by the declaration, PL/I either pads or truncates the string, as necessary. For example, consid-er the following Declare statement: DECLARE B CHARACTER (7); Now, if B='VANbGOGH', then, the stored value for B =VANbGOGIf B = 'KUGEL', then, the stored value for B =KUGELbb. ii) BIT STRINGS: -Whena variable is declared with a statement as follows, viz., DECLARE name BIT (length); then, that variable is treated as a bit string, consisting of only ones and zeros. Bit strings are handled in the same way as character strings. Multiple Declarations: -Itis not necessary to allo-cate storage for- eachvariable in separate Declare statements. For example, consider the followingfour separate declare statements: DCL A FLOAT (6); DCL B FIXED BINARY (18); DCL C CHARACTER (12); DCL X FIXED (7,2); The above can be combined together as illustrated below: DCL AFLOAT(6), B FIXED(18), C CHARACTER(12), XFIXED(7,2); Factored Declarations: -Whenstorage is to be allocated for severalvariables that have a common set of attributes, it is possible to condensethem, as shown below: DCL AFIXED(5, 2), B CHAR(5), C FIXED(5, 2), D FLOAT(7); can be writtenas follows: DCL (A, C)FIXED(5, 2), B CHAR(5), D FLOAT(7); In the above, theFIXED(5, 2) attribute of both A and C has been factored out, leading to a condensed form of the de-clare statement. Note that no more factoring is possible now in the above statement.

Question:

How are the legs of a horse adapted for running? Compare the method of walking of a bear, a cat and a deer.

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Solution:

Animals differ with respect to which part of the foot they put on the ground when walking and running. Men and bears walk on the entire sole of the foot. This method of locomotion is known asplatingrade. Animals such as dogs and cats, to increase their effective limb length and thus their running speed, have become adapted to running on their digits, or fingers. This type of locomotion is known asdigitigrade. Speed is increased still further in the horses, deer, and cattle, which walk on their hoofs, or nails. In these animals the lower limb bones are lengthened, raising the wrist and ankle farther off the ground. This type of locomotion is known asunguligrade. In the case of the horse, the leg is supported by one digit, terminating in a hoof, whereas in cattle, two digits are used for support, terminating in two hoofs. Those ungulates, such as the horse, which walk on one digit are known as thePerisso-dactyla. Those walking on two digits are known as theArtiodactyla.

Question:

Write a CSMP111 program for the following inventory control system: ẏ = u - V ẋ = V - S V = (Y/T_1) u = S + K(I - X) where X - Current inventory level Y - Outstanding level of orders placed with the suppliers u - Rate of ordering from suppliers V - Delivery rate from suppliers S - Sales rate I - Planned inventory level T_1- Average delivery time (= 3 days) Let the initial conditions be I= X = 0 Y = 12 S = 4 items/day.

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Solution:

The control of inventory requires a feedback system. A retailer tries to keep inventory at a level where he maintains what he views as a reasonable balance between the cost of holding goods in inventory and the penalty of losing sales if the inventory should become empty. A simple model of the inventory control system would need two levels and three rates defined as follows: X - Current inventory level Y - Outstanding level of orders placed with the supplier u - Rate of ordering from supplier V - Rate of delivery by supplier S - Rate of sales Also, assume that planned inventory level and average lead timeI andT_1respectively, are constant. A systems diagram for the above levels diagram for the above levels is: The rate of change of the level of orders is Ẏ = u - V while the time derivative of X is Ẋ - V - S. The rate of delivery is inversely proportional to the outstanding level of orders placed with the suppliers. The order rate is u = S + K(I - X) where K is the ordering constant. The systems diagram acts like a flowchart for the CSMPIII program. Assume that time is measured in days and let the initial conditions be, at t = 0: I= X = 20 Y= 12 S = 4 items/day The program looks as follows: TITLEINVENTORY CONTROL SYSTEM \textasteriskcentered PARAMK=(1.0, 0.5, 0.25, 0.125, 0.0625) \textasteriskcentered S=4.0+2.0\textasteriskcenteredSTEP(4.0) V=Y/T_1 u=S+K\textasteriskcentered(I-X) YDOT=U-V Y=INTGRL(YO,YDOT) XDOT=V-S X=INTGRL(XO,XDOT) \textasteriskcentered CONST I=20.0, YO=12.0, XO=20.0, T1=3. 0 \textasteriskcentered TIMER DELT=0.1 FINTIM=50.0 PROEL=1.0 PRINT u, v, X, Y END STOP

Question:

Write the micro-operations for the execute cycles of the Memory reference instructions of Figure 2 using these registers and control signals. M - memory PC - program counter MAR - memory address register MBR - memory buffer register OPR - op-code register F, R - cycle identification registers E - carry bit register The cycle code is shown in Figure 1. Cycle ID Registers FR Cyclecode Cycledescription 00 C_0 Fetch cycle 01 C_1 Indirect cycle 10 C_2 Execute cycle 11 C_3 Interrupt cycle Fig. 1 Each cycle consists of four steps t_0, t_1, t_2, t_3. Instruction Symbol OP-Code Control Signal Effective Address Memory Word Symbolic Designation AND 000 q_0 M M AC\leftarrow[AC]\wedgeM \wedge ADD 001 q_1 m M EAC\leftarrow[AC]+M LDA 010 q_2 m M AC\leftarrow[M] STA 011 q_3 m M M\leftarrow[AC] BUN 100 q_4 m - PC\leftarrowm BSA 101 q_5 m M M\leftarrow[PC] PC+m+1 ISZ 110 q_6 m M M\leftarrowM+1, if M+1=0, then PC\leftarrow[PC]+1 Fig.2

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Solution:

At the beginning of the execute cycle the registers contain the following information: [PC] = The address of the instruction about to be exe-cuted; incremented by one. [MBR]_4-15 = The address of the operand (effective address). [OPR] = The op-code part of the instruction format (this was decoded into one of eight control signals q_0 - q_7) [F]= 1 }decoded to excite c_2 [R] = 0 } A step is specified if the control signal expression to the left of the colon is true (logic 1). AND:This instruction logically "ands" the contents of the accumulator and the operand. The result is stored back in the accumulator. Note:[] means "contents of"; < > means "addressed by." q_0c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_0c_2t_1 :MBR \leftarrow [M]\mid q_0c_2t_2 :AC \leftarrow [AC] \wedge [MBR]AND with AC, store result in AC q_0c_2t_3 :F\leftarrow0Go to Fetch cycle (c_0 is excited) ADD:This instruction adds the contents of the AC and the operand. The result is stored in the AC. If there is a carry, it is stored in E. q_1c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_1c_2t_1 :MBR \leftarrow [M]\mid q_1c_2t_2 :EAC \leftarrow [AC] + [MBR]Add to AC, store result in AC and E q_1c_2t_3 :F\leftarrow0Go to Fetch cycle LDA:This instruction loads the operand into the AC. q_2c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand, clear AC q_2c_2t_1 :MBR \leftarrow [M]; AC\leftarrow0\mid q_2c_2t_2 :AC \leftarrow [AC] + [MBR]Add to AC q_2c_2t_3 :F\leftarrow0Go to Fetch cycle STA:This instruction stores the contents of the AC into the memory location specified by [MBR]_4-15. q_3c_2t_0 :MAR \leftarrow [MBR]_4-15Store address in MAR q_3c_2t_1 :MBR \leftarrow [AC]Store [AC] in MBR q_3c_2t_2 :M\leftarrow[MBR]Load into memory q_3c_2t_3 :F\leftarrow0Go to Fetch cycle BUN(branch unconditionally):This instruction is similar to the JUMP or GO TO statement in FORTRAN. The address of the next instruction is specified by the address portion of the MBR. The execute cycle is: q_4c_2t_0 :PC \leftarrow [MBR]_4-15Transfer next address to PC q_4c_2t_1 :Do nothing q_4c_2t_2 :Do nothing q_4c_2t_3 :F\leftarrow0Go to Fetch cycle BSA(branch and save return address):This instruction stores the contents of the PC in the memory location specified by the effective address and then control branches to the next higher location of the effective address. q_5c_2t_0 :MAR \leftarrow [MBR]_4-15\mid MBR_4-15 \leftarrow [PC]\midLoad effective address into PC PC\leftarrow[MBR]4-15\mid q_5c_2t_1 :[M]\leftarrowMBRLoad old PC into effective address q_5c_2t_2 :PC\leftarrow[PC]+1Increment PC q_5c_2t_3 :F\leftarrow0Go to Fetch cycle The BUN and BSA instructions are very useful when branch-ing to and from subroutines. ISZ(increment and skip-if-zero):This instruction incre-ments the operand and then tests to see if the operand is zero. If the operand is not zero then the program flows normally. If the operand is zero then the next' instruction is skipped and control resumes at the instruction after the skipped instruction. q_6c_2t_0 :MAR \leftarrow [MBR]_4-15\midFetch operand q_6c_2t_1 :MBR \leftarrow [M]\mid q_6c_2t_2 :MAR\leftarrow[MBR]+1Increment operand q_6c_2t_3 :M\leftarrow[MAR]\midStore new operand; if [MBR] = 0 then PC=PC+1\midIncrement PC if \midoperand is zero; F\leftarrow0\midGo to Fetch cycle.

Question:

A typical bacterial cell has a cell wall and plasma membrane. If the cell wall is removed by treatment withlysozyme, an enzyme that selectively dissolves the cell wall material, will the bacteria live? Explain. What is the biochemicaldifference between the cell walls of bacteria, and plants?

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Solution:

The main function of the bacterial cell wall is to provide a rigid framework or casing which supports and protects the bacterial cell from osmotic disruption. Most bacteria live in a medium which is hypotonic relative to the bacterial protoplasm: i.e., the bacterial protoplasm is more concentrated than the medium, and water tends to enter the cell. If intact, the cell wall provides a rigid casing, which prevents the cell from swelling and bursting. If the cell wall were destroyed, water would enter the cell and cause osmoticlysis(bursting). If the bacterium is placed in a medium which has the same osmotic pressure as the bacterial cell contents, osmotic disruption would not occur when the cell wall was dissolved. A bacterium devoid of its cell wall is called a protoplast. Pro-toplasts can live in an isotonic medium, (medium having equal osmotic pressure as the bacterial protoplast).If the bacterium was placed in a hypertonic medium, waterwould leave the cell andplasmolysis(shrinkage) would ensue. The cell wall varies in thickness from 100 to 250 \AA (an angstrom, \AA, is 10^\rule{1em}{1pt}10 meter) and may account for as much as 40% of the dry weight ofthe cell. While the cell wall ineucaryotesis composed of cellulose, in bacteria , the cell wall is composed ofinsoluablepep-tidoglycan. Peptidoglycan consists of sugars (N-aceylglucosamineand N- acetylmuramic acid) and amino acids, includingdiaminopinelicacid, an amino acid unique to bacteria. In gram-negative bacteria, the peptidoglycan constitutes a much smaller fraction of the wall component than it does in gram-positive bacteria. The higher lipid content in the cell walls of gram-negative bacteria accounts for the differences in gram- staining . The plasma membrane, a thin covering immediately beneath the cell wall, is too fragile to provide the support needed by the cell. Instead, this semipermeablemembrane controls the passage of nutrients and waste products into and out of the cell.

Question:

A ship is being towed by a tug by means of a steel wire. If the drag on the ship is equivalent to 2 × 10^6 lb, and if the breaking strain of the wire is 0.025, what is the smallest permissible diameter of wire that may be used?

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Solution:

If the ship is being towed, it eventually settles down at a steady speed so that the force being exerted by the tow wire is equal and opposite to the drag on the ship. Now, the formula for Young's modulus is Y = (stress)/(strain) = (F_n/A)/(∆1/1_0) where the symbols have their usual significance. Thus A = F_n1_0/Y∆1, where F_n the normal force on the wire, is equal to the force exerted by the wire on the ship, which balances the drag force. (See the fig.) Further, the strain ∆1/1_0 must be less than 0.025, the strain at which the wire breaks. A table of Young's modulus for different materials shows that for steel, Y = 29 × 10^6 lb\bulletin^-2 Hence A \geq [(2 × 10^6 lb)/(29 × 10^6lb\bulletin^-2 × 0.025)] The cross sectional area of the wire (A) is \piR^2, where R is the radius of the wire. Since R = D/2 where D is the diameter of the wire, then A = \pi[(D^2)/4] and [(\piD^2)/4] \geq 80/29 in^2. \therefore D \geq \surd(320/29\pi) in. = 1.87 in., Hence, 1.87 in. is the smallest possible diameter for the tow wire.

Question:

Write a PL/I procedure that reads an integer N > 1 and calculatesthe sequence Y =[ ^N\sum_i= 1{i(N -i)^2} ] / N The program should output N and the corresponding Y on eachline.

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Solution:

We will set up a DO loop to calculate the terms as in-dicated by the summationsymbol, (\sum). Program control passes to the end of the program whenthe value of N is less than or equal to zero. In this example, we will input N = 2, N = 10, and N = 25. SEQ: PROC OPTIONS (MAIN); DCL Y FIXED (10,3); LOOP: GET DATA (N); / \textasteriskcentered TEST FOR END OF DATA \textasteriskcentered / IF N < = 0 THEN GO TO FINISH; / \textasteriskcentered INITIALIZE Y AND COMPUTE THE SUM \textasteriskcentered / Y = 0; SUMLOOP: DO J = 1 TO N; Y = Y + J \textasteriskcentered (N - J) \textasteriskcentered (N - J); END SUMLOOP; Y = Y/N; / \textasteriskcentered PRINT ANSWERS \textasteriskcentered / PUT SKIP DATA (N,Y) ; GO TO LOOP; FINISH: END SEQ; Sample input: N = 2; N = 10; N = 25; N = 0; Sample output: N =2Y =0.500; N =10Y =82.500; N =25Y =1300.000;

Question:

A 1,000-horsepower steam engine produces about 750,000 W of mechanical power. If this engine operates a generator with an efficiency of 95 per cent, what is the electrical power produced ?

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Solution:

Efficiency \epsilon is defined by \epsilon = (P_output/P_input)×100% The steam engine delivers 750,000 W to the generator. The generator operates at an efficiency of 95% whereP_outputis its Power output. For the generator , {(P_output)/(7.5 × 10^5 W)} = 95% = 9.5 × 10^\rule{1em}{1pt}1 The electric power delivered in this energy-transformation process is, P = (7.5 × 10^5 W) (9.5 × 10^\rule{1em}{1pt}1) = 7.125 × 10^5 W The 5 per cent difference (37,500 W) appears as heat that must be dissipated by either passing air through cooling fins or circulating water.

Question:

A Celsius thermometer indicates a temperature of 36.6\textdegreeC. What would a Fahrenheit thermometer read atthat temperature ?

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Solution:

The relationship between the Celsius and Fahrenheit scales can be derived from a knowledge of their cor-responding values at the freezing and boiling points of water. These are 0\textdegreeC and 32\textdegreeF for freezing and 100\textdegreeC and 212\textdegreeF for boiling. The temperature change between the two points is equivalent for the two scales and a temperature difference of 100 Celsius degrees equals 180 Fahrenheit degrees. Therefore one Celsius degree is (9/5) as large as one Fahrenheit degree. We can then say \textdegreeF = (9/5)\textdegreeC + B where B is a constant. To find it, substitute the values known for the freezing point of water: 32\textdegree = (9/5) × 0\textdegree + B = B. We therefore have \textdegreeF = (9/5)\textdegreeC + 32\textdegree. For a Celsius temperature of 36.6\textdegree, the Fahrenheit temperature is F = (9/5) × 36.6\textdegree + 32\textdegree = 65.9\textdegree + 32\textdegree = 97.9\textdegree.

Question:

Write a PL/I program to integrate a general function D(X) overA \leq X \leq B. Illustrate for D(X) = (1 - X)-2 .

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Solution:

We slice the interval [a, b] into n equal parts. The size of the slices ish = (b - a)/n . The area under the curve is given ac-cording to Simpson's ruleby S = (h/3) [f (a) + 4f (a+h) +2f(a+2h) + 4f(a+3h) + 2f(a+4h)+...+ f(b)] /\textasteriskcentered SIMPSON RULE INTEGRATION \textasteriskcentered/ RPT: GET LIST (A,B,N); H = (B - A)/N; X = A; S = 0; DO I = 1 TO N/2; S = S + H/3\textasteriskcentered (D(X) + 4\textasteriskcenteredD(X+H) + D(X+2\textasteriskcenteredH)); X = X + 2\textasteriskcenteredH; END; PUT SKIP EDIT (A,B,N,S) (2F(10,4)), F(10), F(10,4)); GO TO RPT; D:PROCEDURE (X); RETURN (1/1 - X)\textasteriskcentered\textasteriskcentered2); END D;

Question:

Why did Mendel succeed in discovering the principlesof transmissions of hereditary traits, whereas others who had conducted such investigations had failed?

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Solution:

Mendelssuccess was a combination of good experimental technique and luck. He chose the garden pea for his studies because it existed in a number of clearly defined varieties; the different phenotypes were thus evident and distinguishable. In addition, he only studied the transmission of one trait at a time, and crossed plants differing with respect to just that one trait. Previous investigators had studied entire organisms at a time, and had crossed individuals differing in many traits. Their results were a conglomeration of mixed traits and interacting factors, making it impossible to figure out what was happening. Mendel was also careful to begin with pure line plants developed through many generations of natural self - fertilization. However, luck was involved in that the traits which Mendel randomly chose were coded for by genes that were located on different chromosomes, and thus assorted independently . If he had chosen traits whose genes were linked, his ratios would not have been understandable in terms of what he knew then.

Question:

With what speed is an electron moving if its kinetic energy is 1% larger than the value that would be calculated if relativistic effects were not present? What is the value of the kinetic energy of the electron?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0947.htm

Solution:

The relativistic expression for the kinetic energy of a particle of rest mass m_0 and mass m traveling with speed v is T = E - m_0c^2= {m_0c^2 / (\surd1 - v^2c^2)}- m_0c^2 m_0c^2[{1 / \surd(1 - (v2/ c^2)} - 1 ] = m_0c^2 [ {1 - (v2/ c^2)}^-1/2 - 1] If the error involved in using the classical formula for T is only 1%, the particle is traveling at a speed such that v^2/ c2<< 1 Hence, using a Taylor expansion for {1 - (v2/ c^2)}^-1/2 = (1 + v2/2c2+ 3v^4/ 8c^4 + .....) T = m_0c^2 {(1/2) ( v2/ c^2) (3/8) ( v4/ c^4) + .......} = [(1/2)m_0v2{1 + (3/4) ( v2/ c^2)}] , when we ignore terms of higher order than those in v^2 / c^2 . But (1/2)m^0v^2is the nonrelativistic expression for the kinetic energy of the particle, and thus, for the electron in the problem, (1/2)m_0v^2 {1 + (3/4) ( v2/ c^2)}] = (1/2)m_0v2+ (1%) [(1/2)m_0v^2] = 1.01[(1/2)m_0v^2] or 1 + ( 3v2/ 4c^2) = 1.01 ( 3v2/ 4c2) = .01 v^2 = .04c^2 / 3 = 4c^2 / 300 v = 0.1155c. T = (101/100) × (1/2)m_0v2 =(101/100) × (1/2) × 9.108 × 10^-31 kg × (4 / 300) × 2.998^2 × 10^16 m^2 \bullet s^2 = 5.512 × 10^-16 J Since 1 eV = 1.602 × 10^-19 J T = (5.512 × 10-16J) / (1.602 × 10^-19 J / eV) = 3442 eV .

Question:

Write a BASIC program to calculate the hypotenuse of a right triangle. Your program should read in values of the two sides A and B, and calculate the hypotenuse C according to the Pythagorean theorem. Illustrate the results when A = 3, B = 4; A = 6, B = 8; A = B = 1.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0228.htm

Solution:

This program illustrates the use of Library Functions in BASIC. It executes function SQR (Square Root) for determining the hypotenuse. The hypotenuse C is calculated from the formula C = \surd(A^2 + B^2). The program follows: 1\O REM CALCULATES THE HYPOTENUSE 2\O REM OF A RIGHT TRIANGLE 3\O PRINT "SIDE A", "SIDE B", "HYPOTENUSE C" 4\O READ A, B 5\O LET C = SQR (A _\uparrow 2 + B _\uparrow 2) 6\O PRINT A,B,C 7\O GO TO 4\O 8\O DATA 3, 4, 6, 8, 1, 1 9\O END OUTPUT: SIDE A SIDE B HYPOTENUSE C 3 4 5 6 8 1\O 1 1 1.41...

Question:

Name the major organs of the human digestive tract and explain their functions.

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0422.htm

Solution:

The human digestive system begins at the oral cavity. The teeth break up food by mechanical means, in-creasing the substrates' surface area available to the action of digestive enzymes. There are four types of teeth. The chisel-shaped incisors are used for biting, while the pointed canines function in tearing, and the flattened, ridged premolars and molars are used for grinding and crushing food. In addition to tasting, the tongue mani-pulates food and forms it into a semi-spherical ball (bolus) with the aid of saliva. The salivary glands consist of three pairs of glands. The parotid glands, located in the cheek in front of the ear, produce only watery saliva which dissolves dry foods. The submaxillary and sublingual glands, located at the base of the jaw and under the tongue, produce watery and mucous saliva which coagulates food particles and also lubricates the throat for the passage of the bolus. Saliva also contains amylases which break down starches. The tongue pushes the bolus into the pharynx which is the cavity where the esophagus and trachea (windpipe) meet (see Fig. 2). The larynx is raised against the epiglottis and the glottis is closed, preventing food from passing into the trachea. The act of swallowing initiates the movement of food down through a tube con-necting the mouth to the stomach called the esophagus. Once inside the esophagus, the food is moved by involuntary peristaltic waves towards the stomach. At the junc-tion of the stomach and the esophagus, is a special ring of muscles called the lower esophageal sphincter. These muscles are normally contracted, but when food from the esophagus reaches the sphincter, it opens reflexively and allows food to enter the stomach. The stomach is a thick muscular sac positioned on the left side of the body just beneath the ribs. The upper region of the stomach, closest to the heart, is called the cardiac region. Below that is the crescent part of the sac called the fundus. The pyloric region is tubular and connects the stomach to the small intestines. The wall of the stomach is made up of three thick layers of muscle. One layer is composed of longitudinal, one of circular, and the other of oblique (diagonal) fibers. The powerful contractions of these muscles break up the food, mix it with gastric juice and move it down the tract. Gastric juice is a mixture of hydrochloric acid and enzymes that further digest the food. Gastric juice and mucus are se-creted by the small gastric glands in the lining of the stomach. The mucus helps protect the stomach from its own digestive enzymes and acid. The partially digested food, called chyme, is pushed through the pyloric sphincter in-to the small intestine. The pyloric sphincter is similar in structure and function to the lower esophageal sphincter. The first part of the small intestine, called the duodenum, is held in a fixed position. The rest of the intestine is held loosely in place by a thin membrane called mesentery which is attached to the back of the body wall. In the duodenum, bile from the liver that has been stored in the gallbladder is mixed with pancreatic juice from the pancreas. The secretions of the pancreas and glandular cells of the intestinal tract contain enzymes that finish digesting the food. As digestion continues in the lower small intestine, muscular contractions mix the food and move it along. Small finger-like protrusions, called villi, line the small intestine facing the lumen. They greatly increase the intestinal surface area and it is through the villi that most of the nutrient absorption takes place. The small intestine joins the large intestine (colon) at the cecum. The cecum is a blind sac that has the appendix protruding from one side. Neither the appendix nor the cecum are functional. At the junction of the cecum and the lower small intestine (ileum.) is a sphincter called the ileocecal valve. Undigested food passes through this valve into the large intestine. The large intestine has the function of removing water from the unabsorbed material. At times there is an excretion of certain calcium and iron salts when their concentrations in the blood are too high. Large numbers of bacteria exist in the colon. Their function in man is not fully understood, but some can synthesize vitamin K which is of great importance to man's blood clotting mechanism. The last section of the colon stores feces until it is excreted through the anal sphincter.

Question:

Discuss the adaptations for land life evident in the bryophytes.

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0176.htm

Solution:

The bryophytes are considered one of the early invaders of the land. In order to survive on land, a plant must have certain structures which will enable it to exploit a terrestrial environment for food, water, gases and to afford it protection against environ-mental hazards. Although the bryophytes are far from being truly terrestrial and retain a strong dependency on a moist surrounding, they have accumulated important adaptations enabling them to live successfully on land. Because land plants are removed from a water environment, they face the potential danger of dehydration due to evaporation. To reduce water loss, most bryophytes have an epidermis. In the mosses, it is thickened and waxy, forming a cuticle. Unlike aquatic plants, which obtain and excrete gases dissolved in water, land plants require an efficient means to exchange gases with the atmosphere. The epidermis of the bryophytes is provided with numerous pores for the diffusion of carbon dioxide and oxygen. Diffusion through pores is a much faster and more efficient process for gaseous exchange than simple diffusion through membranes. Besides gases, land plants need to obtain water, which may be a limiting factor in a terrestrial environ-ment. Most bryophytes absorb water and minerals directly and rapidly through their leaves and plant axis, which may have a central strand of thin-walled conducting cells. Generally, the plant is attached to the substrate by means of elongated single cells or filaments of cells called rhizoids. In those mosses having a cuticle on their leaves, the rhizoids may function to some extent in water absorption. In most bryophytes, however, the rhizoids are not true roots and serve only in anchoring the plant. In addition to structures, the bryophytes have evolved a life cycle in which the developing zygote is protected within the female gametophyte. Here the zygote obtains food and water from the surrounding gametophytic tissue and is protected from drought and other physical hazards present in a terrestrial environment. The bryophyte sperm cell is flagellated and requires a moist medium for its transport. Fertilization, however, can occur after a rain or in heavy dew. The sperm cell swims, in a film of moisture, to the female gametophyte (archegonia) wheregametic union occurs. Because the bryophytes have acquired rhizoids, cutinized epidermis, a porous surface, and a reproductive style in which the embryo is protected within the female gametophyte, they are able to succeed in a terrestrial environment.

Question:

Define the following symbols as they are used in flowcharting:

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0166.htm

Solution:

These symbols have very specific meanings for flowcharting procedures. The following explanations come from the American National Standards Institute: Input or output using a line printer or a terminal Two or more actions may follow from this point, based upon the response to the question raised Input or output using the standard 80- column card Commentary on a particular section of the algorithm, used to explain ambiguous actions Usually with a number inside the circle, it connects flowlines when going to or coming from another spot on the same page or another page

Question:

In peas, yellow color is dominant to green. What will be the colors of the offspring of the following crosses? a) homozygous yellow × green b) heterozygous yellow × green c) heterozygous yellow × homozygous yellow d) heteroygous yellow × heterozygous yellow

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0649.htm

Solution:

This problem offers practice in genetic crosses. First we must determine the alphabetic repre-sentation of the genes involved. Let Y be the gene for yellow color in peas and y be the gene for green color in peas. For each cross, we write down the genotypes of the parents, the possible types of gametes, and de-termine the F_1 offspring. If it helps we may use the Punnett Square to obtain the first generation. In the crosses then: All the progeny of F_1 are yellow because Y is dominant. Half the progeny is yellow (Yy) and half is green (yy). All the progeny will be phenotypically alike (yellow). Genotypically, half will be homozygous and half will be heterozygous. A ratio is obtained of 3 yellow progeny (1 YY + 2 Yy) to 1 green (yy).

Question:

If the gold foil in a Rutherford type experiment is 1/10,000 in. thick, what is a probable minimum number of gold atoms an alpha particle passed through before hitting the fluorescent screen? 1 in = 2.54 cm, 1 cm = 10^8 \AA. The radius of a gold atom is 1.5 \AA.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0665.htm

Solution:

In Rutherford's experiment, a beam of alpha particles is aimed at a gold foil of given thickness. Most of the alpha particles pass through undeflected. The deflections that occur suggest that the alpha particles collided with a solid body. This body is the gold nucleus. This experiment elucidated some important properties of the nucleus. One of these properties is the positive charge of the nucleus. If one knows the thickness of the foil and the atomic radius of the particles that compose the foil, one can calculate the number of particles which make up the width of the foil. If one assumes cubic packing in the foil, the sum of the lengths of the diameters of the gold atoms will be equal to the thickness of the foil. Proceed as follows: The radius, and therefore the diameter, of the gold atom is given in terms of angstroms. The thickness of the foil must be converted to the same units. The thickness = (10^-4 in) (2.54 cm/in) (1 × 10^8 \AA/cm) = 2.54 × 10^4 \AA. Diameter of gold nucleus = 2 × Radius = 2 × 1.5 \AA= 3 \AA. The minimum number of gold atoms passed by alpha particles is equal to [(2.54 × 10^4 \AA)/(3 \AA)] = 8.466 × 10^3

Question:

It is desired to produce penetrating x-rays of wavelengths about 0.20 A. What is the minimum voltage at which the x- ray tube can be operated?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1034.htm

Solution:

An x-ray tube accelerates electrons through a potential difference. In order to see how this occurs, let us refer to the figure. The net force on an electron emitted from plate A is, by definition of electric field F^\ding{217} = eE^\ding{217} where e is the signed charge of an electron, and E^\ding{217} is the electric field intensity. Hence, the work done by this force (since it is the resultant force on the electron) is equal to the change in kinetic energy of the electron (1/2)m_e v2_b -(1/2)m_e v2_a = ^b\int_a F^\ding{217} \bullet dr^\ding{217}. Assuming that the initial velocity, v_a, of the electron is zero, (1/2)m_e v2_b = ^b\int_a eE^\ding{217} \bullet dr^\ding{217} (1/2)m_e v2_b = e ^b\int_a E^\ding{217} \bullet dr^\ding{217}. Replacing the signed number e by - \verte\vert (since e < 0), we may write (1/2)m_e v2_b = - \verte\vert ^b\int_a E^\ding{217} \bullet dr^\ding{217} = \verte\vert(V_b - V_a) Hence (1/2)m_e v2_b = \verte\vert (V_b - V_a). Assuming that V_a = 0 volts (1/2)m_e v2_b = \verte\vert V_b . If all the energy of the electron when it reaches point b is changed into the energy of an x-ray photon, we obtain (1/2)m_e v2_b = \verte\vert V_b = hѵ HenceV_bmin = hѵ/\verte\vert . V_b is the minimum voltage needed to produce x-rays of fre-quency ѵ. (Only those electrons which give up all their energy to the x-ray photons will produce x-rays). V_bmin = hѵ/\verte\vert Butѵ = c/\lambda where c = speed of light and \lambda is the x-ray wavelength. V_bmin = (hc)/(\verte\vert\lambda) V_bmin = (6.63 × 10^-34 joule-sec3.0 × 10^8 m/sec)/(1.60 × 10^-19 coul0.20 × 10^-10m) = 62,000 volts

Question:

What is the possible function of nonsense codons in the genes?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0619.htm

Solution:

Nonsense codons are those which do not code. There are three such nonsense triplets for an amino acid : UAA, UAG, and UGA. They are found at the end of nucleotide sequences coding for given polypeptides, and code specifically for the termination of the chain. They signify that peptide elongation at the ribosome should stop. These codons are read by specific proteins known as release factors. These release factors are necessary to release the chain from the ribosome. Two release factors have so far been identified, each of which recognizes two codons. One is specific for UAA and UAG and the other for UAA and UGA. Nonsense codons may be employed in the following manner: the nucleotide sequence etc.- CUU AGG UAU AUA UAG GCC ACG - etc. on the mRNA might represent two genes, with the UAG triplet acting as a terminating codon between them :

Question:

Explain the TRAILER CARD method of stopping a computer job, i.e., a program.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0333.htm

Solution:

In this method, we use a certain unique number on the last card in thedata section of a program card deck. This last card is called the Trailer Card. The unique number selected is such that it is different in value, or numberof digits, etc., from all the other authen-tic data on the other precedingdata cards of the program deck. For example, if we know that allour data consists of positive numbers, we could use a negative number onthe Trail-er Card. Or, if we know that all the data is non-zero, we could usea value of zero on the Trailer Card. Then, in the body of the program weinsert a statement which tells the computer to end the job when the valuewhich we have put on the Trailer Card is encountered. The Trailer Card method is most useful in cases where different re- runsof a given program will have different amounts of data, which cannot beknown in advance. Hence, instead of modifying the program each time thedata changes, we use the same program, but take advantage of the Trailer Card to tell the computer when to end the program. We give below a program to illustrate the use of the Trailer Card method. SUM: PROCOPTIONS(MAIN); DCL(X,SUM)FIXED(4)INIT(0); LOOP:GET LIST (X) ;/\textasteriskcentered WE READ ONE DATA VALUE \textasteriskcentered/ SUM = SUM + X; IF X = 0000 THEN GO TO FINISH; /\textasteriskcentered WE READ A VALUE OF 0000 FOR X, WHICH IS ON THE TRAILER CARD, AND, AS THE TRAILER CARD IS THE LAST DATA CARD, WE GO OUT OF THE LOOP TO LABEL FINISH. HERE WE ASSUME NONE OF THE DATA NORMALLY HAS A VALUE '0000' \textasteriskcentered/ GOTO LOOP; /\textasteriskcentered IF THE TRAILER CARD IS NOT REACHED, WE REPEAT THE LOOP \textasteriskcentered/ FINISH: PUTLIST(SUM); END SUM;

Question:

A + B \rightarrow products. Consider the following reaction rate data and then determine the rate law for each case: (a) When [A] is doubled, the initial rates doubles; when [B] is doubled, the initial rate doubles, (b) When [A] doubles, initial rate doubles; when [B] doubles, the initial rate is halved, (c) Doubling [A], doubles initial rate; but doubling [B] leaves the rate unchanged.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0452.htm

Solution:

The given reaction is one with two different molecules being converted to products. The rate of any such reaction is: r = rate = k [A]^x [B]^y,where k = a rate constant and [A] = concentration of A and [B] = concentration of B. x and y are the number of each A and B molecules, re-spectively, that are present in the activated complex or transition state. The determination of these exponents is the key to writing the rate law for each case. You proceed as follows: (a) You are told that when A or B is doubled, the rate is doubled. Thus, when A \rightarrow 2a,r \rightarrow 2r, which means 2r = k [2A]^x[B]^y. Now, if x = 1, then [2A]^x = [2A] and dividing by 2 you come back to r \rightarrow k[A][B]^y. In other words, r \rightarrow 2r if A \rightarrow 2A, implies rvaries directly as A; that is, x = 1. The same type of situation exists with B, so that y = 1. Thus, the rate law becomes r = k[A][B]. (b)Here, when A \rightarrow 2A, r \rightarrow2a, which means, again, A varies directly as r, and so x = 1. When B \rightarrow 2B, r \rightarrow (1/2)r , however, in this case, (1/2)r = [A][2B]^y. This can only come about if y = - 1. If y = - 1, then [2B]^-1 = 1/[2B], so that (1/2)r = [A] 1/[2B]. The halves can be cancelled to return a directionproportion. Thus , the rate law becomes r = k[A][B]. (c)Here, when [A] \rightarrow [2A], r \rightarrow 2r but when [B] \rightarrow[2B], r \rightarrow r. For A: you have, again, a direction proportion between A and r, which requires x = 1. For B: when B's concentration is doubled, nothing happens to r; it remains constant. This means r does not depend upon the [B]. For r not to depend upon [B]^y, y must equal zero, thus, r = k [A] [B]\textdegree = k [A] (1). As such, the rate law becomes r = k [A].

Question:

Figure 1 shows a flat strip of copper of width a and negligi-ble thickness carrying a current i. Find the magnetic field B at point P, at a distance R from the center of the strip, at right angles to the strip. .

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1079.htm

Solution:

Subdivide the strip into long infinitesimal fila-ments of width dx, each of which may be treated as a wire carrying a current di given by i(dx/a). The field contribu-tion dB at point P in Fig. 1 is given, for the element shown, by the differential formula for a long cylindrical wire, which is dB = (\mu_0 / 2\pi)(di / r) = (\mu_0 / 2\pi) [{i(dx/a)} / (R sec \texttheta)]. Note that the vector dB is at right angles to the line marked r, Only the horizontal component of dB, dB cos \texttheta, is effec-tive, the vertical component being canceled by the contribution associated with a symmetrically located filament on the other side of the orgin. Thus B at point P is given by the (scalar) integral B = \intdB cos \texttheta = \int [{\mu_0i (dx / a)} / (2\piR sec \texttheta)] cos \texttheta = [\mu_0i / (2\piaR)] \int[(dx) / (sec^2\texttheta)] . The variables x and \texttheta are not independent, being related by x = R tan \texttheta or dx = R sec^2\texttheta d\texttheta. Bearing in mind that the limits on \texttheta are \pmtan^-1 (a/2R) and eliminating dx from this expression for B, B = [\mu_0i / (2\piaR)] \int[(R sec^2\texttheta d\texttheta) / (sec^2\texttheta)] = [\mu_0i / (2\pia)]^+tan-1(a/2R)\int^ _-tan-1(a/2R) d\texttheta = [\mu_0i / (\pia)] tan^-1(a/2R). At points far from the strip, (a/2R) is a small angle, for which tan^-1\alpha \cong \alpha. Thus, as an approximate result, B \cong [\mu_0i / (\pia)] (a/2R) = [\mu_0 / (2\pi)] (i / R). This result is expected because at distant points the strip cannot be distinguished from a cylindrical wire.

Question:

A source emits sound waves at 1000 cycles/second (frequency f_0) and the speed of the wave with respect to the source is 1000 ft/second(v_w) rest with respect to the source (\lambda_0)? towards the listener, what frequency (f_sl) and wavelength (\lambda_sI) does he observe? listener, what frequency (f_s2) and wavelength (\lambda_s2) does he observe?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0840.htm

Solution:

Everyone has heard the drop in pitch of a passing train's whistle. The Doppler effect is the apparent change of wave frequency observed by a listener when he and the wave source are in relative motion. In this problem, we deal with sound waves and derive the change in observed frequency and wavelength for two kinds of relative motion: motion of the observer and motion of the source. For any observer, v = \lambdaf , which means that the velocity at which he observes the wave to move is equal to the product of its observed frequency and wavelength. This is no more than an application of the conventional velocity de-finition v = ∆s/ ∆t if we remember f = 1/T where T is the wave's period. Then v = \lambdaf = \lambda/T where \lambda is the observed displacement in time T. (a) Using v = \lambdaf, v_w = \lambda_0f_0, \lambda_0 = v_w/f_0 = (1000 ft/second)/(1000/second) = 1 ft (b) It is known that the velocity of waves in a medium depend only on the mechanical properties of that medium. When the source moves towards the listener (see L_1 in figure) a "bunching up" of waves is noted by the listener, as shown in the figure. We seek a mathematical relation to determine by what amount their wavelength is effectively reduced, and how the frequency is affected. In a time, t, the source emits t/T = f_0t waves. Along the line between listener and source these waves are spread over a distance v_wt - v_st. Thus, if the wavelength is smaller by a constant amount, it must now be \lambda_s1 = [(v_wt - v_st)/(f_0t)] = [(v_w - v_s)/(f_0)] v_wt - v_st f_0t v_w - v_s)/(f_0)] = [(1000 ft/sec - 100 ft/sec) / (1000/sec)] = [(900 ft/sec) / (1000/sec)] = 0.9 ft. As the speed of the wave is unchanged, the listener will observe an in-crease in the wave's frequency, according to v_w = \lambda_s1f_s1; f_s1 = v_w/\lambda_s1 = (1000 ft/sec)/(.9 ft) = 1111 sec^-1 (c) As the source recedes a listener observes the waves to occupy more space than those of a stationary source, as can be seen by looking at L_2 in the figure. In a time, t, f_0t waves are emitted, spread over a distance v_wt + v_st. As before, we find the wavelength \lambda_s2 = [(v_wt + v_st)/( f_0t)] = [(v_w + v_s)/(f_0)] = [(1000 ft/sec + 100 ft/sec) / (1000/sec)] = 1.1 ft And, as in (b) we have for the observed frequency, f_s2 = v_w/\lambda_s2 = (1000 ft/sec)/(1.1 ft) = 909/sec.

Question:

2.000picogram(pg) of ^33P decays by ^0_-1\beta emission to 0.250 pg in 75.9 days. Find the half-life of ^33P.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0719.htm

Solution:

The half-life is defined as the time it takes for 1/2 of the amount of a certain compound present to de-compose. For example, if a substance has half-life of day, after one day there will only be 1/2 of the original amount left. When given the original and final amount of a substance, after a given time has elapsed, the number of half-lives that have passed can be found. his is done by dividing the original amount by the final amount and determ-ining how many factors of 2 are present in the quotient. The half-life is then found by dividing the time elapsed by the number of half-lives. Solving for the half-life of ^33P: (original amount) / (final amount) = (2.000 pg) / (0.250) = 8 8 = 2 × 2 × 2, therefore 3 half-lives have passed. One is given that these half-lives elapse in 75.9 days. half-life = (75.9 days) / (3) = 25.3 days.

Question:

Find all three-digit numbers with non-zero first digit that are equal to the sum of the cubes of their digits.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0537.htm

Solution:

The solution is based upon having the computer check each three-digit number HTU = 100\textasteriskcenteredH + 10\textasteriskcenteredT + U to see if the condition HTU = H3 + T3 + U3 is satisfied. [Note that H stands for hundreds, T stands for tens, and U-for units]. Since we want all possible combinations of H,T, and U where U = 0,1,2,...,9; T = 0,1,2,...,9; and H = 1,2,3,...,9 - there are 10 × 10 × 9 = 900 numbers to check. We do this checking with three nested DO loops. The DO LOOP parameters take on values 1,2,...,10; 1,2,...,10; and 1,2,...,9, progressing from the inner to the outer loops. Since 1 is the lowest possible Integer to be assigned to any DO-LOOP index, the program subtracts 1 from indexes TP1 and UP1 to initiate them at zero. Meanwhile, the final values of TP1 and UP1 are set to 10, which become 9's after the same subtraction. The program looks as follows: INTEGER H,T,U,TP1, UP1 DO 10 H = 1,9 DO 10 TP1 = 1,10 T = TP1 - 1 DO 10 UP1 = 1,10 U = UP1 - 1 IF(100\textasteriskcenteredH + 10\textasteriskcenteredT + U.EQ. H\textasteriskcentered\textasteriskcentered3 + T\textasteriskcentered\textasteriskcentered3 + U\textasteriskcentered\textasteriskcentered3) GO TO 1 GO TO 10 1WRITE (6,100) H,T,U 10CONTINUE 100FORMAT (3I1) STOP END

Question:

Explain how character strings are declared and manipulated in Pascal.Give examples.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0406.htm

Solution:

Unlike some other programming languages, Pascal has no facilities for explicit declaration of character strings. Instead, Packed arrays of characters are used to define character strings. Since the length of an array must be known at compilation time, only strings of fixed length are allowed. Thus, if we define TYPE STRING = PACKED ARRAY[1..10] of char; VAR WORD: string; then the variable WORD can be assigned any value consisting of exactly ten characters. The following assignments are valid: WORD: =' HELLObbbbb'; WORD: =' GGODbBYEbb' ; WORD: = 'CHARACTERS'; WORD: = '#123456789' and the following are not: WORD: = 'HELLO' WORD: = 'WORD' WORD: = 'AVERYLONGWORD' When a variable declared as packed array of characters ap-pears in a WRITE or WRITELN statement, the entire array is written out. For example, the effect of these two state-ments WORD: =' HELLObbbbb'; write1n(word); will be HELLObbbbb However, to read in a character string, the only possibility is character by character, and then only if the length of the string is exactly equal to the length of the array hold-ing it. Thus, to read in a ten-character word, we would write FOR I:= 1 to 10 DO READ(word[I]); Character strings can also be compared. For example, the Boolean variable P will have value 'true' after the execu-tion of any of these statements. P: = 'ALEX' < 'JILL' P: =' JOHNbbb' < ' JOHNSON ' P: ='ZITO' > 'MARY' P: = 'JACK' ='JACK'

Question:

A sample of blood has freezing point of - 0.560\textdegreeC. What is the effective molal concentration of the solute in the serum? According to this concentration, what is the osmotic pressure in the blood serum at body temperature (37\textdegreeC)?

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Solution:

The molal freezing point depression and boiling point elevation for the ideal 1 molal solution are also called the cryoscopic and ebullioscopic constants. The mathematical expressions for the freezing point depression is ∆T_f = m K_f where K_f is the freezing point constant for water, 1.86\textdegreeC, m is the effective molality of the solute, and ∆T_f is the change in the freezing point. For this particular problem ∆T_f = 0.560\textdegreeC and K_f = 1.86\textdegreeC, so that m = ∆T_f / K_f = 0.560\textdegreeC / 1.86\textdegreeC = 0.30 m. Hence, 0.30 m is the effective molal concentration of solute in serum. Osmosis is the diffusion of a solvent through a semi- permeable membrane into a more concentrated solution. The osmotic pressure of a solution is the minimum pressure that must be applied to the solution, in excess of the pressure on the solvent, to prevent the flow of solvent from pure solvent into the solution. An empirical osmotic pressure equation for solutions is \pi = CRT, where \pi is the osmotic pressure, C is the concentration in molality or molarity, R is the gas constant (0.082 atm- liter/\textdegreeK-mole), and T is the temperature (in \textdegreeK). Thus, at 37\textdegreeC = 310\textdegreeK and at 0.30 m, blood serum has an osmotic pressure of \pi = CRT = (0. 30 m) (0.082 atm-liter/\textdegreeK mole) (310\textdegreeK) = 7.6 atm.

Question:

Balance the equation for the following reaction taking place in aqueous basic solution: MnO^-_4 + H_2O_2 \rightarrow MnO_2 + O_2

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Solution:

The equation in this problem involves both oxidation and reduction. When balancing it, you can use the following rules. Separate the net reaction into its two major components, the oxidation reaction (the loss of electrons) and the reduction reaction (the gain of electrons). For each half- reaction, balance the charges with H^+, if the medium is acidic, or OH^-, if the medium is basic. Next, balance the oxygens by the addition of H_2O. Balance total atoms by the addition of H atoms. Finally, combine the two half reactions, so that all charges from electron transfer cancel out. You employ these rules as follows: The net reaction is MnO^-_4 + H_2O_2 \rightarrow MnO_2 + O_2 The oxidation process is H_2O_2 \rightarrow O_2 + 2e^- The oxygen atoms in H_2O_2 go from -1 to zero in O_2. Thus, you have a loss of two electrons. To balance charges, add 2 OH^- to the left side, since there exist 2 negative charges on the right side. You obtain H_2O_2 + 2OH^- \rightarrow O_2 + 2e^- You now have 4 oxygens on the left, but only 2 on the right. Thus, add 2 water molecular ro the right, obtaining H_2O_2 + 2OH^- \rightarrow O_2 + 2e-+ 2H_2O There are the same number of H's on each side. Proceed, now, to reduction half-reaction. Here, MnO^-_4 + 3e^- \rightarrow MnO_2 Mn begins with a + 7 oxidation number and ends up with + 4 in MnO_2. Therefore, 3 electrons must be added to the left side of the equation. To balance the charges, add 4 OH^- ions to right, since you have a total of 4 negative charges on left. Rewriting the equation MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. Add 2 water molecules to the left, so that oxygen atoms can be balanced, obtaining 2H_2O + MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. The hydrogens are balanced. Thus oxidation:H_2O_2 + 2OH^- \rightarrow O_2 + 2e-+ 2H_2O reduction:2H_2O + MnO^-_4 + 3e^- \rightarrow MnO_2 + 4OH^-. To combine these two so that electrons cancel out. Select a multiple of 3 and 2, since these are the number of electrons involved in the half- reactions. This multiple is six. Multiply the oxidation by 3 and the reduction by 2, obtaining oxidation:3H_2O_2 + 6OH^- \rightarrow 3O_2 + 6e-+ 6H_2O reduction:4H_2O + 2MnO^-_4 + 6e^- \rightarrow 2MnO_2 + 8OH^- The net reaction is the total. Thus, adding you obtain: 3H_2O_2 + 4H_2O + 6OH^- + 2MnO^-_4 + 6e^- \rightarrow 3O_2 + 6e^- + 6H_2O + 8OH^- + 2MnO_2 Cancel the electrons, subtract OH^- ions and H_2O's to obtain: 2MnO^-_4 + 3H_2O_2 \rightarrow 2MnO_2 + 3O_2 + 2H_2O + 2OH^- which is the balanced equation.

Question:

A car starts from rest and reaches a speed of 30 miles per hour in 8 seconds. What is its acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0058.htm

Solution:

v = at for constant acceleration. We shall convert the velocity in miles per hour into feet per second. A useful conversion factor to remember is that 60 mph is about 88 ft. per second. Therefore, 30 mph is about 44 ft. per second. Substituting we have: a =v_final/t= (44 ft)/(sec × 8 sec) = 5.5 ft per sec per sec.

Question:

The difference of potential V_1 between the terminals of a resistor is 120 volts when there is a current I_1 of 8.00 amp in the resistor. What current will be maintained in the re-sistor if the difference of potential is increased to 180 volts?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0646.htm

Solution:

Ohm's law indicates that the resistance R will remain the same when the potential difference is increased; hence we can write R = [(V_1 )/(I_1 )] = [(120 volts)/(8.00 amp)] = 15.0 ohms I_2 = [(V_2 )/(R)] = [(180 volts)/(15.0 ohms)] = 12.0 amp .

Question:

Write an assembler language program in BAL to add 20 numbers from main storage.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0200.htm

Solution:

Let AA denote the starting location of the num-bers A_i to be added together. We assume these numbers oc-cupy consecutive full words in main storage (i.e. 4 bytes per word). If the address of the A_i just added to the sum is kept in a register, the number 4 can be added to the con-tents of that register; the result is the address of the next number to be added. Suppose register 7 is initialized to AA, the address of A_0. Each time the adjustment step is executed, the quantity in register 7 is increased by 4. Hence the second time the instruction A 8,\varphi(7) is executed, the address \varphi in register 7 is 4 greater than the address of A_0; thus it is the address of A_1. The third time the body of the loop is executed, the address of register 7 has been increased again, so A_2 is growing to sum. The process of adding on A_i and adjusting the address continues until exe-cution of the loop is terminated. We exit from the loop when the contents of register 9 is 20. \textasteriskcenteredA SEGMENT TO ADD 20 NUMBERS FROM \textasteriskcenteredCONSECUTIVE FULL WORDS IN MAIN STORAGE \textasteriskcenteredINITIALIZATION LA8,\varphiSUM =\varphi \varphi \varphi LA9,\varphiI =\varphi \varphi \varphi LA7, AA ADDRESS OF FIRST A \textasteriskcenteredBODY: ADD A (SUBI) TO SUM LOOPA8,\varphi(7) \varphi LA9, 1 (9) INCREASE I LA7, 4 (7) GET ADDRESS OF NEXT A C9, = F '2\varphi' \varphi BNE LOOP BRANCH BACK IF I NOT YET 20, LV LOOP IF I = 2\varphi \varphi

Question:

Explain how a conditional branch instruction is specified in APL programs.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0371.htm

Solution:

A conditional branch instruction is specified in APL by means of a right-pointing arrow followed on its right by an expression. For example: i) [4] \rightarrow (I \leq N) × 2 In the above, a test for I is performed. Is I \leq N? If I is indeed less than or equal to N, then the value of the test is true, i.e., 1. But if I is not less than or equal to N then the value of the test is false, i.e., 0. Now, if the result of the test is 1, the value of the expression (I \leq N) × 2 reduces to 1 × 2, equal to 2. This means that in the given example the statement [4] of the program re-duces to the following: [4] \rightarrow (I \leq N) × 2 i.e.,[4] \rightarrow 1 × 2 i.e.,[4] \rightarrow 2. Thus, a branching to statement 2 is performed. However, if the result of the test is 0, then the statement [4] of the example reduces to the following: [4] \rightarrow (I \leq N) × 2 i.e.,[4] \rightarrow 0 × 2 i.e.,[4] \rightarrow 0. This means that a branching to statement 0 is to be per-formed. However, there is no statement 0 in the program. Hence, the program execution halts. ii) [4] \rightarrow (I \leq N) × BOB. The above is similar to parti) explained before, except that instead of the statement number of the statement to which a branch off is required, a label is specified. La-bels have an advantage over numbers. If statements are added or removed from the program, the statement numbers change. This necessitates corrections to be made within the statements too, in order to branch to the proper state-ment according to the new numbering. But if labels are used, corrections of the statements are not necessary. A label takes on a value equal to the statement number of the statement which has the given label. In the example given above, the label is BOB. This means that if the test is true, the program will jump to a statementlabelledBOB. The program could be something like the following: [2] BOB: I \leftarrow I + 1 [4] \rightarrow (I \leq N) × BOBor,[4] \rightarrow (I \leq N) × BOB [7] BOB: I \leftarrow I + 1 In the above, if the test is false, statement [4] becomes: [4] \rightarrow 0 × BOB, i.e.,[4] \rightarrow 0. and the program halts. iii) In the above two cases, the program went to a halt if the test was false. However, at times it is not proper for the program to halt when the test is false. It is just required that the program goes to the next sequen-tial instruction if the test is false, and branch to the statement indicated if the test is true. Consider the following statement, for example: [4] \rightarrow 2 × ʅ(I \leq N) In the above, if the test of I \leq N is true, the value of (I \leq N) is equal to 1. Therefore the statement [4] reduces to 2 × ʅ(1) = 2 × 1=2. Thus, the program branches to statement 2. But, if the test is false, the statement [4] reduces to 2 × ʅ(0) Now, iota zero, i.e., ʅ 0, generates an empty vector, and therefore the statement [4] becomes meaningless. Hence, the program goes to the next sequential statement. iv) A label can be used, instead of a statement num-ber, with the iota function. For example, [4] \rightarrow BOB × ʅ(I \leq N). v) Another way to specify a branch instruction is by using the compression function, which has the symbol /. It is not necessary to know the meaning of the compression function in order to understand the branching. Consider the following illustration: [4] \rightarrow (I \leq N) / 2. In the above, if the test, that is, the quantity to the left of the compression function symbol is true, a transfer is made to the statement whose number is specified on the right of the compression function symbol. Thus, if I \leq N is true, the program jumps to statement 2. But, if the test is false, the program goes to the next sequential statement, i.e., to statement 5. vi) A label can also be used on the right side of the compression function symbol instead of a number. For ex-ample, [4] \rightarrow (I \leq N) / BOB.

Question:

Consider an elementary inventory control system where there is no delay between the ordering of goods and their receipt into inventory. Suppose the order rate is direct-ly proportional to the difference between desired inventory and actual inventory, with proportionality constant 1/A, where A is the time that would be required to correct the inventory if the order rate were constant. Write a FORTRAN program to simulate this system, using modified Euler's method (the method is fully explained, and the subroutine for it is given in the SIMULATION chapter) from time t=0 to t=t_f if I(t) is the inventory level at time t and I(0)=I_0.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0464.htm

Solution:

Clearly, the goal of this system is to maintain the desired inventory, D. If \.{I} is the order rate (and hence, the rate of change in inventory, since there is no delay), we can write: \.{I} = 1/A(D - I)(1) with block diagram as follows: Note that this model "fills to capacity" using negative feedback control. That is, knowledge of the present system state is used to determine the future system state. If change in inventory is done continuously, the exact solution to equation (1) would be: I(t) = D - (D - I_\O)^e-t/A Note, that as t \rightarrow \infty, l(t) approaches D, which is the steady state solution, the program looks as follows: DATA T/_\O , \O/ COMMON D, A INPUT N, TFIN, ACCUR, D, A, I\O OUTPUT T, I SPACE = D - I\O REALN = N DT = TFIN/REALN I = I\O DO 1\O J = 1, N T = T + DT CALL MEULER (T, I, ACCUR, DT) EXACT = D - SPACE \textasteriskcentered EXP(-T/A) ERROR = ABS( (EXACT - I) /EXACT) \textasteriskcentered1\O\O. OUTPUT T, I, EXACT, ERROR 1\OCONTINUE STOP END FUNCTION G (W) COMMON, D, A G = (D - W)/A RETURN END

Question:

CO_2 + H_2O +hv(light energy) (CH_2O) + O_2. It was generally assumed in the past that the carbohydrate came from a combination of the carbon atoms and water molecules and that the oxygen was released from the carbon dioxide molecule. Do you agree with this assumption? Explain your answer.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0216.htm

Solution:

Superficially, the assumption appears sound, and was believed for years to be true. It is however, quite wrong, and was refuted in the 1930's by C.B. vanNielof Stanford University. VanNielwas studying photosynthesis in different types of photosynthetic bacteria. He observed that in their photosynthetic reactions, bacteria reduce carbon dioxide to carbohydrate, but do not release oxygen. The purple sulfur bacteria, which require hydrogen sulfide for photosynthesis, were found to accumulate globules of sulfur within the cells. Based on these observations VanNiel then wrote the following equation for photosynthesis in purple sulfur bacteria: CO_2 + 2H_2S \rightarrow CH_2O + H_2O + 2S This equation shows that it is unlikely that carbon dioxide is split during photosynthesis; if it were, oxygen would be formed. Moreover, carbohydrate could not have come from the combination of carbon atoms and hydrogen sulfide molecules. It would be reasonable to propose that hydrogen sulfide was the molecule to be cleaved rather than carbon dioxide. This led VanNielto propose the following general equation for photosynthesis, CO_2 + 2H_2A \rightarrow (CH_2O) + H_2O + 2A Here, H_2A represents anoxidizablespecies capable of donating its electrons upon being split. According to this hypothesis, in the photosynthesis of green plants, it is the water molecule, not carbon dioxide, that is split to release O_2. This brilliant speculation was not proven until many years later when advancement of radio - isotopic techniques made it possible to reveal the mechanism of the actual reaction. The use of isotopic oxygen, ^18O, allowed the successful tracing of the path of oxygen from water to free molecular oxygen: CO_2 + 2H_2^18O - CH_2O + H_2O + ^18O_2 Thus, it can readily be seen that molecular oxygen comes actually from water and not from carbon dioxide. It is now known that the splitting of water occurs during the light reactions of photosynthesis, releasing O_2 gas, electrons, which are energized by light to provide for ATP formation, and protons, which are used to form NADPH, the reducing power necessary for carbohydrate formation. Carbon dioxide is then reduced in the dark reactions of photosynthesis using both ATP and NADPH to form carbohy-drates.

Question:

What are the implications of the Hardy-Weinberg Law?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0694.htm

Solution:

The Hardy-Weinberg Law states that in a population at equilibrium both gene and genotype frequen-cies remain constant from generation to generation. An equilibrium population refers to a large interbreeding population in which mating is random and no selection or other factor which tends to change gene frequencies occurs. The Hardy-Weinberg Law is a mathematical formulation which resolves the puzzle of why recessive genes do not disappear in a population over time. To illustrate the principle, let us look at the distribution in a population of a single gene pair, A and a. Any member of the popu-lation will have the genotype AA,Aa, oraa. If these genotypes are present in the population in the ratio of 1/4 AA : 1/2Aa: 1/4aa, we can show that, given random mating and comparable viability of progeny in each cross, the genotypes and gene frequencies should remain the same in the next generation. Table 1 below shows how the genotypic frequencies of AA,Aa, andaacompare in the population and among the offspring. Table 1 The Offspring of the Random Mating of a Population Composed of 1/4 AA, 1/2Aaand 1/4aaIndividuals Mating Frequency Offspring MaleFemale AA×AA 1/4 × 1/4 1/16 AA AA×Aa 1/4 × 1/2 1/16 AA + 1/16Aa AA×aa 1/4 × 1/4 1/16Aa Aa ×AA 1/2 × 1/4 1/16 AA + 1/16Aa Aa ×Aa 1/2 × 1/2 1/16 AA + 1/8Aa+ 1/16aa Aa ×aa 1/2 × 1/4 1/16Aa+ 1/16aa aa ×AA 1/4 × 1/4 1/16Aa aa ×Aa 1/4 × 1/2 1/16Aa+ 1/16aa aa ×aa 1/4 × 1/4 1/16aa Sum : 4/16 AA + 8/16Aa+ 4/16aa Since the genotype frequencies are identical, it follows that the gene frequencies are also the same. It is very important to realize that the Hardy-Weinberg law is theoretical in nature and holds true only when factors which tend to change gene frequencies are absent. Examples of such factors are natural selection, mutation, migration, and genetic drift.

Question:

What is the mole fraction of H_2SO_4 in a 7.0 molar solution of H_2SO_4 which has a density of 1.39 g/ml?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0292.htm

Solution:

The mole fraction of H_2SO_4, is equal to the number of moles of H_2SO_4 divided by the sum of the number of moles of. H_2SO_4 and of H_2O. mole fraction of H_2SO_4 = [( moles of H_2SO_4 )/(moles of H_2SO_4 + moles of H_2O)] Since the solution is 7.0 molar in H_2SO_4, you have 7 moles of H_2SO_4 per liter of solution. If one knows how much 7 moles of H_2SO_4 weighs, and how much one liter of the solution weighs, the weight of water can be determined by taking the difference between the quantities. The weight of one liter of the solution can be calculated by multiply-ing the density by the conversion factor 1000 ml/1 liter. weight of 1 liter of the solution = 1.39 g/ml × 1000 ml/liter = 1390 g/l. weight of the water in a one liter solution = weight of the total solution - weight of 7 moles of H_2SO_4 7 moles of H_2SO_4 weigh 7 times the molecular weight of H_2SO_4 . weight of 7 moles of H_2SO_4 = 7 × 98 g = 686 g. Hence, weight of the water = 1390 g - 686 g = 704 g. The number of moles of H_2O is found by dividing its weight by the molecular weight of H_2O. no. of moles of H_2O = [(704 g)/(18 g/mol] = 39 moles The number of moles of both components of the system is now known, therefore, the mole fraction of H_2SO_4 can be obtained. mole fraction H_2SO_4 = [(moles of H_2SO_4)/(moles of H_2SO_4 + moles H_2O)] mole fraction of H_2SO_4 = [7 / (7 + 39)] = 0.15.

Question:

A proton, starting from rest, falls through a potential differenceof 10^6 V. What is its final kinetic energy andfinal velocity?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0620.htm

Solution:

At its rest position, the proton has only potential energy and no kineticenergy. As it falls, it loses potential energy but gains kinetic energy. In its final position, the proton has only kinetic energy and no potential energy. However, the law of conser-vation of energy demands that at all times, the sum of the kinetic and potential energies remain constant. Consequently, the sum of the two energies at the rest position must equal thesum of the two energies at the final position. Or PE + 0 = 0 + KE By definition of potential V, V = Energy/Charge=E/q Then PE =eV wheree is the charge of the proton and V is the potential dif-ference throughwhich it falls. Therefore 10^6 elec. volts = 1/2m_pv^2 , wherean electron volt is the energy required to move a charge e througha potential of 1 volt. The final kinetic energy is then 10^6eV= 10^6eV×(1.602 X 10\rule{1em}{1pt}12erg/1eV) =1.602 × 10^\rule{1em}{1pt}6 erg = 1/2 m_pV2 To compute the final velocity, we use this relation, then v=\surd[{2×(1.60×10^\rule{1em}{1pt}6 erg)} / (1.67×10^\rule{1em}{1pt}24g)]. wherem_p = 1.67 ×10^\rule{1em}{1pt}24 g. Consequently, v =1.38 × 10^9 cm/sec

Question:

What is the elemental percent composition (by weight) of a mixture that contains 20.0 g of KAl(SO_4)2and 60.0 of K_2 SO_4 ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0151.htm

Solution:

The elemental percent composition by weight is equal to the weight of the element present divided by the total weight of the mixture multiplied by 100. percent composition = {(wt of elements) / (total wt) }× 100 The weight of the total mixture is equal to the sum of the weights of the KAl(SO_4)_2 and the K_2 SO_4 . total weight = 20.0 g + 60.0 g = 80.0 g The weights of the various elements present is found by dividing the molecular weight of the element multiplied by the number of atoms of that particular atom present in the molecule by the molecular weight of the compound, and then multiplying the quotient by the number of grams of the compound present. Solving for the weights of the elements in KAl(SO_4)2: MW = 258.12. For K = MW = 39.1 weight of K = {(30.1 g/mole × 1)/(258.12 g/mole)} × 20.0g = 3.03g For Al: MW = 26.98 weight of Al = {(26.98 g/mole × 1)/(258.12 g/mole)} × 20.0g = 2.09g. For S: MW = 32.06 weight of S = {(32.06 g/mole × 2 moles)/(258.12 g/mole)} × 20.0g = 4.97g. For O: MW = 16 weight of O = {(16 g/mole × 8moles)/(258.12 g/mole)} × 20.0g = 9.92g. Solving for the weight of the elements in KAl(SO_4)_2 : MW = 174.26 For K: MW = 39.1 weight of K = {(39.1g/mole × 2 moles)/(174.26 g/mole)} × 60.0g = 26.93 g For S; MW = 32.06 weight of S = {(32.06 g/mole × 1moles)/(174.26 g/mole)} × 60g = 11.04 g For O: MW = 16 weight of O = {(16 g/mole × 4 moles) / (174.26 g/mole)}× 60g = 22.04 g. One can find the total weights of the various elements by taking the sum of their weights from the two compounds. Total weight of K = 3.03 g + 26.93 g = 29.96 g Total weight of Al = 2.09 g Total weight of S = 4.97 g + 11.04 g = 16.01 g Total weight of O = 9.92 g + 22.04 g = 31.06 g. One can now determine the elemental percent composition of the mixture. % = {(weight of elements) / (total weight of mixture)} × 100 % of K = {(29.96g) / (80.0g)} × 100 = 37.45% % of Al = {(2.09g) / (80.0g)} × 100 = 2.61% % of S = {(16.01g) / (80.0g)} × 100 = 20.01% % of O = {(31.96g) / (80.0g)} × 100 = 39.95% .

Question:

You have 1 liter of an ideal gas at 0\textdegreeC and 10 atm press-ure. You allow the gas to expand against a constant ex-ternal pressure of 1 atm, while the temperature remains constant. Assuming, 24.217 cal/liter-atm, find q, w, ∆E and ∆H in calories, (a) in these values, if the expansion took place in a vacuum and (b) if the gas were expanded to 1 atm pressure.

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Solution:

The solutions to the parts of this problem require a combination of thermodynamics and ideal gas law theory. The gas expands because its pressure is greater than the external pressure. When the pressure of the gas falls to 1 atm, which is the pressure being applied ex-ternally, gas expansion will terminate. This allows for determination of the volume change using Boyle's Law. You want ∆V (volume change) , since w = work = P∆V. Boyle's law states that PV = constant for ideal gas. Thus, P_1V_1 = P_2 V_2. Originally, P_1 = 10 atm with V_1 = 1 liter. You end up with P_2 = 1 atm. Thus, V_2 = (P_1 V_1) / P_2 = [{(10 atm) (1 liter)} / {1 atm }] = 10 llters. ∆V = 10 liters (final) - 1 liter (original) = 9 liters. Therefore, w = P∆V = (1 atm)(9 liters) = 9 liters-atm. But, there are 24.217 cal per liter-atm, so that, in calories, w = 9 liter-atm × 24.217 cal/liter-atm = 218 cal. q = the heat absorbed by the system. To find its value, you employ the first law of thermodynamics, which says ∆E = q - w, where ∆E = change in energy of system and w = work performed. Since the gas is ideal, E = energy is only a function of temperature. As such, ∆E = 0, since the temperature is constant. You have, therefore 0 = q - w or q = w. But, you just found w, which means q = 218 cal. To find ∆H, the enthalpy change, remember that ∆H = ∆E + ∆(PV). You know that ∆E = 0. To find ∆(PV) , re-call that ∆(PV) = P_2 V_2 - P_1 V_1, as derived from Boyle's law. However, P_1 V_1 = P_2 V_2 thus; P_2 V_2 - P_1 V1= 0. This means ∆H = 0. In a vacuum, there is no external pressure. This means that no work can be done, so that w = 0. This is derived from the fact that work is defined as performing an action against a surrounding environment. If there is no environment, there cannot be any work. Since ∆E remains 0 and q = ∆E + w, then q = 0. Since there is zero pressure, PV = 0. Since ∆ (PV) = 0, ∆E = 0 and ∆H = ∆E + ∆(PV), then ∆H = 0. You now find the work (w) , required in going from zero pressure to 1 atm pressure; use the equation for the reversible expansion of a gas in terms of work. w = 2.303 n RT log V_2/V_1, where V_1 = initial volume, V_2 = final volume, R = universal gas constant, T = temperature in Kelvin (Celsius plus 273\textdegree) and n - number of moles. V_2 was calculated from Boyle's law. The rest is given. Thus, you substitute to obtain w = 2.303 (1 mole) (1.987) (273)log (10/1) = 1249 cal. ∆E\textdegree and ∆H\textdegree are still zero, but since q = ∆E + w, q = 0 + 1249 = 1249 cal.

Question:

One gram of water (1 cm^3) becomes 1671 cm^3 of steam when boiled at a pressure of 1 atm. The heat of vaporization at this pressure is 539 cal/gm. Compute the external work and the increase in internal energy.

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0531.htm

Solution:

The work done by the water in expanding from volume V_1 to volume V_2 is W = ^V2\int_V1 pdV where p is the pressure exerted by the water on its container. In our problem, p is constant, whence W = p(V_2 - V_1) = p(V_V - V_L) where V_V is the volume occupied by the steam, and V_L is the volume occupied by the water. Hence, W = p(V_V - V_L) = 1.013 × 10^6 dynes/cm^2 (1671 - 1)cm^3 Here we used the fact that 1atm= 1.013 × 10^6 dynes/cm^2 = 1.695 × 10^9 ergs Since 1 erg = 2.389 × 10^-8 cal W= (1.695 × 10^9 ergs)(2.389 × 10^-8 cal/erg) W= 41 cal The Law of Thermodynamics is ∆U = ∆Q - ∆W where ∆U is the change in internal energy of the water- steam system during the stated process. Also, ∆Q and ∆W are the heat added to and the work done by the system during the process, respectively. Denoting the gaseous state by the subscript V, and the liquid state by the subscript L, we obtain U_V - U_L = ∆Q - ∆W The heat, ∆Q, added to the system is the amount needed to vaporize the water, or ∆Q =mL= (1 g)(539 cal/g) = 539 cal ThenU_V - U_L= 539 cal - 41 cal = 498 cal. Hence the external work, or the external part of the heat of vaporization, equals 41 cal, and the in-crease in internal energy, or the internal part of the heat of vaporization, is 498 cal.

Question:

For the following PL/I program, CONDITION' arises. b) Show what will be printed out if the program is run. c) Write a corrected program to avoid errors due to this condition. d) Show what the print out will be from the corrected program . EXAMPLE:PROCOPTIONS(MAIN); DCL (B(6),C) CHAR(4), Z FIXED (2); GET LIST (Z,C); DO I=1 TO Z; B(I)=SUBSTR(C,1,I); PUTLIST(I,B(I)); END; END EXAMPLE; The DATA CARD holds data as follows:6,ROVE.

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Solution:

a) The purpose of the above program is to illus-trate a type of conditionknown as 'STRINGRANGE CONDITION'. In the program, the array B is declared as having 6ele-ments . The GET LIST statement gets the value of Z from the DATA CARD asZ = 6. Also, the value of C is obtained from the DATA CARD as C = 'ROVE'. In the next statement, viz., DO I = 1 TO Z, the value of Z = 6 gets substituted. Hence, the DO loop statement becomes DO I = 1 TO 6;. This means that the DO loop will be performed 6 times. Each time., thevalue of an array element B(I) is calculated. This value ofB(I) = SUBSTR(C,1,I) means that from C, i.e., from 'ROVE', starting from the first character, select I characters. These selectedI characters are to be assigned to B(I). Thus, for example, if 1 = 2, then SUBSTR(C,1,2) gives B (2) = 'RO'. In the given program, as Z = 6, the loop will be perform-ed 6 times. But, during the 5th loop, on substituting values ofIand C in the equation: B(I) = SUBSTR(C,1,I) , gives: B(5) = SUBSTR ('ROVE',1,5). The above statement tries to find a substring of 'ROVE', starting fromthe first character up to (and including) the fifth character. However, 'ROVE' has only 4 characters. Hence, the computer will just print the available characters. The programmermust be informed. Hence, the computer is pro-grammed to printout a message stating that the STRINGRANGE CONDITION had occurredin a certain program statement of the program. b) The printout of the program will be as follows: 1'R ' 2'RO ' 3'ROV ' 4'ROVE ' 5'ROVE ' 6'ROVE ' c) The program can be corrected as follows: The computer must be told thatit must look out for the STRINGRANGE CON-DITION. This is done by usinga word within parentheses, viz., (STRINGRANGE):, as a prefix to thefirst program state-ment, as shown below. Also, in the body of the program, a statement such as:. On STRINGRANGE GOTO .....,must be in-cluded. A good place for inserting this statement is just after the DCL statementso that the computer reads it be-fore it enters the loop. The corrected program can now be written as shown below: (STRINGRANGE) :EXAMPLE:PROC OPTIONS(MAIN); DCL(B(6),C) CHAR(4), Z FIXED(2) ; ON STRINGRANGEB(I) = '0000' ; GETLIST(Z ,C) ; DO I = 1 TO Z; B (I) = SUBSTR(C, 1,I) ; PUTLIST(I,B(I)); END; FINISH:END EXAMPLE; In the above program the program is made to go to the FINISH label, whichis the end of the program when the STRINGRANGE CONDITION arises. But as a matter of fact any valid instruc-tion could be given to the computerinstead of the 'GOTO FINISH' instruction. d) The print-out of the corrected program is as follows: 1'R ' 2'RO ' 3'ROV ' 4'ROVE ' 5'0000 ' 6'0000 ' The computer will also print out a STRINGRANGE CONDITION mes-sage. It is assumed that the user of the program knows that whenever '0000' is printedout it means that a STRINGRANGE CONDITION has occurred at that stage.

Question:

Write a BASIC program which calls for the input of an arbitrary number, and then prints out whether the number read in is positive, negative, or zero.

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Solution:

This problem represents an aspect of elementary decision making. The implementation of these decisions is best handled by use of several IF ... THEN clauses. 1\O\O REM DETERMINE IF X IS POSITIVE, NEGATIVE 11\O REM OR ZERO 12\O PRINT "WHEN I ASK, YOU ENTER A NUMBER AND"; 13\O PRINT "I WILL TELL YOU WHETHER YOUR NUMBER"; 14\O PRINT "IS POSITIVE, NEGATIVE, OR ZERO." 15\O PRINT 16\O PRINT "WHAT IS YOUR NUMBER?"; 17\O INPUT X 18\O IF X > \O THEN 19\O 181 IF X < \O THEN 21\O 182 IF X = \O THEN 23\O 19\O PRINT X; "IS POSITIVE" 2\O\O GO TO 15\O 21\O PRINT X; "IS NEGATIVE" 22\O GO TO 15\O 23\O PRINT X; "IS ZERO" 24\O GO TO 15\O 999 END

Question:

A biology student places one bacterium in a suitable medium and incubates it under appropriate conditions. After 3 hours and 18 minutes, he determines that the number of bacteria now present is 1000. What is the generation time of this bacteria?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0134.htm

Solution:

Bacterial growth usually occurs by means of binary fission with one cell dividing into two and these two cells dividing into four, etc. The population at any time can be represented by a geometric progression (1, 2, 4, 8...). The time interval required for each division (or for the entire population to double) is called the generation time. Different bacteria have different generation times, ranging from 20 minutes forEscherichiacoli to 33 hours forTreponemapallidum(the bacterium responsible for syphilis). The generation time also differs for the same bacterial species under different environmental conditions. To determine the generation time (G) of a bacterial population, we must know the number of bacteria present initially (N\textdegree_0), the number of bacteria present at the end of a given time period (N), and the time period (t). We can determine the generation time by using some simple mathemati-cal expressions. If we start with a single bacterium , the total population (N) after the nth generation is 2^n: N = 1 × 2^n For example, at the end of three generations, we would have eight bacteria, (2^3 = 8) . However, since we usually start with many bacteria, we must modify the formula to account for more than one parental bacterium: N = N_0 × 2^n. Solving for n, the number of generations, we get log N = log N_0 + n log 2. By substituting .301 for the log 2 and rearranging, n = (log N - log N_0) / (.301) = 3.3 (log N - log N_0). Since the difference of the logarithms of two numbers is the logarithm of the quotient of the two numbers, n = 3.3 log (N / N_0) The generation time G is simply equal to the time elapsed between N_0 and N divided by the number of generations: G = (t / n) = [t] / [3.3 log (N / N_0)] In our example: N = 1000, N_0 = 1 and t = 198 minutes. Substituting into the general formula, G = (198) / (3.3 × log 1000) . G = (198) / (3.3 × 3) = 20 minutes The time interval required for this particular bacterial species to divide is twenty minutes. If we plot the number of bacterial cells against time, we get a bacterial growth curve. In our case, with a generation time of 20 minutes, we would obtain this curve: This curve does not represent the complete pattern of growth, but rather one selected portion of the normal growth curve. This portion is called the exponential or logarithmic phase of growth. Only at this phase does the population double at regular time intervals. Eventually the reproduction rate is checked because of the lack of nutrients or the accumulation of waste products. In our example, we assumed that the bacteria were growing at a logarithmic rate where both N_0 and N should be measured during the exponential growth phase.

Question:

Describe the curve one would obtain by plotting pressure versus volume for an ideal gas in which the temperature and number of moles of gas are held constant.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0053.htm

Solution:

This problem requires a plot of the ideal gas equation. This equation reads PV = nRT, where P = pressure of gas, V = volume of gas, n = number of moles of gas, R = gas constant, and T = absolute tem-perature of gas. Since R is a constant and n and T are held constant, the product nRT may be combined into a single constant, call it k. Then, PV = nRT = k, or PV = k which is the equation of a hyperbola. Plotting P on the abscissa axis and V on the ordinate axis, we obtain the following curve: Note: Even if we did not know PV = k is the graph of a hyperbola, we could still obtain this graph. For ex-ample, if k is a constant, P and V must vary inversely with each other. In other words, if P is large, v must be small for k to remain constant. Likewise, if V is large, P must be small.

Question:

A racing car traveling in a circular path maintains a constant speed of 120 miles per hour. If the radius of the circular track is 1000 ft, what if any acceleration does the center of gravity of the car exhibit?

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Solution:

Since the car is traveling a circular path at constant speed, v, its acceleration is radial, and given by a = v^2/R Here, R is the radius of the circular path. Using the given data a = [(120 miles/hr)^2]/(1000 ft) = (14400 miles^2)/(1000 ft\bullethr^2) a = 14.4 miles^2/ft\bullethr^2 In order to keep units consistent, note that 1 hr = 3600 s and 1 mile = 5280 ft Then a = [14.4(5280 ft)^2]/[(1 ft)(3600 s)^2] a = 30.98 ft/s^2

Question:

Write a program in BASIC to find and print the sum of two complex numbers, inputted from the terminal. You may treat a single complex number as an ordered pair of real numbers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0223.htm

Solution:

If Z_1 and Z_2 are two complex numbers given by a + bi and c +di respectively wherei= \surd-1 , we define Z_1 + Z_2 = B (a + c) + (b + d)i. In terms of ordered pairs of real numbers, Z_1 = (a, b), Z_2 = (c, d), and Z_1 + Z_2 = (a+c,b+d), where a, b, c and d are the coordinates of the two points representing given numbers: a and c are on the real axis, b and d are on the imaginary axis The program below used the input of a particular number 999 to ter-minate the program. This is done by means of an IF .., THEN statement, which has the form IF variable = expression THEN label AND works as fol-lows: If the value of given variable equals the given expression, then the next statement to be executed is the one with the label following the word THEN. Otherwise, the computer goes to the next statement. 1\O PRINT "THIS PROGRAM ADDS TWO COMPLEX 15 PRINT "NUMBERS IN A, B FORM" 2\O PRINT 3\O PRINT "TYPE IN THE FIRST NUMBER" 4\O INPUT A, B 5\O IF A = 999 THEN 999 6\O PRINT "TYPE IN THE SECOND NUMBER" 7\O INPUT C,D 8\O PRINT "THE SUM Z1 + Z2 = (";A + C; " , " ;B + D; ")" 9\O GO TO 20 999 END

Question:

A capacitor is found to offer 25 ohms of capacitive reactance when connected in a 400-cps circuit. What is its capacitance?

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Solution:

Capacitive resistance, X_C, is given by X_C = [1/(2\pifC)] Here X_C = 25 ohms and f = 400 cps. Then X_C = 25 ohms = {1/(2\pi × 400 cps × C)} orC = [1/(2\pi × 400 cps × 25 ohms)] Since 1 farad = 1 (coul/volt) = {1coul/(amp \textbullet ohm)} = 1 (sec/ohm) C = [1/(2\pi × 400 × 25) farad = 1.6 × 10 ^-5 farad = 16\muf

Question:

Lemon juice is very acidic, having a pH of 2.1. If we assume that the only acid in lemon juice is citric acid (represented by HCit), thatHCitismonoproticand that no citrate salts are present, what is the concentration of citric acid in lemon juice? The dissociation constant for citric acid HCit+ H_2O \rightleftarrows H_3O^+ + Cit^- is K_a = 8.4 × 10^-4 mole/liter.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0417.htm

Solution:

The solution to this problem involves deter-mination of [HCit] from the expression for K_a of the reaction HCit+ H_2O \rightleftarrows H_3O^+ + Cit^-. By definition, K_a = {[H_3O^+] [Cit^-]} / [HCit], or,[Hcit] = {[H_3O^+] [Cit^-]} / K_a Since one mole of Cit^- is produced per mole of H_3O^+ produced, [Cit^-] = [H_3O^+]. [H_3O^+] can be determined from thepH.By definition, pH = - log [H_3O^+], hence [H_3O^+] = 10^-pH = 10^-2.1 mole/liter. Therefore, [H_3O^+] = [Cit^-] = 10^-2.1 mole/literand we have [HCit]= {[H_3O^+] [Cit^-]} / K_a = (10^-2.1 mole/liter×10^-2.1 mole/liter) / (8.4 × 10^-4 mole/liter) = [(10^-2.1)^2 / (8.4 × 10^-4)] mole/liter = [(10^-4.2) / (8.4 × 10^-4)] mole/liter \allequal [(6.3 × 10^-5) / (8.4 × 10^-4)] mole/liter = 7.5 × 10^-2 mole/liter = 7.5 × 10^-2 M.

Question:

A billiard ball is struck by a cue as in figure (a). The line of action of the applied impulse is horizont-al and passes through the center of the ball. The initial velocity v_0^\ding{217} of the ball after impact, its radius R, its mass M, and the coefficient of friction \mu between the ball and the table are all known. (a) How far will the ball move before it ceases to slip on the table and starts to roll? (b) What will its angular velocity be at this point?

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Solution:

(a) Between the time the ball is struck by the cue, and the time it begins pure rolling, friction with the table decelerates it linearly, but simultaneously exerts a torque upon it about its center of mass. This causes the ball to undergo an angular acceleration. The ball begins pure rolling when its linear velocity and its angular velocity have been decreased and increased respectively to the point at which the relation v = R\omega holds. We recognize this as the definition of linear velocity with respect to angular velocity for pure rolling. The force of friction on the ball is by definition: F_f = \muN = \mumg where N = mg is the normal force between the ball and the table. The negative sign indicates that F_f is directed opposite to V_0. The ball's linear acceleration is: a = Ff/ M =- \mug Thus, its linear velocity at time t is given by: v(t) = v_0 + at = v_0 - \mugt The torque on the ball is (see figure (b)) : \cyrchar\cyrt = F_fR = \muMgR Since we know that the moment of inertia of a solid sphere about an axis passing through the center is I as 2/5 MR^2, we can calculate its angular accelera-tion: \cyrchar\cyrt = I\alpha,\alpha = \cyrchar\cyrt/I = (\muMgR)/[(2/5)MR^2] = (5/2)(\mug/R) The ball's angular velocity at time t is given by: \omega(t) = \omega_0 + \alphat,\omega_0 = 0 \omega(t) = \alphat = (5/2)(\mug/R) t To calculate the distance the ball will move before it begins pure rolling, we must first calculate how long it is after the ball has been struck that this occurs. Rolling begins when v(t) = R\omega(t) v_0 - \mugt = R[(5/2)(\mug/R)] t = (5/2) \mugt (7/2) \mugt= v_0 , t = (2/7)(v_0/\mug) The distance the ball travels is therefore, since the acceleration is constant, s = v_0t + (1/2)at^2 = v_0t - (1/2)\mugt^2 s = v_0[(2/7)(v_0/\mug)] - (1/2)\mug[(2/7)(v_0/\mug)]^2 = (2/7)[(v_0^2)/(\mug)] - (2/49) [(v^2_0)/(\mug)] = (12/49) [(v^2_0)/(\mug)] Here v_0 is the ball's initial velocity. (b) Its angular velocity at this point is: \omega(t) = \omega_0 + \alphat \omega(t) = \alphat = (5/2)(\mug/R) [(2/7) (v_0/\mug)] = (5/7)(v_0/R) Since the ball's initial angular velocity \omega_0- = 0.

Question:

Given that 1 gram of sugar, starch, or other dry carbo-hydrate requires one liter of oxygen for complete oxi-dation and one gram of vegetable oil or other dry fat requires about 2 liters, answer the following: (a) Why is there a difference in the volume of oxygen required, (b) If mineral oil, paraffin wax, and hydrocarbons are foods would they have fewer calories per unit weight than vegetable oils? (c) Can hydrocarbons be a food source for man?

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Solution:

Consider what is meant by oxidation and the general chemical composition of the substances mentioned. Complete oxidation reactions may be defined as the addition of oxygen to an organic compound to produce water (H_2O) and carbon dioxide (CO_2) plus heat. The general formula of a fat may be written (RCO_2)_3C_3H_5 , where the R's can be the same or different molecules. In other words, a fat may be defined as an ester of glycerol and three long chain carboxylic acids. Carbo-hydrates are simple sugars or the substances that hydrolyze to yield simple sugars of general formula C(H_2O) x, where x may range from 3 to many thousands. Thus, you can proceed as follows: (a) In fats, from the general formula, you see the percentage of C and H in fats is greater than that in carbohydrates. The percentage of O is less in fats than in carbohydrates. The amount of oxygen required in combustion is a direct function of the C, H, and O percentages in the organic compound. The greater the C, H percentage, the more oxygen is required. Thus, fats should need much more oxygen for complete combustion than carbohydrates. (b) Mineral oil, paraffin wax, or hydrocarbons should yield more Calories per unit weight than vegetable oil. From (a) , you know that the amount of oxygen re-quired is proportional to the amount of heat produced upon oxidation; this is a direct function of C, H, and O percentage in the organic compound. Mineral oil, paraffin wax, and hydrocarbons have a higher C and H content than fats. (Hydrocarbons are 100% H and C whereas fats are part oxygen.) Therefore, mineral oil, paraffin and hydrocarbons yield more calories (or heat). (c) Hydrocarbonscan notbe used as a food source. Food is not useful to the body until the food particles have been degraded and oxidized by enzymes to yield heat. The human body possesses no enzyme systems that can perform the degradation of hydrocarbons.

Question:

Write a FORTRAN program to implement Gauss's elimination method for solving the systems of equations.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0506.htm

Solution:

Consider the system of n linear equations with n unknowns: a_11x_1 + a_12x_2 + ... + a_1nx_n = b_1 a_21x_1 + a_22x_2 + ... + a_2nx_n = b_2(1) .. .. .. a_n1x_1 + a_n2x_2 + ... + a_2nx_n =b_n Gauss's elimination method is used to find a solution of (1) i.e., a set [x_1,...,x_n} such that when it is substituted into (1), all the equations are satisfied. The method is as follows: 1) Divide the first equation by a_11 to obtain x_1 + (a12/ a_11) x_2 + ... + (a1n/ a_11)x_n= b_1 / a_11.(2) 2) Now subtract a_21 times first equation from row 2, a_31 times the first equation from row 3,... , a_n1 times the first equation from row n to obtain x_1 + (a_12 / a_11) x_2 + ... + (a_1n / a_11)x_n= b_1 / a_11(3) w_22x_2 + ... + w_2nx_n = c_2 . . . w_n2x_2 + ... +w_nnx_n=c_n. where the result of adding - (a_j1 (a1j/ a_11)x_j) froma_ijx_ihas been written as w_ijx_j and the result of adding - (a_i1b_1 /a_11) to b_1 has been written asc_i (i= 2,...,n). The method is applied again to eliminate x_2 from the third, fourth,...,n-thequations. Repeated application yields thetridiagonal system: x_1 + v_12x_2 + ... + v_1nx_n = d_1 x_2 + v_23x_3 + ... + v_2nx_n = d_2(4) ...... x_n-1 + vn-1,nxn= d_n-1 x x_n=d_n Now the system can be solved by back-substituting. The X, Y and Z's need not be presented to solve such systems on a computer. The coefficients can be loaded into an array A(N,N+1) and solved in a similar manner. DIMENSION A (N, N+1) CAN EXTRA COLUMN IS NEEDED FIRST TO STORE. CTHE CONTINUOUSLY CHANGING VALUES (b_1....b_n) OF THE CEQUATIONS, AND THEN THE ROOTS OF THE SYSTEN READ (5,10)N DO 40 I = 1, N DO 40 J = 1, N+1 READ (5, 20) A (I, J) 40CONTINUE M = N - 1 L = N + 1 DO 60 K = 1, M K1 = K + 1 DO 60 J = K1, L DO 60 I = K1, N A (I, J) = A (I, J) - A(I, K) / A (K, K)\textasteriskcenteredA (K, J) 60CONTINUE I = N J = I + 1 DO 70 I1=1, N DO 70 J1 = 1, J A (I1, N + 1) = A (I1, N + 1) / A (I1, I1) 70CONTINUE 100I = I - 1 110J = J - 1 B (I, N + 1) = B (I, N + 1) + A (I, J)\textasteriskcenteredA (J, N + 1) IF (J.EQ.0) GO TO 120 GO TO 100 120I = N I = I - 1 J = J - 1 A (I, N + 1) = I / A (I, I)\textasteriskcentered(A (I1, N + 1) - B (I, N + 1)) 130IF (I.EQ.0) GO TO 140 GO TO 120 140WRITE (3, 30) (A (I, N +1), I = 1, N) STOP END Note: The formats 10, 20, 30 are not included because they depend on actual data.

Question:

What are some practical applications of plant hormones?

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Solution:

Auxins have a variety of practical uses and are of tremendous economic importance. Because auxins stimulate root growth, it is possible to propagate a plant vegetatively by obtaining a cut stem containing an apical meristem which can produce auxin. Root formation will occur on the cut stem because of the presence of auxin. A new plant will grow from the cut stem. By treating the female flower part with auxin, it is possible to produce parthenocarpic fruits (fruits formed without pollination and hence without seed, such as seedless oranges and grapes). Auxin-treated fruits grow larger, ripen faster, and remain on the plant until they can be harvested. Plants sprayed with synthetic auxins retain their leaves, branches, and flowers, as well as their fruits, for longer periods of time. The synthetic auxin 2, 4-D (2, 4-dichlorophenoxy acetic acid) is widely used as a weed killer. Most of the common weeds are dicotyledonous plants which are more susceptible to the auxin than are the monocots. The dicots take up the 2, 4-D and are stimulated to metabolize at such a high rate that they consume their own cellular constitu-ents and eventually die. Many of the monocots are important food sources, such as corn, oats, rye, barley and wheat. Auxins are especially useful since they destroy the dicotyledonous weeds, but leave the monocotyledonous crop untouched. Like auxins, gibberellins can cause the development of parthenocarpic fruits, including apples, currants, cucumbers, and eggplants. When treated with both auxin and gibberellin, these plants produce fruits more than twice as large as those obtainable by the application of either hormone alone. Application of gibberellin to some species of long-day (or short-night) plants induces flowering without appropriate long-day exposure. Gibberellins will also substitute for the dormancy-breaking requirement and induce germination by promoting the growth of the embryo and the emergence of the seedling. Treatment of a plant with cytokinin causes the loss of apical dominance and the plant grows bushier. It also prevents leaves from yellowing, and thus has particular commercial value in indoor, decorative plants.

Question:

Given the equilibrium H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbarand HS\Elzbar+ H_2 O \rightleftarrows H_3 O^+ + S\Elzbar2, find the final concentration of S\Elzbar2, if the final concentrations of H_3 O^+ and H_2 S are 0.3 M and 0.1 M, respectively. k_1 = 6.3 × 10\Elzbar8for H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbarand k_2 = 1 × 10\Elzbar14for HS\Elzbar+ H_2 O \rightleftarrows H_3 0^+ + S\Elzbar2.

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Solution:

This problem can be solved by writing the equilibrium constant expressions for the equilibrium. These expressions give the ratio of the concentrations of products to reactants, each raised to the power of its coefficient in the equilibrium equation. Therefore, for H_2 S + H_2 O \rightleftarrows H_3 0^+ + HS\Elzbar, we can write k_1 = 6.3 × 10\Elzbar8= {[H_3 0^+] [HS\Elzbar] } / [H_2 S] For HS\Elzbar+ H_2 O \rightleftarrows H_3 0^+ + S\Elzbar2, k_2 = 1 × 10\Elzbar14= {[H_3 0^+] [S\Elzbar2]} / [HS^\rule{1em}{1pt}] Note: water concentration is not included in the equi-librium expression since its concentration is assumed to be constant. One is not given any information concerning [HS\Elzbar]. [HS\Elzbar] is common to both k_1 and k_2 , so that if one solves one equation for [HS\Elzbar] and substitutes it into the other equation, [HS\Elzbar] is eliminated. Proceed as follows: If one solves for [HS\Elzbar] in k_1 one obtains [HS\Elzbar] = {k_1 [H_2 S] } / [H_3 0^+ ] = { 6.3 × 10\Elzbar8[H_2 S] } / [H_3 0^+ ] Substituting this into k_2 , one obtains k_2 = 1 × 10\Elzbar14= {[H_3 0^+] [S\Elzbar2]} / [HS^\rule{1em}{1pt}] = {[H_3 O^+] [S\Elzbar2]} / ({6.3 × 10\Elzbar8[H_2 S]} / [H_3 0^+ ]) Rewriting in terms of [S\Elzbar2] , [S\Elzbar2] = { (1 × 10\Elzbar14) (6.3 × 10\Elzbar8) [H_2 S] } / [H_3 0^+ ]^2 One is given that [H_2 S] and [H_3 O^+] equal 0.1 M and 0.3 M, respectively. Therefore, one can substitute these values into this equation to solve for [S\Elzbar2] . [S\Elzbar2] = {(1 × 10\Elzbar14) (6.3 × 10\Elzbar8) (0.1)} / [0.3]^2 = 7 × 10\Elzbar22M.

Question:

Differentiate clearly between "breathing" and "respiration".

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/Users/wenhuchen/Documents/Crawler/Biology/F16-0391.htm

Solution:

Respiration has two distinct meanings. It refers to the oxidative degradation of nutrients such as glucose through metabolic reactions within the cell, resulting in the production of carbon dioxide, water and energy . Respiration also refers to the exchange of gases between the cells of an organism and the external environment. Many different methods for exchange are utilized by different organisms. In man, respiration can be categorized by three phases: ventilation (breathing), external respiration, and internal respiration. Breathing may be defined as the mechanical process of taking air into the lungs (inspiration) and expelling it (expiration). It does not include the exchange of gases between the bloodstream and the alveoli. Breathing must occur in order for respiration to occur; that is, air must be brought to the alveolar cells before exchange can be effective. One distinction that can be made between respiration and breathing is that the former ultimately results in energy production in the cells. Breathing, on the other hand, is solely an energy consum-ing process because of the muscular activity required to move the diaphragm.

Question:

Using conditioning, how can one determine the sensitivity of a dog's hearing to sounds of different frequencies?

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Solution:

Dogs can usually discriminate between sounds of different frequencies or pitches (measured in cycles per second, cps) . To determine how well they discriminate, one can use operant conditioning. By rewarding or punish-ing a particular act when it occurs, the probability that it will occur again is either increased or decreased. For example, one can expose the dog to a sound frequency of 1000 cps. The dog is then rewarded, whenever it responds to a sound of that frequency by barking. Soon the dog barks whenever it hears a sound of 1000 cps. In the same manner, the dog can be conditioned to raise its paw when-ever it hears sounds of another frequency. By making this frequency closer and closer to 1000 cps, and observing the dog's response, either a bark or a paw lift, one can determine the smallest differences in frequency between which the dog can discriminate. A point is eventually reached beyond, which the dog is unable to discriminate between a "bark\textquotedblright pitch and a "lift-paw" pitch. Dogs have been found to be capable of discriminating between two sounds that differ by only two cycles per second. Generalization is the response that is contradictory to discrimination. This occurs when an animal conditioned to one stimulus responds in the same manner to a similar, though different stimulus. For example, a bird conditioned to peck at a dark blue spot may peck at a violet spot when the blue one is unavailable.

Question:

Perform the following conversions (synthesis) using any inorganic reagents you require:

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Solution:

In any synthesis problem you can use a step by step approach. Try to find reactions that lead to inter-mediates from which the product can be obtained. In (A) you want to shift the hydroxyl group from the 1\textdegree carbon to the 2\textdegree carbon. You can do this through the correspond-ing alkene. Once the alkene is obtained, the addition of acid and water will generate the desired product. To obtain the intermediate, add acid to the starting material. Because it is an alcohol, it will dehydrate to the alkene. You have, then, In (B) you want to switch the CI from the 2\textdegree carbon to the 1\textdegree carbon. Convert the starting material to an alkene and from this the product can be obtained. You have, therefore, the following process: Potassium hydroxide has the ability to dehalogenate an alkyl halide to an alkene. With peroxides, the hydrogen from the HCI adds to the carbon with the least number of carbons attached. You are performing an addition reaction across a double bond. In (C), you are converting the OH group to a NH_2 group. You can convert the OH to CI, first, since upon addition of NH_3, the NH_2 group is obtained. To obtain the chlorine, dehydrate the alcohol with acid to form the alkene, then add HCI. You have the following sequence:

Question:

In 1932, many cities showed higher sunlight levels than in the previous year. Explain this observation.

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Solution:

To answer this problem, consider the effect of atmospheric particles on the amount of sunlight received and what happened in 1932. By 1932 three years of slackened industrial activity- had passed. Then as now, industries released massive a- mounts of particulate matter, i.e. solid and liquid aerosols, into the atmosphere. These particles have the ability to scatter and absorb light. Because industrial activity was drastically reduced, fewer particles were released that could scatter or absorb the sunlight. The cities, therefore, received more sunlight in 1932 than in years when industry was thriving. It is interesting to note that because of particulate matter, cities receive about 15 to 20 % less solar radiation than do rural areas.

Question:

Gas expanding in a gas engine moves a piston in a cylinder and does 2000 ft-lb of work at a constant pressure of 3000 lb/sq ft. How much does the volume change?

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Solution:

The work done by a gas in expanding a piston a distancedsis dW= F^\ding{217} \textbulletds^\ding{217} where F^\ding{217} is the force exerted by the gas on the piston of area A. Since F^\ding{217} andds^\ding{217} are in the same direction, dW= Fds By definition of the gas pressure, p, we have p = F/A orF =pA Using this in the expression fordW, dW = p Ads= pdV wheredVis the change in volume of the gas during the expansion. The net work done by the gas, in ex-panding at constant pressure from volume V_1 to volume V_2, is W = ^V2\int_V1 p dV = p ^V2\int_V1 dV = p(V_2 - V_1) Solving for V_2 - V_1, we obtain V_2 \rule{1em}{1pt} V_1 = W/P = (2000 ft lb)/(3000 lb/ft^2) = (2/3) ft3 for the volume change of the gas.

Question:

Given the balanced equation 4NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2 0 (g), how many grams of NH_3 will be required to react with 80 g of O_2 ?

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Solution:

In this problem, one is asked to find the number of grams of NH_3 required to react with 80 g of O_2. From the equation, one can see that 4 moles of NH_3 react with 5 moles of O_2. This means that one must first determine the number of moles of O_2 in 80 g. One can then use the following ratio to determine the number of moles of NH_3 that will react with this number. This ratio holds because the ratio of NH_3 to O_2 is 4 : 5. (4/5) = [(number of moles of NH_3 ) / (number of moles of O_2 )] After one knows the number of moles of NH_3 required, one can determine the weight of the NH_3 by multiplying the number of moles by the molecular weight. To solve this problem: (1)calculate the number of O_2 in 80 g (2)calculate the number of moles of NH_3 that react with 80 g O_2 (3)find the number of grams of NH_3. Solving: (1)The number of moles of O_2 in 80 g is found by dividing 80 g by the molecular weight of O_2. The molecular weight of O is 16, which means the molecular weight of O_2 is twice that amount, or 32. number of moles = [(number of grams present) / (molecular weight)] number of moles = [(80 g) / (32 g/mole)] = 2.5 moles (2)The number of moles of NH_3 reacting with a certain number of moles of O_2 is in the ratio 4 : 5, as shown previously. (4/5) = [(number of moles of NH_3) / (number of moles of O_2)] (4/5) = [(number of moles of NH_3) / (2.5moles)] number of moles of NH_3 = [(4 × 5 moles) / (5)] = 2 moles 2 moles of NH_3 are necessary to react with 2.5 moles or 80 g of O_2. (3)The number of grams of NH_3 in 2.0 moles is found by multiplying 2.0 moles by the molecular weight of NH_3 MW of NH_3 = 17. number of grams = number of moles × molecular weight number of grams = 2.0 moles × 17 g/mole = 34 g.

Question:

what is the energy content of 1 gm of water?

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Solution:

If the mass of the gram of water was completely converted to energy, the amount of energy released would be E = mc^2 = 1 × 10^-3 × (3 × 10^8)^2 = 9 × 10^13 joules

Question:

Calculate the interatomic nonbonded distance between 2 bromine nuclei in a molecule of carbontetrabromide (CBr_4) . The C-Br distance is 1.94 \AA.

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Solution:

From diagram A, one can see that CBr_4 is in a tetrahedral conformation. Thus, one knows that the Br-C-Br angle is equal to 109.5\textdegree. As such, diagram B is then true. One is solving for the length ofBr_1Br_2. Angle Br_1-C-A is 109.5\textdegree / 2 or 54.75\textdegree, since one can let lineCAbe an angle bisector. To findBr_1\AA, note that sin 54.75 = B_1A / 1.94 A^o. Therefore, Br_1A= 1.94 \AA × sin 54.75 Br_1A= 1.94 \AA × .8166 = 1.58 \AA B_1B_2= 2 ×Br_1A= 2 × 1.58 \AA = 3.17 \AA, sinceBr_1A=Br_2AforCAis also a median.

Question:

500 liters of a gas at 27\textdegreeC and 700torrwould occupy what volume at STP?

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Solution:

STP (Standard Temperature and Pressure) is defined as being 0\textdegreeC and 760torr, thus, in this problem, one is asked to find the new volume of a gas when the temperature and pressure are changed. One refers to the combined gas law in such a case. This law can be stated: For a given mass of gas, the volume is in-versely proportional to the pressure and directly pro-portional to the absolute temperature. Stated algebrai-cally (PV/T) = K where P is the pressure, V is the volume, T is the absolute temperature, and K is a constant. This means that if two of the variables are changed, the third changes so that the relation PV/T = K remains true. This means that one can now state that (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original temperature, P_2 is the new pressure, V_2 is the new volume, and T_2 is the new absolute temperature. In this problem, one is given the temperature on the Celsius scale, it must be converted to the absolute scale before using the combined gas law. This can be done by adding 273 to the temperature in \textdegreeC. T_1 = 27 + 273 = 300\textdegreeK T_2 =0 + 273 = 273\textdegreeK One knowsP_1, V_1, T_1, P_2, and T_2. One is asked to find V_2. Using the Combined Gas Laws: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2P_1 = 700torr V_1 = 500 liters T_1 = 300\textdegreeK P_2 = 760torr V_2 = ? T_2 = 273\textdegreeK [(700torr× 500 liters)/(300\textdegreeK)] = [(760torr× V_2)/(273\textdegreeK)] V_2 = [(700torr× 500 liters × 273\textdegreeK)/(760torr× 300\textdegreeK)] V_2 = 419 liters.

Question:

Write Declare statements for the following variables or con-stants, making use of the rules of PL/I regarding attributes. Also explain the reasons why you selected the particularat-tributes. The variables are given as follows: a) A variable named PERCENT which takes values from 0\textbullet00 to 100\textbullet00.PERCENT is an array of 8 elements. b) A variable named AMOUNT which represents dollars and cents, from 0\textbullet00 to 1 million dollars. c) A variable named NUMBER which represents Avogadro's number= 6\textbullet022×10^23 d) A variable named TOTAL which represents integer numbersfrom 1 to 4000, to be stored in a binary form.

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Solution:

a) We can Declare PERCENT as follows: DCLPERCENT(1:8) FIXED (5,2) REAL DEC; Here, the parentheses (1:8) shows that PERCENT is an array. FIXED specifies the Scale. (5, 2) specifies the precision, which is of the form XXX\textbulletXX. REAL specifies the Mode DEC specifies the Base. b) We can Declare AMOUNT as follows: DCL AMOUNTFIXED(9, 2) DEC REAL; In the above, the words FIXED, DEC and REAL specify the scale, the Base and the Mode respectively.Also, (9, 2) can represent num-bers from -9,999,999\textbullet99 to +9,999,999\textbullet99.Thus, our range of 0\textbullet00 to 1,000,000\textbullet00 canbe easilyaccomodatedin this range. c) We can Declare NUMBER as follows: DCL NUMBERFLOAT(4) REAL DECIMAL; In the above, the value of the variable NUMBER is equal to 6\textbullet022 × 10^23. Hence, if we tried to use the FIXED Scale, we would have to specify its scaleand precision as follows: FIXED (26, 3). This huge number of almost all digits of zero, is very cum-bersome to use, store, etc. Hence, we use instead the Floating point scale as above. The specificationFloat(4) means that the number is of the form X\textbulletXXX, i.e., 4 digits, with the remaining multiplier of powers of ten implicitly provided. A printoutof the NUMBER would be: 6\textbullet022E + 23. d) We declare TOTAL as follows: DCL TOTALFIXED(12) BINARY REAL; In the above, we have specified the Scale as FIXED, the Mode as REAL, and, the Base as BINARY. The reason for specifying the precision as (12) isas follows: We are given that the variable TOTAL takes values from 0 to 4000. And we observe that a number 2^11 = 2048,2^12 = 4096, 2^13 = 8192. Hence, a precision of 2^12 is just right to represent a range from 0 to 4000.

Question:

An astronaut is to be put into a circular orbit a distance of 1.6 × 10^5 m (about 100 miles) above the surface of the earth. The earth has a radius of 6.37 × 106m. and mass of 5.98 × 10^24 kg. What is the orbital speed?

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Solution:

The force between the astronaut and the earth is: F = G (mM/R^2) = ma where G is the gravitational constant, m the mass of the astronaut and his ship, M the mass of the earth, and the letterathe centripetal acceleration of the ship. The term on the far right of the equation is just a state-ment of Newton's second law. We know that centripetal acceleration a, is equal to V^2/R where V is the instantaneous linear velocity of the orbiting body at any time, thus: G(mM/R^2) = m(V^2/R) V = [G (M/R)]^1/2 = [(6.67 × 10^-11 N - m^2/kg^2){(5.98 × 10^24 kg)/(6.37 × 10^6m)}]^1/2 = 7.91 × 10^3 m/sec [Note 1N = 1 {(kg - m)/(sec^2)}].

Question:

Suppose that an electron bounces back and forth along the x axis in a box of length 10^-10 m. (a) What is its lowest energy level? (b) What wavelength of radiation would be emitted if the electron changed from the n = 2 to the n =1 state?

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Solution:

(a) No external forces act on the electron except when it hits the walls and is reflected elastically. There-fore, it experiences no loss of energy and its momentum while it travels between the walls is constant in magnitude. Wilson and Sommerfeld stated a general rule prescribing the quantization of a system which undergoes periodic motion. This rule is \ointp_qdq = nh(1) where p_q is the momentum associated with the periodic coordinate q. (In our case, the x position of the particle is varying periodically. Hence, q = x.). Also, n is an integer, h is Planck's constant, and the integral in (1) is evaluated over 1 period of the variable q. (For this example, we evaluate (1) over 1 period of x). If we consider the electron to be at the left wall at the beginning of its cycle, then \ointp_xdx = ^L\int_0 mvdx + ^0\int_L (-mv) dx = 2mvL.(2) Note that the integral is split into 2 parts because the momentum changes direction in 1 cycle of motion. Using (2) in (1) 2mvL = nh orp = mv = (nh/2L).(3) Assuming the electron to be nonrelativistic, its energy is E = (1/2)mv^2 = (p^2/2m) = (n^2h^2) / (8mL^2). The lowest energy level occurs in the n = 1 state. E_1 = [h^2/(8mL^2)] = [{(6.625 × 10^-34)^2} / {8(9.11 × 10^-31) × 10^-20}]joule. = 6.02 × 10^-18 joule. Because 1 joule = 6.24 × 10^18 eV E_1 = (6.02 × 10^-18 joule) [6.24 × 10^18 (eV/joule)] E_1 = 37.5 eV where v is the photon frequency. (b) The energy of the photon is hv = E_2 - E_1 = [h^2/(8mL^2)] (2^2 - 1^2) hv = [{(6.625 × 10^-34 J \textbullet s)^2}/{(8) (9.11 × 10^-31 kg) (10^-20 m^2)}](4 - 1) = 18.1 × 10^-18 joule. Since 1 joule = 6.24 × 10^18 eV hv = (18.1 × 10^-18 joule) [6.24 × 10^18 (eV/joule)] hv = 113 eV = 18.1 × 10^-18 joule. From c = v\lambda, the wavelength \lambda is \lambda = (hc/E) = [{6.625 × 10^-34(3.0 × 10^8)} / (18.1 × 10^-18)]m = 1.10 × 10^-8 m = 110 A.

Question:

GivenK_ifor acetic acid is 1.8 × 10^-5, calculate the percentage of ionization of 0.5 M acetic acid. The dis-sociation reaction is HC_2H_3O_2 \rightleftarrows H^+ + C_2H_3O_2^-.

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Solution:

K_iis the ionization constant and indicates to what degree acids and bases will dissociate in solution. Acetic acid is a weak acid becauseK_iis so small. The larger the value ofK_i, the greater the % dissociation and the stronger the acid. A useful range of values forK_iis given below. Strength Range Very strong greater than 1 × 10 ^3 strong 1 × 10 ^3 to 1 × 10^-2 weak 1 × 10^-2 to 1 × 10^-7 Very weak less than 1 × 10^-7 K_iis calculated from the ratio of products to re-actants, each raised to the power of its coefficient in the reaction equation; K_i= ({[products]^reaction ^moles ^(or ^coefficient)}/{[reactants]^ reaction ^moles ^(or ^coefficient)}) From the chemical reaction, K_i= {[H^+][C_2H_3O_2^-]}/[HC_2H_3O_2] = 1.8 × 10^-5 To solve this problem, one uses the following method. Before the reaction begins, no dissociation occurs, so that only acetic acid is present, and in its full concentration of 0.5 M. After dissociation occurs, b number of moles/liter has dissociated to the products in equal amounts. Or b moles/liter of each H^+ and C_2H_3O_2^- are produced since the coefficients of the reaction indicate that they are formed inequimolaramounts. If b moles/liter dissociate, then there are 0.5 - b moles/liter of acetic acid left. This can be summarized as follows: Before: 0.5 moles O O After: 0.5 - b b b This gives aK_iof, K_i= {[b][b]} / [0.5 - b] = 1.8 × 10^-5 . SinceK_iis so small, very few moles of acetic acid dissociate, which causes b to be insignificantly small. Therefore, 0.5 - b is approximately equal to 0.5, giving K_i= {[b][b]} / [0.5] = 1.8 × 10^-5 . Solving for b, one obtains: b^2 = 9.0 × 10^-6 b = 3.0 × 10^-3 mole of acetic acid ionized. Using the following equation to find % ionization: % ionization = [(number of moles ionized)/(original number of moles of acetic acid)] × 100 = [(3.0 × 10^-3) / (0.5)] × 100 = 0.60 %.

Question:

Define a fossil and distinguish between the two major kinds of fossils. Discuss their preservation, dating, and usefulness in tracing human evolution.

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Solution:

Fossils are remains or traces of a plant or animal found in the earth. There are basically two types of fossils: physical (anatomical) fossils and cultural fossils (archaeological artifacts). Physical fossils are those which reveal the form and structure of the animal or plant. Examples of physical fossils are bones, teeth and footprints. Cultural fossils differ in that they reveal information concerning the activities of an organism, or a group of organisms. Cultural fossils include tools and ceremonial artifacts. Successful preservation of any type of fossil depends on the material of which it is composed and the environment in which it was formed. Whereas soft tissues of the body deteriorate rapidly in the air and leave little trace, hard bones may remain preserved for millions of years. The most frequently preserved anatomical fossils are teeth, which are coated with dentin . This is one of the hardest substances found in the body. Also commonly found are the larger and thicker bones of the cranium, femur, jaw, and pelvis. Archaeological artifacts found are made of materials that are difficult to decompose. This explains why the obsidian tools (obsidian is a dark volcanic rock resembling glass) made by early man are found in large quantity, whereas no wooden tools have been uncovered from the same period. Preservation of the body of an organism is most effective when death occurs rapidly and an air supply is immediately eliminated. This may occur when an organism falls into a quicksand pit, or is trapped in an avalanche or in the lava flow of a recently erupted volcano. Once the animal has died and the soft tissues begin to decay, minerals from soil or rocks may start to infiltrate these tissues resulting in a well preserved, petrified specimen. However, this preservation of an entire organism trapped by some natural phenomenon is relatively rare. More often only fragments of fossilized teeth and bones are discovered, from which the nature of the unpreserved body parts must be deduced. Digging for fossils is usually a costly enterprise requiring a great deal of time and precision. It is therefore important that a logical site be chosen before the project has begun. Historic water sites, caves and natural shelters are suspicious sites for a digging. When fossils are found, precise maps are drawn relating their positions relative to one another and to the surroundings. Digging is done in levels since the lower a fossil is found the older it usually is. This may not be true in areas of extreme erosion or regions where mountain formation or other natural phenomena such as earthquakes have upset the ordered layering of sediment. The dating of fossils is essential in fitting together the pieces of the puzzle of man's history. The scheme of the evolutionary process can only be determined by knowing which features appeared first in the evolution leading to modern man. There are two methods of dating fossils; (1) relative dating, and (2) radioactive dating. A fossil is dated relatively when the position in which it is found is used as the reference point in time. In other words, what stratum the fossil was found in, along with the depth of the find and the other fossils occurring with it can be used to determine its relative age. Radioactive dating is more exact than relative dating. There are several different types of radioactive dating, for example, carbon-14, potassium-argon, and uranium-lead. They all work on the principle of half-lives. Fossils are dated by measuring the decrease in the level of radioactivity from the calculated original value. C\rule{1em}{1pt}14 is the only method that directly dates the fossils themselves. The others determine the approximate year the fossil was deposited by calculating the age of the rock the fossil is found in. After enough fossils have been obtained and dated, and structural differences determined, the story of man's evolution can be better understood. The story will never be complete, but some trends in structural changes leading to modern man can be observed. Some forms of pre-historic man are believed to have become extinct due to changing climate, competition with other animals (including hominoid forms), and by disease. In the end, the hominoids who were our ancestors, must have possessed distinctive selection factors that permitted their survival, per-petuation and evolution.

Question:

Give a brief summary of the events occurring in spermatogenesis and oogenesis. Compare the two processes.

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Solution:

Spermatogenesis is the production of sperm (male sex cells). It begins with spermatogonia, which are primitive, unspecialized germ cells lining the walls of the tubules in the testes. Throughout embryonic development and during childhood, the diploid spermato-gonia divide mitotically, giving rise to additional spermatogonia to provide for growth of the testes. After sexual maturity is reached, some of the spermatogonia grow and enlarge into cells known as primary spermato-cytes (see figure), which are still diploid. The primary spermatocytes then undergo the first meiotic division, each producing two equal-sized secondary spermatocytes. Each of these in turn undergoes the second meiotic division, forming two spermatids of equal size. The haploid spermatids then undergo a complicated process of maturation, including development of a tail, to become functional sperm. Oogenesis is the process of ovum (female gamete) formation. The immature germ cells of the female are called oogonia and are located in the ovary. Early in development, they undergo numerous mitotic divisions. Then some oogonia develop and enlarge into primary oocytes. At maturation, the primary oocyte undergoes the first meiotic division. The nuclear events are the same as in spermatogenesis, but the cytoplasm divides unequally to produce two cells of different sizes. The small cell is the first polar body and the large cell is the secondary oocyte. This haploid secondary oocyte undergoes the second meiotic division, which also involves unequal, cytoplasmic distribution, and forms a second polar body and a large ootid. The first polar body divides to form two additional second polar bodies. The ootid matures into the egg, or ovum. The polar bodies disintegrate and do not become functional gametes. Both spermatogenesis and oogenesis involve the production and maturation of the gametes. However, there are a few notable differences intrinsic in each process with the most important being the differences in the amount of cytoplasm and the number of viable gametes produced. The egg cell accumulates much more nutrient material during its development than does the sperm. The sperm actually loses cytoplasm during maturation. In addition, the unequal meiotic division of the oocyte results in a much larger gamete than would normal division. As a result, one viable gamete with a considerable amount of cytoplasm is produced in oogenesis. In spermatogenesis, four viable gametes with very little cytoplasm are produced. These differences are a reflection of the different roles of the egg and sperm in reproduction. It is the fertilized egg which will ultimately develop into the new individual. Large amounts of cytoplasm are necessary to provide adequate nutrition for the developing embryo. The un-equal cytoplasmic division solves the problem of reducing chromosome number without losing cytoplasm. Sperm, on the other hand, are specialized for their role in fertilization. Their small size is necessary for motility. The production of large numbers of sperm, in contrast to eggs, is to ensure the occurrence of fertilization against the odds resulting from physical barriers and environmental factors. Fertilization of the haploid ovum by the haploid sperm restores the diploid number in the resultant zygote.

Question:

The recessive gene s produces shrunken endosperm in corn kernels and its dominant allele S produces full, plump kernels. The recessive gene c produces colorless endosperm and its dominant allele C produces colored endosperm. Two homozygous plants are crossed, producing an F_1 which are all phenotypically plump and colored. The F_1 plants are test-crossed with homozygous recessive plants and produce the following progeny: shrunken, colorless4035 plump, colored4032 shrunken, colored149 plump, colorless152 a) What were the phenotypes and genotypes of the original parents?b) How are the genes linked in the F_1? c) Cal-culate the map distance between the two gene loci.

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Solution:

In solving this problem we will apply principles that have been demonstrated in earlier problems. determined by looking at the F_1 and F_2. We are told that all the F_1 offspring are plump and colored. This indicates that the genotypes of these offspring are either SSCC or SsCC, because they show the dominant trait. Now we must decide which one of the two alternatives is correct. This is where the F_2 comes in. If the F_1 plants were SSCC, only offspring expressing the dominant trait could have been produced in the F_2, since the other parent in the test cross was a homozygous recessive. All offspring would have been heterozygous, and in such a state would have expressed only the plump and colored traits. However, four phenotypes were actually produced in the F_1, including the homozygous recessive (shrunken, colorless). Therefore, the F_1 parent had to have carried a copy of both types of recessive genes. The genotype of the F_1 is then SsCc. Since we are told that both parents of the F_1 off-spring were homozygous, their genotypes must be either SSCC and sscc, with phenotypes plump, colored and shrunken, colorless respectively or SScc and ssCC with phenotypes plump, colorless and shrunken, colored re- spectively. We illustrate this with a diagramatic cross: b)Now, looking at the ratios of offspring in the F_2 generation, we see that Now, looking at the ratios of offspring in the F_2 generation, we see that there is a significant departure from the expected 1 : 1 : 1 : 1 ratio when a test cross of a heterozygote with a recessive homozygote is performed. This is indicative of crossover. (See previous problems for more detailed explanation of this and other concepts of linkage). Looking at the ratios in the F_2 generation, we see that the larger groups, which represent the parental linkage combinations, are plump, colored and shrunken, colorless. Each of these separate combinations represents the combination of genes on the parental chromosomes. There-fore, the parental genes must have been linkedSCandsc. We have determined the F_1 genotype to be SsCc. Therefore, in the F_1, the genes are linked . In order to prove this, let us look at how this configu-ration could indeed give rise to the progency in F_2. In meiosis, can produce four types of gametes if crossover occurs: The homozygous recessive can only produce gametes of the sc type. In the cross, then: To calculate the map distances, we first determine the recombination frequency (RF) between the two genes: RF = [(number of recombinants) / (total number of progeny)] = (149 + 152) / (149 + 152 + 4035 + 4032)] = (301) / (8368) = 0.036 An RF of 0.036 would give a map distance of 0.036 × 100 or 3.6 map units. Hence the map distance between the locus for shape and the locus for color is 3.6 map units apart.

Question:

H_2O_3, hydrogen trioxide, a close relative of hydrogen peroxide, has recently been synthesized. It is extremely unstable and can be isolated only in very small quantities. Write a Lewis electron dot structure for H_2O_3.

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Solution:

A Lewis dot structure of a compound shows the arrangement of valence electrons. Valence electrons are defined as an element's outer electrons which participate in chemical bonding. Thus, to write an electron dot structure for H_2O_3, calculate the total number of valence electrons. This can be done by considering the electronic con-figurations of H (hydrogen) and O (oxygen). Hydrogen possesses one valence electron while oxygen has 6 valence electrons. In H_2O_3, a total of (2) 1 + (3) 6 = 20 valence electrons are involved. With this in mind, the Lewis electron dot structure becomes Notice, you have represented the required 20 valence electrons.

Question:

A 1000-gram metal block fell through a distance of 10 meters . If all the energy of the block went into heat energy, how many units of heat energy appeared?

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Solution:

Work is given by the force acting on the block times the distance travelled by the block, when force and distance are in the same direction. Hence, Work = (980 × 1000)dynes× 1000 cm = 98 × 10^7 erg. Since 1 joule = 10^7 ergs, 98 × 10^7 ergs equals 98 joules. There are 4.19 joules/cal. Therefore, Heat=(98 joules)/(4.19 joules/cal) = 23.5 cal approximately.

Question:

Let the force constant k of the spring in the figure be 24 Let the force constant k of the spring in the figure be 24 newtons /meter, and let the mass of the body be 4 kgm. The body is initially at rest, and the spring is initially unstretched. Suppose that a con-stant force p^\ding{217} of 10 newtons is exerted on the body, and that there is no friction. What will be the speed of the body when it has moved 0.50 meters?

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Solution:

The equations of motion with constant acceleration cannot be used, since the resultant force on the body varies as the spring is stretched. However, the speed can be found using the work-energy theorem. This states that the work done by all the forces acting on a body is equal to the change in kinetic energy of the object. Then, if v_-F is the final speed of the block, work = \intp\ding{217}\bullet dx\ding{217}+ \intF\ding{217}\bullet dx\ding{217}= (1/2) m(v2_F - 0^2) = ∆E_Ksince the initial kinetic energy of the body is zero. P is constant, the angle \ding{217} \ding{217} \ding{217} \ding{217} between P^\ding{217} and dx is 0\textdegree and the angle between F^\ding{217} and dx^\ding{217} is 180\textdegree. Also the magnitude of the restoring force is F = kx, by Hookes law. Then P\intdx - \intkx dx= (1/2) mV2_F Px - 1/2 kx^2= (1/2) mV2F 10n ×0.5m - (1/2) × 24(n/m) × 0.25m^2 = 1/2 ×4kg × V2_F orV_F =1(m/sec)

Question:

Determine the quantity of heat required to convert 10 g of ice at 0\textdegreeC to vapor at 100\textdegreeC, For water, heat of fusion = 80 cal/g, heat of vaporization = 540 cal/g, and specific heat = 1 cal/g -\textdegreeC.

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Solution:

Heat is absorbed when ice is transformed from the solid to the liquid state. The heat absorbed when 1 g of solid melts is called the heat of fusion. The heat of fusion for water is 80 cal/g. In calculating the amount of heat needed to raise the temperature of the liquid water, the specific heat must be used. The specific heat is de-fined as the number of calories of heat necessary to raise the temperature of a gram of a substance 1\textdegreeC. The specific heat of water is 1\textdegree, thus 1 calorie is needed to raise the temperature of 1 gram of water 1\textdegreeC. The quantity of heat necessary to convert 1 g of a liquid into a vapor is termed the heat of vaporization. For water, 540 calories are ne-cessary to change 1 g of liquid water at 100\textdegreeC into vapor (steam) at 100\textdegreeC. In this problem three steps are involved. 1) The ice must be melted. 2) The water must be heated from 0\textdegreeC to 100\textdegreeC. 3) The water must be vaporized to steam. Heat is absorbed in each of these steps, so that each must be accounted for. Melting the ice: Because the heat of fusion is 80 cal/g for water, one knows that for every gram of ice melted, 80 cal of heat must be absorbed. Here, there are 10 grams of ice be-ing melted, so the specific heat must be multiplied by 10 to find the amount of heat absorbed. 10 g × 80 cal/g = 800 cal. Therefore, 800 cal of heat are absorbed when the ice is melted. Heating the liquid: To calculate the amount of heat absorbed when the water is heated, the specific heat should be taken into account. The specific heat of water is 1 cal/g\textdegreeC. This means that for every degree of temperature each gram of water is raised, 1 cal of heat is absorbed. In this case, 10 grams of water are raised 100\textdegree. Therefore, the specific heat is multiplied by 10 to take into account the weight of the water and then multiplied (again) by 100 to account for the fact that the water is being raised 100\textdegree. 1 cal/g-deg × 10 g × 100\textdegree = 1000 cal. Therefore, 1000 calories are absorbed when the liquid is heated to 100\textdegreeC. Vaporization: To find the amount of heat absorbed when the water is vaporized, the heat of vaporization must be used. For water, the heat of vaporization is 540 cal/g, which means that for every gram of water vaporized, 540 calories are absorbed. Here, 10 g of water is vaporized so that the heat of vaporization must be multiplied by 10 g to find the amount of heat absorbed. 10 g × 540 cal/g = 5400 cal. To calculate the total amount of heat absorbed, the heat absorbed in all three steps must be added together. melting800 cal heating1000 cal vaporization5400 cal 7200 cal or 7.20 Kcal. Therefore, 7200 calories are needed to melt 10 g of ice, heat it to 100\textdegreeC and vaporize it to steam.

Question:

A chemist wants to calculate Avogadro's number by the inspection of a solid cube of AgCl. The density of the cube is 5.56 g/cm^3. The spacing between the Ag^+ and Cl^- ions in the cube is 2.773 × 10^-8 cm from their centers. From these data, perform this calculation.

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Solution:

To solve this problem, the following quantities must be determined: the volume of one mole of AgCl, cubic edge length of one mole, the number of ions on an edge, and the total number of ions. The total number of ions divided by 2 will yield Avogadro's number. From the atomic weights of Ag and Cl (107.868 amu and 35.453 amu) one mole of AgCl weighs 143.321 grams. Because density = mass/volume, the volume of one mole = (143.321 g/mole) / (5.56 g / cm^3) =25.78 cm^3 / mole. The volume of a cube = (e)^3, where "e" is the length of the cube edge. Therefore, e = ^3\surd{25.78 cm^3/moles} = 2.954 cm / ^3\surdmole. Because the spacing between ions is 2.77 × 10^-8 cm, the number of ions along the edge is (2.954 cm / ^3\surdmole.)^3 / (2.773 × 10^-8 cm) = 1.065 × 10^8 ions / ^3\surdmole. The total number of ions in 1 mole of AgCl must be {(1.065 × 10^8/ ^3\surdmole)^3} = 1.209 × 10^24mole. This number is not Avogadro's number. Because you have two ions per formula unit, there is one Avogadro's number of Ag^+ and one Avogadro's number of Cl^-. It follows, therefore, that you must divide by two. As such, (1.209 × 10^24 / mole) / 2 = 6.04 × 10^23 / mole = Avogadro's number.

Question:

Nitrogen reacts with hydrogen to form ammonia (NH_3). The weight-percent of nitrogen in ammonia is 82.25. The atomic weight of hydrogen is 1.008. Calculate the atomic weight of nitrogen.

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Solution:

When elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. In a given amount of NH_3, 82.25% of its weight is contributed by nitrogen. The weight-percent of hydrogen in ammonia is, then 100 - 82.25 = 17.75%. One mole of nitrogen and 3 moles of hydrogen combine to form one mole of ammonia. Therefore, one mole of nitrogen constitutes 82.25% and 3 moles of hydrogen constitute 17.75% by weight of one mole of ammonia. The atomic weight of an element is equal to the weight of one mole of that element. The following ratio can be set up to solve for the atomic weight (AW) of nitrogen: AW_N / (3 × AW_H) = weight-percent N / weight-percent H Solving for AW_N AW_N= [(3 × AW_H) (weight-percent N)] / (weight-percent H) = [3 (1.008) (18.25)] / 17.75 = 14.01.

Question:

The captain of a submarine fitted with a directional transmitter finds that he receives bad echoes from the sea bed when he is submerged if the transmission di-rection makes an angle greater than 45\textdegree with the vertical [see Figure (A)]. He wishes to transmit a message to shore using radiation which is completely horizontally polarized. How far from the coast should he surface if the receiving point is a house on top of a coastal cliff 500 ft high and he intends to polarize his radiation by reflection on the sea surface? Take the location of the transmitter as 12 ft above the water when the submarine has surfaced.

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Solution:

If the transmission direction makes a small angle with the vertical, most of the radiation passes through the water surface into the air and only a small fraction is reflected back to the ocean bed to produce echoes. The amount reflected increases with the angle until, when the critical angle is reached, reflection is total, and the echo becomes troublesome. The re-fractive index of water for the radiation used is found from Snell's Law, n sin \texttheta_i = n_air sin \texttheta_r where \texttheta_i and \texttheta_r are the angles of incidence and re-fraction for the transmitted signal. We are told that the critical angle is \texttheta_i = 45\textdegree. At the critical angle, \texttheta_r = 90\textdegree. Hence, n = (1/sin 45\textdegree) = \surd2 = 1.414. When the submarine has surfaced, it can produce a completely plane-polarized beam by reflection of the radiation at the Brewster angle from the surface of the sea. The angle required is tan \textphi = n = 1.414.That is, \textphi = 54.75\textdegree. [See figure (B)]. Thus, the distance of the submarine from the cliff is d = s_1 + s_2 From figure (B), however, (12ft/s_1) = tan (90 \rule{1em}{1pt} \textphi) = tan 35.25\textdegree and(500 ft/s_2) = tan (90 - \textphi) = tan 35.25\textdegree ord = (500 ft/tan 35.25\textdegree) + (12 ft/tan 35.25\textdegree) = (512/0.707) ft = 724 ft.

Question:

If all the chemical components found in a living cell were to beput together in the right proportion, the mixture would not functionas living matter. What characteristics would enable usto consider something living?

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Solution:

Living things are characterized by a specific, complex organization: thesimplest functional unit of organization is the cell. The cell possesses a membrane, called the plasma membrane, which delimits its living substancesfrom the surroundings. It also has a nucleus which is a specializedregion of the cell controlling many important life processes. Living things have to consume energy in order to stay alive. This energymust come from food; plants (with some exceptions) have the abilityto manufacture their own food from sunlight and inorganic materials inthe soil and air. Animals, lacking this ability, have to depend ultimately onplants for food. Complex food substances are taken in and degraded intosimpler ones; and in the process, energy is liberated to carry out life sustainingfunctions. Living things are characterized by their ability to move. The movementof most animals is quite obvious - they crawl, swim, climb, run orfly. The movements of plants are much slower and less apparent, but nonethelessdo occur. A few animals, for example sponges and corals, aresedentary - they do not move from place to place. To procure food and othernecessities of life, these animals rely on the beating of hair-like projections, called cilia, to move their surroundings (from which they obtaintheir food) past their body. The ability to respond to physical or chemical changes in their immediatesurrounding is another charac-teristic of living things. Higher animalshave special-ized cells or complex organs to respond to certain typesof stimuli. Plants also show irritability; the VenusFlytrapis a remarkableexample. The presence of an insect on the leaf of this plant stimulatesthe leaf to fold; the edges come together and ensnare the fly. The fly provides a source of nitrogen for the VenusFlytrap. Living things grow, that is, increase in cellular mass. This growth canbe brought about by an increase in the size of individual cells, by an increasein the number of cells, or both. Some organisms, such as most trees, will grow indefinitely, but most animals have a definite growth period whichterminates in an adult form of a characteristic size or age. Another characteristic of living things is the ability to reproduce. No organismcan live forever; hence there must be some way for it to leave descendantsthat can continue its kind on earth. There are a variety of reproductionmechanisms. Some simple organisms split into two after havingattained a certain size. Higher organisms have complex reproductivepatterns involving partners of the opposite sex. The ability of a plant or animal to adapt to its environment is anothercharacteristic of living things. It enables an organism to survive theexigencies of a changing world. Each kind of organism can adapt by seekingout a favorable habitat or mode of life or by undergoing changes tomake itself better fit to live in its present environment.

Question:

Two different compounds of elements A and B were found to have the following composition: first compound, 1.188 g of A combined with 0.711 g of B; second compound, 0.396 g of A combined with 0.474 g of B. (a) Show that these data are in accord with the law of multiple proportions, (b) If the formula for the first compound is AB_2, what is the formula for the second?

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Solution:

(a) The law of multiple proportions can be stated: When two elements combine to form more than one compound, the different weights of one that combine with a fixed weight of the other are in the ratio of small whole numbers. This means that if one solves for the expected amount of B that is used in forming the second compound from the ratio of A : B in experiment one, the experimental amount should be a multiple of the calculated value. This is seen more clearly after looking at the data. In experiment 1, A combines with B in a ratio of 1.188 g A : 0.711 g B or 1 : .598. In experiment 2, A combined with B in a ratio of 0.396 A : 0.479 B or 1 : 1.20. The law of multiple proportions states that .598 should be a small multiple of 1.20. 1.20/.598 = 2, thus the law is supported. (b) If the formula for the first compound is AB_2, one knows that the proportion of the number of moles of A to B is 1 : 2. Thus 1 unit volume of A weighs 1.188 g and 2 units of B weigh 0.711 g. Therefore, 1 unit of B weighs 0.711 g/2 = 0.356 g. Using this data, one solves for the number of units of A in the second compound by dividing the weight of A in compound 2 by 1.188 g. no. of units of A in compound 2 = (.396g) / (1.188g/unit)= .311 units no. of units of B in compound 2 = (.474) / (.356 g/unit) = 1.331 units The ratio of A to B in compound 2 can now be found. Let x = no. of B atoms A / B = .311 / 1.331 = 1 / x x = 1.331 / .311 = 4.0 Therefore, the second compound is AB_4. This result could also have been obtained by using the data from part (a) . It was determined that twice as much B is present in the second compound. Thus, if the first compound is AB_2, then the second compound must be AB_4.

Question:

Consider the decay of the positively chargedpion, \pi^+ . Pion \ding{217}muon+ neutrino. The energy equivalent of the mass of thepionis about 140 MeV, and themuonand neutrino produced in the decay have rest masses corresponding to energies of 106MeVand zero, respect-ively. How much kinetic energy is available to the particles created in the decay?

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1053.htm

Solution:

The difference of the energy equivalents of the masses of the pion and its decay products is 140 MeV - 106 MeV = 34 MeV. Let us consider the frame of reference in which the pion was at rest. The total energy of the system is given by the energy equivalent of the pion mass. After the decay, the total mass of the decay products is less than thepionmass. The principle of con-servation of energy requires this mass difference to be transformed into another form of energy, hence the two daughter particles (muonand neutrino) share it as kinetic energy.

Question:

Find the speed of acompressionalwave in an iron rod whose specific gravity is 7.7 and whose Young's modulus is 27.5 × 10^6 1b/in^2.

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Solution:

A wave in an iron rod will move with a speed v = \surd(E/\rho) where E is Young's modulus, and \rho is the specific weight of iron. E = 27.5 × 10^6 1b/in ^2 = 27.5 × 10^6 × (1b/in^2) × (144 in^2/ft^2) where we've used the fact that 1^2 1b/in ^2 = 144 1b/ft^2 \rho = (D/g) = {(7.7 × 62.4 1b/ft^3)/(32 ft/sec^2)} = 15.0 slugs/ft^3 Hence, v = \surd{(27.5 × 10^6 × 144 1b/ft^2)/(15.0 slugs/ft^3)} = 1.6 × 10^4 ft/sec.

Question:

Describe the three major structural distinctions between DNA and RNA.

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Solution:

DNA or deoxyribonucleic acid is in the form of a double helix having a deoxyribose sugar and phosphate backbone. The two helices are linked together by hydrogen bonds between nitrogenous bases bound to the sugar moiety of the backbone. In DNA, the bases are adenine, guanine, cytosine and thymine. RNA or ribonucleic acid differs from DNA in three important respects. First, the sugar in the sugar-phosphate backbone of RNA is ribose rather than deoxyribose. Ribose has hydroxyl groups on the number 2 and 3 carbons, whereas deoxyribose has a hydroxyl on the number 3 carbon only (see figure 1). Secondly, RNA has only a single sugar-phosphate backbone with attached single bases. Although hydrogen bonding may occur between the bases of a single RNA strand causing it to fold back on itself, it is not regular like that of DNA, where each base has its complement on the other strand. Thus, RNA is similar to a single strand of DNA. Finally, the pyrimidine base uracil (see figure 2) is found in RNA instead of the pyrimidine base thymine found in DNA. Therefore, the nitrogenous bases present in RNA are adenine, guanine, cytosine and uracil.

Question:

A 20 g chunk of Dry Ice (CO_2) is placed in an "empty\textquotedblright 0.75 liter wine bottle and tightly corked. What would be the final pressure in the bottle after all the CO_2 has evaporated and the temperature has reached 25\textdegreeC?

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Solution:

The final pressure in the bottle will equal the original pressure plus the pressure contributed by the CO_2. Since the bottle was originally open in atmos-pheric pressure, the original pressure is 1 atm. The pressure contributed by the CO_2 is found by using the Ideal Gas Law. This law is stated PV =nRTorP =nRT/ V where P is the pressure of the gas, n is the number of moles, R is the gas constant, 0.082 liter-atm/mole -\textdegreeK, V is the volume, and T is the absolute temperature. One is given R and V in the problem. n is found by dividing the number of grams of CO_2 present by the molecular weight (MW of CO_2 = 44 ). no. of moles = 20 g / (44 g/mole) = .45 moles One can convert \textdegreeC to \textdegreeK by adding 273\textdegreeto the temperature in \textdegreeC. \textdegreeK = \textdegreeC + 273\textdegree T = 25 + 273\textdegree= 298\textdegreeK Solving for P_(CO)2 : P_(CO)2 = [(.45 moles × .082 liter -atm/ mole - \textdegreeK × 298\textdegreeK) / (.75 liter)] = 14.81 atm. P_total= P_(CO)2 +P_original P_total=14.81atm+ 1atm= 15.81 atm.

Question:

What is the structure arid function of the chloroplasts in green plants?

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Solution:

The chloroplasts have the ability to transform the energy of the sun into chemical energy stored in bonds that hold together the atom of such foods (fuel) as glucose. By the process of photosynthesis, carbon dioxide and water react to produce carbohydrates with the simultaneous re-lease of oxygen. Photosynthesis, which occurs in the chloro-plasts, is driven forward by energy obtained from the sun. Photosynthesis involves two major sets of reactions, each consisting of many steps. One set depends on light and cannot occur in the dark; hence, this set is known as the light reaction. It is responsible for the production of ATP from sunlight and the release of oxygen derived from water. This process is known as photophosphorylation. The other set, referred to as the dark reaction, is not dependent on light. In the dark reaction, carbon dioxide is reduced to carbo-hydrates using the energy of ATP from the light reactions. Chlorophyll, the pigment contained in chloroplasts which gives plants their characteristic green color, is the molecule responsible for the initial trapping of light energy. Chlorophyll transforms light energy into chemical energy; then it passes this energy to a chain of other compounds involved in the light and dark reactions. Plastids are organelles which contain pigments and/or function in nutrient storage. Chloroplasts are but one example os a plastid; they give the green color to plants. Each chloroplast is bounded by two layers of membranes; the outer one is not connected with the other membranes of the cell. The inner membrane gives rise to the complex internal system of the chloroplast. Surrounding the in-ternal membranes is a ground substance termed the stroma. The stroma contains the enzymes which carry out the dark reactions of photosynthesis. Granules containing starch are also found in the stroma. The internal membrane system of the chloroplast are usually in the form of flattened sacs called thylakoids (see accompanying diagram) . A stack of thylakoids is called a granum (plural - grana). The lamel-lae contain the chlorophyll molecules and other pigments involved in photosynthesis. The layered structure of the granum is extremely crucial for the efficient transfer of energy from one pigment molecule to another without the dissipation of a great deal of energy in the process.

Question:

It is easy to understand how selfish behavior can be explained by natural selection. But how can one account for altruism?

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Solution:

Selfish behavior is exemplified in the formation of dominance hierarchies. The high-ranking animals eat before the other animals when food is scarce and take preference in the selection of mates and resting places. Consequently, these individuals are generally healthier, live longer and reproduce more frequently. Natural selection favors these strong animals, in that they produce more offspring having their superior traits. Selfish behavior is expected to be transferred to future generations since it leads to the more successful survival of the animal. Altruism is self-sacrificial behavior that benefits other members in a group. An altruistic animal reduces its own fitness, and therefore its chances of survival and reproduction, in order to increase the success of another animal or of the group as a whole. For example, a male baboon remains to fight a leopard while the troop escapes. Natural selection would not favor the continuation of this costly behavior since those who do show it are less likely to survive and reproduce. One would therefore expect the genes for altruism to be lost, but they are not. The reason for this lies in the fact that the male baboon - only risks its life for baboons in his own troop, not for baboons in other troops. His behavior therefore benefits only those baboons related to itself; these baboons probably share common genes, among which are those for altruistic behavior. The al-truistic behavior thus preserves the genes in the other members that are shared with the altruistic member by common descent. For the altruistic genes to be favored in natural selection, the benefit that the related members receive must outweigh the loss of the altruistic animal. Although the baboon might die and lose its ability to pass on its genes, the baboon's brother can compensate for the loss. To do so, it must have more than twice as many offspring as a result of its brother's altruistic behavior. (Since about 1/2 of the genes of a given animal are shared by its brother or sister.) Similarly, since first cousins share 1/4 of their genes, on the average, compensation would involve quadrupling the survivor's offspring. Altruism directed at a variety of relatives, from offspring to cousins, results in a summation of the benefits so that the altruistic gene(s) may be preserved and the related behavior may evolve. Thus, the existence of altruism can be accounted for by natural selection. To demonstrate the compensation by a relative for the loss of an altruistic relative, we will assume that altruism is determined by one allele, A. If one offspring of F_1 generation dies, its brother can compensate by doubling its offspring, (F_2). This will double the frequency of A in F_2 (2 × 1/4) to give 1/2, the frequency that A would have had in F_1 if the brother had not died. Similarly, if one altruistic offspring in the F_2 generation dies, its cousin (also in F_2) can compensate by quadrupling its offspring, F_3. Four times the frequency of A in F_3 (4 × 1/8) gives 1/2, which would be the frequency of A in F_1 if the cousin had not died. It is important to note that compensation must always be referred back to the F_1 generation, since that is the only way to prevent a loss of the altruistic gene.

Question:

A proton is accelerated from rest for 1 nanosecond [= 10^\rule{1em}{1pt}9 s] by an electric field E = 3 x 10^4 volts/meter. What is the final velocity (see figure)?

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Solution:

The general equation of motion involving time, constant acceleration, and velocity is v_f = v_0 + at. The acceleration is given by the electric force, eE, divided by the mass of the proton. v_0, the initial velocity, is zero. Therefore, the expression for v_f is: vf= at = (eE/Mp) t e = 1.6 x 10^\rule{1em}{1pt}19 coulombs E = 3 × 10^4 volts/meter M_p = 1.67 × 10^\rule{1em}{1pt}27 kg t= 1 x 10^\rule{1em}{1pt}9 sec v_f = [(1.6 × 10^\rule{1em}{1pt}19)(3 × 10^\rule{1em}{1pt}4)(1 × 10^\rule{1em}{1pt}9)] / [1.67 × 10\rule{1em}{1pt}27] \approx 2.9 × 10^3 meters/sec

Question:

A sample of gaseous krypton, maintained at constant press-ure , is found to have a volume of 10.5 l at 25\textdegreeC. If the system is heated to 250\textdegreeC, what is the resulting volume?

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Solution:

We need a relationship between volume and temperature. Such a relationship is provided by Charles' Law, which states that volume V and absolute temperature T are proportional, or, as an equality, V =kT, where k is a constant. We can determine k for our system from the initial volume and temperature . Thus, K = (V/T) = [(10.5 l)/(25\textdegreeC)] = [(10.5 l)/(298.15\textdegreeK)] = 0.0352 l-\textdegreeK^-1 The volume corresponding to a temperature of 250\textdegreeC, which is 523.15\textdegreeK, is V =kT= 0.0352l-\textdegreeK^-1 × 523.15\textdegreeK = 18.41 l.

Question:

Describe the data types available in Pascal.

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Solution:

Pascal has 4 standard data types. Used to represent signed and unsigned integers. ex: -25, 600, 37523 Used to represent real numbers. ex: -7.18, 3.1459, 2.718, 1.414, 17,0 Also in scientific notation: 3.14E02, 3.14E-02 3. Char. Used to represent character data. ex: 'a' , ' '(blank), '$' , ')' 4.Boolean. Has only two values - TRUE and FALSE. In addition to these four standard types, there are enumer-ated, or user defined types. ex: TYPE DAYOFWEEK = (sun, mon, tue, wed, thu, fri, sat); color = (red, blue, green, yellow); There are also subrange types, which are some ranges of in-teger, char, or enumerated types. ex: TYPE WORKDAY = (mon . . fri); Letters = ('A' . . 'Z'); DIGITS = (0 . . 9); All of the above types are scalar types. In addition to these, Pascal also has structured data types, composed of a number of scalar types. Examples are sets, arrays and records.

Question:

A steel shaft 12 ft long and 8 in. in diameter is part of a hydraulic press used to raise up cars in a garage. When it is supporting a car weighing 3200 lb, what is the decrease in length of the shaft? Young's modulus for steel is 29 × 10^6 lb\bulletin^-2.

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Solution:

The formula for Young's modulus is Y = (Stress)/(Strain) = (F_n/A)/(∆l/l_0) where F_n is the normal force at the end of the shaft compressing it, A is the cross sectional area of the shaft, l0is the length of the shaft when no stress is present and ∆l is the change in length due to the compression. A is given by \piR^2 = (\piD^2)/4, where R and D are the radius and diameter of the cross section of the shaft. Thus, the decrease in length is ∆l = (F_nl_0)/(YA) = (3200 lb × 144 in.)/(29 × 10^6 lb\bulletin^2 × 16\pi in^2) = 3.16 × 10^-4 in.

Question:

Write the structural formulas for all chlorine derivatives having molecular formula C_5H_11CI.

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Solution:

In hydrocarbon derivatives, the hydrogen atoms are replaced by other atoms or groups of atoms, such as oxygen, chlorine, hydroxyl (-OH), or nitro (-NO_2), which produce chemically active centers in an otherwise less active molecule. To solve this problem, one first writes the structural isomers with five carbons in a continuous chain, in which each different hydrogen atom of the alkane has been replaced by a chlorine atom: Their names are (a) 1-chloropentane, (b) 2-chloropentane, (c) 3-chloropentane. Second, one writes the structural isomers with four carbons as the parent molecule, in which two hydrogen atoms have been replaced by a chlorine atom and a methyl group: Their names are (d) 1-chloro-2-methylbutane, (e) 1- chloro - 3-methylbutane, (f) 2-chloro -2-methylbutane, and (g) 2-chloro -3-methylbutane. Third, one writes the structure of the isomer with three carbons for a parent molecule, in which three hydrogen atoms of the three carbon chain have been re-placed by a chlorine atom and two methyl groups: This compound's name is 1-chloro -2, 2-dimethylpropane or 1-chloro neopentane. Hence, there are eight structural isomers with the molecular formula C_5H_11CI.

Question:

Determine the phase velocity of the de Broglie waves associated with a neutron which has an energy of 25 eV.

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Solution:

If the neutron of mass m has energy 25 eV = 25 × 1.062 × 10^-19 J = 4.00 × 10^-1 8 J, then its speed is given by the relation (1/2) mv^2 = 4.00 × 10-1 8Jor v = \surd[(8.00 ×10-1 8J)/(1.67 ×10-2 7kg)] = 6.92 × 10^4 m \textbullet s^-1 . The phase velocity of the associated de Broglie waves is then v_p = (c^2/v) = [{(3.00 × 10^8)^2m^2 \textbullet s^-2}/(6.92 × 10^4 m \textbullet s^-1)]. = 1.30 × 10^12 m\bullets^-1 The difference between the phase and group velocities can be seen in the figure. When sinusoidal waves of different frequency (or wavelength) are combined, they appear as in the figure, with their amplitude modulated. Clearly, the frequency of variation of the amplitude of this resultant wave is less than the frequency of variation of the wave enclosed by the amplitude envelope. Similarly, the velocity of the envelope (v_group) is different from the velocity of the enclosed wave (v_phase). However, the 2 are related by v_phase = (c^2/v_group) where c is the speed of light in the case of de Broglie waves. Since the envelope locates the approximate po-sition (due to the uncertainty principle) of the particle, the group velocity of the wave must represent the actual velocity of the particle, or the particle's wave would not be able to keep up with the particle.

Question:

Explain the buffering action of a liter of 0.10 M acetic acid containing 0.1 mole of sodium acetate. In the ex-planation, use ionic equations.

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Solution:

By buffering action, one means the ability of a substance to maintain relatively constant conditions of pH in the face of changes that might otherwise affect the acidity orbasicityin solution. For example, a weak acid, such as acetic acid, will dissociate according to the following equation: HOAc+ H_2O \rightleftarrows H_3O^+ +OAc^- . The sodium acetate particle will be completely ionized in solution: NaOAc\rightleftarrows Na^+ +OAc^-. These are the two processes that occur in this buffer. Suppose one increases the acetate concentration. By doing this the equilibrium is shifted to the left, i.e., to acetic acid. k_a = {[H_3O^+][OAc^-]} / [HOAc]or[H_3O^+] = ([HOAc] / [OAc^-]) k_a . The key to the buffering action is this [HOAc] to [OAc^-] ratio. If [HOAc] changes, [OAc^-] will change accordingly so that the same value of the ratio is obtained. Thus, the ratio is a constant. This means, therefore, that [H_3O^+] is maintained as a constant. pH = - log [H^+], it is also constant.

Question:

Write a trapezoidal integration routine in FORTRAN to evaluate^.4\int0 sin2 xdx+ ^.3\int_.1 cos2 xdx.

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Solution:

Let us structure the problem as follows: there shall be a main routinewhich calls a trapezoid function routine, which in turn references twofunction routines defining the functions sin2x and cos2x. First note that accordingto the trapezoid rule ^b\int_af(x)dx\approx (h/2)[f(a) + 2f(a+h) + 2f(a+2h) + ... + 2f(a+(n-1)h)+f(b)], whereh = (b-a)/n is the number of trapezoids. Thus we have FUNCTIONTRAP(A,B,N,FX) H = (B - A)/N SUM = 0 K = N - 1 DO 6 I = 1,K 6SUM = SUM +FX(A + I\textasteriskcenteredH) TRAP = (FX(A) + FX(B) + 2.\textasteriskcenteredSUM)\textasteriskcentered(H/2.) RETURN END The main calling program and function routines are given below. EXTERNAL FX1, FX2 APPROX =TRAP(0.,.4,5,FX1) + TRAP(.1,.3,5,FX2) PRINT, APPROX STOP END FUNCTION FX1(X) FX1 = (SIN X)\textasteriskcentered\textasteriskcentered2 RETURN END FUNCTION FX2(X) FX2 = (COS X)\textasteriskcentered\textasteriskcentered2 RETURN END Notice that the main program uses an EXTERNAL statement. This declara-tionmust be used in every calling program which passes the name ofa subprogram or built-in function to another subprogram. Also rememberthat the main program is entered first. After the END statement inthe main program, all subroutines may be entered. In this program, it is up to you to define the accuracy of the integra-tion. In otherwords, you choose the value of N, which determines the number of iterationsto be done.

Question:

If you administer too much insulin to a diabetic, it may cause a condition called "insulin shock". Explain this condition. Why does it come about?

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Solution:

To solve this problem, you must know how insulin, a hormone, (chemicals that stimulate or depress certain functions of the body) works in the body. The addition of insulin in the body stimulatesglycogenesis, a process which removed glucose from the blood and converts it to glycogen. Insulin also facilitates the transport of glucose across the membranes of muscle cells and adipose (fatty) tissue. If too much insulin is added, too much glucose is removed from the blood. The concentration of glucose in the blood becomes inadequate, a condition called hypoglycemiaresults.This"insulin shock" or hypoglycemia may be characterized by all or any of the following symptoms: weakness, trembling, profuse perspiration, rapidheart beat, delirium, and loss of consciousness.

Question:

Distinguish between viviparous, ovoviviparous, and oviparous reproduction.

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Solution:

The females of all birds, amphibians and bony fish, most insects, and many aquatic invertebrates lay eggs from which the young eventually hatch; such animals are said to be oviparous (egg-bearing). In such cases, the major part of embryonic development takes place outside the female body, even though fertilization may be internal. The eggs of oviparous animals therefore contain relatively large amounts of yolk, which serves as a nutrient source for the developing embryo. Mammals, on the other hand, have small eggs with comparatively little yolk. The mammalian embryo develops within the female's body, deriving nutrients via the maternal bloodstream, until its development has proceeded to the stage where it can survive independently. Such animals are termed viviparous (live-bearing) . The third type of reproduction is intermediate in character. This is ovoviviparous reproduction and it in-volves the production of large, yolk- filled eggs which remain in the female reproductive tract for considerable periods following fertilization. The yolk of the egg is the nutrient source for the developing embryo, which in this case, usually forms no close connection with the wall of the oviduct or uterus and does not receive nourishment from the maternal blood. A diverse group of animals, including snails, trichina worms, flesh flies, sharks, and rattle-snakes are in this category.

Question:

Give a list of the built-in single-precision mathematical functions that can be used in FORTRAN language.

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Solution:

NAME FUNCTION TYPE OF ARGUMENT TYPE OF RESULT ABS(X) IABS(I) SQRT(X) SIN(X) COS(X) ATAN(X) ATAN(X,Y) ASIN(X) ACOS(X) TAN(X) COT(X) ALOG(X) AL0G 10(X) EXP (X) GAMMA(X) INT(X) FLOAT(I) IFIX(X) AMOD(X,Y) MOD(I, J) SIGN(X,Y) ISIGN(I,J) DIM(X,Y) IDIM(I,J) ERF(X) ERFC(X) AMAX 1(U,V,X...) MAX 0(I,J,K...) AMIN 1(U,V,X...) MIN 0(I,J,K...) Absolute value of X Absolute value of I Square root of X Sine of X Cosine of X Principal value of arc Tangent of X Arc-tangent of X/Y Principal value of arc sine of X Principal value of arc cosine of X Tangent of X Cotangent of X Natural logarithm of X Common logarithm of X e^X Gamma function of X Integer part of X Converts integer to real Converts real to integer Remainder of X/Y Remainder of I/J Sign of Y times ABS(X) Sign of J times IABS(I) Difference between X and the smaller of X and Y Difference between I and the smaller of I and J Error function of X Complement error function of X Largest of real constants Largest of integer constants Smallest of real constants Smallest of integer constants real integer real real real real real real real real real real real real real real integer real real integer real integer real integer real real real integer real integer real integer real real real real real real real real real real real real real integer real integer real integer real integer real integer real real real integer real integer

Question:

Through several successive reactions, a chemist uses carbon,CaO,HCland H_2 O to produce C_6 H_4, Cl_2. Assuming an efficiency of 65 %, how much C_6 H_4 Cl_2 can be produced from 500 grams of carbon? Assume that 1/3 of the carbon is lost as 3 moles CO.

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Solution:

In solving this problem, you must account for all carbon atoms and employ the mole concept. You need not be concerned with the actual sequence of reactions nor the roles ofCaOand H_2 O. Dichlorobenzene, C_6 H_4 Cl_2, consists of 6 carbon atoms. You can determine that there were originally 9 moles of carbon present since 3 moles of CO are produced and the carbon present in the CO represents 1/3 of the original amount of carbon present. A mole is defined as weight in grams/molecular weight. The molecular weight of carbon is 12. You started with 500 grams. Therefore, the number of moles of carbon is 500/12. It is stated above, however, that for every 9molesof C, 1 mole of C_6 H_4 Cl_2 was produced. Therefore, the number of moles of C_6 H_4 Cl_2 is 1/9 of the moles of carbon that you started with. Namely, the number of moles of C_6 H_4 Cl_2 is 1/9 × 500/12. The problem calls for an efficiency of 65 %. Therefore, we must multiply this number of moles by (65/100)or,(1/9) × (500/12) × (65/100). The molecular weight of C_6 H_4 Cl_2 is 147. Recalling the definition of a mole, the weight of C_6 H_4 Cl_2 produced is 147 × (1/9) × (500/12) × (65/100) = 442 grams.

Question:

A sealed glass bulb contained helium at a pressure of 750 mm Hg and 27\textdegreeC. The bulb was packed in dry ice at - 73\textdegreeC. What was the resultant pressure of the helium?

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Solution:

The only parameters mentioned in this problem are pressure (P) and temperature (T) . One is given the initial pressure and temperature and the final temperature. One is asked to determine the pressure at - 73\textdegreeC. The Law of Gay-Lussac relates temperature and pressure. It can be stated mathematically (P_1 /T_1 ) = (P_2 /T_2 ) where P_1 is the initial pressure, T_2 the initial absolute temperature, P_2 the final pressure and T_2 the final absolute temperature. The temperature in \textdegreeC is converted to \textdegreeK by adding 273. T_1 = 27\textdegreeC + 273 = 300\textdegreeK T_2 = - 73\textdegreeC + 273 = 200\textdegreeK Solving for P2 [(750 mm Hg)/(300\textdegreeK)] = [(P_2 )/200\textdegreeK)] P_2 = [{(750 mm Hg) (200\textdegreeK)}/(300\textdegreeK)] = 500 mm Hg.

Question:

A brass rod of density 8.5 g\textbulletcm ^\rule{1em}{1pt}3 and length 100 cm is clamped at the center. When set into longitudinal vibration it emits a note two octaves above the fundamental note emitted by a wire also of 100-cm length weighing 0.295 g and under a tension of 20 kg weight which is vibrating transversely. What is Young's modulus for brass?

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Solution:

The rod vibrates with its center clamped (see figure). Its fundamental frequency of vibration is thus such that the center of the rod is a node and each of the ends an antinode. The length of the rod, L, is half a wavelength (\lambda). Thus \lambda = 2L. Further, the speed of sound in the rod is c = \surdY/\rho, where Y is Young's modulus for brass and \rho its density. Hence the frequency of vibration is f = (c/\lambda) = (1/2L) \surd(Y/\rho) For the vibrating wire the length is the same and, if \mu is the mass per unit length of the wire (\mu = m/L) the frequency of the fundamental vibration is f^1 = (1/2L) \surd(S/\mu) = (1/2L) \surd(SL/m), where m is the mass of the wire. But f = 4f_1, since one frequency is two octaves above the other. Hence (1/2L) \surd(Y/\rho) = (4/2L) \surd(SL/m) orY = (16\rhoSL)/m = [(16 × 8.5 g\textbulletcm ^\rule{1em}{1pt}3 × 2 × 10^4 kg × 981 cm\textbullets ^\rule{1em}{1pt}2 × 100 cm)/(0.295 g)] = 9.04 × 101 1dynes\textbulletcm^\rule{1em}{1pt}2 .

Question:

A current of 30 amp Is maintained in a thin, tightly wound coil of 15 turns, with a radius of 20 cm. What is the magnetic induction at the center of the coil?

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Solution:

The magnetic induction at the center of a coil of radius R with one turn can be found by using the Biot-Savart Law. dB^\ding{217} = [(\mu_0 i)/(4\pi)] [(dl^\ding{217} × R^\ding{217})/(R^3)] Where the permeability constant \mu_0 = 4\pi × 10^[-7\omega/(a-m)] , and dl^\ding{217} is an element of the coil in the direction of the current i. From the figure we have dB = [(\mu_0 i)/(4\pi)] [(dlR sin 90 o )/(R^3 )] = [(\mu_0 i)/(4\pi)] where R^\ding{217} and dl^\ding{217} are 90 o apart. We have used the magnitude of dB^\ding{217} because all the infinitesimal contributions to the magnetic induction from the infinitesimal lengths dl^\ding{217} are in a direction perpendicular to the plane of the coil. Therefore the total magnetic field at the center is the sum of the infinitesimal contributions dB or B = \intdB = [(\mu_0i)/(4\pi)] (1/R^2) \intdl = [(\mu_0i)/(4\pi)] [(2\piR)/(R^2) = (\mu_0/2) (i/R) The magnetic induction at the center of a coil containing N coils will be equal to the sum of the contributions due to each of the coils, or B_T = NB = [(\mu_0N^i)/(2R )] = [(4\pi × 10^-7 weber)/(amp-m)] [(15 × 30 amp)/(2 × 0.02m)] = 1.4 × 10^-3 weber/m^2 The direction of B_T is perpendicular to the plane of the orbit.

Question:

The hydrated K^+ and Cl^- ions are approximately spherical in shape, with a diameter of 6.0 \AA. What fraction of the total volume of the water phase of an E. coli cell is occupied by these ions?

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Solution:

A hydrated ion is an ion that is completely surrounded by water molecules. Water gathers around ions because of the strong electrostatic attraction between the water dipoles and the ions. The O atom of a water molecule is slightly negative and is attracted to the positive K^+ ion and the H atom of the water is slightly positive and is attracted to the negative Cl^- ion. The concentration of KCl in an E. coli cell is (150 × 10^-3) M. Because KCl dissociated, into K^+ and Cl^\rule{1em}{1pt} ions, the ion concentration in the cell is 2(150 × 10^-3 M) . If one knows the volume of the aqueous phase of an E. coli cell, one can determine the number of moles of the ions present. An E. coli is cylindrical; it is 2.0\mu long and has a diameter of 1.0\mu. The volume of a cylinder is equal to l\pir^2, where l is the length of the cylinder and r is the radius of its cross-section. 1\mu = 10^-4 cm. Volume of a single E. coli cell = (2.0 × 10^-4 cm) (\pi) × (0.5 × 10^-4 cm)^2 = 1.57 × 10^-12 cm^3 = 1.57 × 10^-15 l 80% of the E. coli cell is in the aqueous phase. and the K^+ and Cl^- ions will only exist in this aqueous phase. Volume of the aqueous phase = .80 × (1.57 × 10^-15 l) = 1.26 × 10-15l . The number of moles of ions can now be found. Moles of ions = (2) (150 × 10^-3 moles / liter) (1.26 × 10^-15 l) = 3.78 × 10^-16 moles There are 6.02 × 10^23 ions per mole. No. of ions = (3.78 × 10^-16 moles) (6.02 × 10^23 ions / mole) = 2.28 × 10^8 ions. Each ion is taken to be spherical with a diameter of \AA. The volume of a sphere is equal to 4/3 \pir^3 , where r is the radius. 1 \AA = 10^-8 cm. Volume of 1 ion = (4 / 3) \pi(3.0 × 10^-8 cm)^3 = 1.131 × 10^-22 cm^3 The total volume occupied by the ions is 2.28 × 10^8 ions times the volume of 1 ion. Total volume of ions = (2.28 × 10^8 ions) × (1.131 × 10^-22 cm^3 / ion) = 2.58 × 10^-14 cm^3 The fraction of the total volume of the cell occupied by the ions is equal to the volume of the ions divided by the volume of the cell. fraction of volume occupied by the ions = (2.58 × 10^-14 cm^3) / (1.57 × 10^-12 cm^3) = 1.65 × 10^-2 or about 2%.

Question:

Calculate and print out the area of a rectangle whose width and length vary in integral steps from 1 to 4.

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Solution:

In this problem a "FOR - NEXT\textquotedblright loop is being introduced. It has a general form of: FOR variable = expr_1 TO expr_2 STEP expr_3 ..................... NEXT variable where expr_1 is the initial value of the variable, expr_2 - the final value, and expr_3 indicates the value by which the variable has to be incremented after each pass through the loop. If the STEP part is omitted, it is assumed to be 1. We use two nested "FOR ... NEXT" loops as shown below. 1\O REM CALCULATES RECTANGLE'S AREA 2\O FOR W = 1 TO 4 3\O FOR L = 1 TO 4 4\O PRINT "L ="; L; "W = "; W; "A ="; L\textasteriskcenteredW 5\O NEXT L 6\O NEXT W 7\O END

Question:

The molar absorptivity of phytochome (a light-sensitive pigment thought to control the process of flowering in plants) is 76,000. What will be the absorbance of a 5.0 × 10^-6 molar aqueous solution of phytochome if the pathlength of the cell is 1.0 cm?

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Solution:

The fraction of the incident light absorbed by solution at a given wavelength is related to the thickness of the absorbing layer and to the concentration of the absorbing species. These two relationships are combined into the Lambert-Beer Law, given in integrated form as log (I_0 / I) = acl, where I_0 is the intensity of the incident light. I is the intensity of the transmitted light, a is the molar absorb-ing index (the molar extinction coefficient), c is the concentration of the absorbing species in moles per liter, and l is the thickness of the light absorbing sample. Log (I_0 / I) is also known as the absorbance A. Solving for A, A = acl A = (76,000) (5.0 × 10^-6 moles / liter) (1 cm) = 0.38.

Question:

Find the pH of a 0.1 M solution of ammonia, NH_3.pK_b= 1.76 × 10^-5.

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Solution:

This problem involves the dissociation of a weak base, which proceeds by the following general equation: B + H_2O \rightleftarrows BH^+ + OH^-, where B is the base and BH^+ and OH^- are the dissociation products, x moles of B dissociate into x moles of BH^+ and x moles of OH^-. This is more clearly seen below: NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- Before:0.1 MOO After:0.1 - xxx If x moles of NH_3 dissociate, then, at equilibrium, one has 0.1 - x left, since 0.1 is the initial concen-tration. NH_3 is a weak base and very littledissociation occurs, which means x is extremely small. This allows one to approximate 0.1 - x as 0.1. By now, it should be evident that [H^+] does not appear in the equilibrium equation. The strategy will be to first find [OH^-], then calculating thepOH, and then using the equationpK=pOH+ pH = 14 to findpH. In order to calculate the value of [OH^-], set up the equilibrium constant equation for which K_b = 1.76 × 10^-5. Thus,K_b = {[NH_4^+] [OH^-]} / [NH_3]. Substituting, K_b = [(x)(x)] / [0.1] = 1.76 × 10^-5, so that x = [OH^-] = 1.33 × 10^-3. Then, one can calculatepOHby: pOH = - log [OH^-] = - log [1.33 × 10^-3] = 3 - log 1.33 = 2.88. Therefore, pH = 14 -pOH = 14 - 2.88 = 11.12.

Question:

Compute the electrostatic force of repulsion between two \alpha\rule{1em}{1pt}particles at a separation of 10^\rule{1em}{1pt}11 cm, and compare with the force of gravitational attraction between them.

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Solution:

For this problem, we use Coulomb's Law which states that the electrostatic force between two charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance by which the charges are separated. The constant of proportionality is written as 1/4\pi\epsilon_0,where \epsilon_0 has the value of 8.85 ×10^\rule{1em}{1pt}12coul^2 /nt \rule{1em}{1pt} m^2 . Each \alpha-particle has a charge of +2e, or 2 × 1.60 × 10^\rule{1em}{1pt}19 = 3.20 × 10^-19 coul. The force of repulsion at a separation of 10^\rule{1em}{1pt}11 cm or 10^\rule{1em}{1pt}13 m is F = [1 / (4\pi\epsilon_0)] (qq' ) / (r^2) = 9 × 10^9[(3.20 ×10^\rule{1em}{1pt}19)^2 / (10^\rule{1em}{1pt}13)^2 ] = 9.18 × 10^-2 newton and since 1 newton = 10^5 dynes, F = 9180 dynes. This is a sizable force, equal to the weight of nearly 10 grams ! To find the force of gravitational attraction we use Newton's Law of Universal Gravitation. This has the same form as Coulomb's Law(an inverse square law) but instead of charges we have masses and the constant of proportionality is called G. The mass of an \alpha-particle (2 protons + 2 neutrons) is 4 × 1.67 x 10^\rule{1em}{1pt}24 = 6.68 x 10^\rule{1em}{1pt}24 gm = 6.68 x 10^\rule{1em}{1pt}27 kgm. The gravitational constant G is G = 6.67 × 10^\rule{1em}{1pt}11 (newton\rule{1em}{1pt}m^2 / kgm^2). The force of gravitational attraction is F = G(mm' / r^2 ) = 6.67 × 10^\rule{1em}{1pt}11 [(6.68 × 10^\rule{1em}{1pt}27)^2 / (10^\rule{1em}{1pt}13)^2 ] = 2.97 × 10^\rule{1em}{1pt}37 newton. The gravitational force is evidently negligible in comparison with the electrostatic force.

Question:

An object is placed 12 centimeters from a convex mirror whose focal length is 6 centimeters. Where will the image be formed?

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Solution:

First, we construct a ray-diagram. Here we trace the path of two rays - one parallel to the axis, and one through the center of curvature. The image is virtual and smaller than the object (for the convex mirror this is true for all positions of the object). We may now solve mathematically. D_0 is 12 centimeters, f is - 6 centimeters. Sub-stituting in (1/D_0) + (1/D_I) = (1/f) (1/12 cm) + (1/D_I) = (1/- 6 cm) whence D_I = - 4 cm D_I is negative implies that the image is on the same side of the mirror as the object and is virtual.

Question:

Describe, by both formal and ionic equations, what takes placewhen 0.10 mole of sodium sulfite is added to 1 liter of 1 M hydrochloric acid.

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Solution:

The formal equation for the reaction is Na_2SO_3 + 2HCI \rightarrow 2NaCl + H_2O + SO_2. The ionic equation gives more detailed information than this. The soluble sodiumsulfite salt will react with excess hydrochloric acid according to the ionicreaction (2Na^+, SO_3^2-)+ 2(H_3O^+, CI^-) \rightarrow 2(Na^+, CI^-) + H_2SO_3 (ag) + H_2O. Sulfurous acid, H_2SO_3, exists in equilibrium with both sulfur dioxide and hydroniumions: H_2SO_3 \rightleftarrowsSO_2 (g) +H_2O , SO_3^2- + 2H_3O^+ \rightleftarrowsH_2SO_3 . SO_2 gas will escape from solution, thus pulling the re-action towards completion, to give the products 2(Na^+ , CI^- ), SO_2, and 2H_2O.

Question:

How do transpiration and the theory of water cohesion explain the process of water transport in plants? When does root pressure become important in water transport?

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Solution:

Water transport in plants can be explained by transpiration and the theory of cohesion-tension. When a leaf transpires, water is lost from the leaf cells, molecule by molecule . As a consequence of the water loss, theosmolarityof the leaf cells increases and water molecules enter these cells by osmosis from adjacent cells with a lowerosmolarity. These adjacent leaf cells in turn receive water from neighboring leaf cells and so on, which all ultimately get their water from the xylem cells in the vascular tissue. Since water molecules are linked to each other by hydrogen bonds into continuous columns within the xylem vessels right down to the root tip , a molecule leaving a xylem cell to enter amesophyllcell of the leaf will necessarily tug this entire column of water along behind it. Because water has sufficient tensile strength, the column does not snap when being pulled . The upward movement of this column causes a negative pressure to develop within the xylem. This negative pressure is similar to the suction produced while sucking a drink through a straw. This draws water molecules from the root tip cells into the vessels. This theory of water movement in the xylem is known as the cohesion-tension theory. In the root tip, water molecules from the soil pass down a gradient through the root hairs to the cells that have just lost water to the xylem. In this way, the upward stream of water is continuous. In short, while transpiration provides a pull at the top of the plant, the strong tendency of water molecules to stick together attributed to hydrogen bonds (cohesive force of water molecules) transmits this force through the length of the stem and roots and results in the movement of the entire water column up the xylem . In the spring, before leaves are formed and trans-piration becomes important , root pressure is probably the major force bringing about the rise of sap. In addition, under conditions of high humidity, still air currents, or extremely low temperatures, root pressure may become important in raising water to the leaves of some plants. Thus root pressure may contribute to the upward movement of water under some conditions, but it probably is not the principal force causing water to rise in the xylem of most plants most of the time.

Question:

What are the distinguishing characteristics of cardiac and smoothmuscle.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0464.htm

Solution:

Smooth muscle fibers are considerably smaller than skeletal musclefibers. Each smooth muscle fiber has a single nucleus located in thecentral portion of the cell. By contrast, each skeletal muscle fiber is multi-nucleatedwith the nuclei located peripherally. The most noticeable morphologicalfactordistingushingsmooth from either skeletal or cardiac muscleis the absence of striated banding patterns in the cytoplasm of smoothmuscle. Smooth muscle does not contain myofibrils. However, myosinthick filaments andactinthin filaments can be seen distributed throughoutthe cytoplasm, oriented parallel to the muscle fiber, but not organizedinto regular units of filaments as in skeletal and cardiac muscle. It is believed that the molecular events of force generation in Smooth muscle cells are similar to those in skeletal muscle. Smooth muscleexhibits wide varia-tions in tonus; it may remain almost entirely relaxedor tightly contracted. Also, it apparently can maintain the contractedcondition of tonus without the expenditure of energy, perhaps owingto a reorganization of the protein chains making up the fibers. Smooth muscle cells are not under voluntary control. Their activity is underthe regulation of the autonomic nervous system and they have an abilityto perform work for long periods of time, since their contraction is slowand sustained. Smooth muscle is often referred to as a visceral musclesince it regulates the internal environment of a great many systemsand organs. It is found in the walls of hollow organs such as the intestinaltract, the bronchioles, the urinary blad-der, and the uterus. Vascular smooth muscle lines the walls of blood vessels. It may also be foundas single cells dis-tributed throughout an organ such as the spleen or insmall groups of cells attached to the hairs in the skin. Cardiac muscle has properties similar to those of skeletal muscle. It is multinucleated and striated, with its thick and thinmyofilaments organizedinto myofibrils. The sliding-filament type of contraction is also foundin cardiac muscle. Unlike skeletal muscle but like smooth muscle, cardiacmuscle is involuntary. Each beat of the heart represents a single twitch. Cardiac muscle has a long refractory period. Consequently, it is unableto contracttetanically, since one twitch cannot follow another quicklyenough to maintain a contracted state. A unique feature of cardiac muscleis its inherentrhyth-micity; it contracts at a rate of about 72 beats perminute. The muscle is innervated by nerves, but these nerves only serveto speed up or slow down the inherent cardiac rhythm. In addition, cardiacmuscle is unique in having intercalated discs, or tight junctions, betweencells; these aid in the transmission of electrical impulses throughout the heart.

Question:

Calculate the length (in \AA) of a polypeptide chain con-taining 105 amino acid residues if (a) it exists entirely in \alpha-helical form, or (b) it is fully extended.

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Solution:

(a) Peptide chains may assume the \alpha-helix con-figuration spontaneously, because this form is stable and has the least free energy, providing there are no opposing inter-actions of the R groups or of the solvent. The \alpha-helix is pictured in Figure A. As shown, when the peptide chain assume this configuration each amino acid residue contributes 1.5 A^\textdegree to the length of the chain. If a chain in the \alpha-helical con-figuration contains 105 amino acid residues, its length is 1.5 \AA × 105 = 157.5 A. (b) If the chain is fully extended, each amino acid residue contributes its full length to the chain. Figure B shows that the length of an average amino acid is 3.6 \AA. The length of the chain in this conformation is 3.6 \AA × 105 = 378.0 \AA.

Question:

Give an example of a distributed lag model for the national economy, using the national accounting scheme. Is the computer essential for dealing with distributed lag models?

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Solution:

In simulation studies, a problem in the real world is solved using a simple mathematical model in conjunction with the computer. Distributed lag models belong to the class of discrete simulation models. Such models change only at fixed intervals of time, and current values of a given variable are determined by previous values of the same variable. Using Keynesian macroeconomic analysis we can construct a model of the national economy. Let C be consumption, I be investment, T be taxes, G be government expenditure and Y be national income. Then, a general macroeconomic model would be: C = a + c(Y-T) I = b + rY(1) T = tY Y = c + I + G Here, "a" represents consumption that is independent of income while "c" is the marginal propensity to consume. "c" represents the amount per dollar that is spent on con-sumption. "b" is investment that is independent of income, and "r" is the marginal propensity to invest. Finally, "t" is the tax rate per dollar of income. Note that o \leq c, r, t \leq 1. Since simulation deals with concrete situations, we should give values to the parameters in (1). Thus, C = 30 + 0.7(Y - T) I= 3+ 0.1(Y)(2) T = . 2Y Y = C + I + G. The above static model can be made dynamic by picking a fixed time interval and then expressing the current values of the variables in terms of values at previous intervals, or lagging them. Thus, we can lag all the variables in (2) by a year to obtain C = 30 + 0.7(Y-1- T_-1) I = 3 + 0.1(Y_-1)(3) T = 0.2Y_-1 Y = C_-1 + I_-1 + G_-1 Model (3) is to be understood as follows: Current consump-tion depends on disposable income from the previous year. Investment is a function of the previous year's national income, as is taxes. In large-scale models, like (3), most calculations can be done without the aid of the computer. But at lower levels of aggregation (say where investment in different industries is considered separately), the computer is indispensable.

Question:

A copper kettle, the circular bottom of which is 6.0 in. in diameter and 0.062 in. thick, is placed over a gas flame. On assuming that the average temperature of the outer surface of the copper is 214\textdegreeF and that the water in the kettle is at its normal boiling point, how much heat is conducted through the bottom in 5.0 sec? The thermal conductivity may be taken as 2480 Btu/(ft^2 hr F\textdegree/in).

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Solution:

The heat Q conducted through the bottom of the kettle in time t is given by Q = K At (∆T/∆x) where ∆T/∆x is the temperature gradient in F\textdegree/in, K is the thermal conductivity and A is the area of the bottom in ft^2. We have A = \pir^2 = \pi[(30/12) ft]^2 = 0.20 ft^2 t = 5.0 sec = (5.0/3600) hr = 0.0014 hr The temperature on the inside of the bottom of the kettle is the same as that of boiling water (212\textdegreeF). Since the temperature on the outside of the bottom is 214\textdegreeF and the thickness of the bottom of the kettle is 0.062 in, the temperature gradient across the bottom is ∆T/∆x = (214\textdegreeF - 212\textdegreeF)/(0.62 in) = 32\textdegreeF/in The heat conducted through the bottom is then Q = [2480 {(Btu)/(ft^2 hr \textdegreeF/in)}] (0.20 ft^2)(0.0014 hr)(32\textdegreeF/in) = 22 Btu

Question:

Examine the DATA DIVISION of Fig. 1 and explain the usage and effects of File Description entries.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0256.htm

Solution:

The DATA DIVISION of a COBOL program is divided into sections such as WORKING-STORAGE section, and the FILE Section. The File Section always precedes the Working-Storage Section and includes descriptions of all the files and their associated records. File Section header begins with margin A. In the given example program FD means File Description and it describes 2 files, PARTS-FILE and OUTFILE. Both PARTS-FILE and OUTFILE must have been specified in an ENVIRONMENT DIVISION SELECT Statement. Following this, LABEL and DATA records are also included and are very impor-tant in the FD section. Every FD entry must include a LABEL RECORDS clause. Many files use labels to tell the computer where the particular file begins and ends. These records may by unique to that file, they may be standard, or in some files they may be omitted. For unit record files, such as card files or printer files, the label clause reads LABEL RECORDS ARE OMITTED. The DATA RECORD clause specifies the record name associated with the file being described. The DATA RECORD clause is not required in most computer systems since the description of the associated data record must directly follow each FD entry. However, the LABEL RECORDS clause is always required for every FD entry. The data records P_RECORD and SUMMARY are used in the PROCEDURE DIVISION to store the data read into or out of their asso-ciated files.

Question:

Discuss the structural and hormonal aspects of insect metamorphosis.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0316.htm

Solution:

Basically, there are two kinds of metamorphosis: incomplete metamorphosis which results in a similiar but larger form, and complete metamorphosis, which gives rise to strikingly different forms. Many insects, such as moths, butterflies and flies undergo metamorphosis. Generally, a wormlike larva \rule{1em}{1pt} called a caterpillar in moths, maggot in flies, and grub in bees - hatches from the egg. The larva is a relatively active form; it crawls about, eats voraciously, and molts several times, each time becoming larger in size. The last larval molt gives rise to a pupa, an inactive form which typically does not move or eat. Moth and butterfly larvae spin a cocoon around their bodies and molt within the cocoon to form the pupae. During this process all the structures of the larva are broken down and used as raw materials for the development of the adult. Each part of the adult body develops from a group of cells called a disc. The discs are embryonic cells derived directly from the egg and remain quiescent during larval stages. During the pupal stage they grow and differentiate into adult structures such as wings, legs, and eyes. These structures remain collapsed, folded and thus nonfunctional at first; when the pupa molts into the adult form, blood is pumped in to inflate them, and chitin is deposited to make them hard. Metamorphosis in insects is characterized not only by sharp changes in appearance from the larval to the adult form, but also by striking changes in their modes of life. The butterfly larva eats leaves whereas the adult feeds on nectar from flowers. The mosquito larva lives in ponds and eats algae and protozoa, while the adult sucks the blood of humans and other mammals. The process of insect metamorphosis involves a brain hormone, a molting hormone called ecdysone, and a regulatory hormone called the juvenile hormone. The brain hormone stimulates glands in the insect's prothorax (the part of the body immediately behind the head, to which the first pair of legs is attached). The prothoracic glands respond by secreting ecdysone which induces molting. Ecdysone is believed to be involved also in many growth and developmental processes in insects. Juvenile hormone is produced by a pair of glands, called corpora allata, which is located just behind the brain and closely associated with it. High concentrations of juvenile hormone at the time of molting result in another immature stage following the molt, while low concentrations lead to a more advanced stage after molting. This is supported by the demonstration that removal of the corpora allata from insects in a highly immature stage resulted in pupation at the next molt, followed by a molt that resulted in a midget adult. Conversely, implantation of an active corpora allata into an advanced pupa resulted in another immature stage after molting rather than an adult. In the larva, juvenile hormone is present in a high concentration, hence the larva molts to form a pupa. In the pupa, juvenile hormone is absent and thus when the pupa molts an adult results. Notice that insect metamorphosis involves the nervous system as well as endocrine glands. Of the three hormones discussed, one is released directly by the brain (the brain hormone), one is secreted under the influence of the brain hormone (ecdysone),and the third is secreted by glands closely associated with the brain (juvenile hormone). The interactions between these three hormones are shown in the accompanying figure.

Question:

Briefly discuss the major functional unit of the Central Processing Unit (CPU).

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0011.htm

Solution:

The CPU (Central Processing Units), or simply PU or PE (Processing Element) since there may be multiple cooperating units, is the engine that executes the instructions (software). The conventional CPU, also referred to as the Von Neumann type processor is sequential. It fetches , decodes, and executes in-structions one at a time. Generally, the CPU (or simply the processor) consists of three major functional units: Control Unit (CU): Controls the fetch, decode, execute cycles for instructions stored in memory. Arithmetic Control Unit (ALU): Accomplishes arithmetic operations and logical comparisons of words of data (a word of data is as long as the length of register in bits). Internal Storage Unit (ISU): This is the fast internal memory, not to be confused with primary memory that is external to a processor chip, that temporarily stores and manipulates data. Typically, it consists of a well defined set of regis-ters. The registers hold small chunks of data (8 bits, 16 bits , 32 bits, 64 bits) to be manipulated. In addition, this unit contains the address and data busses, in a way analogous to highways, for data to move data to and from memory. Finally, this unit contains busses (a bunch of wires) for I/O, as when a signal needs to be sent to CPU (e.g. and I/O interrupt ) regard-ing the status of an attached I/O device (also referred to as peripheral).

Question:

Let (X_i Y_i) be a set of points in the X,Y plane. Devise a flow-chart for a program that will find the regression line \^{y} =\alpha˄+ \beta˄ x_i that minimizes the sum of the squared \alpha˄ deviations from the line.

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Solution:

The simple linear regression model is as shown in Fig.1. It is required to find the equation of the straight line, \^{y} =\alpha˄+ \beta˄ x_i. The \alpha˄ unknowns are\alpha˄, the y- intercept and \beta˄- the slop of the line or the \alpha˄ regression coefficient. Assuming that such a line exists, it is required to minimize ^n\sum_i=1 (y_i - \^{y})^2 , the sum of squared y_i deviations. But (y_i - \^{y}) = y_i - (\alpha˄+ \beta˄ x_i). \alpha˄ Thus, the problem is to Minimize^n\sum_i=1 (y_i -\alpha˄+ \beta˄ x_i)^2(1) \alpha˄ where x_i ,y_i are known data points and\alpha˄, \beta˄ are the variables. \alpha˄ Taking the partial derivatives of (1) with respect to\alpha˄, \beta˄, setting them \alpha˄ equal to zero to find the minimum of (1) and solving the result-ing set of linear equations yields \beta˄ = [^n\sum_i=1 (y_i -y) (x_i -x)] / [^n\sum_i=1 (x_i -x)^2] \alpha˄=y- \beta˄x \alpha˄ \alpha˄ where y= (1/n)^n\sum_i=1 y_iandx= ^n\sum_i=1 x_i. In order to convert the above program from algebraic to computer form, the following steps must be performed: 1.The points xi,y_i are read and stored in arrays X and Y. 2.X,Yare calculated. 3.\sum(x_i -x)^2 , \sum(y_i -y)^2, \sum(y_i -y) (x_i -x) are calculated. 4.\alpha˄, \beta˄ and \sum(y_i - (\alpha˄+ \beta˄ x_i)^2 are computed. \alpha˄ \alpha˄ \alpha˄ The flow-chart of the procedure is as shown in Fig. 2

Question:

A 1200-lb sled is pulled along a horizontal surface at uniform speed by means of a rope that makes an angle of 30\textdegree above the horizontal (see figure (a)). If the tension in the rope is 100 lb, what is the coefficient of friction?

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Solution:

Since the sled is being pulled at constant velo-city, there are no unbalanced forces. We break up the ten-sion in the rope into components parallel and perpendicular to the horizontal (see figure (a)). By Newton's second law \sumF_x = 0 therefore F_friction = \muN = T cos 30\textdegree(1) \sumF_y = 0 therefore N + T sin 30\textdegree = 1200 N = 1200 lb - T sin 30\textdegree(2) From (1), \mu = (T cos 30\textdegree)/N Substituting (2) into this expression, \mu = (T cos 30\textdegree)/(1200 lb - T sin 30\textdegree) = [(100 lb)(.866)]/[1200 lb - (100)(1/2)lb] = (86.6)/(1150) = 0.0753.

Question:

Define the term mutation. What types of mutations can be distinguished and how may each be produced?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0634.htm

Solution:

A mutation can be defined as any inheritable change in a gene not due to segregation or to the normal recombination of unchanged genetic material. There are two major types of mutation: chromosomal mutation, which can involve an extensive chemical change in the structure of a chromosome, and point mutation (or gene mutation) which involves a single change in molecular structure at a given locus. There are a variety of types of chromosomal muta-tions (see fig. 1). A deletion is a mutation in which a segment of the chromosome is missing. In duplication, a portion of the chromosome is represented twice. When a segment of one chromosome is transferred to another non- homologous chromosome, the mutation is known as a trans-location. An inversion results when a segment is removed and reinserted in the same location, but in the opposite direction. (See Figure 1). Chromosomal mutations can also involve only one nucleotide. The insertion or deletion of a single nucleotide can have extensive effects. Such an occurrence results in a frame-shift by throwing the entire message out of register. (Recall that the gene message is read continuously from triplet to triplet, without "punctuation" between the codons.) Frame-shift mutations usually lead to the production of completely non-functional gene products. Point mutations involve some change in a nucleotide of the DNA molecule, usually the substitution of one nucleotide for another. Point mutations can result from exposure to x-rays, gamma rays, ultraviolet rays and other types of radiation, from errors in base pairing during replication, and from interaction with chemical mutagens. How radiation leads to changes in base pairs is not clear, but the radiant energy may react with water molecules to release short-lived, highly reactive radicals that attack and react with specific bases to cause chemical changes. These changes can block normal replication or cause errors is base pairing. Errors arising from DNA replication are better understood. There exist analogs of nitrogenous bases, such as azoguanine and bromouracil. These analogs possess structures similar to purines or pyrimidines, but have different chemical properties. For example, bromouracil, an analog of thymine, can pair with adenine during repli-cation. However, bromouracil can also pair with guanine and if it does that, subsequent replication results in a strand having a G-C base pair, replacing the A-T base pair of the original strand. (See Fig. 2) Mutagenic chemicals - that is, chemicals which induce mutations - include nitrogen mustards, epoxides, nitrous acids and alkylating agents. These are chemicals that react with specific nucleotide bases in DNA and can convert one base to another or cause mispairing of bases during replication. If a mutated gene is transcribed, and results in changes at or near the active site of an enzyme, the altered enzyme may have markedly decreased or altered enzymatic properties. However, if the mutation occurs elsewhere in the enzyme molecule, it may have little or no effect on the catalytic activities of the enzyme and may go undetected. Therefore, the true number of mutations may indeed be much greater than the number observed.

Question:

Write the micro-instructions for the interrupt cycle of the computer. What changes must be made in the execute cycle of the previous instruction to accommodate the interrupt cycle?

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Solution:

Previously, after executing the instruction the F register would be cleared and control would then go to the fetch cycle. To facilitate the interrupt capability, however, the IEN flag must be checked at the end of each execute instruction. If IEN = 0 the interrupt capability is not used and control then goes to the fetch cycle. If IEN = 1 the FGO and FGI flags must be checked to see if the teletype needs service. If FGO and FGI are both clear (zero), then control goes to the fetch cycle, but if FGO or FGI are set, then con-trol must go to the interrupt cycle. Hence, the end of the execute cycle must be changed from this: c_2t_3 :F\leftarrow0 to this:c_2t_3 { [IEN] \bullet [ [FGI] + [FGO] ] } :R\leftarrow1 c_2t_3 { [IEN] \textbullet [ [FGI] + [FGO] ] } :F\leftarrow0 When the computer enters the interrupt cycle it goes to a subroutine specifically designed for interrupts. The return address is stored in a special location in memory (location 0), that can be recalled at the end of the sub-routine. The IEN is cleared so another interrupt will not occur until the present one is dealt with. c_3t_0 :MBR_4-15 \leftarrow[PC]\mid PC\leftarrow0\midGo to Interrupt Subroutine c_3t_1 :MAR\leftarrow [PC]\midlocation1, Save return address PC \leftarrow [PC] + 1\midin location O, Clear IEN. c_3t_2 :M\leftarrow[MBR]\mid IEN \leftarrow 0\mid c_3t_3 :F\leftarrow0\midGo to fetch cycle R\leftarrow0\mid The FGO and FGI flags will be checked in the interrupt subroutine to determine if data must enter or leave the AC. After the teletype is serviced the IEN is set to 1 and control leaves the subroutine at the BUN instruction. The computer then fetches the instruction addressed by the PC contents saved in location O.

Question:

Use the quantum aspect of light to derive the formula for the Doppler effect. Assume that the light source moves with a nonrelativistic velocity with respect to the observer.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1012.htm

Solution:

Let M be the mass of the source and v^\ding{217} its velocity. In order to emit photons, the atoms of the source must initially be in excited states. If the inter-nal energy of the excited source is E, the total energy of the source before emission is E_t = (1/2) Mv^2 + E. When a photon is emitted, the internal energy changes by an amount hv_0 where v_0 is the frequency of the photon observed by the source. The source suffers a recoil as a result of emission and acquires a new velocity v'. By the law of conservation of energy, (1/2) Mv^2 + E = (1/2) Mv'^2 + E' + hv(1) where v is the frequency of the photon as measured by the observer. From the figure, we write the conservation of the momentum in directions parallel and perpendicular tovas Mv = Mv' cos\alpha + (hv/c) cos\texttheta.(2) 0 = mv' sin\alpha + (hv/c) sin\texttheta.(3) Since E' = E - hv_0' (1) can be written as Mv^2 = Mv'^2 + 2h(v - v_0).(4) Rewriting (2) and (3) as Mv' cos \alpha = Mv - (hv/c) cos \texttheta Mv' sin \alpha = (hv/c) sin \texttheta and squaring both sides M^2v'^2 cos^2 \alpha = M^2v^2 + (h^2v^2/c^2) cos^2 \texttheta - 2Mv (hv/c) cos \texttheta M^2v'^2 sin^2 \alpha = (h^2v^2/c^2) sin^2 \texttheta and adding the two, we get M^2v'^2 = M^2v^2 + [(h^2v^2)/c^2] - 2Mv (hv/c) cos \texttheta.(5) Here we made use of the trigonometric identity cos^2 \texttheta + sin^2 \texttheta =1. Combining (4) and (5), we have M^2v'^2 = M^2v2- 2hM(v - v_0) = M^2v^2 + [(h^2v^2)/c^2] - 2Mv (hv/c) cos \texttheta or2hM(v - v_0) - 2Mv (hv/c) cos \texttheta + [(h^2v^2)/c^2] = 0. If the mass M is sufficiently large, we may neglect the last term in the preceding equation, 2hM(v - v_0) = 2Mv (hv/c) cos \texttheta givingv = [v_0/{1 - (v/c) cos \texttheta}]. Therefore, the frequency of the emitted light gets smaller as the source moves faster with respect to the observer.

Question:

A microbiologist takes a stool specimen from you and a few days later tells you that Escherichia coli andSalmonella typhi are growing in your intestinal tract. Which typeshould you be concerned about?

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Solution:

Escherichiacoli, orE.coli, is a species of bacteria which normally inhabits the intestine of man and other animals. Only rarely canE.colibe pathogenic . For example, they may cause diarrheal disease in infants, and are occasionally found in infections of theurogenitaltract. The presence of E .coliin the stool should not cause any alarm, since it is one of the many types of bacteria that normally inhabit our intestine. The genusSalmonella, however, includes several species which are pathogenic to man and other animals.Salmonellatyphi, for example, causes the acute infectious disease known as typhoid fever. Typhoid fever is character-ized by a fever, inflammation of the intestine, intestinal ulcers, and a toxemia (presence of toxins in the blood). The presence of Salmonella is therefore ample reason for concern. Typhoid fever, like most intestinal infections, is transmitted from one person to another through food and water. Transmission may be indirect . For example, wastes from an infected person can pollute drinking water or food, or the infected person may handle food at some point in its processing or distribution, and contaminate it, af-fecting the consumer. The common housefly can also trans-mitSalmonellafrom wastes to food. Typhoid fever occurs in all parts of the world. In locations where good sanita-tion (proper disposal and treatment of biological waste, and water purification ) is practiced, typhoid fever inci-dence is very low. Carriers are people who are infected withSalmonellatyphi, but who have had only a slight intestinal infection, and hence do not knowthayharbor the pathogenic organism. Carriers should not be allowed to handle or prepare food .

Question:

Consider a satellite in a circular orbit concentric and coplanar with the equator of the earth. At what radius r of the orbit will the satellite appear to remain stationary when viewed by observers fixed on the earth? We suppose the sense of rotation of the orbit is the same as that of the earth.

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Solution:

The satellite is being pulled towards the earth by a gravitational force F\ding{217}_g. By Newton's second law, this force equals ma where m is the \ding{217} mass of the satellite and a is the linear acceleration along the direction parallel to the force F\ding{217}_g (which acts as the centripetal force in this case \ding{217} since the motion is circular). Therefore F_g = ma Furthermore, we know that in circular motion, linear acceleration is given by a = \omega^2r where \omega is the angular velocity of m. The force of gravity (F_g) is given by GmM/r^2 where G is the gravitational constant and M is the mass of the earth. Substituting in the equation above we have GmM/r^2 = m\omega^2r Solving for r we have r^3 = GM/\omega^2 In this equation we note that G is a constant as is M (the mass of the earth). Therefore, r varies as a function of \omega (or vice versa). If we fix \omega, we necess-arily fix r. For the satellite to appear to remain stationary when viewed by observers fixed on the earth, the satellite must have the same angular velocity as the earth. The angular velocity of the earth (\omega_e = ∆\texttheta/∆t, or change of angle per unit time) is 2\pi/1 day= 2\pi/[(60 sec/min)(60 min/hr)(24 hr/day)] = 2\pi/(8.64 × 10^4 sec^-1) = 7.3 × 10^-5 sec^-1 Substituting in the above equation we have: r^3= [(6.67×10^-11N\bulletm^2/kg^2)(5.98×10^24kg)]/(7.3×10^-5 sec^-1)^2 r^3\approx 7.49×10^22 m^3 r\approx 4.2×10^7 m r\approx 4.2×10^9 cm The radius of the earth is 6.38×10^8 cm. The distance calculated is roughly one-tenth of the distance to the moon.

Question:

Write a FORTRAN program which uses the modified Euler method to simulate the competition of two species of population, N_1(t) and N_2(t), isolated from the environment, from time t = 0 to t =t_f, if it is observed that: Ṅ_1 = (A_1 - K_11N_1 - K_12N_2) N_1(1) Ṅ_2 = (A_2 - K_21N_1 - K_22N_2) N_1(2) where Ṅ_1Ṅ_2 are the time rates of change of N_1(t) and N_2(t), respectively. A_1 and A_2 are positive constants involving natural birth/death rates for each species when isolated. The K_ij-are positive constants involving cross-effects between species. Initially, N_1(0) = N_10, and N_2(0) = N_20 .

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Solution:

First, note that equations (1) and (2) are coupled, non-linear, first-order differential equations. Since exact solutions of such equations are rare, a numerical method, such as the modified Euler's method, is essential. The program first applies the modified method to equation (1), then to equation (2). By putting N1 and N2 in a COMMON block, the new computed value of N1 will be used in determining N2. A logical flag, TWO, is used to determine which equation is being integrated: REAL T/0.0/,TFIN,DT,REALN,ACCUR,N1, 1N2,N10,N20,A1,A2,K11,K12, 1K21,K22 INTEGER I,N COMMON TW0,N1 ,N2 ,A1 ,A2, K11, K12, K21, K22 LOGICAL TWO/.FALSE./ READ, N,TFIN,ACCUR, N10,N20,A1,A2, K11, K12, K21,K22 PRINT, T,N10,N20 REALN = N DT = TFIN/REALN N1 = N10 N2 = N20 DO 10 I = 1,N T = T + DT CALL MEULER(T,N1,ACCUR,DT) TWO = .TRUE. CALL MEULER(T,N2,ACCUR,DT) TWO = .FALSE. PRINT,T,N1,N2 10CONTINUE STOP END FUNCTION G(W) COMMON TWO,N1, N2,A1,A2, K11, KL2, K21, K22 REAL G, W,N1 ,N2 ,A1 ,A2, K11, K12, K21, K22 LOGICAL TWO G = (A1 - K11\textasteriskcenteredW - K12\textasteriskcenteredN2)\textasteriskcenteredW IF (TWO) G = (A2 - K21\textasteriskcenteredN1 - K22\textasteriskcenteredW)\textasteriskcenteredW RETURN END

Question:

When heated to 600\textdegreeC, acetone (CH_3COCH_3) decomposes to give CO and various hydrocarbons. The reaction is found to be first order in acetone concentration with a half-life of 81 sec. Given at 600\textdegreeCa 1-liter container into which acetone is injected at 0.48 atm, approximate how long would it take for the acetone pressure to drop to 0.45 atm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0466.htm

Solution:

When a reaction is said to be first order in a particular reactant it means that the rate of the red-action is proportional to the concentration of that re-actant. Thus the rate law for this reaction can be written : Rate = k [CH_3COCH_3] where k is the rate constant and[ ] indicate concentration. The rate constant is related to the half-life (t_1/2) by the following equation: t_1/2 = 0.693 / k Solving for k:t_1/2 = 81 sec 81 sec = 0.693 / k k = 8.56 × 10^-3 sec^-1 To solve for the time it takes for a reaction to proceed to a certain degree one uses the equation: dln [A] = - k dt where dln[A] is the natural logarithm of the original con-centration subtracted from the natural log of the final concentration of acetone, k is the rate constant and dt is the time elapsed. Because acetone is a gas in this ex-periment the pressure is proportional to the concentration. Therefore one can let [A]_ original = 0.48 and[A]_final =0.45. Solving for dt: ln 0.48- ln 0.45 = (- 8.56 × 10^-3 sec^-1)(dt) (-.734) - (-.799) = (- 8.56 × 10^-3 sec^-1)(dt) (-.734) - (-.799) =dt -8.56 × 10^-3 sec^-1 -7.5 sec = dt Therefore 7.5 sec have elapsed since the beginning of the reaction.

Question:

Develop a program to find rare prime decades. Consider a decade to be a series of numbers between and including 10a + 1 and 10a + 10, where a = 0,1,2,3... Rare decades are those which contain four primes. Output such decades less than 5000.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0548.htm

Solution:

The rarity of this occurrence is due to the fact that in any decade, there can be at most four primes. If you consider that in each decade there are five even numbers and one odd number that is a multiple of five, you should recognize that the remaining four possible primes must end with 1,3,7, or 9. The only exception to this rule is the first decade, the primes of which are 2,3,5 and 7. The needed algorithm must check the odd integers between 3 and 5000 to establish the existence of primes. To do this task, a FUNCTION subprogram IPRIME is used. This subprogram may use the Sieve of Eratosthenes (developed earlier) or some other method. It is assumed that the subprogram returns either of the two values: a 1 if the integer is a prime, a 0 if it is composite. The main program makes use of a four element queue. Each time a prime is found, it is stored in the array K. When four consecutive primes are in the queue, a check is made to see if they are included in one decade. If so, the entire queue is printed out. If not, the first prime in the queue is removed, the remaining three elements move up one position, and the next prime is inserted at the end of the queue. This process continues until I reaches 5000. The program looks as follows: DIMENSION K(4) K(1) = 2 ISUM = 1 J = 2 I = 3 II = 1 CDO WHILE I IS LESS THAN 5000 20IF (I.GE. 5000) GO TO 60 CASSIGN A NEW DECADE TO ISUM IF II = 5 IF (II.LT. 5) GO TO 5 ISUM = ISUM + 10 II = 0 CIF I IS A PRIME, ASSIGN IT TO K(J) 5CALL IPRIME (I) IF (IPRIME.EQ. 0) GO TO 57 K(J) = I J = J + 1 I = 1+2 II = II + 1 IF (J.LE.4) GO TO 20 CCHECK IF ALL 4 PRIMES IN ARRAY K ARE WITHIN CTHE SAME DECADE IF (K(1).LT.ISUM) GO TO 30 IF (K(4). GE. ISUM + 10) GO TO 30 CIF, IN FACT, THEY ARE - OUTPUT THE ARRAY K WRITE (5,100)(K(J), J = 1,4) 100FORMAT (4(2X,I4)/) J = 1 GO TO 20 CIF THEY ARE NOT WITHIN ONE DECADE 30DO 35 N = 2,4 35K(N-1) = K(N) J = 4 GO TO 20 57I = I + 2 II = II + 1 GO TO 20 60STOP END

Question:

How does asexual reproduction in plants differ from sexual reproduction? What are the evolutionary advantages of the latter?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0209.htm

Solution:

Asexual reproduction in plants takes place by the separation of any portion, or specialized portion, of a parent plant to form new individuals, or by the for-mation and germination of spores or single cells specialized in the replication of new individuals. Both these methods preserve the genetic makeup of the parent plants. Sexual reproduction involves the union of two gametes. The combination of two different sets of chromosomes from the gametes results in a new individual with a new genetic composition. Sexual reproduction therefore gives rise to new genotypes while asexual reproduction perpetuates the same ones. Sexual reproduction is an important process in the evolution of a species, and is considered much more ad-vantageous than asexual reproduction. The fusion of gametes from two genetically different parents makes possible new combinations of genetic material and thus increases the variation among individuals in the population. Genetic variation provides a pool of diversified phenotypes upon which natural selection can act. By this process, natural forces determine which individuals of the species will survive and which will become extinct, ultimately improving the species by sorting out the advantageous gene in the pool. An asexually reproducing species offers little genetic variation for selection and evolution is minimized. In addition, the natural forces acting upon any species are always changing. The more a species is able to respond to these forces with evolutionary changes that maintain or increase its fitness, the more likely it is to survive and multiply. Such evolutionary changes can occur through genetic mutations, expressed in phenotypes which are then acted upon by natural selection. Because a mutated gene is usually recessive, sexual reproduction increases its chance of expression by bringing two mutated alleles together in the process ofgameticfusion. Asexual reproduction will need a double mutation, which occurs in the order of 10^-12 per generation, to result in an altered phenotype. Therefore, without sexual reproduction there could be little genotypic and thus phenotypic changes. Evolution could not progress, as there could be no response to inevitable environmental changes, and extinction of the species would soon follow.

Question:

In an earlier question, we learned about the various factors which are involved in the clotting mechanism. Show how these factors are affected in conditions that cause excessive bleeding in humans, and in conditions that cause intravascular clotting.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0348.htm

Solution:

Excessive bleeding can result from a deficiency of any one of the blood clotting factors. An insufficiency ofprothrombin, one of the intermediates in the clotting mechanism, can cause a patient to develop a severe tendency to bleed. Both hepatitis and cirrhosis (diseases of the liver) can depress the formation ofprothrombinin the liver. Vitamin K deficiency also depresses the levels ofprothrombin. Vitamin K deficiency does not result from the absence of the vitamin from the diet, since it is continually synthesized in the gastrointestinal tract by bacteria. The deficiency results from poor absorption of fats (vitamin K is fat soluble) from the tract due to a lack of bile, which is secreted by the liver. Hemophilia, or "bleeder's disease," is a term loosely applied to several different hereditary deficiencies in coagulation, resulting in bleeding tendencies. Mosthemo-philiasresult from a deficiency in one of the protein factors called Factor VIII or theantihemophilicfactor necessary for production ofprothrombinaseby the platelets. Thrombocytopenia, the presence of a very low quantity of platelets in the blood, also causes excessive bleeding. With adefinciencyof platelets, not enoughprothrombinasecan be synthesized. One major type of thrombocytopenia results from the development of immunity to one's own plate-lets. [Normally, the immune system does not develop immunity against the body's own proteins, but sometimes one can deve-lop this "autoiInmunity".] The antibodies (specific proteins in the blood) attack and destroy the platelets in the person's own blood. Thrombocytopenia also results from pernicious anemia and certain drug therapies. These - abnormalities cause excessive bleeding; but there are also pathological conditions caused by clotting when it should not normally occur. An abnormal clot that develops in a blood vessel is called a thrombus. If the thrombus breaks away from its attachment and flows freely in the bloodstream, it is termed an embolus. Should an embolus block an important blood vessel (to the heart, lungs, or brain), death could occur. What causes intravascular (within a blood vessel) clotting? Any roughened inner surface of a blood vessel, which may result from arteriosclerosis, bac-terial infection, or physical injury, can initiate the clotting mechanism by releasingthromboplastinfrom the platelets. Blood which flows too slowly may also cause clotting. Since small amounts of thrombin are always being produced, a hampered flow can increase the concentration of thrombin in a specific area, so that a thrombus results. The immobility of bed patients presents this problem. Heparin is a strong anticoagulant normally produced in small amounts by special cells in the body. Heparin can be used as an intravenous anticoagulant in the prevention of thrombi formation.

Question:

For the reaction PbSO_4 (s) \rightarrow Pb^2+ + S0^2-_4, ∆H = + 2990 cal/mole. Will the solubility of PbSO_4 increase or decrease with increasing temperature?K_sp=1.8 × 10^-8 at 25\textdegreeC. Find itsk_spat 55\textdegreeC.

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Solution:

Whether the solubility of a salt increases or decreases with an increase in temperature can be determined by an investigation of its ∆H, the enthalpy change. The formula for the temperature dependence on k is log k = [(∆H\textdegree)/(2.303 RT)] + [(∆S\textdegree)/(2.303 R)]. A positive ∆H\textdegree suggests that solubility increases with increasing temperature. When ∆H\textdegree is positive, the first term remains a negative value. When T increases, the first term becomes a smaller negative number. Thus, k increases. Since k measures solubility, the solubility increases. To find thek_spof PbSO_4 at 55\textdegreeC, use the fact that log (k_2 /k_1) = [(- ∆H\textdegree) / (2.303 R)] [(1/T_2) - (1/T_1)], where k_2 = solubility constant at the temperature in Kelvin (Celsius plus 273\textdegree) of T_2; k_1 = solubility constant at a temperature of T_1. Thus, if thek_sp at one temperature is known, thek_spat another temperature can be found given ∆H\textdegree. Let T_2 = 55\textdegreeC or 328\textdegreeK and T_1 = 298\textdegreeK. (25\textdegreeC).Thus log [(k_328)/(k_298)] = [(- ∆H\textdegree)/(2.303 R)] [(1/328) - (1/298)] log [(k_328 )/(1.8 × 10^-8)] = [{- (2990 cal/mole) (- 3.07 × 10^-4/k)} /{(2.303) 1.987 cal/mole K)}] = 0.200 = 0.200 k_328 = (1.8 × 10^-8) (10^0.200) = (1.8 × 10^-8) (1.58) = 2.8 × 10^-8.

Question:

Write a program to solve for integral zero of a third-degree polynomial

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0519.htm

Solution:

The flowchart illustrates the basic program. Now we give an explanation of the theory behind our search. Which in-tegers do we try for Z to test P(z) for 0? Let us assume that there are N complex zeros denoted by. Z_N, Z_N-1 ,..., Z_2, Z_1. The fac-tor theorem for polynomials states that if f(z1) = 0, i.e., z1 is a root of f(z), then f(z) = (z - z1)g(z) where g(z) is a polynomial one degree lower than f(z). Then, by repeated applications of the factor theorem, we obtain (X - Z_N) (X - Z_N-1) ...(X - Z_2) (X - Z_1) = A_NX^N + A_N-1X^N-1 +... +A_1X + A_0.(1) Multiplying out the left hand side of (1), we see that the only constant term in the product is (-1)^n (Z1)\bullet(Z2)...(Z_N) = A_0. Hence, we can conclude that if a polynomial has any integral zeros, they must be factors of the constant term A_0 . Note that the converse need not be true, i.e., a factor of A_0 need not be a root. Note, however, that any factor of P(1) = A_0 will lie in the interval -\vertP(1)\vert to \vertP(1) \vert. The actual program makes use of the sign function SGN to assure stopping in the right direction. 10DEF FNP(X) = P(4)\textasteriskcenteredX\uparrow3 + P(3)\textasteriskcenteredX\uparrow2 + P(2)\textasteriskcenteredX + P(1) 20PRINT 25FOR S = 4 TO 1 STEP - 1 30READ P(S) 40PRINT P(S); 50NEXT S 60PRINT "INTEGRAL ZERO(S): " ; 68REM TURN SWITCH OFF 70LET K = 0 78REM STUDY LINE 80 CAREFULLY 80FOR X = -P(1) TO P(1) STEP SGN (P(1)) 88REM LINE 90 PREVENTS ERROR MSG CAUSED BY 89REM DIVIDING BY ZERO 90IF X = 0 THEN 140 98REM IS X A FACTOR, OF P(1)? 100IF P(1)/X = INT(P(1)/X) THEN 140 108REM IS THE REMAINDER ZERO? 110IF FNP(X) < > 0 THEN 140 118REM IF THE COMPUTER GETS THROUGH HERE, THE 119REM VALUE OF X IS A ZERO OF THE FUNCTION 120PRINT X; 128REM TURN THE SWITCH ON - WE HAVE A ZERO 130LET K = 1 140NEXT X 150IF K = 1 THEN 20 160PRINT "NONE FOUND"; 165GO TO 20 170DATA 1, -2, -11, 12 180DATA. 1, 1, -5, -2 190DATA 1, -2, 3, -4 200DATA 2, -3, -10, 3 210END

Question:

What is the resultant force on a body of mass 48 kg when its ac-celeration is 6 m/sec^2?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0096.htm

Solution:

The relationship between a body's acceleration and the net force on it is given by Newton's Second Law. The mass of the body is given, hence the net force on the body is \sumF = ma = 48 kg × 6(m/sec^2) = 288newtons.

Question:

Write a program in BASIC to find the sum and number of factors for the integers between 1000 and 1050, not including 1 and the integer itself.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0550.htm

Solution:

As an example, consider the integer 36, It has 7 factors: 2,3,4,6,9,12,18, the sum of which is 54. The presented program outputs the considered integer, sum of its factors, and the number of factors. The program terminates when the considered number is greater than 1050. The program looks as follows: 15LET F = 1133 20LET F = F + 1 25IF F < 1175 THEN 30 28STOP 30LET M = 0 32LET N = 0 34LET D = 2 35LET Y = F/D 40LET W = INT(Y) 45IF W < > Y THEN 70 50LET M = M + D 55LET N = N + 1 60IF D = Y THEN 70 65LET M = M + Y 68LET N = N + 1 70LET D = D + 1 75IF D\textasteriskcenteredD < = F THEN 35 80PRINT F,M,N 90GO TO 20 95END

Question:

A uniform rod of mass m and length 2a stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has not occurred, show that, when the rod makes an angle \texttheta with the vertical, \omega^2 = (3g/2a)(1 - cos \texttheta) where \omega is the rod's angular velocity. Also find the normal force exerted by the floor on the rod in this position, and the coefficient of static friction involved if slipping occurs when \texttheta = 30\textdegree.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0195.htm

Solution:

The forces acting on the rod are the weight mg^\ding{217} acting downward and the normal force N^\ding{217} and the frictional force F^\ding{217} of magnitude \muN exerted by the floor at the end 0 in contact with the floor. In order to find \omega, we re-late the net torque \cyrchar\cyrt on the rod to the rod's angular acceleration a by using \cyrchar\cyrt = I\alpha Here, I is the rod's moment of inertia. We will then be able to solve for \omega. When one takes moments about 0, the only force producing rotation about 0 is the weight of the rod. Hence \cyrchar\CYRT = mga sin \texttheta = I_0 \alpha = (4/3)ma^2 \alpha Here, I_0 is the rod's moment of inertia about 0. Now, \alpha = (d\omega/dt) = (d\omega/d\texttheta) × (d\texttheta/dt) = \omega(d\omega/d\texttheta) = (3/4)(g/a) sin \texttheta. ^\omega\int_0 \omega d\omega = ^\texttheta\int_0(3/4)(g/a) sin \texttheta d\texttheta. [(1/2) \omega^2]^\omega_0 = [-(3/4)(g/a) cos \texttheta]^\texttheta_0 or \omega^2 = [(3g)/(2a)] ( 1 - cos \texttheta). The center of gravity G has an angular acceleration \alpha about 0, and thus a linear acceleration a\alpha at right angles to the direction of the rod. This linear acceler-ation can be split into two components, a\alpha cos \texttheta horiz-ontally and a\alpha sin \texttheta vertically downward. The horizontal acceleration of the center of gravity is due to the force \muN and the vertical acceleration is due to the net effect of the forces mg and N. Thus, using Newton's Second Law, and taking the positive direction downward, mg - N = ma\alpha sin \texttheta = (3/4) mg sin^2 \texttheta(1) andF_max = \muN = ma\alpha cos \texttheta = (3/4)mg sin \texttheta cos \texttheta.(2) From (1) N = mg - (3/4) mg sin^2 \texttheta= [(mg)/(4)] (4 - 3 sin^2 \texttheta) . But when \texttheta = 30\textdegree, slipping just commences. At this angle F has its limiting, maximum value of F_max. We have \mu_s= (F_max)/N = [(3/4)mg sin \texttheta cos \texttheta]/[(mg)/(4) (4 - 3 sin^2 \texttheta)] = (3 sin \texttheta cos \texttheta)/(4 - 3 sin^2 \texttheta) = [3 × (1/2)× (\surd3/2)]/[4 - (3/4)] = (3\surd3)/(13) = 0.400.

Question:

What is the least radiation frequency v, capable ofpositron- electronpair production?

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Solution:

The process in which a photon disappears and an electron-positron pairappears, is known as pair production. The energy of the photon shouldat least be large enough to provide for the rest masses of the electronand proton. Therefore E_photon= m_electronc^2 + m_positronc^2 hv= 2m_ec^2 wherev is the photon frequency. Hence v = (2m_ec^2/h) = [{2 x (9.11 x 10^-31 kg) x (3 x 10^8m/s)^2} / (6.63 x 10^-34 J.S)] = 2.47 x 10^20 (1/sec).

Question:

Given a one-dimensional array Y with 50 elements, and numbersU and I, write statements to compute the value of S fromthe following equation: S =y_i+ U {(y_i+ 1-y_i- 1) / 2} + (U^2 / 2) (y_i+ 1- 2y_i +y_i- 1) S =y_i+ U {(y_i+ 1-y_i- 1) / 2} + (U^2 / 2) (y_i+ 1- 2y_i +y_i- 1)

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Solution:

Note that since the subscripts on Y can be i+1,i, or i-1 where the Y arrayis dimensioned 50, then we must restrict I to being an integer from 2 to49, i.e., there are 48 possible values of I. With this restriction in mind we cansimply write: DIMENSIONY(50) DO 100 I = 2,49 S =Y(I) + U \textasteriskcentered (Y(I + 1) - Y(I - 1)/2.0) 1+ U \textasteriskcentered U \textasteriskcentered (Y(I + 1) - 2\textasteriskcenteredY (I) 1+Y(I - 1))/2.0 100CONTINUE STOP END Further remarks on interpretation: This formula is calledStirling's interpolationformula [S stands forStirling!] through second differences andis described below. We have three points on a curve: (Xi -1,Yi - 1), (X_i, Y_i), and (Xi + 1, Yi + 1) such that Xi + 1- X_i = X_i - Xi - 1= h , (i.e., the x coordinates are equally spaced) and have a value of x for which wewant to guess Y(x). We write U = (x -x_i )/h. Then the formula stated givesthe interpolated value of Y corresponding to x, found by passing a quadraticthrough the three given points.

Question:

If a 1,000\rule{1em}{1pt}V battery is connected to two parallel plates separated by 1mm (10^\rule{1em}{1pt}3m), what is the electric field?

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Solution:

The potential of the battery, V = 1,000 V, and the distance between the plates, d = 10^\rule{1em}{1pt}3 m, are given. By definition, the difference in potential experienced by moving a charge from point A to point B is V_B \rule{1em}{1pt} V_A = \rule{1em}{1pt} ^B\int_A E \bullet dl(1) where E is the electric field and d is an element of the path traversed in moving the charge. Now, looking at the figure, we see that, for the plates of a battery, E is perpendicular to the plates. If we evaluate (1) over a straight line path parallel to E, we find V_B \rule{1em}{1pt} V_A = \rule{1em}{1pt}B\int_A Edl = \rule{1em}{1pt}d\int_0 Edl = \rule{1em}{1pt} E d where d is the plate separation. Then │E│ = │(v_b \rule{1em}{1pt} v_a)│ / d = (10^3 v) / (10^\rule{1em}{1pt}3m) = 10^6 v / m.

Question:

Artificial kidneys have been devised for patients with kidney disease. How do these artificial kidneys work? What is the basic principle involved?

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Solution:

The artificial kidney is used to replace the diseased kidney in eliminating the excess ions and wastes which would accumulate in the blood as a result of failure of the latter. In an artificial kidney, the patient's blood is passed through a system of very fine tubes bounded by thin membranes. The other side of the membrane is bathed by a dialysis fluid into which waste products can pass from the blood. The fine tubes con-verge into a tubing which then conducts the blood back into the patient's body through a vein. The fine tubes of the artificial kidney are made of cellophane. This material is used because it has much the same characteristics as the endothelium of blood capillaries; it is highly permeable to most small solutes but relatively impermeable to protein. The dialysis fluid which bathes the cellophane tubes is a salt solution with ionic concentrations similar to those of blood plasma. The function of the artificial kidney can be ex-plained by the principle of diffusion. Since the cel-lophane membrane is permeable to most small solutes, the concentrations of solutes in the blood, as it flows through the tubes, tend to equal those in the dialysis fluid. However, if there is an above normal level of a certain solute in the blood, this solute will diffuse out of the blood into the surrounding fluid, which has a lower concentration of the solute. In this way, waste products and other fluid substances in excess will leave the blood and pass into the dialysis fluid since their concentrations in the fluid are very low or non-existent. To prevent waste solutes from building up in the fluid and inter-fering with diffusion, the fluid is continually being re-placed by fresh supply. Great care must be taken to maintain the sterility of the artificial kidney. The total amount of blood in an artificial kidney at any one moment is roughly 400-500 ml. Usually heparin, an anticoagulant, is added to the blood as it enters the artificial kidney to prevent coagulation; an antiheparinis added to the blood before returning it to the patient's body in order to allow normal blood clotting to take place. An artificial kidney can clear urea from the blood at a rate of about 200 ml. of plasma per minute, - that is, 200 ml of plasma can be completely cleared of urea in one minute - this being almost three times as fast as the clearance rate of both normal kidneys working together. Thus we can see that the artificial kidney can be very effective, but it can be used for only 12 hours every 3 or 4 days, because of the danger to the blood's clotting mechanisms due to the addition of heparin.

Question:

Show that the area of a parallelogram, whose sides are formed by the vectors A\ding{217} and B\ding{217} (see figure) is given by Area = \vertA\ding{217} × B\ding{217}\vert .

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Solution:

The area of the parallelogram shown in the figure is Area = bh But h = \vertA\ding{217}\vert sin \texttheta and b = \vertB\ding{217}\vert Area = \midA\ding{217}\mid \midB\ding{217}\mid sin \texttheta(1) The left side of (1) is the magnitude of A\ding{217} × B\ding{217}, hence Area = \midA\ding{217} × B\ding{217}\mid = \midA\ding{217}\mid \midB\ding{217}\mid sin \texttheta If we are interested in obtaining a vector area, we may write Area = A\ding{217} × B\ding{217} where the direction of the area is the direction of A\ding{217} × B\ding{217} . Such vector areas are useful in defining certain surface integrals used in physics.

Question:

How is the human eye regulated for far and near vision? In what respects does the frog eye differ in these regulatory mechanisms ?

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Solution:

The human eye can focus near or distant images by changing the curvature of the lens. The lens is bound tociliarymuscles via the suspensory ligaments. When inverted to focus on a distant object, the ciliary muscles contract, stretching thesuspensoryligaments as well as the flexible lens. A flat lens correctly focuses the image of a distant object on the retina.Ciliarymuscles relax when one focuses on a near object. This allows the lens to contract into a shape that will correctly focus the closer object. Lens regulation in frogs differs from man in one impor-tant aspect. The frog focuses objects by moving the eye lens forward or backward, whereas inmanaccommodation is achieved by changing the shape of the lens , without any change in its position.

Question:

Calculate the pH of a 0.10 M solution of sodium acetate, NaOAc. The equilibrium constant for the dissociation of acetic acid is 1.8 × 10^-5.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0438.htm

Solution:

The pH will be governed by two processes: the ionization of NaOAc, and the hydrolysis of acetate ion,OAc^-. NaOAcdissociates according to the equation NaOAc\rightleftarrows Na^+ +OAc^-. If we assume that dissociation is complete (a reasonable assumption for dilute salt solutions), the concentrations of Na^+ andOAc^- are equal to the initial concentration ofNaOAc: [Na^+] = [OAc^-] = 0.10 M. Hydrolysis of acetate ion proceeds according to the equation OAc^- + H_2O \rightleftarrowsHOAc+ OH^-. Let x denote the concentration ofundissociatedHOAcformed by hydrolysis. Then, since one mole of OH^- is formed and one mole ofOAc^- is consumed per mole ofHOAcformed, the concentration of hydroxide ion formed is x, and the concentration of acetate ion remaining is 0.10 - x. That is, [OH^-] = x,[OAc^-] = 0.10 - x. These concentrations can be substituted into the expression for the hydrolysis constant, K_h= {[HOAc] [OH^-]} / [OAc^-] to find x = [OH^-].K_hcan be found by using the water constant,K_w = 1.0 × 10^-14 = [H_3O+] [OH^-] and the equilibrium constant for dissociation of acetic acid (HOAc), K_a = 1.8 × 10^-5 = {[H_3O^+] [OAc^-]} / [HOAc]. Then, K_h= {[HOAc] [OH^-]} / [OAc^-] = [H_3O+] [0H^-] × {[HOAc] / ([H_3O^+] [OAc^-])} =K_w/ K_a = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10. Substituting the unknown concentrations into the expression gives K_h= 5.6 × 10^-10 = {[HOAc] [OH^-]} / [OAc^-] = {(x) (x)} / (0.10 - x) \allequal x^2 / 0.10 = 10 x^2, where we have assumed that x is much smaller than 0.10 (hence 0.10 - x \allequal 0.10). Then 10 x^2 = 5.6 × 10^-10 x = (5.6 × 10^-11)^1/2 = 7.5 × 10^-6 Therefore, [OH^-] = x = 7.5 × 10^-6. Using the water constant, [H_3O+] [OH^-] = 1.0 × 10^-14 [H_3O^+] = [1.0 × 10^-14] / [OH^-] = [1.0 × 10^-14] / [7.5 × 10^-6] = 1.3 × 10^-9 and the pH of the solution is pH = - log [H_3O^+] = - log (1.3 × 10^-9) = 8.9.

Question:

What is the average velocity of the molecules of the air at 27\textdegree C?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0490.htm

Solution:

From a simple atomic model in which we consider the atom as a hard spherical body subject to completely elastic collisions we can develop an equation for the average kinetic energy of a molecule which is: (1/2)mv^2_a = (3kT)/2 where m is the mass of one molecule,v_ais its average velocity, k is Boltzmann's constant, and T is the absolute temperature of the environment of the molecule. Multiply both sides by Avogadro's number N_A and 2 N_Amv^2_a = 3N_AkT ButN_Am= M_m andN_Ak= R since M_m is the mass of one molecules, and R is the gas constant which is Boltzmann's constant times Avogadro's number. Substitut-ing and rearranging we have: v^2_a = (3RT)/(M_m). Air consists mainly of nitrogen, which is diatomic, and its effective molecular mass is approximately twice the atomic mass of nitrogen. So M_m \approx 2 × 14 = 28 gm/mole v^2_a = [(3)(8.32 Joule/mole \textdegreeK)(273 + 27)\textdegreeK]/[28 gm/mole] where we have used the fact that 27\textdegree C = (273 + 27)\textdegreeK. Hence v^2a= 267.43 Joule/gm But 1 Joule = 1nt\textbullet m = 1 kg \textbullet m^2/s^2 =10^7 gm \textbullet cm^2/s^2 and v^2_a^ = 267.43 × 10^7 [(gm \bullet cm^2)/(gm \bullet s^2)]^ = 267.43 × 10^7 cm^2/s2 Thereforev_a= 5.17 × 10^4 cm/s. This is equivalent to 1,160 miles per hour!

Question:

The freezing point of silver is 960.8\textdegreeC and the freezing point of gold is 1063.0\textdegreeC. Convert these two readings to Kelvin (\textdegreeK), Fahrenheit (\textdegreeF), and Rankine (\textdegreeR).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0021.htm

Solution:

Kelvin: Temperatures measured in Celsius (\textdegreeC) are converted to \textdegreeK by adding 273.15 to the original measurement. freezing point of silver = 960.8\textdegreeC + 273.15 = 1234\textdegreeK freezing point of gold = 1063.0\textdegreeC + 273.15 = 1336.2\textdegreeK Fahrenheit: \textdegreeC are converted to \textdegreeF by using the equation \textdegreeF = 9/5 (\textdegreeC) + 32. freezing point of silver = 9/5 (960.8\textdegreeC) + 32 = 1761\textdegreeF freezing point of gold = 9/5 (1063.0\textdegreeC) + 32 = 1945\textdegreeF Rankine: The Rankine scale is an absolute scale used by engineers. Its unit is the Fahrenheit degree. Absolute zero is equal to zero degrees Rankine. Convert \textdegreeK to \textdegreeR by using the equation \textdegreeR = 9/5 (\textdegreeK) freezing point of silver = 9/5 (1234\textdegreeK) = 2221\textdegreeR freezing point of gold = 9/5 (1336.2\textdegreeK) = 2405\textdegreeR.

Question:

If you mix 50ml of 5.0 × 10-^4 M Ca(NO_3)_2 and 50ml of 2.0 × 10-^4 M NaF to give 100ml of solution, will precipitation occur? The K_sp of CaF_2 is 1.7 × 10-^10 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0401.htm

Solution:

Whether or not precipitation will occur, when these two solutions are mixed, depends upon the ion product. If two solutions containing the ions of a salt are mixed, and if the ion product ex-ceeds the K_sp , then precipitation will occur. You need to determine the concentrations of Ca^2+ and F- and to see if the product exceeds the given K_sp of 1.7 × 10-^10 for CaF_2 . The equation for this reaction is CaF_2 (s) \rightleftarrows Ca^2+ + 2F- . Thus, the K_sp = [Ca^2+] [F-]^2 . You are told that a 50ml solution of Ca(NO_3 )_2 has a molarity of 5 × 10-^4 M. Each Ca(NO_3 )_2 dissociates to yield one Ca++ ion which means the molarity of Ca++ is also 5 × 10-^4 M. However , it is 5 × 10-^4 M in 50ml , and the mixture is 100ml . Therefore , the molarity must be divided by two . The molarity of NaF, 2.0 × 10-^4 , and therefore Na+ , because of its dissociation, must also be divided by two for the same reasons . As such , in the mixture, [Ca^2+ ] = 2.5 × 10-^4 M and [F-] = 1.0 × 10-^4 M. Recalling that K_sp = [Ca^2+ ] [F-]^2 , you have K_sp = (2.5 × 10-^4 ) (1.0 × 10-^4 )^2 = 2.5 × 10-^12 . You were given that the K_sp for a saturated solution of CaF_2 is 1.7 × 10-^10 . Since 2.5 × 10-^12 is less than 1.7 × 10-^10 , the mixture is unsaturated, and precipitation should NOT occur .

Question:

A plane can travel 100 miles per hour (mph) without any wind. Its fuel supply is 3 hours. (This means that it can fly 300 miles without a wind). The pilot now proposes to fly east with a tail wind of 10 mph and to return with, of course, a head wind of 10 mph. How far out can he fly and return without running out of fuel? (The student should first reflect on whether or not the answer is the same as it is without a wind, namely, 150 miles.)

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0086.htm

Solution:

Our basic equation is (time out) + (time back) = 3 hours We now use the fact that time = (distance)/(speed) so that if d represents the distance out (= distance back) in miles we have [d/(110 miles/hr)] + [d/(90 miles/hr)] = 3 hours since the speed out is (100 + 10)and the speed back is (100 - 10)mph. Thus: (90 d + 110 d)/(9900 miles/hr) = (200 d)/(9900 miles/hr) =3 hours d = 148.5 miles.

Question:

An electric motor, rated at 1 hp, operates from a 100-V line for 1 hr. How much current does the motor require and how much electrical energy was expended?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0785.htm

Solution:

The power expended in an electric circuit is given by the product of the voltage across its terminals and the current passing through the terminals. Hence, the current in the motor is I = (P_E/V) = (1 hp/100 V) = (746 W/100 V) = 7.46 A The power of the motor indicates the amount of electrical energy it draws from the electrical source. per unit time and converts to mechani-cal energy. Then, the electrical energy expended in one hour is E_E = P_E × t = 1 hp × 1 hr = 746 W-hr = 0.746 kW-hr or, in CGS units, E_E = (746 W-hr) × {(10^7 ergs/sec)/1W} × (3600 sec/1 hr) = 2.68 × 10^13 ergs

Question:

Explain how the sense organ of equilibrium functions in the crayfish.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0309.htm

Solution:

The sense organ of equilibrium in the crayfish is the antenna. All crustaceans have two pairs of antennae. The second pair in lobsters and crayfish is extremely long; this pair functions in touch and taste. At each base of the first pair, astatocystis present, which functions in the maintenance of equilibrium. Thestatocystsare sacs that open to the exterior. Inside the sac is astatolith, com-posed of fine sand grains cemented together by secretions from thestatocystwall. Along the floor of the sac are a number of rows of sensory hairs. These hairs arise from sensory receptor cells and are innervated by theantennularnerve. When the crayfish is in an upright po-sition, continual impulses from both right and leftstatocystscounterbalance each other and the crayfish does not lean to either side. If the animal is rotated onto its right side, thestatolith of the left side exerts a gravi-tational pull on the sensory hairs, which send impulses to the brain. Thestatolithon the right side does not exert a pull on the sensory hairs, and the crayfish as a result rotates to the left, thus balancing itself. If the animal is turned so that bothstatolithsare exerting forces, and impulses are sent from bothstatocysts, the crayfish rotates in the direction of whicheverstatocystis most stimulated. Thestatocystis aninvaginationof the ectoderm, which forms the exoskeleton, and is shed with each molt. The animal must gather new sand grains to form thestatolith. This is done by inserting sand into the statocyst sac or by burying the head in sand. Studies of thestatocyst's function were done by inserting' iron filings into sand when crayfish molted, and using a magnet to directstatolithforces on the sensory hairs. Thestatocystmay also function to maintain position during movement. For this purpose there are free sensory hairs, which are not in contact with thestatolith. These hairs are stimulated by the motion of fluid in the chamber when the animal moves.

Question:

Each liter of human blood serum contains about 3.4 g of sodium ions (Na^+, ionic weight = 23 g/mole) . What is the molarityof Na^+ in human blood serum?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E08-0282.htm

Solution:

We are given that the concentration of Na^+ is 3.4 g/liter. To convert this concentration tomolarity, we need to convert 3.4 g to the corresponding number of moles. This is accomplished by dividing 3.4 g by the ionic weight of Na^+ (which is essentially the atomic weight, since the missing electron does not detract much from the weight) to obtain 3.4 g/23 g/mole = 0.15 mole. The con-centration is thus 0.15 mole/liter = 0.15 molar = 0.15 M.

Question:

When 100 g of water at 0\textdegreeC are mixed with 50 g of water at 50\textdegreeC, what is the change of entropy on mixing?

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/Users/wenhuchen/Documents/Crawler/Physics/D15-0525.htm

Solution:

The 100 g of water at 0\textdegreeC are arbitrarily said to have zero entropy. The 50 g of water at 50\textdegreeC have a greater entropy than the same quantity of water at 0\textdegreeC, since it contains more heat energy. Entropy S is defined as ds ds =dQ/T(1) =dQ/T(1) wheredsis the infinitesimal change in the entropy due to an infinitesimal quantity of heatdQin the system. T is the instantaneous temperature in Kelvin degrees. Integrating both sides of equation (1) gives S_2 - S_1 = ^Q2\int_Q1dQ/T.(2) In raising its temperature from a temperature t_1 to t_2 a substance absorbs heatdQgiven by dQ=mcdT(3) where m is the mass of the substance, c is its specific heat, anddTis its change in temperature in Kelvin degrees (same as change in Celsius degrees). Substitution of equa-tion (3) into equation (2) yields S_2 - S_1 = ^T2\int_T1 mc (dT/T)(4) In degree Kelvin, 0\textdegreeC = (0 + 273)\textdegreeK = 273\textdegreeK and 50\textdegreeC = (50 + 273) \textdegreeK = 323\textdegreeK. Let m_2 = 50 gm and m_1 = 100 g. The specific heat of water is given by c = 1 cal \textbullet g^-1 \textbullet K deg^-1 . S_2 - S_1 = ^323\textdegreeK\int_273\textdegreeK m_2c (dT/T) = m_2c 1n(323/273) = 50 g × 1 cal \textbullet g^-1 \textbullet K deg^-1 × 2.303 × 0.0730 = 8.4 cal \textbullet K deg^-1. Since S_1 = 0, it follows that S_2 = 8.4 cal \textbullet K deg^-1. When the water is mixed, the heat gained by the cold water is equal to the heat lost by the hot water. There-fore, m_1c(t_3 - t_1) = m^2c(t_2 - t_3), where t_1 is the original temperature (0\textdegreeC) of the 100 g of water, t_2 is the original temperature (50\textdegreeC) of the 50 g of water and t_3 is the final, intermediate temperature of the system. 100 g × (t^3 - 0\textdegreec) = 50 g × (50\textdegreeC - t_3).\thereforet_3 = (2500\textdegreeC)/150 = 16.67\textdegreeC. Converting to Kelvin degrees in order to be able to use equation (4), we have t_3 = 16.67\textdegreeC = (16.67 + 273)\textdegreeK = 289.67\textdegreeK. The entropy of the final mixture is S_3 = ^t3\int_t1 dQ dQ /T /T = ^289.67\textdegreeK\int273\textdegreeK(m_1 + m_2)c (dT/T) = ^289.67\textdegreeK\int273\textdegreeK(m_1 + m_2)c (dT/T) = = (m_1 + m_2)cln(289.67/273) = 150 g × 1 cal \textbullet g^-1 \textbullet K deg^-1 × 2.303 × 0.0257 = 8.9 cal \textbullet K deg^-1. The increase in entropy is thus 0.5 cal \textbullet K deg^-1.

Question:

In an inertial frame a body moves freely with a trajectory given by x_I = v_Ot;y_I = 0;z_I = 0. What is the trajectory in a frame rotating with constant angular velocity \omega counterclockwise about the z_I axis?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0218.htm

Solution:

In the figure the particle is travelling at velocity v_0 in the inertial coordinate system. After time t, the body has traveled a distance v_0t along the x-axis. Now, we consider a coordinate system which rotates with angular velocity \omega about the z_I axis. Therefore the z- axis of the new system coincides with the inertial z_I axis and the angle between the rotating (x_R, y_R) axes and the inertial (x_I, y_I) axes is by definition of \omega, \texttheta = \omegat. The coordinates of the position of the particle in the rotating frame in terms of the coordinates of the particle in the inertial frame can be read off the figure as x_R = v_0t cos \omegat y_R = -v_0t sin \omegat z_R = 0.

Question:

Wave number (ѵ̅) are the reciprocals of wavelengths, \lambda, and are given by the expression ѵ̅ = 1/\lambda. For the hydrogen atom, the Bohr theory predicts that the wave number for the emission line associated with an electronic transition from the energy level having principal quantum number n_2 to that with principal quantum number n_1 is ѵ̅ = R_H [(1/n_1 ^2) - (1/n_2 ^2)], where R_H is theRydbergconstant. In what region of the electromagnetic spectrum would there appear a spectral line resulting from the transition from the tenth to the fifth electronic level in hydrogen?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0696.htm

Solution:

This problem can be solved by applying theRydbergformula used for determining thewavenumbersof the spectral lines. TheRydbergconstant is 1.10 × 10^5 cm^-1. n_2 = 10 and n_1 = 5. Substituting: ѵ̅ = R_H [(1/n_1 ^2) - (1/n_2 ^2)] = 1.10 × 10^5 cm^-1 [(1/5^2) - (1/10^2)] = 1.10 × 10^5 cm^-1 [(1/25) - (1/100)] = 1.10 × 10^5 cm^-1 (0.04 - 0.01) = 1.10 × 10^5 cm^-1 × 0.03 = 3.3 × 10^3 cm^-1. This line appears in the infrared region of the spectrum (approximately 20 cm^-1 to 10^4 cm^-1).

Question:

Write a FORTRAN program which uses the modified Euler method to simulate the growth of an isolated species from time t = 0 to t = t_f, if the number of individuals at time t is N(t), N(0) = N_0 and the birth rate B and the mortality rate M (per unit of population per unit of time) are given by: Case (1): B = b_1; M = m_1 where b_1 and m_2 are positive constants. Case (2): B = b_2 > 0. Competition for food causes the death rate to increase in direct proportion to the ratio N(t)/2N_0 with the con-stant of proportionality equal to m_2 where 0 < m_2 < 1, and N_0 > 0. Compare the approximations with the exact values.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0583.htm

Solution:

An initial schematic diagram of the system looks as shown in figure #1. The rate of flow of individuals into and out of the system (per unit of time) is given by BN and MN respectively. Thus, B and M act as valves (see Fig. 02). Thus, the net rate of change in population, N, is the difference between the entrance and exit rates: Ṅ = BN - MN = (B - M)N, and the block diagram of the process is shown in figure #3. Case (1):If B = b_1 and M = m_1 are positive constants, B-M = b_1 - m_1 = r Case (1):If B = b_1 and M = m_1 are positive constants, B-M = b_1 - m_1 = r will also be a constant. If b_1 > m_1, then r > 0 and the population will rise indefinitely. If b_1 < m_1, then r < 0 and the population will continuously decrease to zero. If b_1 = m_1, r = 0 and there will be no change in population. In any case, one can write: Ṅ = (B - M)N = rN(2) The exact solution to equation (2)is N(t) = N_0e^rt Note that when r > 0, t \ding{217} \infty \Rightarrow N(t) \Rightarrow \infty . When r < 0, t\ding{217} \infty \Rightarrow N(t) \Rightarrow 0 When r = 0, N(t) = N_0e^0 = N_0(1) = N_0. The program uses POP for N(t). REAL T/0.0/, Ml COMMON R READ, N,TFIN,ACCUR,B1,Ml, POPO PRINT, T,POPO REALN = N DT = TFIN/REALN POP = POPO R = B1 - Ml DO 10 I = 1,N T = T + DT CALL MEULER(T,P0P,ACCUR,DT) EXACT = POPO\textasteriskcenteredEXP(R\textasteriskcenteredT) ERROR - ABS((EXACT-POP)/EXACT)\textasteriskcentered100 PRINT,T,POP,EXACT,ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON R G = R\textasteriskcenteredW RETURN END Case (2): In this case, equation (1) becomes: Ṅ = (b_2 - m_2N/2N_0)N = b_2N - (m_2N^2/2N_0)(3) Equation (3) is a nonlinear differential equation (because of the N^2 term) . Though, generally, solutions for nonlinear equations are not easy to find, equation (3) is known as the Bernoulli equation with the solution: N(t) = (2b_2N_0) / (m2+ (2b_2 - m_2)e^-b2t) There are three possible cases: (i)m_2 = 2b_2 . Substituting for m_2 in equation (3) at time t = 0 when N = N_0 gives: Ṅ = b_2N_2 - (2b_2N_0^2 / 2N_0) = b_2N_2 - b_2N_0 = 0 which means population remains at initial level. This can also be seen by substituting for m_2 in equation (4): N(t) = 2b_2N_0/ {2b_2 (2b_2 - 2b_2 )e^-(b)2t } = 2b_2N_2 / 2b_2 = N_0 . (5) Since t does not appear in equation (5), in this case, N(t) is the constant function, N(t) = N_0. What this means to the programmer is that for the pair of parameters (b2 ,m_2), where m_2 = 2b_2 any initial population will be an equilibrium point; and the system is stable at this point. (ii)m_2 < 2b_2 . m_2 < 2b_2 \Rightarrow m_2 > 0 \Rightarrow as t \ding{217} \infty : (2b_2 - m_2)e^-bt \ding{217} 0^+ m_2 + (2b2- m_2)e^-bt 2b_2 / {m_2 + (2b_2 - m_2)e^-bt} \ding{217} [2b_2 / m_2] > 1 (since m_2 < 2b_2) N(t) \ding{217} (2b_2 / m_2)N_0 > N_0(6) From equation (6), we see that population will increase to the equilibrium value 2b_2N_0/m_2 . (iii)m_2> 2b_2 . An analysis similar to that used in case (ii) yields: N(t) \ding{217} (2b_2 / m_2 )N_0 < N0= (since m_2 > 2b_2 \Rightarrow {2b_2 / m_2} < 1)(7) as t \ding{217} \infty. From equation (7), the population will decrease to the equilibrium level . It follows that the system is asymptotically stable, since it converges back to its equilibrium point as a result of any arbitrary disturbance: For this first-order case, the equilibrium point could also have been found by the graphical method, plotting B and M as functions of N: The equilibrium point is the point of intersection, where the birth rate just equals the death rate, since the net rate of change In population, N = (B - M)N, will be zero. But B - M = 0 \Rightarrow b_2N - (m_2 / 2N_0)N^2 = 0 N{b_2 - (m_2/2N_0)N} = 0 \Rightarrow N = 0 or If N > 0, b_2 - (m_2/ 2N_0) N = 0 (m_2/2N_0)N = b_2 \Rightarrow N = (2b_2/m_2) N_0 which is the same result obtained in equations (6) and (7), using the exact solution. Some sort of analysis of the system under study is an invaluable aid to the programmer in distinguishing between peculiar system behavior due to system stability or instability and peculiar results due to programming or logical errors. The program uses POP for N(t), POPO for N_0 . REAL T/0.0/, NUMBER,M2 COMMON B2,M2,POPO READ,N,TFIN,ACCUR,B2,M2,POPO PRINT ,T, POPO REALN = N DT = TFIN/REALN POP = POPO CCONSTANT TERMS IN EXPRESSION FOR EXACT NUMER = 2.\textasteriskcenteredB2\textasteriskcenteredPOPO DENOM = 2.\textasteriskcenteredB2 - M2 DO 10 I = 1,N T = T + DT CALL MEULER(T,POP,ACCUR,DT) EXACT = NUMER/ (M2+(DEN0M\textasteriskcenteredEXP(-B2\textasteriskcenteredT))) ERROR = ABS((EXACT-POP)/EXACT)\textasteriskcentered100. PRINT,T,POP,EXACT,ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON B2,M2,POPO REAL M2 G = (B2-(M2\textasteriskcenteredW)/(2.\textasteriskcenteredPOPO))\textasteriskcenteredW RETURN END

Question:

The angular velocity of a body is 4 rad/sec at time t = 0, and its angular acceleration is constant and equal to 2 rad/sec^2. A line OP in the body is horizontal ( \texttheta_0 = 0) at time t = 0. (a) What angle does this line make with the horizontal at time t = 3 sec? (b) What is the angular velocity at this time?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0172.htm

Solution:

The angular kinematics equations for constant angular acceleration are identical in form to the linear kinematic equations with a corresponding to a, \omega to v, and \texttheta to x. (a) Comparable to x = x_0 + \omega_0t + (1/2)at^2, we have \texttheta = \texttheta_0 + \omega_0t + (1/2)\alphat^2 where \texttheta_0, \omega_0 are the initial angular position and velocity of the body. Since \texttheta_0 = 0, we have \texttheta_-= \omega_0t + (1/2)\alphat^2 = 4 (rad/sec) × 3 sec + (1/2) × 2 (rad/sec^2) × (3 sec)^2 =21 radians = 21 radians [(1 revolution)/(2\pi radians)] = 3.34 revolutions. The angle \texttheta is then \texttheta= 0.34 × one revolution = 0.34 × 360\textdegree \approx 122\textdegree. (b)\omega= \omega_0 + \alphat = 4(rad/sec) + 2(rad/sec) × 3 sec = 10(rad/sec). Alternatively, \omega^2= \omega^2_0 + 2\alpha\texttheta = 4(rad/sec)^2 + 2 × 2(rad/sec^2) × 21 rad = 100(rad^2/sec^2)\omega = 10(rad/sec).

Question:

The mass of an electron is 9.11 × 10^-31 and Planck's constant = 6.63 × 10^-34 J sec. How much energy does an electron possess in its five lowest states, if it is placed in a cube of dimension 50 nm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0680.htm

Solution:

Wave function theory states that the permitted values of the energy of a particle in a box is given as E = [(N^2 h^2)/(8m L^2)], where E = energy, N = principal quantum number, h = Planck's constant, m = mass of particle, and L = length of box. Substitute in the known values and solve for E. The principal quantum number N corresponds to the energy levels of the electrons. The five lowest energy states are N = 1, N = 2, N = 3, N = 4, and N = 5, where N = 1 is the lowest and N = 5 the highest energy level. Substituting: E = [(N^2 h^2)/(8m L^2)] = [{N^2 (6.63 × 10^-34 J sec)^2} / {8 (9.11 × 10^-31 kg) (5 × 10^-8 M)^2}] = N^2 (2.41 × 10^-23) (Note: 50 nm = 5.0 × 10^-8 M) 1st energy level (lowest): N^2 = 1, E = 2.41 × 10^-23 J 2nd energy level: N^2 = 4, E = 9.64 × 10^-23 J 3rd energy level: N^2 = 9, E = 2.16 × 10^-22 J 4th energy level: N^2 = 16, E = 3.85 × 10^-22 J 5th energy level: N^2 = 25, E = 6.025 × 10^-22 J.

Question:

Construct a galvanic cell based on the reaction3Fe(s) Construct a galvanic cell based on the reaction3Fe(s) + 2Au^+3 \rightarrow 3Fe^+2 + 2Au(s) with ∆H negative.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0591.htm

Solution:

A galvanic cell may be defined as one in which spontaneous chemical reactions occur at each of the electrodes of the cell to produce an electric current. Thus, to construct this cell, you need to know those reactions that occur at the electrodes. The two electrodes are the anode and cathode. At the anode oxidation occurs, which means you have the loss of electrons. At the cathode, reduction occurs, which means you have the gain of electrons. The oxida-tion and reduction reactions are each called half-reactions. These are the reactions that occur at the electrodes. From the given reaction, it becomes apparent that oxidation is given by Fe(s) \rightarrow Fe^+2 + 2e^- and re-duction by Au^+3 + 3e^- \rightarrow Au(s). Thus, the anode half-cell would consist of an iron electrode in a solution containing Fe^+2 ions, while the cathode half-cell would consist of a gold electrode in a solution con-taining Au^+3 ions. A compound such as FeSO_4 could serve as a source of Fe^2+ ions, and AuCl_3 as a source of Au^+3 ions. The cell might appear as in the figure shown.

Question:

Both myoglobin and hemoglobin are oxygen-carrying molecules. Why is hemoglobin the molecule of choice to carry oxygen in the blood?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0351.htm

Solution:

In vertebrates, both myoglobin and hemoglobin act as oxygen carriers. Hemoglobin, however, is the only one that acts in the blood, carrying oxygen from the lungs to the tissues where it is needed for metabolic processes. Myoglobin is located in muscle, and serves as a reserve supply of oxygen. What accounts for the functional differ-ences between these two proteins? Myoglobin is a single-chain protein containing only one heme group, which serves to bind an oxygen molecule. Hemoglobin is composed of four polypeptide chains (each chain similar to the one of myoglobin) and thus has four binding sites for oxygen. It is the interaction of these four polypeptide chains that confers special properties to the hemoglobin molecule, making it a better oxygen-carrier in the blood. One of these important properties is the ability to transport CO_2 and H^+ in addition to O_2. Myoglobin does not have this ability. The binding of oxygen by hemoglobin is regulated by specific substances in its environment such as H^+, CO_2, and organic phosphate compounds. These regu-latory substances bind to sites on hemoglobin that are far from the heme groups. The binding of these substances affects the binding of oxygen by producing conformational changes in the protein, that lower hemoglobin's affinity for oxygen. A change at the regulatory site (where these regulatory substances bind) is translated to the heme site by changes in the way the four polypeptide chains are spatially arranged. This results in a structural change in the protein. Such interactions between spatially distinct sites are termed allosteric interactions. Hemoglobin is an allosteric protein (whereas myoglobin is not) and there fore exhibits certain allosteric effects which account for the functional differences between the molecules. This essential difference between myoglobin and hemoglo-bin is reflected in their respective oxygen dissociation curves. The curve is a plot of the amount of saturation of the oxygen-binding sites as a function of the partial pressure of oxygen: The shape of the oxygen dissociation curve reflects the crucial difference: the shape of the hemoglobin curve is sigmoidal (S-shaped) whereas that of myoglobin is hyper-bolic. The sigmoidal shape is ideally suited to hemo globin's role as an oxygen carrier in the blood. When the partial pressure of the oxygen is high, hemoglobin tends to bind oxygen; that is, many of the binding sites are filled with oxygen. There is a high partial pres-sure of oxygen (\sim 100 torrs) in the alveoli of the lungs, so that hemoglobin tends to pick up oxygen there. When the partial pressure of oxygen is low, hemoglobin tends to release oxygen. There is a low partial pres-sure of oxygen (\sim 20 torrs) in the capillaries in active muscle, where oxygen is being rapidly consumed by cellular respiration. CO_2 is being rapidly produced, and its concentration is correspondingly high. Hemo-globin tends to release oxygen. Hemoglobin thus acts to pick up oxygen where it is available and release it where it is needed. Myoglobin, on the other hand, has a higher affinity for oxygen than does hemoglobin. Thus, even at relatively low partial pressures, myoglobin still tends to bind oxygen. Only when the pressure is very low does myoglobin release oxygen. This behavior makes it unsuitable to carry oxygen in the blood. An oxygen carrier in the blood must be able to readily release oxygen at places where oxygen is needed. On a molecular level , the sigmoidal shape means that the binding of oxygen to hemoglobin is cooperative- that is, the binding of oxygen at one heme facilitates the binding of oxygen at another heme on the same hemoglobin molecule. The binding of the first oxygen molecule is thought to be most thermodynamically unfavorable. Sub-sequent oxygen molecules can bind more easily because the first molecule alters the structure (disrupt certain electostatic interactions) upon binding. The increasing affinity of the hemoglobin molecule for oxygen thus accounts for its sigmoidal dissociation curve.

Question:

An object, starting from rest, is given an acceleration of 16 feet per second^2 for 3 seconds. What is its speed at the end of 3 seconds?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0060.htm

Solution:

Since the acceleration is constant, we have a = (v_final-v_initial)/t orv_final= v_initial + at Butv_initial= 0 for the object started from rest. Therefore v_final= a × t = [(16 ft)/(sec^2)] × 3sec = 48 ft per sec.

Question:

Summarize the complete oxidation of glucose to CO2 and H2O.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0100.htm

Solution:

See accompanying figure. The complete oxidation is written as: Glucose + 6 O2 \textemdash\textemdash> 6 CO2 + 6 H2O + 36 ATP Glucose enters the cell. In the cell cytoplasm,glycolysisbreaks glucose downtopyruvate.Pyruvateenters the mitochon-drion. In the matrix, pyruvateis converted to acetyl-CoA. Acetyl-CoAenters the Krebs cycle by combiningwithoxaloa-ceticacid (OAA), a four carbon molecule. Acetyl- CoAis 2 carbons, so when it joins with OAA it forms a 6-c citrate molecule. By a series of oxidation-reduction steps, OAA is regenerated to keepthe cycle spinning,2 CO2s are produced per turn of the cycle, and the reducingequivalents NADH and FADH2 are produced. The reducing equivalentsgo to the in-ner mitochondrial membrane and pass their electronsalong. At three sites, a pair of protons are extruded into the in- termembranespace (only two pairs for FADH2). The last en-zyme of the respiratorychain iscytochromeoxidase: oxygen accepts the electrons, picksup a pair of H+s, and forms wat-er. Concomitant with the flow of electronsis the synthesis of ATP: The extruded H+s flow into the matrix viachannels on the "lollypop" enzymes. One ATP is produced per pair of H+s. Note on the figure that the only substances entering the cell are glucose andoxygen (circled) and the only sub-stances leaving the cell are water, carbondioxide and ATP (circled). This is in agreement with the equation above. When glucose is oxidized completely, 36 ATPs are produced. Fatty acids andamino acids also feed into this scheme at acetyl-CoA, but different amountsof ATP are produced de-pending on the molecule.

Question:

How much silver is deposited on a spoon which is plated for 20 minutes at 0.50 ampere?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0634.htm

Solution:

In order to check this problem, we must under-stand what happens during the electroplating process. As each silver ion migrates to the spoon, it is reduced to a metallic silver atom. All the silver ions are in a plus 1 ionized state. This means that for every electron pass-ing through the circuit, one silver ion will be turned into metallic silver and deposited. The problem then reduces to one of charge transport - as many atoms (or moles of atoms) of silver will be deposited as electrons (or moles of electrons) that pass through the circuit. We merely have to equate the two. The number of moles of deposited silver is given by the mass of the liberated silver (an unknown quantity) divided by the atomic mass of silver (107.9): Mass liberated in grams / (107.9gm / mole) To discover the amount of charge transported, and hence the number of moles transported, we use the relationship I = q / torq = It where I is current, q is charge, and t is time. Since we know that there are 96,500 coulombs of charge per mole of electrons, we can calculate the number of moles of electrons by using q / (96,500 coulombs /mole)orIt / (96,500 coulombs/mole) Equating the number of moles of silver plated, and the number of electrons which have passed through the circuit during the given time period, we have [Mass liberated in grams / (107.9gm / mole)] = [It / (96,500 coulombs / mole)] Since I = 0.50 amp and t = 20 min = 1200 sec. we have Mass / (107.9gm / mole) = (0.50 amp × 1200 sec) / (69,500 coulombs / mole) Solving, we obtain Mass = 0.671 gm, approximately

Question:

The animals having radial symmetry are considered to be at a lower evolutionary stage than those having bilateral symmetry. Why is this true? What is meant by the term secondarily radial?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0290.htm

Solution:

The bodies of most animals are symmetrical, that is, the body can be cut into two equivalent halves. In radial symmetry any plane which runs through the central axis from top to bottom divides the body into two equal halves. In bilateral symmetry, only one plane passes through the central body axis, that can divide the body into equal halves. In a bilateral body plant, six sides are distinguished: front (ventral), back (dorsal), head-end (anterior),tailend(posterior), left, and right. (Most animals are not perfectly symmetrical; for example, in man, the heart is located more to the left, the right lung is larger than the left lung, and the liver is found on the right side of the body.) Coelenterata(including the hydras, the true jelly-fishes, the sea anemones, sea fans and corals) andCtenophora(including the comb jellies and sea walnuts) are two radiate phyla. Their members have radially sym-metrical bodies that are at a relatively simple level of construction. They have no distinct internal organs, no head, and no central nervous systems though they possess nerve nets. There is a digestive tract with only one opening serving as both mouth and anus, and there is no internal space orcoelombetween the wall of the digestive cavity and the outer body wall. These animals have two distinct tissue layers: an outer epidermis (derived from embryonic ectoderm) and an innergastrodermis(derived from em-bryonic endoderm). A third tissue layer,mesoglea(mesoderm) is usually also present between the epidermis andgastrodermis. Often gelatinous in nature, it is not a well- developed layer, and has only a few scattered cells, which may be amoeboid or fibrous. Coelenterates and ctenophores are mostly sedentary organisms. Some are sessile at some stage of their life cycles, while others are completely sessile throughout their lives. The higher animals are usually bilateral in symmetry. The flatworms and the proboscis worms are regarded as the most primitive bilaterally symmetrical animals. They are however far more advanced than the coelenterates and ctenophores. Both have bodies composed of three well-developed tissue layers - ectoderm, mesoderm, and endoderm. Their body structures show a greater degree of organization than theradially symmetrical animals. Although there are no respiratory and circulatory systems, there is a flame- cell excretory system, and well-developed reproductive organs (usually both male and female in each individual). Mesodermal muscles show an advance in construction; cir-cular and longitudinal muscle layers are developed for purposes of locomotion and/or alteration of body shape. Several longitudinal nerve cords running the length of the body and a tiny "brain" ganglion located in the head are present which together constitute a central nervous system. In higher bilaterally symmetrical animals, there is observed a trend toward more complicated construction of the body. Separate organ systems are developed, specializing in dif-ferent functions. There is also a separation of sexes in individuals so that each individual produces only one kind of gametes (male or female but not both). The echinoderms areradiallysymmetrical animals. The phylum Echinodermata includes sea stars (starfish), sea urchins, sea cucumbers, sand dollars, brittle stars, and sealillies. These animals are fairly complex; they have a digestive organ system, a nervous system, and a reproductive system. The echinoderms are believed to have evolved from a bilaterally symmetrical ancestor. Also, whereas the adults exhibit radial symmetry, echinoderm larvae are bilaterally symmetrical. For these reasons, the echinoderms are considered to be secondarilyradiallysymmetrical. (The echinoderms will be discussed in greater detail in Chapter 12.)

Question:

Design an experiment to demonstrate that florigen, the flowering hormone, is indeed produced in the leaves and travels to the flower bud. 12 hrs per day - change to flower - inducing conditions 18 hrs per day - change to flower - inhibiting conditions A \rightarrow a B \rightarrow b C \rightarrow c D \rightarrow d

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0239.htm

Solution:

To demonstrate that florigen is indeed produced in the leaves, a flowering plant is defoliated and then exposed to appropriate conditions under which flowering normally occurs. The observation that the plant fails to flower indicates that florigen must have been produced in the leaves, and that florigen has been removed as a result of defoliation. A control experiment must of course be set up to show that defoliation of the plant produces no adverse effect on the plant. It is postulated that in order to exert its effect on the buds, florigen must be transported from the leaves to the flower buds. This can be tested by defoliating the lower part of a plant and exposing it to flower inhibiting factors, while maintaning the upper, leafy part under flower-promoting conditions. The leafy part is found to flower as expected. Flowering in the lower part, however, can best be explained by the diffusion of florigen from the leaves of the upper part to the buds of the lower part of the plant. The defoliated lower part of the plant can never flower on its own because of the absence of florigen. Because flowers do appear on the defoliated part of the plant, it is concluded that florigen is a diffusible sub-stance made in leaves. Even more compelling evidence for this conclusion comes from grafting experiments. One can grow two plants, one plant exposed to flowering conditions, the other not. The plant exposed to flowering conditions will flower (fig. 1a). The flowering plant is now grafted onto the non-flowering plant, and a partition is used so that the grafted portion continues to be exposed to flowering conditions and the non-flowering plant to flower-inhibiting conditions (fig. 1b). Gradually, the non-flowering plant will begin to produce flowers, usually at the point nearest the graft (fig. 1c) and eventually throughout the entire plant (fig. 1d). If one were to graft a short-day plant onto a long-day plant, and expose both to a short-day period, the short-day graft induces flowering in the long-day plant (fig. 2a). Likewise, one can expose both to a long- day period, and both portions will flower, the short-day portion having been induced by the long-day portions (fig. 2b) . Similar results can be obtained using day- neutral plants. Thus florigen appears to be physiologically equivalent in the three groups of flowering plants. Graft-induced flowering is successful only if there is a living tissue connection between the two plants. In addition, flower-induction is inhibited if one removes the phloem containing tissue in the graft. This supports the conclusion that florigen is transported via the phloem system of the plant. Some plants cannot be induced to flower by the above methods unless they are first defoliated. This suggests that, under flower- inhibiting conditions, some inhibiting substance may be produced by the leaves. Flowering may thus involve the interaction of both inducing and inhibiting factors.

Question:

What is an antipyretic ? Explain how it functions in humans.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E32-0938.htm

Solution:

Antipyretics are fever-reducing substances that relieve headaches and minor pains. They are also important in the treatment of arthritis and rheumatic fever. They belong to the class of medicinal compounds called sali-cylates, which, in turn, are termed analgesics or pain-killers. The antipyretic action of these drugs appears to be accomplished by the influence of the drugs on a center in the hypothalamus of the brain. The hypothalamus causes small blood vessels in the skin to dilate (to swell or ex-pand) thus enabling the body to lose heat.

Question:

The following table gives the number of items sold daily at a super-market. Week 1 Week 2 Sunday 900 800 Monday 400 500 Tuesday 500 300 Wednesday 600 300 Thursday 300 400 Friday 700 600 Saturday 1100 900 Write a FORTRAN program to compute a seven day moving average.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0486.htm

Solution:

Observations taken over periods of time typically show seasonal variability. The purpose of moving averages is to reduce these variations to enable us to spot the trend. First, find the average daily sales from Sunday to Saturday of the first week. This figure is placed in the table next to Wednesday - the midpoint of the examined period. Then find the average from Monday of the first week to Sunday of the second week, etc. Proceed-ing in this way we obtain the seven-day moving average for the given data. 7 day M.A. 7 day M.A. Sunday 571 Monday 586 Tuesday 571 Wednesday 643 543 Thursday 629 Friday 643 Saturday 614 From the table, it can be seen that the sales are falling. Perhaps the supermarket should start an advertising campaign; The program looks as follows: DIMENSION X(14), AV(7) L = 1 M = 7 SUM = 0 DO 2 I = 1,14 READ X(I) 2CONTINUE 10DO L J = L,M SUM = X(J) + SUM 1CONTINUE AV(L) = SUM/7 L = L + 1 M = M + 1 IF (M.EQ. 15) GO TO 11 GO TO 10 11DO 12 I = 1,7 PRINT AV(I) 12CONTINUE STOP END

Question:

How many cubic feet of life preserver of specific gravity .3, when worn by a boy of weight 125 lb and having a specific gravity .9, will just support him 8/10 submerged in fresh water of which 1 cu ft weighs 62.4 lb?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0401.htm

Solution:

In this problem the boy b is 8/10 submerged while the life preserver p must be completely sub-merged to give the maximum buoyancy. p must be completely sub-merged to give the maximum buoyancy. The weight of the boy W_b and the weight of the preserver W_p acting downward are just balanced by the buoyant force of the preserver B_p and the buoyant force of the boy B_b. B_b + B_p = W_b + W_p(1) B_b and B_p are equal to the weight of fluid displaced by the boy and the preserver respectively. Hence B_p = Mg where M is the mass of the water displaced by the boy. Since density d = M/V and the volume of the displaced water V = 8/10 V_b where V_b is the volume of the boy, then B_b = [(8/10) V_b] d_w g where d_w is the density of water. The weight of the boy is W_b = (V_bd_b)g =125 lb W_b = (V_bd_b)g =125 lb where V_bd_b is the mass of the boy and d_b is the density of the boy. Similarly, B_p = V_pd_w gandW_p = V_pd_pg where V_p is the volume of fluid displaced by the preserver, and d_p is the density of the preserver. V_b= W_b/gd_b Now,dw = (mass of 1 cu ft of water) / (1 cu ft of water) = [(weight of 1 cu ft of water) / g] / (1 cu ft of water) = [(62.4 lb) / (1 cu ft of water)] × (1/g) Also, the specific gravity (or relative density) of the preserver with respect to water S_p = d_p/d_w = 3/10 ; and s_boy = (d_b/d_w) = 9/10 S_p = d_p/d_w = 3/10 ; and s_boy = (d_b/d_w) = 9/10 Therefore, equation (1) becomes \therefore(8/10) V_bd_wg_- + V_pd_wg = W_b + V_pd_pg (8/10)(125/gd_b) d_wg_- + V_pd_wg = 125 + V_p (3/10) d_wg V_p = [(8/10)(125)(d_w/d_b) - 125] / [(3/10) d_wg - d_wg] = [(8/10)(125)(1/s_b) - 125] / [-(7/10) d_wg] = [(8/10){(125)/(9/10)} - 125] / [-(7/10)(62.4/g)g] = [125 - (8/9)(125)] / [(7/10)(62.4)] = [(1/9)(125)] / [(7/10)(62.4)] = 13.9/43.7 = .32 cu ft (approx).

Question:

What is the force due to the liquid only acting on a circular plate 2 in. in diameter which covers a hole in the bottom of a tank of oil 4 ft high, if the specific gravity of oil is .5?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0414.htm

Solution:

Since the liquid in the tank is in equilibrium, it can exert only a vertical force on the bottom of the tank, as shown in the figure. Furthermore, the net force on the bottom surface of the tank is due only to the weight of the water above this surface. Hence, F_net = Mg(1) where M is the mass of fluid in the tank. Since the density of the fluid is \rho = M/V where V is the tank volume, we obtain M = \rhoV andF_net = \rhogV(2) The force on the circular plate of radius r is, using (2) F_plate = F_net (\pir^2/A) = \rhogV(\pir^2/A)(3) where A is the area of the tank bottom. We may write the volume of the tank asV = hA(4) Here, h is the depth of fluid in the tank. Combining (4) and (3) F_plate = \rhoghA(\pir^2/A) = \pi\rhoghr^2 F_plate = \pi\rhoghr^2 Now, the specific gravity of a substance, \sigma, is defined as \sigma = \rho/\rho_w where \rho' is the density of the substance and \rho_w is water's density. Hence F_plate = \pi\rho_w\sigmaghr^2 F_plate = (3.14) (1.94 sl/ft^3) (.5) (32 ft/s^2) (4 ft) (1 in^2) In order to keep our units consistent, 1 in = 1/12 ft 1 in^2 = 1/144 ft^2 whenceF_plate = (3.14) (1.94 sl/ft^3) (.5) (32 ft/s^2) (4 ft) (1/144 ft^2) = 2.71 lb.

Question:

How is DNA replicated ?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0615.htm

Solution:

By studying the Watson-Crick model of DNA, Kornberg and his colleagues determined the mechanism of DNA replication. Since DNA is the genetic material of the cell, it must have the information to replicate itself built into it. The Watson-Crick model of DNA seems to offer a good explanation of how DNA is replicated. There are only four different nucleotides found in DNA. Chargaff's rule of. specific base pairing states that these nucleotides (adenine, guanine, thymine and cytosine) are ordered on both strands of the DNA helix so that adenine is always paired to thymine and cytosine to guanine. Thus, in order for DNA to be replicated, the helix need only be unwound to form a template. The nucleotide building blocks can line up in a sequence complementary to the order presented. In this way, two double stranded helices identical to the original molecule are synthesized in the nucleus. This is a general overview of the mechanism of DNA replication. Actually, the first step of DNA replication occurs when unwinding proteins uncoil a part of the DNA helix (see figure - step A). At this time, RNA polymerase combines with the sigma factor to initiate RNA synthesis. This is an essential preliminary step because the RNA that is synthesized serves as a primer which is necessary before DNA replication can occur (Figure-step B). The RNA that is synthesized is not a final product because before DNA replication is completed, it is destroyed by endonucleases and exonucleases. After the primer is made, DNA polymerase initiates DNA replication. There are free nucleotides in the nucleus which complementarily join to form a new strand along the 5' to 3' direction of the unwound portion, however, the DNA strand itself is synthesized in a 3' to 5' manner (Figure-step C). DNA replication is thought to occur in short, uncontinuous bursts allowing the DNA helix to unwind farther (Figure-step D). DNA polymerase then fills in the gaps between the fragments synthesized. DNA ligase joins the 5' end of one fragment to the 3' end of the adjacent fragment. In this way, a strand of DNA is replicated accor-ding to the semi-conservative model, a parental DNA replica-tes to give two first generation hybrids containing one strand of the parental DNA and one newly synthesized strand. (Figure-step E).

Question:

Given that theelectronegativitiesof F,Cl, Br, and I are, respectively, 4.0, 3.0, 2.8, and 2.5, account for the fact that the dipole moment decreases in the sequence HF,HCl,HBr, and HI, even though bond length and the number of electrons increase.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0645.htm

Solution:

A dipole is defined as a molecule which is electrically unsymmetrical - that is, the centers of positive and negative charges are not located at the same point within the molecule. Quantitatively, a dipole is described by its dipole moment, which is equal to the charge times the distance between the positive and negative centers. The decrease in dipole moment will reflect a decrease in either the value of the charges or the distance between them. It is given that the bond length increases in this sequence, which means that distance increases. This suggests an in-crease in dipole moment within the sequence. But the dipole moments decrease, this means that the charge must be de-creasing. This can be explained by noting that the electro-negativity values are decreasing within the sequence. Electronegativity measures the tendency of an atom to attract shared electrons in a molecule. By attracting these electrons, it develops a negative charge. If theelectronegativitydecreases, so does the charge, which would then account for the decrease in dipole moment.

Question:

An electron travels from one to the other of two plates, between which is maintained a potential difference of 1000V. With what speed and with what energy does the electron arrive at the positive plate? A positively charged particle with equal and oppos-ite charge but 3680 times the mass is then released at the positive plate. With what velocity and what energy does it reach the negative plate?

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Solution:

The electron travels through a potential difference of 1000 V. The energy acquired is thus E = eV, where e is the charge of the electron ( = 1.60 × 10^\rule{1em}{1pt}19 C) and V is the potential difference between the plates. Thus E (1.60 × 10^\rule{1em}{1pt}19 c) (1000 V)= 1.60 × 10^\rule{1em}{1pt}16 J. However, the energy required to move an electron through a potential difference of 1 volt is an electron volt (eV). Hence, the energy required to move the electron through a potential difference of 1000 volt is 1000 eV, whence 1000 eV = 1.60 × 10^\rule{1em}{1pt}16 J. As required by the conservation of energy, all this potential energy is converted to kinetic energy on arrival of the electron at the positive plate. Hence (1 /2) mv^2 = 1.60 × 10^\rule{1em}{1pt}16 Jor v^2 = (3.20 × 10^\rule{1em}{1pt}16 J .) / (9.1 × 10^\rule{1em}{1pt}31 kg) \therefore v = 1.88 × 10^7 m \textbullet s^\rule{1em}{1pt}1. For the positively charged particle, the charge it possesses is the same in magnitude as that of the electron and it moves through the same potential differ-ence in the opposite direction. Hence, it acquires the same energy of 1.60 × 10^\rule{1em}{1pt}16 J. The velocity, however, is smaller because of the much larger mass. Thus (1 /2) m_1v^2_1 = 1.60 × 10^\rule{1em}{1pt}16 Jor V^2_1 = (3.20 × 10^\rule{1em}{1pt}16 J.) / (9.1 × 10^\rule{1em}{1pt}31 × 3680 kg) \thereforev_1 = 3.10 × 10^5 m \bullet s^\rule{1em}{1pt}1.

Question:

In Figure 1, a small object of mass m is attached to a light string which passes through a hollow tube. The tube is held by one hand and the string by the other. The object is set into rotation in a circle of radius r_1 with a speed v_1. The string is then pulled down, shortening the radius of the path to r_2. Find the new linear speed v_2 and the new angular speed \omega_2 of the object in terms of the initial values v_1 and \omega_1 and the two radii.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1069.htm

Solution:

The downward pull on the string is transmitted as a radial force on the object. Such a force exerts a zero torque on the object about the center of rotation. Since no torque acts on the object about its axis of rotation, its angular momentum in that direction is constant. Hence initial angular momentum = final angular momentum, mv_1r_1 = mv_2r_2, and v_2 = v_1(r_1 / r_2). Since r_1 > r_2, the object speeds up on being pulled in. In terms of angular speed, since v_1 equals \omega_1r_1 and v_2 equals \omega_2 r_2, mr_1 ^2\omega_1 = mr 2 2 \omega_2 and \omega_2 = (r_1 / r_2) ^2\omega_1, so that there is an even greater increase in angular speed over the initial value.

Question:

In what ways are slime molds like true fungi? In what ways do they resemble animals?

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Solution:

Both cellular slime molds (acrasiomycota) and true slime molds (myxomycota) have unusual life cycles con-taining fungus-like and animal like stages. Slime molds have membrane-bound nuclei, are heterotrophic, ingest food, lack cell walls and produce fruiting bodies. They belong to the kingdom Protista in the phylum Gymnomycota. The true slime mold's adult vegetative stage is decid-edly animal like. At this stage, the slime mold is a large, diploid, multinucleated amoeboid mass called a plasmodium. It moves about slowly and feeds on organic material by phagocytosis. Plasmodium growth continues as long as an adequate-food supply and moisture are available. When these run short, the plasmodium becomes stationary and de-velops organs specialized to produce haploid spores, known as the fruiting bodies. At this stage, the true slime mold is similar to the fungi. Meiosis occurs in the fruiting body, and spores with cellulose cell walls are released. Spores are a resistant and dormant form of a slime mold. When thw spores germinate, they lose their cell walls and become flagellated gametes. Gametes fuse to become zygotes. The zygotes lose their flagella, become amoeboid, and grow into multinucleated plasmodial slime molds. Cellular slim molds differ from the true (acellular) slime molds in being haploid and in that the amoeboid cells, on swarming together, retain their identity as individual cells. The cellular slime molds resemble amoebas through-out most of their life cycle: they lack cell walls, move about and ingest particulate matter. Under certain con-ditions, the amoebas aggregate to form a multicellular pseudoplasmodium, called a slug. The pseudoplasmodium becomes stationary and fruiting bodies are formed. Spores are not produced by cellular division, but by the formation of cell walls around the individual amoeboid cells. Each spore becomes a new amoeboid cell when it germinates. Cellular slime molds are haploid throughout their life cycles. The formation of a fruiting, spore-forming body is characteristic of fungi.

Question:

Near the absolute zero of temperature, the specific heats of solids obey the Debye equation, c = kT^3, where T is measured in \textdegreeK. For a particular solid k has the value 2.85 × 10^-2 cal/g\bulletK deg^4. Calculate the heat that must be supplied to raise 50 g of the solid from 10\textdegreeK to 20\textdegreeK and the mean specific heat capacity of the solid in this interval.

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0469.htm

Solution:

The specific heat capacity varies markedly with temperature. The mean value of the specific heat is defined by c= [1/(T_2 - T_1)] ^(T)2\int_(T)1 cdT Over the range 10\textdegreeK to 20\textdegreeK its mean value will be c= [1/(20 - 10)0K] ^200K\int_100K cdT = 1/10\textdegreeK ^200K\int_100K kT^3 dT = [1/100K] [(1/4) kT^4]200K100K= [1/400K] × 2.85 × 10^-2 cal/g\bulletK deg^4 [20^4 - 10^4] K deg^4 = 106.9 cal/g\bulletK deg. This compares with a magnitude for c of 28.5 cal/g\bulletK deg at 10\textdegreeC and 228 cal/g\bulletK deg at 20\textdegreeC. The heat that must be applied to raise the temper-ature of 50 g through the range of temperature is Q = mc(T_2 - T_1) = 50 g × 106.9 cal/g\bulletK deg × 10 K deg = 53,450 cal.

Question:

What is meant by "genetic drift"? What role may this playin evolution?

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/Users/wenhuchen/Documents/Crawler/Biology/F27-0706.htm

Solution:

The random fluctuation of gene frequencies due to chance processesin a finite population is known as genetic drift. Genetic drift may occurby what is commonly referred to as the Founder Principle. This principleoperates by chance factors alone. A group of individuals may leavetheir dwelling and establish a new feeding or breeding ground in an unexploitedarea. The individuals that find the new area may not represent thesame genetic makeup as the original population, since the individuals thatleft were a random group. This exhibits how one population may establishitself with a new genetic composition by chance alone. Genetic driftmay also affect a population without the migration of individuals. Catastrophes in one area of a certain popu-lation, such as a flood, may randomlykill individuals of a population. Those individuals which survive, doso by chance alone and not by possessing any special selective advantage. Thus genetic drift may result from natural catastrophes where arandom group of individuals may survive by chance alone and change theoriginal population's allelic frequencies. The alteration of allelic frequenciesby chance due to genetic drift is unlike the directional movementscaused by the systematic pressures of mutation, selection, anddifferential migration. The situation is quite different in a large population. In large populations, the relatively small numbers of chance variationsin gene frequencies are absorbed into the population in succeedinggenerations and the overall effect is negligible. In small interbreedingpo-pulations, however, where there are limited numbers of progenyin a given generation, random variation has a significant effect andgenetic drift can become an important factor. Random fluctuation can mostoften lead to fixation of a given allele in a small population. That is, one allele is slowly replacing the other as a result of chance, and the former becomes fixed in the population as the latter is lost. This process of reduction ofheterozygositythrough loss and fixation atvarious loci is also known as the decay of variability. The fixation of certain genes by means of genetic drift may explain theappearance of seemingly unimportant and useless structures in some organisms. These structures may be the expression of homozygous gene pairswhich have accumulated in the population by chance alone. Such a lossof variability can also inactivate natural selection, which can act only whena certain degree of phenotypic variation is present. If there is a lack ofvariation, any adverse situation, such as a period of extreme cold or the arrivalof a group of predators, could terminate an entire population. In sucha homogeneous population, only genetic mutation could hopefully preservethe population. This happens as some mutants arise which are resistantto the unfavorable force. However, since the frequency of an allelicmutation is in the area of 10-6, a homogeneous population could find itselfin great danger of extinction.

Question:

Natural chlorine is a mixture of isotopes. Determine its atomic weight if 75.53% of the naturally occurring element is chlorine 35, which has a mass of 34.968, and 24.47% is chlorine 37, which has a mass of 36.956.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0708.htm

Solution:

If naturally occurring chlorine contains 75.53% chlorine 35 and 24.47% chlorine 37, one mole of chlorine contains .7553 mole chlorine 35 and .2447 mole chlorine 37. The weight of one mole of chlorine is then, equal to the sum of the weight of .7553 mole ^35CL and .2447 mole ^37CL. atomic weight of CL = (.7553) (34.968) + (.2447) (36 .956) = 26.34 + 9.11 = 35.45.

Question:

Most bacteria are heterotrophic, requiring an organic form of carbon, such as glucose, which they oxidize to obtain energy. How do chemotrophic bacteria which can utilize carbon dioxide as a sole carbon source acquire energy?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0129.htm

Solution:

Autotrophs are organisms which require only carbon dioxide as a carbon source, from which they can construct the carbon skeletons of all their organic biomole-cules. The carbon dioxide is reduced to glyceraldehyde-3-4 phosphate in order to synthesize carbohydrates. This reductive process requires much energy. Since glucose is not available to be oxidized for energy, these autotrophic bacteria must oxidize inorganic compounds. They are called chemoautotrophs, since they obtain their energy by oxidizing chemical compounds (as opposed to photoautotrophs, which obtain their energy from light). The inorganic compounds oxidized by the various bacteria include molecular hydrogen, ammonia, nitrite, sulfur (sulfide ions), and iron (ferrous ions). These oxidations result in electrons which enter the respiratory chain, with the concomitant production of ATP. ATP is then used as a source of energy in the reduction of CO_2. Bacteria of theHydrogenomonasgenus obtain energy through the oxidation of hydrogen gas. They possess an enzyme, hydrogenase, which catalyzes the following reaction: H_2 + (1/2) O_2\rightarrow H_2O + 2e^\rule{1em}{1pt} The electrons are transferred to NAD to form NADH_2, which is then oxidized in the respiratory chain, yielding ATP. The bacteriumNitrosomonasobtains energy by the oxidation of the ammonium ion: (2NH_4^+)^ + + 3O_2 \rightarrow 2NO_2^\rule{1em}{1pt} + 2H_2O + 4H^+ + 2e^\rule{1em}{1pt} The electrons produced enter the respiratory chain where they are passed down to O_2, forming ATP along the chain. Bacteria of theNitrobactergenus obtain their energy through the oxidation of nitrite ions into nitrate ions: 2NO_2^\rule{1em}{1pt} + O_2 \rightarrow 2NO_3^\rule{1em}{1pt} + 2e^\rule{1em}{1pt} Again the electrons produced are used in the formation of ATP. The oxidation of ammonia into nitrate is called nitrification. It is one of the most important activities of autotrophic bacteria since it provides the form of nitrogen most available to plants. Nitrification is carried out in two stages, with the first stage involvingNitrosomonasand the second stage featuring the oxidation of nitrites byNitrobacter. The opposite process, called nitrate respiration, is the reduction of nitrate to ammonia. This process is carried out by several heterotrophic bacteria under anaerobic conditions. The oxygen of the nitrate serves as the hydrogen acceptor (under aerobic conditions, molecular oxygen would normally serve as the final electron or hydrogen acceptor). The overall reaction is as follows: HNO_3 + 4H_2 \rightarrow NH_3 + 3H_2O Nitrification should not be confused with nitrogen fixation or denitrification. Nitrogen-fixing microorganisms in the soil use molecular nitrogen in the atmosphere as their source of nitrogen and convert it into ammonia. The ammonia is then used in the synthesis of proteins and other nitro-genous substances. Denitrification is the reduction of nitrates to molecular nitrogen carried out by certain bacteria, such asPseudomonas. The different kinds of reactions are outlined as follows:

Question:

A and B are the players. A tells B to pick a number between 1 and 100. He then asks B to tell him the remainders when the secret chosen number is divided successively by 3, 5 and 7. A then tells B the number he originally chose.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0560.htm

Solution:

The game is centered on the notion of congruence. Let x be the unknown number chosen by B, 1 \leq x \leq 100. Now x divided by 3 can give a remainder of 0, 1 or 2. The numbers between 1 and 100 can be arranged in three sets S_0 = {3,6,9,...,90,93,96,99} S_1 = {1,4,7,...,94, 97,100} S_2 = {2,5,8,...,92,95,98} Note that {S_0,S_1,S_2} forms a partition of 1 - 100. Similarly, on division by 5, the possible remainders 0,1,2,3,4 give rise to the disjoint sets S_0 = { ,5,10,...,90,95,100} S_1 = {1, 6,11,...,91,96} S_2 = {2,7,12,...,92,97} S_3 = {3,8,13,...93,98} S_4 = {4,9,14,...,94,99} Finally division by 7 yields the following Sets:S_0 = {7,14,...,91,98} S_1 = {1,8,15,...,92,99}S_2 = {2,9,16,...,93,100} S_3 = {3,10,17,...,87,94}S_4 = {4,11,...,88,95} S_5 = {5,12,...,89,96}S_6 = {6,13,...,90,97}. To be concrete, suppose x = 41. Then the remainders are 2,1, 6 when 41 is divided by 3,5 and 7 respectively. Thus, x = 41 lies in S_2 = {2,5,8,...,95,98} S_1 = {1,6,11,...,91,96} and S_6 = {6,13,...,90,97}. In fact, 41 is the only number that simultaneously lies in these three sets ! Now memorizing the above sets is one way of mastering this puzzle but fortunately, a much simpler algorithm exists. It uses the fact that 3 × 5 × 7 = 105. Let the secret number be x and let A, B, C be the remainders. Then x \equiv D = 70A + 21B + 15C (mod 105)(1) Thus, for x = 41, 70(2) + 21(1) + 15(6) = 140 + 21 + 90 = 251. But we need x \leq 100. Note that 251 - 105 = 146 and 146 - 105 = 41. Thus, 251 when divided by an integral multiple of 105 leaves a remainder of 41. We say 251 is congruent to 41 modu-lo 105 and write 41 \equiv 251 (mod 105). The program follows: 10PRINT \textquotedblleftTHINK OF A NUMBER BETWEEN 1 and 100." 20PRINT "YOUR NUMBER DIVIDED BY 3 HAS A REMAINDER OF\textquotedblright; 30INPUT A 40PRINT "YOUR NUMBER DIVIDED BY 5 HAS A REMAINDER OF\textquotedblright; 50INPUT B 60PRINT "YOUR NUMBER DIVIDED BY 7 HAS A REMAINDER OF\textquotedblright; 70INPUT C 80D = 70\textasteriskcenteredA + 21\textasteriskcenteredB + 15\textasteriskcenteredC 90IF D < = 105 THEN 120 100D = D - 105 110GOTO 90 120PRINT "YOUR NUMBER WAS\textquotedblright; D;", RIGHT?"; 130INPUT A$ 140IF A$ = "YES" THEN 170 150REM SOMETIMES HUMANS MAKE ERRORS 160IF A$ = "NO" THEN 190 165PRINT "VERY FUNNY! NOW TRY 'YES' OR 'NO'." 166GO TO 120 170PRINT "HAL HAS SPOKEN TRULY" 180GOTO 200 190PRINT "CHECK YOUR ARITHMETIC" 200PRINT "LET'S TRY AGAIN." 210GOTO 10 220END

Question:

(1)Obtain the SUM and CARRY functions for a FULL ADDER and give its circuit implementation. (2)Obtain the Difference and Borrow functions for the FULL SUBTRACTOR and give its circuit implementation.

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Solution:

In a FULL-ADDER we have to add three bits, the Augend bit, Addend bit and previous carry bit. Using the rules for addition in binary, create the truth table shown in Figure 1. A B C_i-1 SUM CARRY 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Fig. 1 Now use the Karnaugh Maps of Figure 2 to find the logic expressions. SUM = AB'C' +A'BC' +ABC + A'B'CCARRY = AC+AB+BC = (AB' +A'B) C' + (AB+A'B') C= AB + (A 'B+AB') C (Distributive Law) = (A+B)C' + (A+B)' C= AB + (A+B) C (NOR Definition) = A+B+C(NOR definition) The Circuit Implementation of a full adder is shown in Figure 3. Figure 3. FULL SUBTRACTOR:A full subtractor is a combinational circuit that performs a subtraction between two bits while taking into account the fact that a 1 may have been borrowed by a lower significant position. The three inputs A,B,C denote the minuend, subtrahend, and previous borrow, respectively. The two outputs D and K represent the difference and next borrow, respectively. The truth table for the circuit is shown in Figure 4. A B C Diff Borrow 0 0 0 0 0 0 0 1 1 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 Fig. 4 Now use the Karnaugh's Maps of Figure 5 to find the logic equations. Difference = AB'C' + A 'BC'Borrow = A 'B'C + A 'BC' + ABC + A'B'C+ BC = A + B + C= A' [B'C + BC'] (same as full-adder)+ BC = A'(B+C) + BC The Circuit Implementation of a full subtractor is shown in Figure 6. in Figure 6.

Question:

An automobile of mass 50 slugs accelerates from rest. During the first 10 sec, the resultant force acting on it is given by \sumF = F_0 -kt, where F_0 = 200 lb, k = 10 lb/sec, and t is the time in seconds after the start. Find the velocity at the end of 10 sec, and the distance covered in this time.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0131.htm

Solution:

The resultant force decreases with time, since F(t) = F_0 -ktwhere t is the elapsed time in seconds. Newton's Second Law tells us that at any time F = ma, where F is the net force on a mass m, and a is its acceleration. Hence F(t) = m a(t);a(t) = [{F(t)} /m] = (F_0 -kt)/m We integrate to find v(t), the instantaneous velocity at time t. v(t) = \inta(t)dt= \int{(F_0/m) - (kt/m)}dt= (F_0/m)t - (k/m) (t^2/2) + c_1 To evaluate the constant of integration, C_1 we use our knowledge that the car accelerates from rest. Hence, v(0) = 0 v(0) = (F_0/m)(0) - (k/2m)(0)^2 + c_1 = 0 \therefore c_1 = 0. The velocity function is therefore v(t) = (F_0/m)(t) - (k/2m)(t)^2 and, at t = 10s, v = (200 lb/50sl)(10 s) - {(10 lb/s)/100sl} 100 s^2 = Again, we integrate v(t) to find s(t), the instantaneous position at time t. s(t) = \intv(t)dt= \int(F_0/m)t - (k/2m)t^2 = (F_0/2m)t^2 - (k/6m)t^3 + c_2 We assume that the car starts at position s(0) = 0. Evaluating c_2, s(0) = (F_0/2m)(0)^2 - (k/6m)(0)^3 + c_2 = 0 \therefore c_2 = 0 The position (and displacement) function is given by s(t) = (F_0/2m)^2 - (k/6m)t^3 Att = 10, the displacement from the origin is s(10 s) = (200 lb/100sl)(100 s^2) - {(10 lb/s)/300sl}(1000 s^3) s(10 s) = 200 f - 33 (1/3) f = 166 (2/3) f.

Question:

There is a disease of tobacco plants called tobacco mosaic. Explain how one can demonstrate the causative agent.

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0148.htm

Solution:

Tobacco mosaic disease causes the leaves of the tobacco plant to become wrinkled and mottled. By grinding these leaves, one can extract the juice from the infected plant. If this juice is rubbed onto the leaves of a healthy plant, it becomes infected. The agent can therefore be transmitted in the juice. If the juice is boiled before being rubbed on a healthy plant, no disease develops. The agent might therefore be a bacterium, since most bacteria cannot sur-vive temperatures above 70\textdegreeC. To isolate the bacteria, we could use filters with known pore diameters. Since most bacteria are larger than .5 \mu (a micron, \mu, is equal to 10^-6 meter), we could use a filter with a pore size slightly less than this. The liquid is passed through this very fine filter in order to remove the bacteria, and the filtrate is checked for the absence of bacteria. When the filtrate is rubbed on the leaves of a healthy plant, it still causes infection. The causative agent could therefore be a toxin produced by some bacteria or a bacterium smaller than any known . It can be demonstrated that the agent is not a toxin by showing that the agent can re- produce. If the filtrate is used to infect a healthy plant , and this plant is then ground up to obtain a new filtrate, this new filtrate could be used to infect another plant. If this procedure is repeated, and if the extent of mottling becomes decreased with subsequent infections , the agent is most probably a toxin. The attenuation of the disease can be attributed to the dilution of the toxin. However, in tobacco mosaic disease the extent of the mottling does not decrease with repetitions of filtration. One can assume that the causative agent does not become diluted. Another process must be occurring, since the agent is reproducing . Further experiments can show that the causative agent reproduces itself only inside the living plant; it cannot grow on artificial media . It is therefore not a bacterium, since bacteria do not require living host cells to reproduce. The microbial agent of this disease is termed a virus: the tobacco mosaic virus, or TMV. It can be iso-lated and crystallized and observed using the electron microscope. It is a rod-shaped virus, much smaller than any known group of bacteria: .28 \mu in length, .015 \mu in diameter.

Question:

Give an example, in pseudocode, of the REPEAT-UNTIL construct. Compare it to the DO-WHILE construct. Are there any similarities? differences?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0184.htm

Solution:

In pseudocode, the REPEAT-UNTIL construct is rendered as repeat tasks A - Z until(X) The condition X can be either logical or numerical. When condition X is satisfied, program control passes out of the loop. Notice also that no matter what the status of the condition X, the enclosed statements (tasks A - Z) are executed at least once. Let us write out the REPEAT-UNTIL construct again, but this time we will include the DO-WHILE construct alongside for comparison. repeatdo while (X) tasks A - Ztasks A - Z until (X)end do while When program control passes into a DO-WHILE construct, the condition, or predicate, is tested before the enclosed statements are executed. The REPEAT-UNTIL construct, as previously mentioned, allows the enclosed statements to be executed at least once. A flowchart representation classifies the structure: The DO-WHILE construct, on the other hand, has this flowchart The DO-WHILE construct, on the other hand, has this flowchart rep-resentation: A general rule can be derived from the similarity of the two constructs: any REPEAT-UNTIL construct can be coded using the DO-WHILE construct plus a Boolean variable. For example, suppose that condition X in the earlier pseudocode is a logical variable TEST. Before entering the loop, TEST = FALSE. To exit, the loop TEST must become TRUE. However, tasks A through Z must be executed at least once, a requirement of the REPEAT- UNTIL construct. To render these notions into a pseudocoded DO-WHILE, we can write: TEST = FALSE do while \textasciitilde (TEST) tasks A - Z TEST = Q end do while The variable Q is dependent upon the execution of tasks A through Z. After evaluating tasks A through Z, Q can become either TRUE or FALSE. The loop is terminated when Q is true; the condition \textasciitilde (TEST) is not satisfied.

Question:

The combined effects of pollution have been known to bring about the phenomena of synergism and antagonism. Explain each of these terms. Provide specific examples of each.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0857.htm

Solution:

The effects of individual pollutants can be disastrous to health, while the combined action of two or more pollutants can be completely different on health and/or the environment. When the combined effects of two or more pollutants are more severe or qualitatively different from the indi-vidual effects, a condition called synergism exists. For example, in 1966 and 1967, there was injury to peanut crops due to the synergistic action of ozone and sulfur dioxide. A condition termed antagonism exists when the combined effects are less severe. For example, cyanides in industrial wastes are quite poisonous to aquatic life, however, in the presence of nickel, a nickel cyanide complex is formed. This nickel-cyanide complex is not as toxic as the cyanides originally produced.

Question:

What is the molarity of ethanol in 90-proof whiskey? Density of alcohol = 0.8 g/ml.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E26-0876.htm

Solution:

Molarity is defined as the number of moles of solute per liter of solution. By definition, 90-proof whiskey is 45 % ethanol by volume. Thus in 1 l (= 1000 ml) of whiskey there is (45 %) (1000 ml) = 450 ml of ethanol. To determine the number of moles of ethanol in 1 l of whiskey first calculate the weight of ethanol in 1 l by multiplying the density of ethanol by the volume of ethanol in 1 l. Then, divide the weight of ethanol in 1 l by its molecular weight (MW ethanol = 46 g/mole). Therefore, (450 ml ethanol) (0.8 g/ml) = 360 g ethanol molarity = [(number moles ethanol) / (1 l whiskey)] = [(360 g/46 g/ mole) / (1 l)] = 7.83 M.

Question:

The engine of a jet aircraft develops a thrust of 3000 lb. What horsepower does it develop at a velocity of 600 mi/hr = 800 ft/sec?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0281.htm

Solution:

Power = Work/Time = Fs/T(1) where F is the force acting on a body and s is the displacement of the object in the direction of F in time t. By definition of velo-city v = s/t(2) v = s/t(2) Therefore, combining equations (1) and (2) P = Fv In this example, P = Fv = 3000 lb × 880 ft/sec = 2,640,000 ft\bulletlb/sec Since 1 hp = 550 ft\bulletlb/sec = [2.64 × 10^6 ft\bulletlb/sec]/[550 (ft\textbulletlb/sec)/hp] = 4800 hp.

Question:

The density, \rho of air at STP is 0.00129 g/cm^3 , the specific heat capacity at constant pressure is 0.238 cal/g - 0K, and the ratio of the principal specific heats is 1.40. what is the mechanical equivalent of heat?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0462.htm

Solution:

The equation of state of an ideal gas for one mole is PV = RT, where R is the gas constant, P is the pressure of the gas, T is its temperature in 0K and V is the molar volume. This equation may be written as, P(M/ \rho) = RT where M is the molecular weight, and \rho is the density of air. Thus, [R/M] = [P/\rhoT] = [(1.013 x 10^6 dynes cm^-2)/(0.00129 g\bulletcm^-3 × 237\textdegreeK)] = [(1.013 × 10^9)/(1.29 × 273)] ergs/g -\textdegreeK Also C_P - C_V = R, where C_P and C_V are the molar heat capacities at constant pressure and constant volume respective-ly. Thus C_p -C_v= R/M, where C_p andC_vare the corres-ponding specific heat capacities per unit mass. Further, C_P /C_V = \Upsilon. Therefore, R/M = (C_p - C_v C_v ) = C_p[1 - (1/\Upsilon)] = 0.238 cal/g - \textdegreeK × [1 -(1/1.40) ) = C_p[1 - (1/\Upsilon)] = 0.238 cal/g - \textdegreeK × [1 -(1/1.40) ] =[(0.2380.40)/ 1.40] cal/g - \textdegreeK. The value of R/M is thus given in two systems of units, one mechanical and the other thermal. The mechanical equivalent of heat is thus obtained by dividing one by the other. Hence J = [(1.013 × 10^9)/(1.29 × 273)]ergs/g -0K × [1.40/(0.238 × 0.40 cal/g -0K)] = 4.23 × 10^7 ergs/cal.

Question:

A lump of clay of mass 30 gm traveling with a velocity of 25 cm/sec collides head on with another lump of clay of mass 50 gm traveling with a velocity of 40 cm/sec in exactly the opposite direction. If the two lumps coalesce (a) what is the velocity of the combined lump after the collision, assuming that no external forces act on the system? (See figure.) (b) What is the energy lost due to collision?

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Solution:

(a) Although all the velocities are in the same straight line, it is important to remember that momentum is really a vector and to distinguish care-fully between positive and negative directions. Choose the positive direction to be to the right, in the same direction as the initial velocity of the 30 gm lump of clay. Then, Initial momentum of 30 gm lump = (30 gm) (25 cm/sec) = 750 gm - cm/sec Initial momentum of 50 gm lump = (50 gm) (- 40 cm/sec) = - 2,000 gm-cm/sec Therefore, total initial momentum =750 gm-cm/sec - 2,000 gm-cm/sec = - 1250 gm-cm/sec. If V is the velocity of the combined 80 gm lump after the collision, Momentum after collision = (80 gm)V The law of conservation of linear momentum assures that the total momentum of the system before collision is equal to the system's momentum after collision. Equating the total momentum before to the total momentum after the collision, - 1250 gm-cm/sec = (80 gm)V V = -15.6 cm/sec. The negative sign indicates that the combined lump is really traveling to the left in the diagram. There-fore, after the collision the combined lump has a velocity of 15.6 cm/sec in the same direction as the initial velocity of the 50 gm lump. The energy lost in the collision is just the difference between the kinetic energy before collision and the kinetic energy after collision: \DeltaE = Ei- E_f = [(1/2) (30 gm) (25 cm/sec)^2 + (1/2) (50 gm) (- 40 cm/sec)^2] - (1/2) (80 gm) (- 15.6 cm/sec)^2 = 49,375 ergs - 9,734 ergs = 39,641 ergs

Question:

Find the total resistance in the circuit in the figure.

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0667.htm

Solution:

This problem is simplified by observing that the circuit can be resolved into its component parts. Since resistors R_1 and R_2 are in series, we may take their sum and consider it as a single resistor (see diagram) with resistance R. Combining the first two resistances, R = R_1 + R_2 = 4 ohms + 2 ohms = 6 ohms Now we have a simple parallel circuit and can easily solve for the total resistance of the circuit. Letting R' be the total resistance of the circuit we obtain: (1/R') = (1/R) + (1/R_3) = (1/6 ohms) + (1/6 ohms) = (1/3 ohms) whence R' = 3 ohms.

Question:

When files are used in a COBOL program, the physical envi-ronment of the program changes. Write a complete ENVIRON-MENT DIVISION for a COBOL program which will make the usage of the file called MEDICAL-DATA possible, using a card reader.

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Solution:

When files are used in COBOL programming the INPUT-OUTPUT SECTION must be included in the ENVIRONMENT DIVISION. INPUT-OUTPUT SECTION follows the CONFIGURATION SECTION, it also begins with Margin A. FILE-CONTROL, which is the first statement and paragraph header in INPUT-OUTPUT SECTION begins with margin A. The two clauses that make the FILE-CONTROL paragraph are the words SELECT and ASSIGN, which must begin on margin B. The file-name specified in the SELECT clause must be the one the programmer will call in the program. The ASSIGN clause describes to the computer exactly what physical equipment is required for that file. The general formats for ASSIGN, and SELECT are, ASSIGN system-name SELECT file-name Fig. 1 shows the complete ENVIRONMENT DIVISION with INPUT- OUTPUT SECTION entries included Variables UR-S-CARDX. following the ASSIGN clause indicate that the file is used for card input (which is a reading device) with external name CARDX. These variables change from system to system and each device has its own code name.

Question:

On a hot day, the pressure in an automobile tire increases. Assuming that the air in a tire at 59\textdegreeF increases in pressure from 28.0 lbs/in^2 to 30.0 lbs/in^2, (a)what is the temperature of the air in the tire, assuming no change in volume? (b) what will the pressure be if the temperature rises to 106\textdegreeF?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0043.htm

Solution:

(a) Because pressure, volume, and temperature are involved, one thinks of the combined gas law. This law can be stated: For a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. This law can be stated algebraically as: (PV/T) = K, where P is the pressure, V is the volume, T is the absolute temperature, and K is a constant. This means that if two of the parameters are changed, the third will adjust itself so that PV/T = K is still true. The following equation is thus true: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original temperature, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the new pressure, V_2 is the new volume, and T_2 is the new absolute temperature, when V_1 = V_2, as in this problem, it follows from the formula that pressure is directly proportional to absolute temperature. In this case, the law can be written (P_1 /T_2 ) = (P_2 /T_2 ) In this problem, the temperature is given in \textdegreeF, and must be converted to \textdegreeK before it is used in the combined gas law. This is done by first converting it to \textdegreeC and then adding 273.\textdegreeP are converted to \textdegreeC by use of the following equation: \textdegreeC = 5/9 (\textdegreeF - 32) T_1 in \textdegreeC = 5/9 (59\textdegree - 32\textdegree) = 15\textdegreeC T_1 = 15 + 273 = 288\textdegreeK. One is given P_1, P_2, and T_1 and asked to find T_2. (P_1 /T_1 ) = (P_2 /T_2 )P_1 = 28.0 1bs/in^2 T_1 = 288\textdegreeK P_2 = 30.0 1bs/in^2 T^2 = ? [(28.0 1bs/in^2 )/( 288\textdegreeK )] = [(30.0 1bs/in^2 /(T^2 )] T_2 = [(30.0 1bs/in^2 × 288\textdegreeK)/(28.0 1bs/in^2 )] = 309\textdegreeK The absolute temperature can now be converted to \textdegreeF by first subtracting 273\textdegree, and then using the equation for conversion from \textdegreeC to \textdegreeF. \textdegreeF = 9/5 \textdegreeC + 32 T_2 in \textdegreeC = 309\textdegree - 273 = 36\textdegreeC T_2 in F = 9/5 (36\textdegree) + 32 = 97\textdegreeF. (b)One can again use the shortened form of the combined gas law, her (P_1 /T_1 ) = (P_2 /T_2 ) Here, one is given T_1, T_2, and P_1 and asked to find P_2. The temperatures must be converted to the absolute scale before use. The same method as used in part (a) will be applied. T_1 in \textdegreeC = 5/9 (59 - 32) = 15\textdegreeC T_1 in \textdegreeK = 15 + 273 = 288\textdegreeK T_2 in \textdegreeC = 5/9 (106 - 32) = 41\textdegreeC T_2 = 41 + 273 = 314\textdegreeK The P_2 can now be found (P_1 /T_1 ) = (P_2 /T_2 )P_1 = 28.0 1bs/in^2 T_1 = 288\textdegreeK P^2 = ? T_2 = 314\textdegreeK [(28.0 1bs/in^2 )/(288\textdegreeK)] = (P_2 /314\textdegreeK) P_2 = [(314\textdegreeK × 28.0 1bs/in^2 )/(288\textdegreeK)] = 30.5 1bs/in^2 .

Question:

A string 100 centimeters long has a mass of 4 grams and emits a frequency of 250 vibrations per second. Another string 100 centimeter long has a mass of 1 gram. What is the frequency of the second string if it is put under the same tension as the first?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0814.htm

Solution:

The second string has one-fourth the mass of the first. The square root of 1/4 is 1/2 . Since the frequency varies inversely as the square root of the mass per unit length, the frequency of the second string is 2 × 250, or 500vps.

Question:

An army captain wants to fire an artillery shell deep into the enemy's flank. However, he knows that there are strong winds blowing above at height H that would blow his shells off course. If his artillery fires shells with muzzle velocity v_0, what is the farthest that he can fire them without their going off course?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0092.htm

Solution:

The equations of motion for such a projectile are the kinematical equations for constant acceleration, with v_0x = v_0 cos \texttheta and v_0y = v_0 sin \texttheta. (Here, v_0 is the initial velocity). x = (v_0 cos \texttheta)t y = v_0 sin \texttheta)t - (1/2) gt^2 where we take the +x-direction as pointing toward the enemy, the +y direction as going straight up, -g is the acceleration of gravity and \texttheta is the angle the shell makes with the horizontal axis as it is fired. We can solve for the time t' at which the shell is at its maximum height since at that point the shell's velocity in the y-direction is zero: v_y = (dy/dt) = v_0 sin \texttheta - gt = 0; t' = (v_0 sin \texttheta)/g The maximum height that the shell reaches is: y_max = (v_0 sin \texttheta)t' - (1/2) gt'^2 = (v_0 sin \texttheta) [(v_0 sin \texttheta)/g] - (1/2) g [(v_0 sin \texttheta)/g]^2 = [(v^2_0 sin^2 \texttheta)/g] - [(v^2_0 sin^2 \texttheta)/2g] = [(v^2_0 sin^2 \texttheta)/2g] But we know that the captain must aim his artillery at an angle \texttheta such that y_max = H, so that the shells just pass under the wind that would blow them off course. Thus we can solve for \texttheta: y_max =[(v^2_0 sin^2 \texttheta)/2g] = H sin \texttheta = \surd[(2gH)/(v_0)] Since the 'shell follows a parabolic path through the air, at time t', when it reaches its maximum height, it has traveled half of its maximum horizontal distance or range. Thus we can solve for its range R: (1/2)R = (v_0 cos \texttheta)t' R = 2(v_0 cos \texttheta)[(v_0 cos \texttheta)/g] = [(2v^2_0 sin \texttheta cos \texttheta)/(g)] Before we can calculate R we must determine cos \texttheta (see figure (b)). To find cos \texttheta, we must find the value of side x in the right triangle, since cos \texttheta = x/v_0. From the Pythagorean theorem: x^2 + 2gH = v^2_0 x = \surd(v^2_0 - 2gH) cos \texttheta = [\surd(v^2_0 - 2gH)]/v_0 Finally, R = [2 v^2_0 {\surd(2gH)/v_0} {\surd(v^2_0 - 2gH)/(v_0)}] / g = [2 \surd{2gH(v^2_0 - 2gH)} ]/g

Question:

At a temperature of 18\textdegree C a beam of diffracted mono- chromatic x-rays is observed at an angle of 150.8\textdegree to the incident beam after being diffracted by a crystal with cubic structure. At a temperature of 318\textdegreeC the corresponding beam makes an angle of 141.6\textdegree with the incident beam. What is the mean co-efficient of linear expansion of the crystal in the given temperature range?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1039.htm

Solution:

If a beam of monochromatic x-rays is deviated by an angle \varphi after diffraction from a set of crystal planes, it is seen from the diagram that, if \texttheta is the glancing angle of the x-rays on this set of planes, 2\texttheta + (180 - \varphi) = 180\textdegreeor\texttheta = \varphi/2 . When the temperature is 18\textdegreeC, the appropriate interplanar spacing is d, and the glancing angle for diffraction from this set of planes is 150.8\textdegree/2 = 75.4\textdegree. Bragg's Law is m\lambda = 2d sin \texttheta where m is the order number, d is the interplanar spacing, \lambda is the x-ray wavelength and \texttheta is the angle the incident x-ray beam makes with the atomic plane from which it is reflected (see figure). Thus 2d sin 75.4\textdegree = m\lambda When the temperature is 318\textdegreeC, the interplanar spacing has increased to d', and the glancing angle has dropped to 141.6\textdegree/2 = 70.8\textdegree. Thus 2d' sin 70.8\textdegree = m\lambda, the wavelength and order of diffraction being unchanged. Hence, d'/d =sin 75.4\textdegree / sin 70.8\textdegree = 1.024. But d' = d(1 + \alpha × 300 C deg), where \alpha is the mean coefficient of linear expansion at right angles to the set of planes considered. Since the crystal is cubic, the coefficient of linear expansion is the same in all directions. \therefore1 + 300 C deg × \alpha = 1.024. \therefore\alpha = 0.024 / 300 C deg = 8.3 × 10^-5 per C deg .

Question:

Describe the operations and built-in functions used for char-acterstring manipulations.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0328.htm

Solution:

One of the most useful operations on character strings is called concatenation. Concatenation is used to join two characterstring operands, giving a single character string as a result. For example, 'JANEb' \vert\vert 'FONDA' = 'JANEbFONDA' The concatenation operator must connect the character string JANEbwith the character string FONDA. The stringJANEbFONDAwill result. The concatenation operator is represented by \vert\vert. The string resulting from the execution of a concatenation oper-ation has a length equal to sum of the lengths of the two operands (character strings). To understand the processing of variable length and fixed length string data, consider the following example. DECLARE[(FIRST_NAME, LAST_NAME) CHAR(10), STUDENT_NAME CHAR(15)] VARYING; GET LIST(FIRST_NAME, LAST.NAME); STUDENT_NAME = FIRST_NAME \vert\vertb\vert\vert LAST_NAME; If on the data card we have 'JANE', 'FONDA' thentheGET statement creates the followings VariableValueCurrent length FIRST_NAMEJANE4 LAST_NAMEFONDA5 but due to the varying attribute, no blanks are added, and STUDENT_NAME =JANEbFONDA butif the varying attribute is not specified thentheGET statement creates thefollowing: VariableValueCurrent length FIRST_NAMEJANEbbbbbb10 LAST_NAMEFONDAbbbbbb10 Hence STUDENT_NAME =JANEbbbbbbFONDAbbbbbb. It has a length of21 characters. But the declared length is 15 so the last 6 characters will betruncated, and hence the following is obtained: STUDENT_NAME =JANEbbbbbbFOND Condition Testing and comparison Operations: All eight comparison operators (>,>=,= ,<=,<,\lnot>,\textquotedblleft\lnot= ,\lnot<) , together withthe condition-testing IF statement can be used with character type operands. Comparison can only be made with character strings of the same length. If the operands are of different lengths, the shorter character string valueis expanded at the right with blank characters. If two character stringsof length zero (i.e., null strings) are compared, the result of the '=' comparisonis true. All other comparison operators re-turn a false when appliedto two null-string operands. Character strings of more than one characterare compared, character by character, in a left-to-right order. STRING BUILT-IN FUNCTIONS: For string manipulations, three built-in functions are provided in PL/I. 1) LENGTH built-in function 2) SUBSTR built-in function 3) INDEX built-in function LENGTH Built-in function:The length of character string variables declaredwith varying attributes may change during the execution of the PL/Iprogrma.It may be necessary to utilize the current length of a characterstring during cal-culations . The built-in function 'LENGTH' returns an integer value which is the currentlength of the character string. e.g.STUDENT_NAME = 'JERRY' N = LENGTH (STUDENT_NAME); LENGTH returns a value of 5 as the length of the character string. The SUBSTR Built-in function:The substring function extracts a segmentof a string (i.e. a substring) from a given character string. The functionreference uses three arguments. e.g.TAR = SUBSTR (CHAR-STRING,i, j) 1. The first argument CHAR-STRING identifies thechar-acterstring fromwhich the selection process is to take place. 2. The second argument 'i' identifies the starting position in the characterstring from where extraction of the substring is to be done. It isalways an integer value. If k is the current length of a character string, then, 0

Question:

What is the wavelength of a 10-eV electron?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0960.htm

Solution:

For an electron of this low energy we can neglect the change of electron mass [^=^mrest/\surd{1 - (v^2/c^2)}] with velocity. Since kinetic energy KE = (1/2)mv^2 and momentum p =mv, we have, because v = \surd[{2(KE)}/m] p = \surd(2mKE). The de Broglie wave theory of matter tells us that every particle has wave-like characteristics. The electron, then, has a wave- length \lambda associated with it. It is given by the de Broglie relation, \lambda = (h/p) = [h/\surd(2m KE)] = (6.62 × 10^-27 erg-sec)/{\surd[2 ×(9.11 × 10^-28g) × (10eV) × (1.60 × 10^-12erg/eV)]} = 3.86 × 10^-8 cm = 3.86 \AA where 1\AA = 10^-8 cm. (The unit of energy was converted from electron volts to ergs, because the CGS System is being used.) A photon with this same wavelength would have an energy \epsilon =hvbut v = (c/\lambda) for a photon. Therefore \epsilon =(hc/\lambda) = [{(6.62 × 10^-27 erg-sec) × (3 × 10^10cm/sec)}/{(3.86 × 10^-8cm) × (1.60 × 10^-12erg/eV)}] = 3.2keV Such an energetic photon is an X-ray.

Question:

There exists a 0.5 M HF solution for whichK_diss= 6.71 × 10^-4. Determine how much dilution is necessary to double the percent dissociation of HF.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0354.htm

Solution:

Percent dissociation means the ratio of [H^+] to original [HF] concentration times 100. To find the amount of dilution necessary to double the percent dis-sociation of an acid, first, establish what the percent dissociation is before dilution. This can be determined fromK_diss_' the equilibrium dissociation constants, which measures the ratio of products to reactants (i.e. their concentrations), each raised to the power of their co-efficients in the balanced chemical equation. The general reaction for the dissociation of an acid, e.g., HA, is HA + H_2O \rightleftarrows H_3O^+ + A^-. Thus, for HF;HF + H_2O \rightleftarrows H_3O^+ + F- for the dissociation reaction. This means that K_diss= 6.71 × 10^-4 = {[H_3O^+][F^-]}/[HF] . It is given that the initial concentration of HF is 0.5. Thus, to find percent dissociation, one needs to know [H_3O^+]. To find this, perform the following operations: Let x = [H_3O^+]. Since H_3O^+ and F^- are formed inequimolar amounts, as can be seen in the chemical re-action, [H_3O^+] = [F^-] = x. If the initial [HF] = 0.5, and x moles/liter dissociate to give [H_3O^+] and [F^-], then, at equilibrium, [HF] = 0.5 - x. Substituting these values: [(x \textbullet x)/(0.5 - x)] = 6.71 × 10^-4 . Solving for x, using the quadratic formula, x = 0.018 = [H_3O^+]. Thus, percentage dissociation becomes (0.018/0.5) × 100 = 3.6 % HF dissociated. Then, to determine the final answer, 7.2 % HF dissociation is needed when one dilutes. Recall that [H^+] = [F^-]. At 7.2 % dissociation, [F^-]/[HF] = [(0.072) / (1 - .072)] = (0.072/0.928). From theK_dissexpression, one has K_diss= 6.71 × 10^-4 = {[H^+][F^-]} / [HF]or (6.71 × 10^-4) / [H^+] = [F^-] / [HF] . [F^-] / [HF] = (0.072/0928) , when percent dissociation is doubled to 7.2 %. After substitution, (6.71 × 10^-4) / [H^+] = (0.072/0928) . Solving for [H^+], [H^+] = 8.65 × 10^-3 = [F^-], Substituting these actual molar concentrations into the equilibrium constant expression, one obtains [(6.71 × 10^-4) / ( 8.65 × 10^-3)] = (8.65 × 10^-3) / [HF] Solving:[HF] = 0.112 at equilibrium [H^+] = 8.65 × 10^-3. Before dissociation [HF] = amount at equilibrium plus amount dissociated = 8.65 × 10^-3 + 0.112 = 0.121 M, Thus, when the percent dissociation is doubled, the initial amount equals 0.121 M for [HF]. The [HF], when the percent dissociation was unchanged, was 0.5 M (given). Therefore, dilute by factor 0.500/0.121 = 4.13. Remember that con-centration ormolarityis a parameter of volume, actually M = moles/liter. One is going from 0.5 M to 0.121 M, which means volume must be increased by a certain factor. When one dilutes a solution, one is adding volume to it and, dividing the solution. This account for using division to obtain the factor of dilution, once initial concentrations are known.

Question:

Spherical particles of pollen are shakenup in water and allowed to stand. The depth of water is 2 cm. What is the diameter of the largest particles remaining in suspension 1 hr later? Density of pollen =1.8 g/cm^3.

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/Users/wenhuchen/Documents/Crawler/Physics/D11-0429.htm

Solution:

The terminal velocity of the particles after they are allowed to settle will very quickly be reached. After 1 hr the only particles left in suspension are those which take longer than 1 hr to fall 2 cm. The larger, heavier particles have already settled. The particles which have just not settled are those which take exactly 1 hr to fall 2 cm. That is, V_T = [2 cm]/[(1 hr)(3600 s/hr)] = 1/1800cm/sec We need another expression for the terminal velocity.Stoke'slaw states that when a sphere moves through a viscous fluid at rest, the resisting force f exerted by the fluid on the sphere is given by f = 6\pi\etarv where \eta is the viscosity of the fluid, r is the radius of the sphere, and v is its velocity with respect to the fluid. The other forces which act on the sphere are its weight mg and the upward buoyant force B of the fluid. Let \rho be the density of the sphere and \rho' the density of the fluid. Then mg = (4/3) \pir^3\rhog B=(4/3) \pir^3 \rho' g(Archimedes' principle) The net force on the sphere equals the product of its mass and acceleration. Taking the downward direction as positive, mg - B - R = ma a = g - [(B + R)/m] Assuming the initial velocity is zero, this net acceler-ation imparts a downward velocity to the sphere. As this velocity increases, so does the retarding force. At some terminalv_T, the retarding force has increased an amount such that the downward acceleration equals zero. At this point, the velocity of the sphere stays constant and is found by setting the acceleration equal to zero. Then mg = B + R (4/3) \pir^3\rhog= (4/3) \pir^3 \rho' g + 6\pi\etarv_T v_T= (2/9) (r^2g/\eta) (\rho - \rho') The radius of the largest particles still just in suspension is thus given by r^2 = (9/2) [(\etav_T)/{g(\rho - \rho')}] = (9/2) [(1 × 10^-2 poise × 1/1800 cm/s)/{980 cm/s^2(1.8 - 1 )g/cm^3}] = [(10^-4)/(64 × 49)] cm^2 d = 2r = [(2 × 10^-2)/(8 × 7)] cm = 3.57 × 10^-4 cm.

Question:

A 2.4 kilowatt generator delivers 10 amperes. At what potential difference does the generator operate?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0781.htm

Solution:

In most situations we cure given the voltage and amperage of a device, and are asked to calculate its power. Here we must work in reverse, calculating the potential difference (voltage) from the power and the current. Nevertheless, we resort to the same formula, namely P = El. P = 2.4 kilowatts = 2400 watts, I = 10 amp P = E × I Then E = (P/I) = (2400 watts/10 amp) = 240 volts.

Question:

An electric heater takes 4 amperes from a 120-volt line. What is its power rating?

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0780.htm

Solution:

E = 120 volts,I = 4 amp P = E × I = 120 volts × 4 amp = 480 watts.

Question:

Compare the bond order of He_2 and He_2^+.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0658.htm

Solution:

The bond order, or number of bonds in a molecule, is equal to the difference in the sum of the number of bonding electrons and the number ofantibondingelectrons divided by two. Bond order = [(no. of bonding electrons-no. ofantibondingelectrons)/(2)] This means that the number of bonding andantibondingelectrons must be determined. There are 2 electrons in He, thus in He_2 there are 4. These electrons are all in the 1s level. For each level, there exists bonding andantibondingorbitals, each of which holds 2 electrons. Thus, in He_2, 2 electrons are bonding and 2 areantibonding. From this, the Bond order = [(2 - 2)/(2)] = 0. Thus, there are no bonds in He_2; and two He atoms will not bond together to form a molecule of He_2. In He_2 +, one electron is removed from He_2, which means that there are now three electrons present. They are all in the 1s level. This is the lowest energy level that an electron can assume. The three electrons are distributed so that two are in bondingorbitalsand one is in anantibonding orbital. Thus, no. of bonds = [(2 - 1)/(2)] = 0.5 = bond order. Because the bond order is not zero, this molecule can form.

Question:

Determine the number of cubic centimeter in one cubic inch.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0005.htm

Solution:

One meter equals 39.37 inches and, since there are 100 centimetersin 1 meter, there are 39.37 inches in 100 cm. Thus, 1 inch is equalto 100/39.37 cm. 1 inch = (100/39.37)cm= 2.54 cm. By cubing both sides of this equation, one can solve for the number of cubiccentimeters in 1 cubic inch. (1 inch)^3= (2.54 cm)^3 1 inch^3 = 16.4 cc.

Question:

Why are magnolias found both in the eastern United States andeastern China, which belong to different realms, and hardlyanywhere else?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0741.htm

Solution:

Early in the Cenozoic Era, the northern hemisphere was much flatterthan it is now and North America was geographically continuous withAsia by means of a land bridge at the Bering Strait. The climate then wasmuch warmer than at present, and magnolias spread over this entire connectedregion. In the later Cenozoic, the Rockies increased in height, andthe western part of North America grew colder and drier. Magnolias, whichwere adapted to a warm moist climate, disappeared from western North America. Subsequentglaciationmoving from the north wiped out anysurviving temperate plants in western North America, and many in Europe,and the western part of China. Because only regions in the southeasternUnited States and eastern China were untouched by the glaciation, magnolias were left growing in these two regions. North America, however, soon broke away from the land mass of Asia, and two separatecontinents were formed. Because of the resulting geographic barrier(ocean) between the magnolias of the two countries, the plants for severalmillion years followed separate evolutionary pathways, and have consequently become slightly different from each other.

Question:

Consider an octahedral complex having a univalent negative ion at each vertex of the octagon. Explain why the d(x)2 - (y)2 orbital of the central atom is less stable relative to the d_xy orbital.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0659.htm

Solution:

Both the d(x)2 - (y)2and the d_xy orbitals are centered about the x - y plane; hence, the difference in the stability of the two orbitals must be related to their different orientations along the x- and y - axes. In the d(x)2 - (y)2orbital, the lobes are directed along the axes; whereas in the d_xy orbital, the lobes are di-rected along the bisectors of the angles between the axes (see Figure A). In the octahedral configuration the ligands (the negatively charged ions electrostatically bound to the central atom) lie on the axes (see Figure B) . Hence, the d(x)2 - (y)2orbital brings the negatively charged electrons in the lobes closer to the negatively charged ligands than does the d_xy orbital. Since the negative charge of the electrons repels that of the ligands, and this electrostatic repulsion falls off as the reciprocal of the square of the distance between the charges, the greater degree of repulsion between the d(x)2 - (y)2orbital and the ligands, relative to the repulsion between the d_xy orbital and the ligands, decreases the stability of the d(x)2 - (y)2orbital.

Question:

Consider the following structure struct element { char keyfield; struct element \textasteriskcenteredback; struct element \textasteriskcenterednext; } } ; ; Now 1) create a doubly linked list of 5 elements. 2) Explain the logic of doubly linked lists.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0437.htm

Solution:

In a doubly linked list the structure contains a) pointer to the next node in the list. b) pointer to the previous node in the list. The logic followed is that the given structure has two pointers i.e. one for the next node and one for the back or previous node. Connecting the pointers in such a way that one single node can refer its previous and its next node is the scheme followed in the program. The first element will have in back NULL value while the next will point to the next node in the list. Similarly, the last element will have a null value in its 'next' and the 'back' will point to the previous node in the list. #define NULL '\textbackslash0' structelement { char keyfieId; struct element \textasteriskcenteredback; struct element \textasteriskcenterednext; }; }; main ( ) { int i; char \textasteriskcenteredcalloc ( ); struct element \textasteriskcenteredstart, \textasteriskcenteredptr, \textasteriskcenterednew _ ptr; ptr = (structelement \textasteriskcentered) calloc(1, sizeof (struct ele-ment)); \textasteriskcentered/ /\textasteriskcenteredfollowing segment creates the 1st element ptr \rightarrow next = NULL; ptr \rightarrow back = NULL; start = pitr; /\textasteriskcenteredfollowing segment creates 4 more elements \textasteriskcentered/ for (i = 1; i < 5; i++) { new _ ptr = (structelement \textasteriskcentered) calloc(1, sizeof (structelement)); new _ ptr \rightarrow back = ptr; new _ ptr \rightarrow next = ptr \rightarrow next; ptr \rightarrow next = new _ ptr ptr = new _ ptr; } } } Diagram Representation:

Question:

A student made three successiveweighingsof an object as follows: 9.17 g, 9.15 g, and 9.20 g. What is the average weight of the object in milligrams?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0011.htm

Solution:

The average of a set of weights is found by adding together all of the weights and then dividing by the number of weighings used. avg. weight = [(9.17 g + 9.15 g + 9.20 g) / 3] = (27.52 g / 3) = 9.17 g. Now that the average weight in grams has been determined, convert it to milligrams using the conversion factor of 1,000 mg/g. 9.17 g × (1000 mg / g) = 9170 mg.

Question:

Design a 4 × 3 random access memory (RAM) using one-bit memory cells, OR gates, and a decoder.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0095.htm

Solution:

Figure 1 is the configuration of a 4 × 3 RAM. Each box labeled M.C. is a one-bit memory cell whose inputs and output are designated in Figure 2. The enable input of the decoder is used as a Memory enable line. When the line is 0 the decoder is disabled and none of the memory words are selected. When the Memory enable line is 1 the decoder is enabled and one of the four words is selected, depending upon the address. The Read/Write control is connected to every M.C. When it is 1, the outputs of the selected M.C.s pass through the three OR gates to the Data-output terminals. The outputs of the non-selected M.C.s are 0, having no effect on the OR gates. When the Read/Write control is 0 the information on the data-input lines is transferred into the flip-flops of the selected word. The M.C.s in the non-selected words are disabled by the selection line so their values remain unchanged.

Question:

C0_2(g) + H_2(g) \rightleftarrows C0(g) + H_20(g) . Initially, 49.3 mole percent CO_2 is mixed with 50.7 mole per cent H_2 . At equilibrium, you find 21.4 mole percent CO_2 , 22.8 mole percent H_2 , and 27.9 mole percent of CO and H_20 . Find K. If you start with a mole percent ratio of 60:40, C0_2 to H_2 , find the equilibrium concentrations of both reactants and products.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0309.htm

Solution:

An equilibrium constant K_eg measures the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, K_eg for this reaction = { [CO] [H_2O] } / { [CO_2] [H_2] } . If you assume each substance occupies the same volume in liters, the concentration can be expressed in moles because concentration = moles/liter , i.e., liters cancel out of the equilibrium constant expression. Thus, to find K , you need to find the number of moles of each of the products and reactants and then to substitute into the equilibrium expression. You are told the final product mole per-cents. The reactants, then, at equilibrium, have mole percents that equal their initial amounts minus the amount that decomposed to pro-duce the products. Thus, at equilibrium, [CO_2] = 49.3\Elzbar 21.4 = 27.9, which was given. Similarly [H_2] = 50.7\Elzbar 22.8 = 27.9, which was given. Thus, by substitution into K = { [CO] [H_2O] } / {[CO_2] [H_2]} ,you obtain K = { (.279) (.279) } / { (.214) (.228) } = 1.60, which is the equilibrium constant for this reaction. The second part follows. You begin with a 60:40 ratio of CO_2 : H_2 , which means that initially you have .600 moles CO_2 and .400 moles H_2 . To find the equilibrium concentrations, let x = moles of CO formed. Thus, x = moles H_2 O formed since coefficients tell us they are formed in equimolar amounts. The fact that moles of a product form, means that x moles of a reactant must have de-composed. Thus, at equilibrium, you have 0.600-x moles of CO_2 and 0.400-x moles of H_2. Recalling, K = {[CO] [H_2O]} / {[CO_2] [H_2]} , you can now substitute these values to give K = [x^2 / {(0.6\Elzbar x)(0.4\Elzbar x)}]. From previous part, K = 1.60, therefore, 1.60 = x^2 / {(0.6\Elzbar x)(0.4\Elzbar x)} . Solving, x = 0.267 . Thus, [CO] = [H_2 O] = x = 0.267M, [CO_2 ] = 0.6\Elzbar x = 0.333M and [H_2 ] = 0.4\Elzbar x = 0.133M.

Question:

(1) Cu(NH_3)^2_4^+ \rightleftarrows Cu^2+ + 4NH_3k = 1.0 × 10^-12 (2) Co(NH_3)^2_6^+ \rightleftarrows Co^2+ + 6NH_3k = 4.0 × 10^-5 (3) Co(NH_3)^3_6^+ \rightleftarrows Co^3+ + 6NH_3k = 6.3 × 10^-36 (4) Ag(NH_3)^+_2 \rightleftarrows Ag^+ + 2NH_3k = 6.0 × 10^-8. Arrange them in order of increasing NH_3 concentration that would be in equilibrium with complex ion, if you start with 1 M of each.

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Solution:

To solve this problem determine the [NH_3] for each complex ion equilibria. This necessitates writing the equilibrium constant expression, which measures the ratio of concentrations of products to reactants, each raised to the power of their coefficients in the chemical reaction. Proceed as follows: k = 1.0 × 10^-12 = {[Cu^2+][NH_3]^4} / [Cu(NH_3)^2_4^+] Start with 1 M of Cu(NH_3)^2_4^+. Let x = [NH_3] at equi-librium. From the chemical reaction one can see that 1 mole of Cu^2+ is formed as 4 moles of NH_3 are formed. Thus, [Cu^2+] = (1/4)[NH_3] = 0.25 x. If x moles/liter of NH_3 form, and the only source is Cu(NH_3)^2_4^+, then, at equi-librium, [Cu(NH_3)^2_4^+] = 1 - 0.25 x. It is initial con-centration minus 1/4 of [NH_3], since they have a mole ratio of 1 : 4, as is seen in the chemical reaction. Substituting into the equilibrium equation: [{(0.25x) (x)^4} / (1 - 0.25 x)] = 1.0 × 10^-12 . Solving for x, x = [NH_3] = 5.3 x 10^-3. To obtain this answer the denominator (1 - 0.025x) is approximated as 1, since the value for x is relatively small compared to 1. To find [NH_3] at equilibrium in (2), (3), and (4), employ the same type of logic and reasoning. The only differences are the values of K and the mole ratios between the substances. For Co^+2 : 1/6 x^7 = 4.0 × 10^-5 [NH_3] = x = 0.3 For Co^+3 : 1/6 x^7 = 6.3 × 10^-36 [NH_3] = x = 9.4 × 10^-6 For Ag^+:1/2 x^3 = 6.0 × 10^-8 [NH_3] = x = 5.0 ×10^-3 In increasing order, rank them as Co^3+, Ag^+, Cu^2+, Co^2+ / increasing [NH_3] .

Question:

If you place HClO_4, HNO_3 orHClin water, you find that they are strong acids. However, they show distinct differences in acidities when dissolved in acetic acid. Such an occurrence is referred to as the leveling effect of the solvent, water: a) Explain the basis for this leveling effect by comparing acid reactions in the water solvent system to the acetic acid solvent system. b) Discuss the leveling effect in terms of basicitiesinstead of acidities.

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Solution:

(a) To explain the leveling effect, you must consider the relative acidic or basic properties of the species involved. In water, the general reaction for the acid is HA + H_2O \rightleftarrows H_3O+ + A^-. in acetic acid, however, the reaction is (assuming HA is a stronger acid than acetic acid (CH_3COOH), and the three acids given are) HA + CH_3COOH \rightleftarrows CH_3COOH_2^+ + A^-. Let us consider the strengths of the acids in these two solvent systems. Water is less acidic, thus more basic and more strongly proton-attracting, than acetic acid. This means that when strong acids are dissolved the equilibrium will be shifted far to the right in water, but not as far in acetic acid. Thus, more products will be produced in the water solution than in the acetic acid. The acidities of the 3 given acids depend upon how much H^+ ion they produce. If the equilibrium is not shifted to the right, less H^+ ion is being produced. Thus, theacidicitiesof the three given acids will be less in acetic acid, since the equi-librium is shifted less to the right, and thus, not much H^+ ion is generated. (b) The leveling effect can also be thought of in terms ofbasicity. For the two types of bases, you have B + H_2O \rightleftarrows BH^+ + OH^- in water and B + CH_3COOH \rightleftarrows BH^+ +CH_3COO^- in acetic acid. Now OH^- is a stronger base than CH_3COO^-, and, therefore, the equilibrium is further to the left in water than in acetic acid. Thus, when a base is added to water it will ionize less than in acetic acid.

Question:

A stream of gas is escaping through a small opening at one end of a large cylinder under the action of an excess pressure (relative to air pressure) ∆P = 10^4 dynes/cm^2. If the density of gas in the cylinder is \varphi = 8 × 10^-4 gm/cm^3, find the escape velocity v.

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Solution:

The excess pressure is about 10^4 dynes/cm^2 \approx 10.2 cm of water, whereas the normal air pressure P_air is about 98 3 cm of water. Since the excess pressure is much smaller than the outside pressure, we can assume that there is no appreciable compression of the gas. Then, we can treat the gas escaping through the hole as incompressible and apply Bernoulli's equation. The pressures inside and outside the cylinder are related by P_inside = P_outside + (1/2) \rhov^2 orv^2 = (2/\rho) (P_a + ∆P - P_a) orv^2 = (2/\rho) (P_a + ∆P - P_a) v= \surd(2∆P/\rho) v= \surd(2∆P/\rho) = \surd[(2 × 10^4 = \surd[(2 × 10^4 dyne/cm^2)/(8 × 10^-4 gm/cm^3)] = 5 × 10^3 cm/sec.

Question:

The trait represented in the pedigree below is inherited through a single dominant gene. Calculate the probability of the trait appearing in the offspring if the following cousins should marry.

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Solution:

Let R represent the dominant allele for the trait, and r the recessive allele. First, we can determine the possible genotypes of the individuals involved. We know that F_1,2 ,F_1,3 , F_1,4 ,F_2,1 ,F_2,3 and F_2,4 are homozygous recessive (rr), since they do not express the trait. P_1 and P_2 must be heterozygous dominant (Rr) since they produced a child F_1,4 who is homozygous recessive. F_1 , could be either RR or Rr. However, F_2 , is homozygous recessive. Therefore F_1,1 must carry the recessive allele and be Rr. F_2,2 is also a heterozygote. We know this because one of his parents (F_1,2) is a homozygous recessive, and the cross (Rr × rr) is incapable of producing a homozygous dominant. In this cross, half the offspring are heterozygous dominant and half are monozygous recessive. Thus we can say that there is a .5 probability that any offspring of the mating will express the dominant trait. Since neither parent carries the dominant allele, it should be obvious even without doing the cross that no offspring expressing the dominant trait will be produced. Thus we can say that there is O probability that any offspring will express the trait.

Question:

What are the basic differences in themechanical arrangement of vertebrate and arthropod joints?

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Solution:

The vertebrates are characterized by having an endoskeleton - a bony or cartilaginous framework lying within the body - surrounded by muscles . The arthropods, on the other hand, have an exoskeleton. The arthropod exoskeleton is achitinousframework on the outside of the body surrounding the muscles. The contraction of muscles in the vertebrates move one bone with respect to another. One end of the muscle is attached to one bone and the other end is attached to another bone. When the muscle contracts, one end - called the origin - remains relatively fixed while the other - called the insertion - moves. The insertion pulls along the bone to which it is attached, and bone movement results. The muscles of the arthropod lie within the skeleton and are attached to the skeleton's inner surface. Regions in which the exoskeleton is thin and flexible allow bending to occur and serve as the joints in the arthropod exoskeleton . A muscle may stretch across the joint so that its contraction will move one part on the next; or a muscle may be located entirely within one section of the body and may be attached at one end to a tough apodeme - a long, thin, firm part of the exoskeleton extending into that section from the adjoining one. (See Figure)

Question:

Using the data from the accompanying figure, calculate the heat of reaction for the following process at 25\textdegreeC: a) CaCl_2 (s) + Na_2CO_3(s) \rightarrow 2NaCl(s) + CaCO_3(s) b) H_2 SO_4 (I) + 2NaCl(s) \rightarrow Na_2SO_4 (s) + 2HCl(g) Standard Heats of Formation, ∆H\textdegree, in kcal/mole at 25\textdegreeC. Standard Heats of Formation, ∆H\textdegree, in kcal/mole at 25\textdegreeC. Substance ∆H\textdegree CaCl_2 (s) - 190.0 Na_2CO_3(s) - 270.3 NaCl(s) - 98.2 CaCO_3(s) - 288.4 H_2 SO_4 (I) - 193.9 HCl (g) - 22.1 Na_2SO_4 (s) - 330.9

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Solution:

Calculate the amount of heat released (or absorbed) from the heats of formation. In general, ∆H for a reaction will be the sum of the heats of formation of products minus the heats of formation of reactants, each of which is multiplied by its coefficient in the equation. Once ∆H is determined, ∆E can be found from ∆E = ∆H - (∆n)RT, where ∆n = change of moles, R = universal gas constant and T = temperature in Kelvin (Celsius plus 273\textdegree) Proceed as follows: (a) CaCl_2 (s) + Na_2CO_3(s) \rightarrow 2NaCl(s) + CaCO_3(s) Thus, ∆H= 2∆H\textdegree_NaCl(s) + ∆H\textdegreeCaCO3 (s)-∆H\textdegree_CaCl2 (s) -∆H\textdegreeNa2CO3 (s) = 2(-98.2) + (-288.4) - (-190.) - (270.3) = - 24.5 Kcal/mole. ∆E= ∆H - ∆nRT. But, in this reaction, no gases appear. This means ∆nRT becomes zero. As such, ∆E = ∆H = -24.5 Kcal/mole. (b) H_2 SO_4 (I) + 2NaCl(s) \rightarrow Na_2SO_4 (s) + 2HCl(g) ∆H= ∆H\textdegreeNa2 SO4_(s) + 2∆H\textdegree_HCl _(g) - ∆H\textdegreeH2 SO4_(l) - 2∆H\textdegree_NaCl(s) = (-330.9) + 2(-22.1) - (-193.9) - 2(-98.2) = +15.2 kcal/mole. ∆E= heat of reaction = ∆H - (∆n)RT. A gas, HCI, is involved in this reaction. ∆n= moles products - moles reactants =2 - 0 = 2 T = 25 + 273 = 298\textdegreeK; R is in terms of cal. and ∆H in terms of Kcal, so that you must use the conversion factor of 1Kcal/1000 cal. Thus, ∆E= -15.2 - (2)(1.987)(298)(1/1000) = - 15.2 - 1.2 = - 16.4 Kcal/mole.

Question:

Genetic variation occurs in bacteria as a result of mutations. This would seem to be the only process giving rise to variation , for bacteria reproduce through binary fission. Design an experiment to support the existence of another process that results in variation.

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Solution:

It was originally thought that all bacterial cells arose from other cells by binary fission, which is the simple division of a parent bacterium. The two daughter cells are genetically identical because the paren-tal chromosome is simply replicated, with each cell getting a copy. Any genetic variation was thought to occur solely from mutations. However, it can be shown that genetic variation also results from a mating process in which genetic information is exchanged. One can show this recombination of genetic traits by using two mutant strains ofE.coli, which lack the ability to synthesize two amino acids . One mutant strain is unable to synthesize amino acids A and B, while the other strain is unable to synthesize amino acids C and D. Both these mutant strains can be grown only on nutrient media, which contain all essential amino acids. When both strains are plated on a selective medium lacking all four amino acids in question, some colonies of prototrophic cells appear which can synthesize all four amino acids (A, B, C, and D) .When the two strains are plated on separate minimal medium plates., no recombination occurs, and noprototrophiccolonies appear. These results may be proposed to be actually a spon-taneous reversion of the mutations back to wild type (normal prototroph) rather than recombination. If a single mutation reverts to wild-type at a frequency of 10^-6 muta-tions per cell per generation, two mutations would simul- taneously revert at a frequency of 10^-6 × 10^-6 = 10^-12 mutations per cell per generation. If one plates about 10^9 bacteria, no mutationalrevertants for both deficiencies should occur. Recombination occurs at a frequency of 10^-7, and thus if 10^9 bacteria are plated,prototrophiccolonies should be found. This recombination of traits from the two parent mutant strains is brought about through a process called conjugation. During conjugation, two bacterial cells lie close to one another and acytoplasmicbridge forms be-tween them. Parts of one bacterium's chromosome are transferred through this tube to the recipient bacterium. The transferred chromosomal piece may or may not get in-corporated into the recipient bacterium's chromosome .

Question:

The easiest method of measuring the refracting angle of a prism is to direct a parallel beam of light on to the angle (vertex A in figure) and measure the angular separation of the beams reflected from the two sides of the prism containing the refracting angle. Show that this angular separation is twice the angle of the prism.

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Solution:

Consider three incoming rays, all parallel and striking the prism at points A, B and C. Erect normals to AB at A and B and to AC at A and C. Designate the angles as in the diagram. The rays striking at B and C are reflected according to the laws of optics, as shown and the angle between the reflected rays is \beta. E is the point at which the normals at B and C meet. The sum of the angles of quadrilateral ABEC is 360\textdegree. Since \angle ABE and \angle ACE are each 90\textdegree, \alpha + \Upsilon = 180\textdegree.(1) In the quadrilateral BDCE, \beta + \Upsilon + \texttheta_1 + \texttheta_2 = 360\textdegree.(2) Since two of the angles surrounding A are right angles, \alpha + \texttheta_1 + \texttheta_2 = 180\textdegree.(3) Add Eqs.(1) and (3) 2\alpha + \Upsilon + \texttheta_1 + \texttheta_2 = 360\textdegree.(4) Subtract (2) from (4) 2\alpha - \beta = 0 or2\alpha = \beta Thus, the angle between the reflected beams, \beta, is twice the refracting angle of the prism, \alpha.

Question:

Describe the growth zones of a typical root.

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Solution:

The tip of each root is covered by a protective root cap, a thimble-shaped group of cells. This root cap protects the rapidly growing meristemic region of the root tip. The outer part of the root cap is rough and uneven because its cells are constantly being worn away as the root pushes its way through the soil. The meristem, protected by the root cap, consists of actively dividing cells from which all the other tissues of the root are formed. It also gives rise to new root cap cells to replace the ones that are sloughed off. Immediately behind the meristem is the zone of elongation. The cells of this zone remain undifferen-tiated but grow rapidly in length by taking in large amounts of water. The meristematic region and the zone of elongation together account for the increase in length In the zone of maturation (or differentiation) of the root cells differentiate into the permanent tissues of the root, for example, the xylem and the phloem tissues. Present in this region too, are slender, elongated, hairlike projections called root hairs that arise laterally. There is one root hair to each epidermal cell. Root hairs greatly increase the surface area of the root and enhance the absorption of the water and minerals into the root. As the root elongates, the delicate, short--lived root hairs wither and die, and are replaced by the ones newly formed by the meristematic region. Root hairs occur only in a short lower segment of the zone of maturation, usually only 1 to 6 cm long.

Question:

An airplane is traveling horizontally at 480 mph at a height of 6400 ft. The airplane drops a bomb aimed at a stationary target on the ground. To an observer on the aircraft, what angle must the target make with the vertical, when the bomb is dropped, if the bomb is to hit the target? (See the figure.) Suppose that the target is a ship which is steaming at 20 mph away from the aircraft along its line of flight. What alterations would need to be made to the previous calculations?

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Solution:

At the moment of release of the bomb, time t = 0, the airplane is at the point which is taken as the origin of the coordinate system, traveling in the positive x-direction with a speed u of 480 mph. 480 mph = 480 (mile/hr) × ( 5280 ft/mile) × (1hr/60 min) × (1min/ 60 sec) = 704 ft/sec. The bomb has the same initial speed. There is no acceleration in the x-direction, for no horizontal force acts on the system. Hence, after time t, when the bomb strikes the target, the distance traveled by the bomb in this direction is given by the kinematic equation for constant velocity, x_0 = ut. The airplane and bomb have no initial speed in the y-direction, but the acceleration g acts in this direction. After time t, the downward distance travel-ed by the released bomb will be, using the kinematic equation for constant acceleration Since ^vo_y = o, this becomes y_0 = 1/2gt^2 \bullet y_o = ^vo_y t + 1/2 gt^2. But y_0 = 6400 ft in this problem, and the time it takes the object to fall this distance is therefore t = \surd[(2y_0)/(g)] = \surd[(2 × 6400 ft)/(32 ft\bullets^-2)] = 20 s. Thus, in the same time, the object moves a horizontal distance x_0 = ut = 704 ft\bullets^-1 × 20 s = 14,080 ft andtan \texttheta = x_0/y_0 = (14080 ft)/(6400 ft) = 2.2 or \texttheta = 65.5\textdegree. The bomb should be released when the target is seen at an angle of 65.5\textdegree to the vertical. If the target is moving, the relative velocity, between plane and ship is the important velocity. For, relative to the ship, the bomb has an initial velocity v_BS\ding{217} = v_BW\ding{217} + v_WS\ding{217}, where v_BW\ding{217} is the initial velocity of the bomb relative to the water, and v_WS\ding{217} the velocity of the water relative to the ship. Since the velocity of the ship relative to water (v_SW\ding{217}) is given as 20 mph, the v_WS\ding{217} = - 20 mph. Thus v_BS = (480 - 20)mph = 460 mph = 450 (mile/hr) × (5280 ft/mile) × (1 hr/60min) × (1 min/60 sec) = 674.6 ft/sec The foregoing analysis can thus be carried out once more, with v_BS in place of u. Thus x_0 = v_BSt' = 674.6 ft/sec × 20 sec = 13,492 ft tan \texttheta' = (13,492 ft)/(6400 ft) = 2.1 and the bomb should now be released when the target is seen at an angle \texttheta' = 64.5\textdegree to the vertical.

Question:

Given the rest masses of the various particles and nuclei involved in the fission _0n^1 +_92U235\ding{217} _56Ba^141 + _36Kr^92 + 3_0n^1 calculate the energy released in ergs and electron volts.

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Solution:

The nuclear masses will be quoted in atomic mass units or u. The atomic mass unit is 1/12 of the mass of a _6C^12 atom. The reaction is _0n^1 +_92U235\ding{217} _56Ba^141 + _36Kr^92 + 3_0n^1 The masses of the reacting particles are Mass of neutron , _0n^1= 1.0087 u Mass of _92U^235=235.0439 u Combined mass of reactants = 236.0526 u The masses of the products are Mass of _56Ba^141= 140.9139 u Mass of_36Kr^92=91.8973 u Mass of three neutrons,3_0n^1 =3.0261 u Combined mass of products = 235.8373 u The mass of the reactants is greater than the mass of the products by 0.2153 u. Hence, If all this mass is converted to energy, we have, by Einstein's mass energy relation, an equivalent amount of energy of E= mc^2 - (0.2153 u)c^2 But,lu. = 1.66 × 10^-24gm E= (0.2153 u) (1.66 × 10^-24gm/u) (9 × 10^20cm^2/s^2) = 3.27 × 10^-4erg But ,1 erg = {1 / (1.6 × 10^-12)} eVand E = (3.21 × 10^-4erg) / (1.6 × 10-12erg / eV) = 2.01 × 10^8eV(1) The energy released in the fission of one uranium 235 nucleus is 201 MeV. Conventional fuels and conventional explosives release their energy by means of chemical reactions. The energy liberated in a chemical reaction is due to a readjustment of the behavior of electrons in atoms and molecules. A good idea of its order of magnitude may be obtained by comparing the E calculated above with the energy needed to re-move an electron from a hydrogen atom(the ionization energy of hydrogen) . Ionization energy of hydrogen = 2.18 × 10^-11erg ={(2.18 × 10^-11) / (1.60 × 10^-12)}eV = 13.6 eV(2) Comparing equations (1) and (2) we see that the fission of one nucleus of uranium 235 releases about ten million times as much energy as a chemical reaction between one or two atoms. One pound of uranium releases as much energy as several thousand tons of a conventional fuel.

Question:

Write the equations for the stepwise dissociation of pyrophosphoricacid, H_4P_2O_7. Identify all conjugate acid- basepairs.

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Solution:

Pyrophosphoricacid is an example of apolyproticacid.Polyprotic acidsfurnish more than one proton per molecule. From its molecular formula, H_4P_2O_7, one can see there exist four hydrogen atoms. This might leadone to suspect that it istetraprotic, i.e. having 4 protons that can be donatedper molecule. This is in fact the case, which means there exist fourdissociation reactions. In general, the equation for a dissociation reactionis, HA + H_2O \rightarrow H_3O^+ + A^- . Polyproticacids follow this pattern. Thus, one can write the followingequations for the step-wise dissociation of H_4P_2O_7. (1) H_4P_2O_7 + H_2O\rightarrowH_3O^+ + H_3P_2O_7^- (2) H_3P_2O_7^- + H_2O\rightarrowH_3O^+ + H_2P_2O_7^-2 (3) H_2P_2O_7^-2 + H_2O\rightarrowH_3O^+ + HP_2O_7^-3 (4) HP_2O_7^-3 + H_2O\rightarrowH_3O^+ + P_2O_7^-4- To identify all conjugate acid-base pairs, note the definition of the term. The base that results when an acid donates its proton is called the conjugatebase. The acid that results when a base accepts a proton is calledthe conjugate acid. From these definitions, one sees that in all casesH_3O^+ is the conjugate acid of H_2O (the base in these reactions) and H_3P_2O_7^-,H_2P_2O_7^-2, HP_2O_7^-3 and P_2O_7^-4- are the conjugate bases of H_4P_2O_7 , H_3P_2O_7^-,H_2P_2O_7^-2 and HP_2O_7^-3 , respectively.

Question:

It is known that, when exposed to air, beryllium does not corrode but barium does. One explanation is that beryllium (Be) forms a tightly protective oxide coat whereas barium (Ba) does not. The density of BeO = 3.01 g/cc and BaO = 5.72 g/cc, find what happens to the volume per atom when the metals become oxides. The density of Be = 1.86 g/cc and of Ba = 3.598 g/cc.

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Solution:

To solve this problem, first calculate the volume per atom of Be and Ba, and then compare it with the volume per atom of their oxides. From this comparison, expansion or shrinkage can be determined. The volume per atom of any element can be found by knowing the atomic weight, the number of atoms per mole and the density of the element. Thus, there are Avogadro's number or 6.02 × 10^23 atoms per mole. You are given the densities of the sub-stances involved in this problem. Thus, by substitution you find volume of Be = MW / {(6.02 × 10^23) (density)} = (9.01 g /mole) / (6.02 × 1023a/molex1.86 g/cc) = 8.05 × 10^-24 cc/atom From similar computations, you find that BeO = 1.38 × 10^-23 cc/atom, Ba = 6.34 × 10^-23 cc/atom, and BaO = 4.45 × 10^-23. One can find the comparative size of BeO to Be by subtracting the volumeper atom of Be from that of BeO. This difference is then divided by the volume of Be and multiplied by 100. This product gives the percentage in-crease in size of the initial and final atoms when Be is oxidized. A similar process is used for Ba. For Be: {(1.38 × 10^-23 cc - 8.05 × 10^-24 cc) / (8.05 × 10^-24 )} × 100 = 71% For Ba: {(4.45 × 10^-23 cc - 6.34 × 10^-23 cc) / (6.34 × 10^-24 )} × 100 = -30% From a comparison of the atomic volumes of Be andBeO, you see that Be expands by 71 % when it forms the oxide. From a comparison of BaandBaO, you see that Ba shrinks by 30 %, when it forms the oxide.

Question:

The mass absorption coefficients for theK_\alphaandK_\betaradiations ofsilver are 13.6 cm^2 \textbulletg\Elzbar1and 58.0cm^2 \textbulletg\Elzbar1when palladium isused as an absorber. What thickness of palladium foil of density11.4g\textbulletcm\Elzbar3reduces the intensity of theK_\alpharadiation toone-tenth of its incid-ent value? What is then the percentagereduction in the intensity of theK_\betaradiation?

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Solution:

The relationwhichdescribes the absorption of x-radiation is I = I_0e\Elzbar^\muz. Solving for z (which locates a point inside the absorber) ln(I / I_0) =\Elzbar\muz orz =\Elzbar (1/\mu)ln(I / I_0) z = (1/\mu) In (I / I_0) = [1/ (\mu_mp)] In (I / I_0), where\mu is the linear absorption coefficient,\mu_mthe mass absorption coefficient, p the density of the absorber and I_0 the intensity of the incident radiation. If theK_\alpharadiation is to be reduced to one-tenth of its incident value, the thickness required is z =[ 1/ (13.6 cm^2 \textbulletg\Elzbar1× 11.4g\textbulletcm\Elzbar3)]ln10 = 1.49 ×10\Elzbar^2 cm. For theK_\betaradiation with this thickness of absorber, (I / I_0) = exp(\Elzbar58.0cm^2 \textbulletg\Elzbar1× 11.4g\textbulletcm\Elzbar3×1.49 × 10\Elzbar2cm ) = exp(\Elzbar9.85)= 5.28 ×10\Elzbar5. Hence, [(I_0 / I) / (I_0)] × 100 = [1\Elzbar(I / I_0) ] × 100 = (1\Elzbar 5.28 × 10\Elzbar5) × 100 [(I_0 / I) / (I_0)] × 100 = 99.995 % whichis the percentage reduction in the intensity of theK_\betaradiation.

Question:

A string 4.0 m long has a mass of 3.0 gm. One end of the string is fastened to a stop, and the other end hangs over a pulley with a 2.0-kg mass attached. What is the speed of a transverse wave in this string?

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Solution:

The string is stretched by a force (the tension), which is in equilibrium with the weight of the mass: T = Mg = 2.0 kg × 9.8 m/sec^2 = 19.6 nt. The linear density \mu (mass per unit length) of the string is \mu = (m/1) = 0.0030 kg/4.0m = 7.5 × 10^\rule{1em}{1pt}4 kg/m A transverse wave can be set up in the string by moving the weight at one end up and down. The speed v of this wave is given by v = \surd(T/\mu) = \surd{(19.6 kg\rule{1em}{1pt}m/sec^2)/(7.5 × 10^\rule{1em}{1pt}4 kg/m)} = \surd(2.6 × 10^4 m^2/sec^2) = 160 m/sec

Question:

A "point-source" unshaded electric lamp of luminous in-tensity 100 candles is 4.0 ft above the top of a table. Find the illuminance of the table (a) at a point directly below the lamp and (b) at a point 3.0 ft from the point directly below the lamp.

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Solution:

a) Illuminance E is the luminousflux F incident per unit area A. Then E = F/A AlsoF = I\cyrchar\cyromega and\cyrchar\cyromega = A / R^2 If the area A is perpendicular to the path of radia-tion from the point source, the illuminance can also be expressed as E = F / A = I\cyrchar\cyromega/A = [{I(A/R^2)}/A] = I / R^2 which is the equation needed since the given quantities are the luminous intensity I and the distance R from the point source. Furthermore, we know that the area is per-pendicular to the radiation being emitted by the lamp. Therefore, E = I / R^2 = 100 candles / (4.0 ft)^2 = 6.25 lu/ft^2 b) For this second point, the area illuminated is a distance R = \surd[(4.0 ft)^2 + (3.0 ft)^2] = 5.0 ft from the lamp, which was found by using the trigono-metric relation for a right triangle. The illuminance is E = F/A whereF = I\cyrchar\cyromega. The solid angle \cyrchar\cyromega is defined as the ratio of the area upon which the source radiates at a radius R to this distance R. The area upon which the source radiates for this case can be seen from the figure to be A cos \texttheta. Therefore, \cyrchar\cyromega = (A cos \texttheta)/R^2 and E = F/A = I\cyrchar\cyromega/A = {I(A cos \texttheta/R^2)} / A = (I cos \texttheta) / R^2 Using trigonometry, cos \texttheta is found to be cos \texttheta = 4.0ft / 5.0ft = 0.80 Then the illuminance on the area A is E = (I cos \texttheta)/R2= {(100 candles)(0.80)} / (5.0ft)^2 = 3.2 lu/ft2

Question:

A current of 10 A is flowing in a long straight wire situated near a rectangular loop, as indicated in the diagram. The current is switched off and falls to zero in 0.02 s. Find the emf induced in the loop and indicate the direction in which the induced current flows.

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Solution:

Consider the diagram, and in particular the shaded portion of width dy situated distance y from the straight current-carrying wire. The magnetic induction at all points of the shaded portion due to the current- carrying wire has the value dB = (\mu_0 /2\pi) (I/y) . where the permeability\mu_0 = 4\pi × 10^-7 N \bullet A^-2 . Therefore the flux through the shaded area is d\varphi = (dB^\ding{217}) \textbullet dA^\ding{217}. Since dB^\ding{217} is constant throughout the shaded portion and dB^\ding{217} is parallel to dA^\ding{217} where dA^\ding{217} is an element of area within the shaded portion, then d\varphi = (dB) \int dA = (dB) (bdy) = (\mu_0 /2\pi) (I/y) b dy. The flux through the whole loop is thus \varphi = \int d\varphi = (\mu_0 /2\pi) Ib0.15 m\int0.05 m(dy/y) = (\mu0 /2\pi) Ib 1n 3.(1) The limits of integration are taken from y = 5cm = 0.05 m to y = 15 cm = 0.15 m to include the flux within the whole loop. The emf induced in the loop is by Faraday's law, \epsilon = - [(∆\varphi)/∆t)]. Thus, the magnitude of the induced emf is (since the flux goes from the value in (1) to 0 in .02 s) \epsilon = [(\mu_0 Ib)/(2\pi) 1n 3 × \epsilon = [(\mu_0 Ib)/(2\pi) 1n 3 × (1/t∆) = [(4\pi × 10^-7 N \bullet A^-2 × 10 A × 0.2m × 1n 3)/(2\pi × 0.02 s)] = 22.0 × 10-6 V. The flux through the loop is decreasing. The induced current must produce effects tending to oppose the change causing it. The current must there-fore be in such a direction as to produce a flux through the loop in the same direction as that produc-ed by the current in the straight wire, i.e., into the plane of the paper. Thus the induced current traverses the loop in a clockwise direction. The negative sign in Faraday's law signifies this opposi-tion of the induced emf to the charge causing it (Lenz's law).

Question:

A ball bearing is released from rest and drops through a viscous medium. The retarding force acting on the ball bearing has magnitude kv, where k is a constant depending on the radius of the ball and the viscosity of the medium, and v is the bearing's velocity. Find the terminal velocity acquired by the ball bearing and the time it takes to reach a speed of half the terminal velocity.

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Solution:

As the ball falls through the medium, it is accelerated by gravity and the viscous force. To find the acceleration of the bear-ing, we use Newton's Second Law to relate the net force on the ball to its acceleration. Taking the positive direction downward (see figure). mg - kv = ma, where a is the acceleration produced at any time. The initial val-ue of a is g, since at the moment of release v = 0. As the value of v increases, the acceleration decreases until, when v = v_0, the terminal velocity, a = 0. Thus mg - kv_0 = 0. Therefore v_0 = (m/k)g. In order to find out at what time v = v_0, we must calculate v as a function of t (or vice versa). At any time t, it will be found that mg - kv = ma = m(dv/dt). or(mg - kv)dt = mdv dt = (mdv) / (mg - kv) = mdv/{m(g - (k/m)v)} dt = dv/{g - (k/m)v};and^t\int_0 dt = ^v\int_0 [{dv} / {g - (k/m)v}]. where, in the integration limits, v = 0 at t = 0 and v = v at t = t. Hence t = ^v\int_0 [{dv} / {g - (k/m)v}] Lettingu = {g - (k/m)v} du = -(k/m) dv To find the new integration limits, we realize that when v = 0, u = g and when v = v, u = g - (k/m)v, whence t = ^g-(k/m)v\int_g [{-(m/k) du} / {u}] = (-m/k) ln (\vertu\vert) ^g-(k/m)v\vert_g t = (-m/k) {ln [\vert{(g - (k/m)v}\vert] - ln(g)} t = -(m/k) ln [\vert{(g - (k/m)v}/g\vert] t = -(m/k) ln \vert1 - (k/m)v\vert(1) The time to acquire half the terminal velocity, T, is thus found by inserting v = v_0/2 in (1) T = -(m/k) ln \vert1 - (k/mg) \bullet (mg/2k)\vert = -(m/k) ln \vert1/2\vert = +m/k ln \vert2\vert = 0.69 (m/k).

Question:

Define the terms haploid, dipolid, gamete, zygote, sporophyte and gametophyte.

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Solution:

These terms refer to the life cycle and reproductive processes of algae, fungi, and higher plants. The chromosome number is termed haploid or diploid. A haploid organism is one which possesses a single complete set of chromosomes while a diploid organism has two sets of chromosomes. A gamete is a cell specialized for sexual repro-duction, and is always haploid. Two gametes fuse to form a diploid cell, which may give rise to a new multi-cellular individual via mitotic division, or which may itself undergo meiosis to form new haploid cells. The diploid cell that results when two gametes fuse is termed a zygote. The zygote is always diploid and is always a single cell. Life cycles of algae and fungi show much more variability than life cycles of higher plants and animals. The different species within a phylum often have very different life cycles, and most algae and fungi are capable of both sexual and asexual reproduction. Sexual reproduction occurs when two haploid gametes fuse to form a zygote. Asexual reproduction occurs by frag-mentation, with each fragment growing into a new plant or thallus. Asexual reproduction also involves mitotic division of a single cell of a thallus to produce daughter cells which are released from the parent thallus to grow into new algae or fungi. Daughter cells which are motile and specialized for asexual reproduction are termed zoospores. In certain species, alternation of generations occurs. This type of life cycle includes both a multi-cellular haploid plant and a multicellular diploid plant, and both sexual and asexual reproduction. In algae, fungi, and most other lower plant forms, the dominant generation is the haploid one, and the reverse is true in the higher plants. Ulva (sea lettuce) , a green alga, and Ectocarpus (a brown alga) are representative of organisms which show alternation of generations. The entire life cycle can be summarized as follows: Haploid zoospores (stage 1) divide mitotically to produce a haploid multicellular thallus (stage 2). These haploid thalli may reproduce asexually by means of zoospores or sexually by means of gametes (stage 3). Gametes fuse to produce a diploid zygote (stage 4). The zygote divides mitotically to produce a diploid multi-cellular thallus (stage 5). Certain reproductive cells of the diploid thallus divide by meiosis to produce zoospores (stage 1). The zoospores are released, and begin the cycle anew. The haploid multicellular stage is called a gametophyte because it is a plant that produces gametes. The diploid multicellular stage is called a sporophyte because it is a plant that produces spores.

Question:

Fibbonacci.numbersare a sequence of numbers such that the first number is 0, the second number is 1, and each subsequent number is the sum of two preceding numbers, e.g. F_1 = 0 Fn- 1 + Fn - 2, n \geq 2 F_2 = 1 Write a Pascal program to print out the first 40Fibbonacci numbers and their sum.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0400.htm

Solution:

We will need five variables for this program: I - to count the number ofiterations .We want only 40 iterations. SUM - to keep the current sum ofnumbers . A, B - the initial two numbers (0 and 1), they will also re-present the two predecessorsof the current number. X - stores the current number, the sum of the above two. PROGRAMFibbonacci(output); VAR A, B, X, I,SUM :integer; BEGIN A : = 0; B : =1; SUM: = 0; {initialize the sum and the 2 initial numbers} FOR I:=1 to 40 DO {Repeat 40 times} BEGIN WRITELN (A); {A is the current number} SUM: = SUM + A; {Update SUM} X: = A + B; {get the next number} A: = B; {A becomes the next predecessor} B: = X {transfer the current number to B} END;{all 40 numbers printed out} WRITEIN ('Thesum of 40Fibbonaccinumbers is', SUM) END.

Question:

Describe and compare the different forms of currently used methods of contraception. Discuss the safety and efficiency of each.

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Solution:

Reproduction in man is an extremely complex process involving many different events. Contraception or birth control acts on the principle that these events are sequential; preventing any given event from occurring will prevent the end result, which is pregnancy. The currently used methods of birth control interfere with the normal reproductive process by (1) suppressing the formation and/or release of gametes; (2) preventing the union of gametes in fertilization; or (3) preventing implantation of the fertilized egg. The oral contraceptive or "pill" acts to prevent ovulation in the female. The pill contains combinations of estrogen and progesterone, which inhibit the release of FSH (follicle stimulating hormone) and LH (luteinizing hormone) by the pituitary via a feedback mechanism; these hormones respectively stimulate follicular growth and trigger ovulation. A woman who is taking the pill in-hibits their release and does not ovulate. The pill is usually taken daily from the fifth to the twenty-fifth day of the female cycle. Withdrawal of the first five days triggers menstruation. Some synthetic hormones used in oral contraception have additional antifertility effects in that they cause a thickening of the cervical mucus and as a result, sperm have more difficulty entering the uterus. The effects of long-term use of the pill are not yet known. There does exist a relationship between increased amounts of estrogen and a higher risk of thromboembolism (the formation of blood clots). Presently, the efficiency of the oral contraceptive is close to 100%. Women using the pill have between zero and one pregnancy per hundred woman-years (a group of one hundred women using that method of contraception for one year). Fifty to eighty pregnancies per hungred woman-years will occur if no attempt is made to restrict conception. Condoms and diaphragms interpose a barrier between the male and female gametes. A condom prevents semen from entering the vagina. A diaphragm blocks the cervix and prevents sperm entry into the uterus. Spermicidal jellies and foams are used with the diaphragm to kill sperm. Condoms and diaphragms are relatively safe, and their effi-ciency results in about 12 to 14 pregnancies per hundred woman-years. The intrauterine device or IUD prevents the implan-tation of the fertilized egg. The IUD is a small object, such as a coil, placed within the uterus, which causes a slight local inflammatory response in the endometrium. This interferes with the endometrial preparation for proper implantation of the blastocyst. In some women, the IUD causes severe cramping and is sometimes spontaneously expelled. Its efficiency is about two pregnancies per hundred woman-years. Restricting intercourse to a "safe" period is known as the rhythm method. Normally, a woman ovulates once a month and abstaining from intercourse during this time can be used as a form of contraception. However, it is extremely diff-icult to precisely determine when ovulation occurs, even when taking body temperature daily (the body temperature rises slighlty following ovulation). Therefore this method is not a very reliable method. About twenty-four pregnancies per hundred woman-years occur using the rhythm method. A more permanent form of birth control is sterilization. In females, this is accomplished by tubal ligation, an operation in which the Fallopian tubes are cut and tied, preventing passage of the ovum from the ovaries to the uterus. This method is 100% effective if performed properly, but involves obvious surgical risks. Male sterilization involves a vasectomy, the cutting and tying off of the vas deferens. Again, if done properly, it is 100% effective and does not involve complicated surgery or anesthesia. The development of the vasectomy marks the beginning of the search for effective male contraceptive techniques.

Question:

The transformation of insect larvae into mature adults is known to be under hormonal control. What hormones are involved and what are their effects?

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Solution:

A hormonal secretion from the brain stimulates the release of the hormone ecdysone by the prothoracic gland. Ecdysone is the hormone that stimulates growth and molting in insects. Another hormone, present in insects prior to adulthood, is secreted by a portion of the brain known as the corpora allata. This hormone, referred to as juvenile hormone, is responsible for the maintenance of larval structures and inhibits metamorphosis. The juvenile hormone can exert its effect only after the molting process has been initiated. It thus must act in conjunction with the prothoracis hormone. When a relatively high level of juvenile hormone is present in the blood, the result is a larva-to-larva molt. When the level of the juvenile hormone is low, the molt is larva-to-pupa. Finally, in the absence of the juvenile hormone, molting gives rise to adults. Ecdysone has been chemically identified as a steroid and it has been obtained in pure form. The pure extract of ecdysone has been used to investigate the functional mechanism of the hormone. In one recent experiment, small amounts of ecdysone were injected into a gnat. Within fifteen minutes, chromosome puffs (enlargements of certain regions of the chromosome) were observed to form at certain sites on the salivary gland. Chromosomes puff formation is taken as an indication of the activation of a certain gene or set of genes. This is one piece of evidence that certain genes may be activated by hormones.

Question:

Find the wavelength of light required for the ionization of sodium atoms to occur. The ionization potential of sodium is 8.17 × 10^-19 J.

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Solution:

The energy necessary to ionize a sodium atom is equal to the ionization potential. Here, one is looking for the wavelength of light, which, when flashed upon sodium, will possess an energy equal to the ionization potential. Wavelength is related to energy by the equation E = hc/\lambda, where E is the energy, h Planck's constant (6.626 × 10^-34 J sec), c the speed of light (3.0 × 10^8 m/sec) and \lambda the wavelength. One can rewrite this equation as \lambda = hc/E and then substitute the given to solve for \lambda. \lambda = [{(6.626 × 10^-34 j sec) (3.0 × 10^8 m/s)} / (8.17 × 10^-19 J)] = 2.43 × 10^-7 m = 2.43 × 10^-7 m × [(1 \AA) / (10^-10 m)] = 2430 \AA.

Question:

Explain bus transfer of data. Show the circuit elements Explain bus transfer of data. Show the circuit elements involved in parallel bus transfer and the symbolic representation of data transfer.

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Solution:

In a system with many registers, the transfer from one register to another requires that lines be con-nected from the output of each flip-flop in one register to input of each flip-flop in another register. Consider for example, the requirement for transfer among three registers as shown in Figure 1. There are six data paths between registers. If each register consists of F flip-flops, there is need of 6F lines for parallel transfer from one register to another. As the number of registers goes to n, there will be (2^n - 2) × F lines (where F is number of flip-flops in each register). This greatly enhances the complexity of connections and cost per bit transfer. The idea of bus transfer is analogous to a central transportation system. Instead of each commuter using his own private car to go from one location to another, a bus system is used with each commuter waiting in line until transportation is available. A bus system is formed with multiplexer and decoder circuit elements. Consider Figure 2 where there are four registers (each register has four flip-flops) and four 4 × 1 multi-plexers. The four inputs to Ml are A_1, B_1, C_1 and D_1. Similarly, for M_2, M_3 and M_4, each multiplexer is con-nected to the same two selection lines x and y. When xy = 00, I_O line of all four multiplexers gets selected and at output we have A_1A_2A_3A_4. Similarly xy = 01 selects B_1B_2B_3B_4. When xy = 10 the selection is C_1C_2C_3C_4, and xy = 11 selects D_1D_2D_3D_4. Figure 2 is symbolically represented as Figure 3. Thus, by proper selection of x and y select lines, we have the desired data on Bus. Now, the question is: How do we transfer the data on bus to the desired register? The technique is to use a Decoder whose output is used to enable the desired register in which the data is to be transferred. As shown in Figure 4, we have a 2 × 4 Decoder whose output is connected to the enable input of the registers. The Bus line carries four output wires from multiplexer. These wires, in the form of a bus, are connected to all four registers R_1, R_2, R_3 and R_4. On selection of xy = 00 we have A_1A_2A_3A_4 on the bus line; all these are going as input to the four registers of Figure 4. Yet the data is not transferred to them until they are enabled. En-abling the register is done by the 2 × 4 decoder, depending on the input v and w. If vw = 00, I_O is on and register R_1 gets enabled. Similarly, if vw = 01, I_1 is on, vw = 10, I_2 is on and if vw = 11, I_3 is on, enabling registers R_2, R_3 and R_4 respectively. Hence the data from register A is transferred to register R_2 by making xy = 00 and vw = 01. Symbolically, this is represented as x̅ y̅:BUS\leftarrowA v̅w:R2\leftarrowBUS where x̅ y̅ or v̅w before the colon (:) represent conditions and the transfer occurs only if the conditions are true.

Question:

A rectangular post 4 in. thick is floating in a pond with three- quarters of its volume immersed. An oil tanker skids off the road and ends up overturned at the edge of the pond with oil of 1.26 slugs \textbullet ft^-3 density leaking from it into the water. When the upper face of the post is just level with the surface of the liquid, what is the depth of the oil layer? What happens if more oil keeps pouring into the pond?

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Solution:

Before the oil is spilled, the post is floating in the water symmetrically. Let its cross- sectional area be A and its density \rho. Then, if three- quarters of the volume is immersed, only 1 in. is above the surface and 3 in. below the surface. The weight downward, F_G and the buoyant force F_B balance, since the post is in equilibrium vertically. Hence if \rho_0 is the density of water, we may write F_G - F_B = 0 \rhoAg × 4" - \rho_0Ag × 3" = 0(1) or\rho/\rho_0 = 3/4 Note that the buoyant force is equal to the weight of water displaced by the post. (We have taken the positive direction downward). When the oil density \rho' pours on, it stays above the water. The water extends up to a height y and the oil fills the other (4 in. - y). Since equilibrium is still achieved, the weight downward must equal the sum of the two buoyant forces due to the water and the oil. Hence \rhoAg × 4" = \rho_0Agy + \rho'Ag(4" - y). Using (1) \rho× 4" = \rho_0 × 3" = \rho_0y + p'(4" - y). \thereforey = [\rho_0 × 3" - p" × 4"]/[\rho_0 - \rho'] = [1.94 slugs \bullet ft^-3 × 3" - 1.26 slugs \bullet ft^-3 × 4"]/[(1.94 - 1.26) slugs \textbullet ft^-3] = (0.78)/(0.68) "= 1.15 " Thus the depth of the oil layer is 2.85". If oil keeps pouring onto the pond, the post must stay as it is with respect to the water-oil interface. While the oil poured on initially, the post rose from the water to compensate for the extra up thrust from the oil by diminishing the up thrust from the water. Once the post is totally immersed, it is at the correct po-sition with respect to the water-oil interface for the sum of the two up thrusts to equal the weight. Adding further oil cannot alter this.

Question:

Distinguish betweenapoenzymesandcofactors .

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Solution:

Some enzymes consist of two parts: a protein constituent called an apoenzymeand a smaller non-protein portion called a cofactor. The apoenzymecannot perform enzymatic functions without its respective cofactor. Some cofactors, however, are able to perform enzymatic reactionswithout anapoenzyme, although the reactions proceed at a muchslower rate than they would if the cofactor andapoenzymewere joinedtogether to form what is called aholoenzyme. The cofactor may be either a metal ion or an organic molecule calleda coenzyme. A coenzyme that is very tightly bound to the apoenzymeis called a prosthetic group. Often, both a metal ion and a coenzymeare required in aholoenzyme. Metalloenzymesare enzymes that require a metal ion;catalase, for example, which rapidly catalyzes the degradation of H_2O_2 to H_2O and O_2, needsFe+^2 or Fe+^3. Iron salts can catalyze the same reaction on their own, butthe reaction proceeds much more quickly when the iron combines with theapoenzyme. Coenzymes usually function as intermediate carriers of thefunctional groups, atoms, or electrons that are transferred in an overall enzymatictransfer reaction.

Question:

How fast does light travel in glass of refractive index 1.5?

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Solution:

By definition, refractive index n is the ratio of the velocity of light in vacuum (3.00 × 10^10 cm/sec) to the velocity of light in the medium in question. Thereforen = [(3 × 10^10 cm/s)/v] = 1.5 Thereforev = [(3 × 10^10 cm/s)/1.5] = 2.00 × 10^10 cm/sec.

Question:

Compare the processes of oxidativephosphorylationand substratelevelphosphorylation. How can certain evolu-tionarychanges be explained by these differences?

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Solution:

In the complete oxidation of glucose to carbon dioxide and water, NADH (reducednicotinamideadeninedinucleotide) is produced. High energyelectrons are transferred from NADH to oxygen via the electron transportsystem. This transfer of electrons involves a series of enzyme catalyzedoxidation - reduction reactions in which the energy released is usedin the production of high energy phosphate bonds. Three high energyphosphate bonds are made for each pair of electrons transferred tooxygen from NADH. This process of harnessing the energy released in theelectron transport chain to synthesize ATP from ADP and inorganic phosphateis termed oxidativephosphorylation. The electron transport chainis not the only system that produces high energy bonds.Atpis pro-ducedin bothglycolysisand the Krebs cycle. Any reaction which synthesizes high energy phosphate bonds but isnot involved in the electron transport system, is called a substrate level phosphorylation. Inglycolysis, a net total of 2 \simP s (high energy phosphate bonds) are produced at the substrate level; in the TCA cycle, one \simP is synthesizedat the substrate level. Most of the ATP produced during aerobicrespiration, however, is synthesized by oxidativephosphorylation. Indeed, therein lies the source of the differences between aerobic and anaerobicorganisms, and the reason that anaerobic organisms must be verysmall, and almost always unicellular. The high energy requirements ofa larger organism could not beaccomodatedthrough the 3 ATP producedvia substrate levelphosphorylation.

Question:

An important application of computers to market research is in computer simulation of continuous models. As an example, there appears to be a positive relationship be-tween the number of new houses built and the sale of wash-ing machines. Give a differential equation connecting the growth of houses and washing machines. Illustrate with graphs.

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Solution:

A mathematical model is said to be continuous when the variables of the system change smoothly over time. Quantities that change continuously over time are best studied with the use of differential equations. Let H be the potential number of families and y be the number of families with houses. H is a constant upper limit on the number of houses and, thus, the number of washing machines that can be sold. Furthermore, assuming that every new house requires a washing machine, the un-filled market is the difference between houses sold and washing machines sold. Finally, as the number of new houses approaches H, the market approaches saturation and the rate of selling houses and, by extension, washing machines drops. The given figure applies: From the figure, the slope of the function between number of houses sold and time is seen to decrease with time, reaching zero at the Time when y(t) = H. Let y = y(t).Then y(0) = 0. If we assume that the rate of new houses sold is a constant function of the remaining potential market, the differen-tial equation is ẏ = k_1 (H - y)y(0) = 0(1)

Question:

An excellent method of making alcohols is the reaction of an aldehydeon a Grignard reagent. If you perform this reaction, you must scrupulously dry thealdehydebefore adding it to the Grignard reagent. Why?

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Solution:

To answer this question, you must know what a Grignard reagent is and some of its properties. The general formula of agrignardreagent is RMgX, where R is an alkyl or aryl group, Mg is magnesium, and X is a halogen atom (usually bromine or chlorine). Grignard reagents are extremely reactive. You can consider it to be the magnesium salt of the weak acid R\ElzbarH. Thus, if there exists in solution a stronger acid, R\ElzbarH will be displaced from its salt. Water is a stronger acid than R\ElzbarH. Thus, if H_2O (water) is present, the following re-action occurs: RMgX+ H_2O \rightarrow RH + Mg(OH)X. Thus, if thealdehydeis not dried, water will be present, and it will then react with thegrignardreagent producing RH. The presence of water reduces or eliminates thegrignardreagent from solution, so that little or none is present to react with thealdehyde.

Question:

Obtain the 1's and 2's complements of the following binary numbers: a)1010101 b)0111000 c)0000001 d)10000 e)0000

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Solution:

Forming complements is useful when one wants to perform subtractionwithout borrowing from another column. This is because subtractionis the same as adding the two's complement. a) The 1's complement of 1010101_2 is found by subtracting it from 1111111 .Thus 1111111 -1010101 0101010 The two's complement is formed by subtracting1010101_2 .from 10000000. Thus, 10000000 - 1010101 00101011 Forming the two's complement by binary subtraction is difficult. b) Since the base, or radix, of the binary systemis2, the one's complementis referred to as the radix minus one complement. Similarly, thetwo's complement is called the radix complement. The radix minus one complementof 0111000_2 is 1111111 - 0111000 1000111 The radix complement of 0111000_2 is 10000000 - 0111000 1001000 While examining a) and b), note that finding the radix minus one complementsof the two given numbers is equivalent to changing 0's to 1's and1's to 0's in the given numbers. c) The radix minus one complement of 0000001 is 1111110. The radix complementis 10000000 - 0000001 1111111 Examining the radix complements of a), b) and c) observe that Radix complement(X)_2 = Radix complement minus one(X)_2 + 1(1) whereX is a binary number. d) The one's complement of 10000 is 01111 or 1111. The two's complement, using (1) is 1111 +1 10000 or, the number itself. e) The radix minus one complement of 0000 is 1111 and the radix complementis 1111 +1 10000 To illustrate the advantage of representing numbers by their radix complements, suppose we wish to store the negative of d), i.e., -10000, in a12 - bit word. The left-most bit has value 1 to represent the minus sign: 1000 0001 0000 Now suppose word size is increased to 16 bits. This is most easily done byadding 4 bits with zero value on the left. But the left-most bit must be changedto 1 and the 1 in the third quartet of bits (third from the left) must bechanged to zero: 1000 0000 0001 0000 Suppose ,instead, that the radix complement of -10000 is in a 12-bit storage. The radix complement of -10000 in a 12-bit storage is 111111111111 -10000 111111101111 +1 111111110000 Now, if word length is to be extended to 16 bits, simply add a quartet of 1's. No special provisions need be made for the sign bit. In other words, whenword size is increased, simply copy the sign bit into all the bits to its left. Finally, the two's complement representation is preferred to the one'scomplement because + 0 \not = - 0 in the one's complement, but + 0 = - 0 in the 2's complement.

Question:

A sample of gold is exposed to a beam of neutrons, and the reaction Au^197 + n = Au^198 + \Upsilon absorbs 10^6 neutrons per second. Au^198 emits \beta-particles and has a half-life of 2.70 days. How many atoms of Au^198 are present after 2 days of continuous irradiation?

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Solution:

At any instant during the irradiation, the increase in the number of Au^198 atoms in unit time will be the number being produced less the number disintegrat-ing. Thus, if \lambda is the decay constant for the \beta decay, dN/dt= 10^6 s\Elzbar1\Elzbar\lambdaN. This follows because the reaction occurs once for every neutron (n) absorbed. If 10^6 neutrons are absorbed per second, 10^6 Au^198 atoms (as well as 10^6 \Upsilon ray photons) are produced per second. dN/(10^6 s\Elzbar1\Elzbar\lambdaN) =dt^N\int_0dN/(10^6 s\Elzbar1\Elzbar\lambdaN) =^t\int_0dt(1) there being no atoms of Au^198 present at time t = 0 when the irradiation starts. Letting u = 10^6 s\Elzbar1\Elzbar\lambdaN du =\Elzbar\lambdadN Since u = 10^6 s\Elzbar1when N = 0, and u = 10^6 s\Elzbar1\Elzbar\lambdaNwhen N = N, (1) becomes \Elzbar(1/\lambda)(10)6(s)\Elzbar1\Elzbar(\lambdaN)\int10^6_s\Elzbar1du/u = ^t\int_0dt or\Elzbar(1/\lambda)In (u)(10)6(s)\Elzbar1\Elzbar(\lambdaN)▏_10^6_s\Elzbar1= t \Elzbar(1/\lambda)In [(10^6s\Elzbar1\Elzbar\lambdaN) / (10^6s\Elzbar1)] = t In[1\Elzbar {(\lambdaN) / (10^6s\Elzbar1)}] =\Elzbar\lambdat and1\Elzbar {(\lambdaN) / (10^6s\Elzbar1)} = exp (\Elzbar\lambdat) N = [(10^6s\Elzbar1) / \lambda] [1\Elzbarexp(\Elzbar\lambdat)](2) Note that in general, the equation of radioactive decay is N = N_0e\Elzbar\lambdat if N_0 is the original number of atoms which are de-caying. After one half life (\tau) , N0/2 atoms are left and N(\tau) = N0/2 = N_0e\Elzbar\lambda\tau or2 = e\lambda\tau Hence\lambda\tau= In 2 and \lambda = 1/\tauIn 2. Using this fact in (2) N = [{(10^6 s\Elzbar1)(\tau)} / (In 2) ] [ 1\Elzbarexp{\Elzbar[(In 2) t} / \tau] } ] where \tau is the half-life period of the \beta-activity. After 2 days, N = [{(10^6 s\Elzbar1) (2.7 days)} / (.6932) ] [ 1\Elzbarexp[\Elzbar[{(In 2)(2 days)} / (2.7days) ] ] ] Since 1 day = 86,400secs N = [{(10^6 ) (86,400)} / (.6932) ] [ 1\Elzbarexp{\Elzbar (2 / 2.7) In 2} ] N = 1.35 x 10^11 atoms.

Question:

A damped harmonic oscillator (DHO) is a system whose behavior can be described by a second-order, linear differential equation of the form: ӱ =-A_1ẏ - A_2y(t)(1) where A_1, A_2 are positive constants. One DHO system is a spring- mass-damper system shown in Figure #1. This could be a simple model of an automobile wheel suspension system (assuming the automobile's body is immobile in a vertical direction). Then the damper acts as a shock absorber. Another DHO system is an RLC electric circuit: The analogy between these two DHO systems is given in Table #1. DHO Y(t) Ẏ = dy/dt A1 A_2 Spring- mass- damper x-displacement V = velocity D/M K/M RLC circuit q-charge in coulombs i = current in amperes R/L 1/LC Given the initial conditions y(0) = y_0 and ẏ(0) = z_0 , write a Given the initial conditions y(0) = y_0 and ẏ(0) = z_0 , write a FORTRAN program that uses the modified Euler method to simulate this system from t = 0 to t = t_f- .

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Solution:

Note that equation (1) is of the form d^2y/dx^2 = g(y,y'). The modified Euler subroutine will be written for this form. First, rewrite equation (1) as a system of simultaneous first-order equations: y = dy/dt = z, z(0) = y' (0) = z_0(2) z = dz/dt = -A_1Z - A_2y , y(0) = y_0(3) Now, use the modified Euler method to find approximate solutions to both (3) and (4). REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2 COMMON A1,A2 READ, N, TFIN, ACCUR, YO, ZO, A1, A2 CALL EULER2(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G2(W,X) REAL G2,W,A1,A2 COMMON A1,A2 G2 = -A1\textasteriskcenteredX-A2\textasteriskcenteredW RETURN END SUBROUTINE EULER2 (N,XN,X,YCOR, ZCOR,ACCUR) REAL N DELTAX = (XN - X)/N 50Y = YCOR Z = ZCOR G2OLD = G2(Y,Z) PRINT,X,Y,Z X = X + DELTAX IF (X.GT.XN) STOP YPRED = Y + Z\textasteriskcenteredDELTAX ZPRED = Z + G2(YPRED,Z)\textasteriskcenteredDELTAX 100YCOR = Y + 0.5\textasteriskcentered(Z + ZPRED)\textasteriskcenteredDELTAX ZCOR = Z + 0.5\textasteriskcentered(G2QLD + G2(YPRED,ZPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) ZDIF = ABS(ZCOR - ZPRED) IF(YDIF.LE.ACCUR.AND.ZDIF.LE.ACCUR) GO TO 50 YPRED = YCOR ZPRED = ZCOR GO TO 100 RETURN END

Question:

A chemist wants the percent dissociation of HC_2H_3O_2 to be 1 %. If theK_dissof HC_2H_3O_2 is 1.8 × 10^-5, what con-centration of HC_2H_3O_2 is required?

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Solution:

To answer this question, consider the dissociation of an acid in H_2O. Then write anequlibriumconstant ex-pression which relates the equlibriumconstant for dis-sociation,K_dissto the ratio of the concentrations of products to reactants, each raised to the power of its coefficients in the reaction. In general, for an acid, there exists the following reaction for dissociation: HA + H_2O \rightarrow H_3O^+ + A^-, where HA is the acid. For this reaction, HC_2H_3O_2 + H_2O \rightarrow H_3O^+ + C_2H_3O_2^- . Theequlibriumconstant expression is K_diss= 1.8 × 10^-5 = {[H_3O^+][C_2H_3O_2^-]} / [HC_2H_3O_2] One wants to find [HC_2H_3O_2] with the knowledge that the percent dissociation is 1 % (note: exclude water from the expression, as it is assumed to be constant). If the percent dissociation is 1 %, then [H_3O^+] / [ HC_2H_3O_2] = 1/99. The H^+ can only come from HC_2H_3O_2. If 1 % of H^+ exists, then 100% (initial percentage) - 1% = 99% of HC_2H_3O_2 must be left at equilibrium. Thus, 1.8 × 10^-5 = (1/99) [C_2H_3O_2^-]. Solving for [C_2H_3O_2^-]; [C_2H_3O_2^-] = 1.78 × 10^-3 M. This concentration of C_2H_3O_2^- must equal the con-centration of H_3O^+, since they are formed inequimolaramounts. One can substitute into the equation 1.8 × 10^-5 = {[H_3O^+][C_2H_3O_2^-]} / [HC_2H_3O_2] to obtain{(1.78 × 10^-3)( 1.78 × 10^-3)} / [HC_2H_3O_2] = 1.8 × 10^-5 . Solving for [HC_2H_3O_2)], 1.76 × 10^-1 M.

Question:

For the following PL/I program, a) Explain how a condition called ENDFILE CONDITION' arises. b) Show what will be printed out if the program is run. c) Write out a corrected program to avoid errors due to this condition . d) Show what the printout from the corrected program will be . EXAMPLE:PROCOPTIONS(MAIN); DCL(X,Y,Z) FIXED(3) INIT(0); GETLIST(Z); DO I = 1 TO Z; GET LIST(X); Y = Y + X; PUTLIST(I,Y); END; PUTLIST(21,Y); END EXAMPLE; The DATA CARD contains the following data: 8, 0, 1, 2, 3, 4, 5, 6

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Solution:

a) The purpose of the above program is to illus-trate a condition knownas ENDFILE CONDITION. In this program, the GETLIST(Z) statement reads a value for Z fromthe DATA CARD. Hence, Z = 8. This value of Z be-comes the upper limitfor the loop of the next statement. Hence, the loop statement becomes: DO I = 1 TO 8. During each loop, a value of X is read from the card and the sum Y = Y + x is calculated. Now, observe that there are only 7 values of X (0 to 6) punched on theDATA CARD. Hence, when the program enters the eighth loop, the computerfinds that the data have run out and terminates the program. This condition of data running out is known as an ENDFILE CONDITION, b) The printout of the computer is as follows: 10 21 33 46 510 615 721 After the seventh loop, Igetsa value of 8. Now the program tries to executeGET LIST(X). But since there is no 8th value ofX,the program haltsgiving out an error message. So there is no chance for executing the otherPUT LIST statement which follows outside the loop. c) The program can be corrected by inserting an ON statement of the generalform ON ENDPILE (FILE NAME) ACTION; The above statement prevents termination of the program abrupt-ly on the occurrenceof the ENDFILE CONDITION. Inside the parentheses, the name of the file,which is to be used, shouldbe specified. For example, if punched cards are being used for input, the name of the input file is (SYSIN). 'ACTION' in the above statement means some sort of in-struction to thecomputer telling it what to do if the data on the file run out. The correctedprogram can therefore be written as follows: EXAMPLE:PROCOPTIONS(MAIN); DCL (X,Y,Z) FIXED(3), INIT(O) ; ONENDFILE(SYSIN) GOTO A ; GETLIST(Z); DO I = 1 TO Z; GET LIST(X) ; Y = Y +X ; PUTLIST(I,Y) ; END ; A:PUTLIST(21,Y); END EXAMPLE; d) The printout from the corrected program will be: 10 21 33 46 510 615 721 1621 Note that after the seventh loop is completed, the computer goes back to theloop head, viz., DO I = 1 TO Z. As Z = 8, therefore I is incremented by 1 from 7 to 8. Now, on trying to executethe next statement, viz., GET LIST(X), the computer finds that an ENDFILE CONDITION has occurred on the SYSIN file. Hence, the programgoes to the statementlabelled'A'. This statement is PUT LIST(2I,Y);. The lat-est values of I and Y lying in the program are 7 and 22 re-spectively; therefore 16 and 22 are printed out. The pro-gram now goes tothe next statement END EXAMPLE which is a normal end of the program.

Question:

Write a BASIC program to reduce a fraction to lowest terms. Test your program with sample data.

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Solution:

We make use of the INT function to find the largest factor of an integer. What we need to do is find the largest common factor of the numerator and denominator. A flowchart illustrating the program logic is given below: 1\O READ N,D 2\O FOR P = N TO 2 STEP - 1 3\O IF N/P = INT(N/P) THEN 7\O 4\O NEXT P 5\O PRINT N'V'D 6\O GO TO 1\O 7\O IF D/P = INT(D/P) THEN 9\O 8\O GO TO 4\O 9\O PRINT N"/"D" ="N/P"/"D/P 1\O\O GO TO 1\O 11\O DATA 5,6 12\O DATA 82, 48 13\O END Note that the absence of commas in both PRINT statements tells the computer to write the numerical values of the variables and the arith-metic operators right after each other, not leaving any blank space in between.

Question:

Calculate, approximately, the relative intensities of the secondary maxima in the single-slitFraunhoferdiffraction pattern.

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Solution:

Theseconderymaxima lie approximately halfway between the minima and are found from \alpha \cong (m + 1/ 2)\pim = 1,2,3,.... Substituting into the formula for intensity in single-slit diffraction yields I_\texttheta=I_m[{sin(m + 1 / 2)\pi} / {(m + 1 / 2)\pi}]^2 , which reduces to (I_\texttheta/I_m) = [1 / {(m + 1 / 2)^2\pi^2}] . This yields, for m = 1,2,3,...,I_\texttheta/I_m= 0.045, 0.016, 0.0083, etc. The successive maxima decrease rapidly in intensity.

Question:

0.1 moles of Cu(NO_3)_2 are added to 1.0molesof NH_3 and then diluted to 1000ml. Find the concentration of Cu^2+ formed. K_eq= 8.5 × 10^-13 .

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Solution:

The first thing to realize is that the complex Cu(NO_3)_2 dissociates to Cu^2+ + 2NO_3^+ in H_2O Cu(NO_3)_2 \rightleftarrows Cu^2+ + 2NO_3^+ when NH is added to the solution the following reaction occurs. Cu^2+ + 4NH_3 \rightleftarrows Cu(NH_3)_4^2+ Cu(NH_3)_4^2+, is called a complex ion. Cu(NH_3)_4^2+, will be in equilibrium with Cu^2+ 4NH_3 . You are given theK_eqfor this reaction which is defined as K = {[Cu^2+][NH_3]^4} / [Cu(NH_3)_4^2+] . In general, if you have the equilibrium: H(A)_X \rightarrow H^+ + XA^- , then K ={[H^+][A^-]^X} / {H(A)_X}]. That is, the coefficients of the balanced equation serve as the ex-ponents in the K expression. The problem stated K = 8.5 × 10^-13 . Therefore, {[Cu^2+][NH_3]^4)} / [Cu(NH_3)_4^2+] = 8.5 × 10^-13 . You are solving for [Cu^2+]. Let x = [Cu^2+], then [Cu(NH_3)_4^2+] = .1-x, since it must be the original amount of Cu(NO_3)_2. present minus the Cu^2+ formed. There is originally 1.0 moles NH_3, at equilibrium 4 moles of NH_3 will have reacted with each mole of Cu^2+ to form Cu(NH_3)_4^2+, thus, [NH_3] may be represented as 1.0-4 × [Cu(NH_3)_4^2+] = 1.0-4(.10-x). Since most of the Cu^2+ present will be in the form of Cu(NH_3)_4^2+ and that there was only .1M of it in the first place, one can make the assumption that x << .1, therefore, .10-x = .1 and 1.0 - 4(.10-x) = 1.0 - 4(.1) = 0.6 = [NH_3] . Using this same assump-tion, [Cu(NH_3)_4^2+] = .1-x = .1 . You now substitute these values into the previous expression. That is, [{x(0.6)^4}/(.1)] = 8.5 × 10^-13 , x = 6.5 × 10^-13 . Therefore, [Cu^2+] = 6.5 × 10^-13 M.

Question:

A shower of protons from outer space deposits equal charges +q on the earth and the moon, and the electrostatic repulsion then exactly counterbalances the gravitational attraction. How large is q?

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Solution:

If R is the distance between the earth and the moon, the electrostatic force in the CGS system is F_e = q^2 / R^2 If M_e and M_i are the masses of the earth and the moon respectively, the gravitationalforce is F_G = (GM_eM_i) / (R^2) Since the two forces are equal, (q^2) / (R^2) = (GM_eM_i) / (R^2) q = \surd(GM_eM_i)

Question:

A small object of weight w^\ding{217} hangs from a string of length l. as shown in the figure. A variable horiz-ontal force P^\ding{217}, which starts at zero and gradually increases, is used to pull the object very slowly (so that equilibrium exists at all times) until the string makes an angle \texttheta with the vertical. Calculate the work of the force P^\ding{217}.

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Solution:

The object is in equilibrium, meaning that its acceleration is zero and the net force acting on the weight is zero. Consider the forces acting on the object, as shown in the diagram. We can say \sumF_x = 0 = P - T sin \texttheta(1) P = T sin \texttheta(2) and\sumF_y- = 0 = T cos \texttheta - W(3) W = T cos \texttheta(4) Dividing eq. (2) by eq. (4), we get P = W tan \texttheta Since P is variable, the work done by it must be found through integration. Recall that work is the integral of the dot product of the force F^\ding{217} and the displacement vector dx^\ding{217}: W = ^x2\int_x1 F^\ding{217} \bullet dx^\ding{217} = ^x2\int_x1 F cos \Upsilon dx where \Upsilon is the angle between F^\ding{217} and dx^\ding{217}. In this case, the force is P, the differential displacement is ld\texttheta and the angle between the two is \texttheta. Substituting these expressions, we have W = ^\texttheta\int_0 P^\ding{217} \bullet ld\texttheta^\ding{217} = ^\texttheta\int_0 (\omega tan \texttheta)(cos \texttheta)(l) d\texttheta = \omegal^\texttheta\int_0 sin \texttheta d\texttheta = - \omegal cos \texttheta\vert^\texttheta_0 = \omegal(1 - cos \texttheta) This result can also be derived using conservation of energy. Since the object's initial and final velocity is zero, kinetic energy is not involved. The change in the object's potential energy must be due completely to the work done on the weight by the force P^\ding{217}. This change in potential energy, ∆PE, is ∆PE = Wh = W(l - x) Butx = l cos \texttheta Therefore, we have ∆PE = \omega(l - l cos \texttheta) = \omegal(1 - cos \texttheta) This is equal to the work: W = \omegal (1 - cos \texttheta)

Question:

A particle of rest mass M moving with speedv_ocollides with and sticks to a stationary particle of rest massm. What is the speed of the composite system after collision?

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Solution:

In fact, we are asked to find the speed ofthe center of mass of the system, since it coin-cides with the composite particle after the collision. The velocity, momentum and total energy of a relativ-istic particle are related to each other as v^\ding{217} / c = (p^\ding{217} c) / E Therefore, we can define the center of mass of a composite system as v^\ding{217}_cm/ c= (\sum_ip_i^\ding{217} c) / (\sum_iE_i) Since there are no forces acting on the system, the velocity of the center of mass does not change during the collision. Before the collision, we have p^\ding{217}_m= 0 ,E_m= mc2 p^\ding{217}_M = M \Upsilon_0 v^\ding{217}_0,E_M = MC^2 \Upsilon0 , where \Upsilon_0 = 1 /\surd{1 ─ (v_0^2 / c^2)} . Hence v^\ding{217}_cm/ c = (p^\ding{217}_Mc) / (mc2+ MC^2 \Upsilon_0 ) v_cm_ / c= (M\Upsilon_0v_0)/ [(m + M\Upsilon_0)c՛] v_cm_ = (M\Upsilon_0)/ [(m + M\Upsilon_0) v_0].

Question:

The heart is removed from the body and placed in an isosmotic solution. Although it is completely separated from nerves, it continues to beat. Explain.

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Solution:

The initiation of the heartbeat and the beat itself are intrinsic properties of the heart and are not dependent upon stimulation from the central nervous system. The heart is stimulated by two sets of nerves (the sympathetic and vagus nerves), but these only partly regulate the rate of the beat, and are not responsible for the beat itself. The initiation of the heartbeat originates from a small strip of specialized muscle in the wall of the right atrium called the sino-atrial (S-A) node, which is also called the pacemaker of the heart. It is the S-A node which generates the rhythmic self-excitatory impulse, causing a wave of con-traction across the walls of the atria. (See figure). This wave of contraction reaches a second mass of nodal muscle called the atrio-ventricular node, or A-V node. The atrio-ventricular node is found in the lower part of the interatrial septum. As the wave of contraction reaches the A-V node, the wave stimulates the node, which produces an excitatory impulse. This impulse is rapidly transmitted to a bundle of nodal fibers called the A-V bundle or bundle of His. This bundle divides into right and left bundle bunches which ramify to the ventricular myocarium (the muscular layer of the heart wall) via the Purkinje fibers. The impulse is then transmitted to all parts of the ventricles causing them to contract as a unit. As the cardiac impulse travels from the S-A node through the atria, the wave of contraction forces blood through the valves and into the ventricles. (In the atria, the impulse is transmit-ted from cell to cell across the intercalated discs) . It is a "wave" of contraction because parts of the atria farthest from the S-A node contract later than do the parts closest to the node. The parts of the atria that contract first are also the first to relax, thus causing a wavelike motion upon contraction. Once the impulse reaches the A-V node, it is delayed (by a fraction of a second) before it is passed on to the ventricles. This transmission delay allows time for the atria to pump the blood into the ventricles before ventricular contraction begins. Unlike the atria, the ventricles contract as a unit. This is due to the rapid trans-mission of the impulse by the Purkinje fibers. For example, in the atria, the impulse travels at a velocity of .3 meter per second while in the Purkinje fibers, the impulse travels at about 2 meters per second. This allows almost immediate transmission of the impulse throughout the ventricular system. The adaptive significance of the contraction of the ventricles as a unit rather than as a wave is clear. The ventricles must exert as great a pressure as possible in order to force the blood throughout a long system of arteries, capillaries and veins. A greater pressure is more easily obtained if the ventricles con-tract as a unit, rather than in waves.

Question:

A resistor R = 10 megohms is connected in series with a capacitor 1 \muf. What is the time constant and half-life of this circuit?

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Solution:

Suppose that we introduce an e.m.f. in the circuit above, for example by connecting a battery to the circuit. (See the figure). In this case a current I will flow in the loop until C is charged up to the battery voltage V. Therefore, by Kirchoff's Voltage Law, V_R and I approach zero as V_C increases. The same current goes through R and C in the loop as we close the switch S; I = (V_R /R) = [(dQ)/(dt)] = (d/dt) (C V_C ) = C [(dV_C )/(dt)] where Q is the charge on the capacitor. As a result of Kirchoff's second rule, V_C + V_R = V and we obtain RC [(dV_C )/(dt)] + (V_C - V) = 0 or[(dV_C )/(dt)] + (1/RC) (V_C - V) = 0 The general solution of this differential equation is V_C (t) = - A e^-t/\cyrchar\cyrt + V(1) where \cyrchar\cyrt = RC is the time constant of the circuit and A is the integration constant. At t = 0, V_C = 0 because the voltage is initially only across R. Thus, substituting (1), A = V, and we get V_C (t) = V (1 - e^-t/\cyrchar\cyrt ). Note that as t becomes very large V_C (t = \infty) = V. The current I is I (t) = C [(dV_C )/(dt)] = C [(d)/dt)] [V (1 - e^-t/\cyrchar\cyrt )] I (t) = [(CV)/(\cyrchar\cyrt)] e^-t/\cyrchar\cyrt = (V/R) e^-t/\cyrchar\cyrt For the circuit under consideration \cyrchar\cyrt = RC = 10 × 10^6 ohms × 10^-6 f = 10 sec. The half-life is the time required for the decay factor (e^-t/\cyrchar\cyrt ) to be (1/2); or e-[{(t)h}/\cyrchar\cyrt] = (1/2) t_h = \cyrchar\cyrtln 2 t_h = 10 sec × ln 2 = 6.9 sec. On the other hand, if R = 10 ohms the time constant is only 10 × 10^-6 sec, or 10 \musec.

Question:

A tomato weighing 60.0 g is analyzed for sulfur content by digestion in concentrated HNO_3 followed by precipitation with Ba to form BaSO_4. 0.466 g of BaSO_4 is obtained. If all of the sulfur is converted to BaSO_4, what is the percentage of sulfur in the tomato?

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Solution:

The percentage of sulfur in the tomato is the weight of sulfur in the tomato divided by the total weight of the tomato multiplied by a factor of 100. To find the percentage, the weight of sulfur must be calculated. This is done by calculating the amount of sulfur precipitated out or, in other words, the amount of sulfur present in 0.466 g of BaSO_4. Using the Law of Definite Proportions, 32 g (1mole) of sulfur are consumed to form 233 g (1mole) of BaSO_4. Thus, once the percentage by weight of S in BaSO_4, is determined, then the weight of sulfur can be calculated. The fraction of S in BaSO_4 is [MW of S / MW of BaSO_4]= [(32 g / mole) / (233 g / mole)] = 0.14 The weight of sulfur in 0.466 g of BaSO_4 is thus, (0.14) (0.466 g)= 0.065 a. Now, that the weight of sulfur is known, the percentage of sulfur in the tomato is, [0.065 g of S / 60.0 g of tomato] [0.065 g of S / 60.0 g of tomato] × 100 = 0.11%.

Question:

An automobile tire whose volume is 1500 in^3. is found to have a pressure of 20.0 lb/in^2. When read on the tire gage. How much air (at standard pressure) must be forced in to bring the pressure to 35.0 lb/in^2. ?

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Solution:

The 1500 in^3. of the air at 20 lb/in^2. is compressed into a smaller volume at 35.0 lb/in.^2 P_1V_1 = P_2V2 We must remember to add, 14.7 lb/in^2. , the atmospheric pres-sure to the value read on the tire gauge. P_1 = 20.0 lb/in.^2 + 14.7 lb/in.^2 = 34.7 lb/in.^2 P_2 = 35.0 lb/in.^2 + 14.7 lb/in.^2 = 49.7 lb/in.^2 (34.7 lb/in.^2)(l500 in.^3) = (49.7 lb/in.^2)v_2 V_2 = 1050 in.^3 The volume of air added to the tire is 1500 in.^3 - 1050 in.^3 = 450 in.^3 when its gage pressure is 35.0 lb/in^2 . The volume at atmospheric pressure will be found from Boyle's law, 14.7 lb/in.^2 × V = 49.7 lb/in.^2 × 450 in.^3 V= (49.7/14.7) 450 in.^3 = 1500 in.^3.

Question:

Assuming theK_spfor radium sulfate is 4 × 10-^11 , what is its solubility in (a) pure water, and (b) ,1M Na_2 SO_4 ?

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Solution:

Whenever an ionic solid is placed in water, an equilibrium is established between its ions and the excess solid phase. A solu-bility constant,K_sp, measures this equilibrium. The concentration of solid is always constant, no matter how much it is in contact with the ions. This means the solid phase will not appear in the equili-brium constant expression. The dissociation is RaSO_4 (s) \rightleftarrows Ra^2+ + SO^2-_4 . TheK_spis, thus, [Ra^2+] [SO^2-_4 ] = 4 × 10-^11 . From the chemical reaction, it becomes evident that the concentration of Ra^2+ must equal that of SO^2- _4 , since they are generated inequimolaramounts. Therefore, let x = the solubility of each. Substituting into [Ra^2+] [SO^2-_4 ] = 4 × 10-^11 , you have x \bullet x = 4 × 10-^11 . Solving for x, you obtain x = 6 × 10-^6 mol/liter. Thus, the solubility of RaSO_4is 6 × 10-^6 mol/liter, the solution is 6 × 10-^6 M Ra^2+ and 6 × 10-^6 SO^2-_4 in water. b) The equation for this reaction in water is RaSO_4 \rightleftarrows Ra^2+ + SO^2-_4 . Na_2 SO_4 is a salt and therefore will ionize completely in water. The concentration of the Na_2 SO_4 solution is .1M, which is equivalent to a solution of .2M of Na+ and .1M SO-_4 . If one adds more SO-_4 to the system, the equilibrium is forced to the left. Therefore, there will be even less Ra^2+ present than in pure water. Thus, [Ra^2+ ] << 6.0 × 10-^6 M when additional SO-_4 is present. Let y = concentration of Ra+. The solubility of RaSO_4 is equal to the product of [Ra+ ] and [SO-_4 ] in this solution. Solving: [Ra+] is equal to y. Because RaSO_4 dissociates into Ra++ and SO-_4 , the RaSO_4 in this con-tributes y to [SO-_4] . There is also SO-_4 already present in this solution, its concentration is .1M, therefore [SO-_4 ] = (y + .1). Writing the equation for the solubility solubility of RaSO_4 = [Ra++] [SO-_4] = y(y + .1). It has been shown that y is much smaller than .1 and the following approximation can be made, y + .10 \cong .1 . Solving for . y: y(y + .1) = 4 × 10-^11 .1y = 4 × 10-^11 y = 4 × 10-^10 .

Question:

Allow exactly 100 ml of 1.5 NNaOHsolution to be mixed with 100 ml 3.0 N H_3PO_4 solution and allow them to reach equilibrium. (1) Determine what species will be present at equilibrium. (2) Find the pH of the solution. k_2 = 6.2 × 10^-8.

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Solution:

To solve this problem, first consider the type of reaction, if any, that occurs and to what extent it has proceeded. Once this is done, the answers to (1) and (2) follow directly. One is adding a base of a given amount to an acid of a given amount. This is a neutralization; i.e., the reaction of an acid and base to produce a salt and water. One is told there are 100 ml of 1.5 NNaOHand 100 ml of 3.0 N H_3PO_4 solution. N = normality, which is defined as equivalents/liter. An equivalent = the number ofhydroniumions that an acid can produce or hydroxide ions that a base can produce per mole of that substance. In a neutral solution the number of equivalents of base is equal to the number of equivalents of acid. Thus, to see to what extent the reaction has proceeded, calculate the equivalents of base and acid. If they are equal, the re-action went to completion. Recall, N = equivalents/liter. Thus, equivalentsNaOH = (1.5 equiv/l)100ml/1000 m/l = 0.15 equiv. Equivalents H_3PO_4 = (3.0 equiv/l) 100ml/1000l = 0.30 equiv. The equivalents of acid do NOT equal those of the base. The H_3PO_4 has 0.30 - 0.15 = 0.15 equivalents that do not react. The rest has been neutralized by the base to form a salt and water. H_3PO_4 is a polyproticacid, which means it can donate 3 protons when a mole of it exists. Since only one-half of the equivalents are neutralized, only one-half of the protons were used. This means that the reaction H_3PO_4 + OH^- \rightleftarrows H_2PO_4^- + H_2O goes to completion, but H_2PO_4- + OH^- \rightleftarrows HPO^2_4^- + H_2O will only be half complete. With this information, (1) can be answered. One started with 1.5 N of 100 mlNaOHsolution, which indicates that 0.15 moles of Na^+ is present.NaOHcan react with only 1 mole ofhydronium ion. Thus, its equi-valents correspond to its moles. 0.30 equivalents of H_3PO_4 correspond to 0.10 moles, since each mole generates 3hydronium ions. In (0.1) H_3PO_4 + (0.15)OH^- \rightleftarrows (0.1)H_2PO_4^- + (0.1)H_2O + (.05)OH^- there is a total 0.1 moles of H_2PO_4^- generated, since H_3PO_4 is the limiting reagent with only 0.1 moles compared with OH^-'s 0.15 moles. This means that for the reaction; (0.1) H_2PO_4^- + (0.05)OH^- \rightleftarrows (0.05) HPO^2_4^- + (0.05)H_2O + (0.05)H_2PO_4^-, 0.05 moles of OH^- is left to react with the 0.1 moles of H_2PO_4^- generated previously. Thus, only 0.05molesof H_2PO_4^- is consumed in producing 0.05 moles of HPO^2_4^-. Thus, anunreactedamount of 0.05 moles of HPO_4^- is left. In summary, the species in solution are 0.05 moles each of H_2PO^2_4^- and HPO^2_4^- with small amounts of H_3O^+, H_3PO_4, PO^3_4^-, OH^- and 0.15 moles Na+. One can now calculate the pH of this. There exists some H_2PO_4^- in solution. This is the source ofhydroniumions, whose pH is measured by pH = - log [H_3O^+]. The reaction is H_2PO_4^- + H_2O \rightleftarrows H_3O^+ + HPO^2_4^-. The equilibrium constant, k_2 = {[H_3O^+] [HPO^2_4^-]} / [H_2PO_4^-] = 6.2 × 10^-8. One can calculate [HPO^2_4^-] and [H_2PO_4^-], since one calculated the mole amounts. The volume of the solution is 0.2 liters, and the concentration in the brackets, are in moles/liter. Substituting these values, 6.2 × 10^-8 = {[H_3O^+] (0.05/.2)} / (0.05/.2). Solving for [H_3O^+], one obtains [H_3O^+] = 6.2 × 10^-8 M. pH = - log [H_3O^+], Thus, pH = - log [6.2 × 10^-8] = 7.21.

Question:

The spherical shell of a Van de Graaff generator is to be charged to a potential of 10^6 V. Calculate the minimum radius the shell can have if the dielectric strength of air is 3 × 10^6 V\bulletm^-1.

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Solution:

The potential and electric intensity at the surface of a sphere of radius R are V = Q/4\pi\epsilon_0RandE = Q/4\pi\epsilon_0R^2 = V/R Note that these relations are the same as the formulae for the potential and electric field of a point charge at a distant R from the charge. Thus, the field due to a sphere of charge Q is the same as that due to a point charge of charge Q. Therefore, R = V/E. But the maximum acceptable value of E is 3 × 10^6 V\bulletm^-1 for, at any higher value of E, the air will break down, and arc discharges through the air will result. Hence, the maximum radius for the spherical shell is R = (10^6 V)/(3 × 10^6 V\bulletm^-1) = 1/3m.

Question:

Design a 5 × 32 decoder using four 3 × 8 decoders (with enable inputs) and one 2 × 4 decoder.

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Solution:

a decoder is a digital device with n input lines and 2^n output lines. A 2 × 4 decoder with inputs X and Y and outputs Z_0 Z_1 Z_2 Z_3 has the following truth table, shown in figure 1. Inputs XY Outputs Z_0Z_1Z_2Z_3 00 1000 01 0100 10 0010 11 0001 Fig. 1 Hence, for each of the four combinations of 1's and 0's there is one and only one output line that assumes the value of 1. Similarly, the truth table for a 3 × 8 de-coder with an enable input is shown in figure 2. Inputs enableWXY Outputs Z_0Z_1Z_2Z_3Z_4Z_5Z_6Z_7 0XXX 00000000 1000 10000000 1001 01000000 1010 00100000 1 011 00010000 1100 00001000 1101 00000100 1110 00000010 1111 00000001 Fig. 2 Note: X denotes "don't care" The 5 × 32 decoder is shown in fig. 3. The 2 × 4 decoder decodes bits A_3 and A_4 and ensures that only one of the 3 × 8 decoders is enabled. Hence bits A_3 and A_4 determine which of the 3 × 8 decoders to enable and bits A_0, A_1, and A_2 determine which output of the enabled decoder will become 1.

Question:

Explain why aldehydes but not ketones can be oxidized to carboxylic acids.

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Solution:

An examination of the structures of aldehydes, ketones, and carboxylic acids will reveal the solution to this problem. In answering this problem, keep in mind that a carbon atom can participate in only four bonds. Aldehydes and ketones have the carbonyl functional group; i.e., The aldehyde has the general formula where R is any carbon group. The ketone's formula is where R and R' are different or the same carbon groups. A carboxylic acid has the formula where R is any carbon group. Now, let us look at the oxidation: But, notice, in the oxidation from the ketone, the resulting compound has a carbon atom with 5 bonds, which is not permissible. Such a situation does not occur with oxidation in aldehydes. This explains why aldehydes, but not ketones,can be oxidized to carboxylic acid.

Question:

Let us estimate the gravitational energy of the Galaxy. We omit from the calculation the gravitational self-energy of the individual stars.

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Solution:

The gravitational energy of an arbitrary system of N parti-cles consists of the sum of the mutual potential energies of each pair of particles. Hence, U = (1/2) [(U_12 + U_13 + U_14 + ... U_1N) + (U_21 + U_23 + U_24 + ... + U_2N) + (U_31 + U_32 + U_34 +...+ U_3N) +...+ ( U_NI + U_N2 + U_N3 +...+ U_NN-1 )](1) The terms U_12, etc., represent the mutual potential energy of particles 1 and 2. By Including sets of terms such as U_12 and U_21 we have double counted, because these represent the mutual potential energies of particles 1 and 2, and particles 2 and 1, respectively. However, these 2 terms are the same. Hence, the factor (1/2) must be included in (1) to negate the process of double counting. Furthermore, terms such as U_11 are omitted because they represent the mutual potential energy of particle 1 with particle 1, i.e., they are self energies. We approximate the gross composition of the Galaxy by N stars, each of mass M, and with each pair of stars at a mutual separation of the order of R. From the definition of potential energy U_ij= (-Gm_im_i)/(r_ij) where m_i andm_jare the masses of particlesiand J, respectively, G = 6.67 × 10^-11 N\bulletm^2/kg^2, is their mutual separation. For our case r_ij= R andm_i =m_j= M for all pairs of particles. Then U_ij= (-GM^2)/R(2) for any 2 particles. Notice that in equation (1), the first parenthesis has N-1 terms of the type (2), the second parenthesis has N-1 terms of this type, and similarly for all Nparenthesises. Altogether, there are N(N-1) terms of type (2) in (1). Therefore, U = (1/2)(N)(N - 1)[(-GM^2)/R] U = -(1/2)[{N(N - 1)GM^2} / (R)] If N \approx 1.6 × 10^11, R \approx 10^21m, and M \approx 2 × 10^30 kg, then U \approx [-(1/2)(1.6 × 10^11)(1.6 × 10^11 - 1)(6.67 × 10^-11N\bulletm^2/kg^2) (2 × 10^30 kg)^2] / [10^21m] U \approx[ (-34.15 × 10^71)/(10^21)] J = -34.15 × 10^50J U \approx -3.42 × 10^51J

Question:

How many moles of hydrogen gas are present in a 50 liter steel cylinder if the pressure is 10 atmospheres and the temperature is 27\textdegreeC? R = .082 liter-atm/mole \textdegreeK.

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Solution:

In this problem, one is asked to find the number of moles of hydrogen gas present where the volume, pressure and temperature are given. This would indicate that the Ideal Gas Law should be used because this law relates these quantities to each other. The Ideal Gas Law can be stated: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (.082 liter-atm/mole \textdegreeK), and T is the absolute temperature. Here, one is given the temperature in \textdegreeC, which means it must be converted to the absolute scale. P, V, and R are also known. To convert a temperature in \textdegreeC to the absolute scale, add 273 to the temperature in \textdegreeC, T = 27 + 273 = 300\textdegreeK Using the Ideal Gas Law: PV = nRTorn = (PV/RT) P = 10 atm V = 50 liters R = .082 liter-atm/mole \textdegreeK T = 300\textdegreeK n = number of moles of H_2 present n = [{(10 atm) (50 liters)/{(.082 liter-atm/mole \textdegreeK) (300\textdegreeK)}] = 20 moles.

Question:

Show that the optical length of a light path, defined as the geometricallength times the refractive index of the medium inwhich the light is moving, is the equivalent distance which thelight would have traveled in a vacuum.

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Solution:

Suppose that light travels a distance l in a medium of refractive indexn. The optical length is then opticallength =nl and, since n = c/v opticallength =cl/v Here c and v are the speeds of light in vacuum and the medium, respectively. But light travels with constant velocity in the medium, and hence l/v = t, where t is the time taken to traverse the light path. nl= (cl/v) = ct = l_0 wherel_0 is the distance the light would have traveled at velocity c, that is, ina vacuum. Thus the optical length is the equivalent distance which the light would have traveled in the same time in a vacuum.

Question:

Explain structures in C. Give an example and show how they can be referenced.

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Solution:

Structures in C are created by astructstatement. It is similar to the typedef statement in the sense that it does not create a variable we can use but provides a format for future variable declarations. Suppose we want to create a structure of students in which we want to store their name, id and theirgpa. The following format is used. #define MAX_NAME 30 #define MAX_ID 10 #define NUMBER _ of _ STUDENTS 100 structstudents { char name [MAX_NAME]; char id [MAX_ID]; float gpa; } stud[NUMBER _ of _ STUDENTS]; In the above declaration stud is the variable created that will use the structure students. Students in this case are called a tag and each subfield within the structure i.e. name, id,gpa, etc., are called a member. We can also define the variableNYCstudof type students for example, structstudentsNYCstud; We can reference individual students in the following manner. stud [O].name; stud [0].id; stud [0].gpa; Thus if we want to printgpaof all students along with their names then for(i= 0;i< NUMBER _ of _ STUDENTS; ++i) printf("%s%1.1f\textbackslashn", stud [i].name, stud [i].gpa); prints the name andgpaof each student.

Question:

A chemist decides to prepare some chlorine gas by the following reaction: MnO_2 + 4HCl \rightarrow MnCl_2 + 2H_2 O + Cl_2 \uparrow If he uses 100 g of MnO_2, what is the maximum volume of chlorine gas that can be obtained at standard temperature and pressure (STP)?

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Solution:

The solution of this problem entails a combination of stoichiometry and gas law theory. The first thing to determine is how many moles of chlorine will be produced. The volume can be determined using the fact that each mole, at STP, occupies 22.4 l. Since the equation is balanced, you must look at the coefficients. The equation indicates that for every mole of MnO_2, 1 mole of chlorine gas is produced. A mole is defined as being equal to weights in grams of material/molecular weight of material. The molecular weight of MnO_2 is 86.9. It follows, then, that the number of moles of MnO_2 is 100/86.9. This will be equated to the number of moles of chlorine gas produced. At STP one mole of gas occupies 22.4 liters. However, we do not have 1 mole, but 100/86.9 moles. To calculate the new volume, you multiply the STP volume by the number of moles of chlorine gas produced. In other words, the volume of the chlorine gas is 22.4 × = [(100) / (86.9)] = 25.8 liters.

Question:

Water flow into a tank is regulated by the position of a valve. The position of the valve is controlled by the height of a float: as the float drops, the valve is turned on to admit the water, while the rising water level causes the float to rise, which gradually shuts off the valve, stopping the flow of water. Write a FORTRAN program to simulate this system, using the modified Euler's method, from time t = 0 to t = tf, if the volume at time t is V(t), V(0) = 0, and the capacity of the tank is C units of volume.

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Solution:

If V̇ is the flow rate, the behavior of the system can be diagrammed as shown in Fig. 2. The present water level is an accumulation of all past flow-rates. The next flow rate is determined by the present water level. Mathe-matically expressed: V̇ = f(V). But note that the flow rate is the greatest when the tank is empty and decreases to zero as the tank fills up. Stating this mathematically: V̇ = K(C - V(t))(1) where 0

Question:

A rifle weighing 7 pounds shoots a bullet weighing 1 ounce, giving the bullet a speed of 1120 feet per second. (a) If the rifle is free to move, what is its recoil speed? (b) If the rifle is held tight against the shoulder of a man weighing 133 pounds and if he were free to move, what would be the recoil speed of the rifle and man? (c) If the bullet imbeds itself in a block of wood weighing 3 pounds and 7 ounces and if the block were free to move, what would be the speed of the block plus bullet?

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Solution:

The law of conservation of momentum may be applied in an isolated system where no external forces are applied. (a) The momentum of the gun plus bullet before firing is zero, and it is therefore also zero after firing. The momentum after firing is M_bullet × vbullet+ M_gun × v_gun = 0(1) Since the law of conservation of momentum involves mass and not weight, we must convert weight into mass by dividing by the acceleration of gravity [M = {(Wt)/g} ].Therefore M_bullet = [(1/16) lb] / [32 ft/s^2] = .001953 slugs and M_gun = (7 lb) / (32 ft/s^2) = 7/32 slugs Equation (1) then becomes .001953 slugs × 1120 ft/sec + 7/32 slugs × v_gun = 0 Whence v_gun = -10 ft/sec, or 10 ft/sec backwards (b) The momentum after firing is M_bullet × v_bullet + M_gun + _man × v_gun _+ _man = 0or .001953 slugs × 1120 ft /sec + [(132/32) slugs + (7/32) slugs] × vgun + man= 0 whence vgun + man= - 0.5 ft/sec,or0.5 ft/sec backwards (c) The momentum of the bullet before the collision with the block is M_bullet × v_bullet = .001953 slugs × 1120 ft/sec × 32 ft/s^2 = 70 lb-ft/sec The momentum after collision is the same, 70 lb-ft/sec. Then 70 lb-ft/sec =(Mbullet + block) × (vbullet + block) = [ (3/32) slugs + (7/2) slugs + 2 slugs] × (v_bullet+block) whence (vbullet + block) = 20 ft/sec forwards.

Question:

An unbanked curve has a radius of 80.0 m. What is the maximum speed at which a car can make the turn if the coefficient of friction \mu_s is 0.81?

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Solution:

We assume that the car is travelling in a circular path (see fig.(a)) at a constant speed. However, its velocity is constantly changing in direction. Hence, the car is accelerating, and, therefore, a force must be acting on the car. This force accelerates the car towards the center of the circular path and is therefore centri-petal. Applying Newton's Second Law, F = ma, to the car, we obtain f = ma(1) where f is the frictional force and a is the acceleration of the car. But, in uniform circular motion a = v^2/R(2) where v is the speed of the car and R is the radius of the circle. Furthermore, the frictional force f is f \leq \musN(3) where N is the normal force exerted by the road on the car. (Note that if f = \mu_s N, the car will begin to slip relative to the road.) Combining (1), (2) and (3) mv^2/R = f \leq \musN mv^2/R \leq \mu_s N(4) Applying Newton's Second Law to the vertical direction of motion (see fig. (b)), we obtain N - mg = 0 because there is no acceleration of the car in this direction. Using this in (4) mv^2/R \leq \mu_s mg Solving for v v \leq \surd(\mu_s gR) v_max = \surd(\mu_s gR) v_max = \surd[(.81) (9.8 m/s^2) (80 m)] v_max = 25 m/s

Question:

(a) Explain what is meant by a paritycheck; (b) Encode the decimal digits 0-9 into their (i)7421code(ii) two-out-of-five code (iii)biquinarycode representations.

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Solution:

(a) A parity check is simply the technique of counting the number of0's or 1' s in a data word. In some systems, the result then determines thevalue assigned to an extra "parity bit", which is carried along as a check. (b) (i) Codes are sometimes identified by their positional values. The binarycode may be referred to as the 8421 code (since a 1 in the leftmost positionhas a value of 8, a 1 in the rightmost position has a value of 1, etc.) The 7421 code is identical to the 8421 code for the first seven digits, butthe high-order bit is used to represent decimal 7 rather than decimal 8 asin 8421 code. The codes are shown in fig. 1. Decimal 8421 7421 0 1 2 3 4 5 6 0000 0001 0010 0011 0100 0101 0110 0000 0001 0010 0011 0100 0101 0110 7 8 9 0111 1000 1001 1000 1001 1010 Note that no legal 7421 word has more than two1' s. Therefore, if a parity checkindicated that a received word had three or four 1' s, an error conditionwould besignalled. (ii) The two-out-of-five code can be obtained from the 7421 code by addinga fifth bit. This new bit, added in the least significant position, is calledthe paritybyt. If the 7421 code representation of a particular decimalinteger has a single 1 bit, the parity bit is 1. If the 7421 code for a decimalinteger has zero or two 1 bits, the parity bit is 0, The two-out-of- fivecode is shown in fig. 2. Decimal 7421 Two-out-of-five 0 0000 00000 1 0001 00011 2 0010 00101 3 0011 00110 4 0100 01001 5 0101 01010 6 0110 01100 7 1000 10001 8 1001 10010 9 1010 10100 Note that every legal two-out-of-five code word has a pair of 1's, except 00000. Thus, the presence of a single 1 bit or more than two 1 bits would signalan error. (iii) The problem with the two-out-of-five code is that there is no parity checkfor zero, since zero is represented by a code consisting only of zeros. If a digit is being transmitted and for some reason the 1's are suppressed, the result will be 00000, and the computer has no way of knowingif this is an error or the number is supposed to be zero. In the biquinarycode, zero is indicated as 01 00001. The value of each succeedinginteger is increased by 1 by moving the 1 left one column until decimal5 is reached. Decimal 5isrepresented as 1000001. The values of decimals6-9 are obtained by moving the 1 to the left, again. Thebiquinary codeis shown in fig. 3. Decimal Biquinary 0 0100001 1 0100010 2 0100100 3 0101000 4 0110000 5 1000001 6 1000010 7 1000100 8 1001000 9 1010000 Every legalbiquinarycode word must contain a pair of 1's. Thus, the presenceof a single 1 bit or more than two 1 bits would signal an error.

Question:

In an experiment, x-rays of frequency f are scattered by the electrons in a block of paraffin. It is found that the x-rays scat-tered at an angle of less than 900, have a wavelength f ', greater than f. Interpret this result in terms of a collision of a photon with an electron.

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Solution:

The collision is shown in the diagram. The photon is scat-tered through an angle \texttheta, and the electron through an angle \varphi. The momentum of the photon is given by the de Broglie relationship. p = h/\lambda = hf/cusing \lambdaf = c As in all collision problems, the law of conservation of momentum may be applied to this collision. The momentum of the photon-electron system, then must remain constant throughout the collision. Since momentum is a vector quantity, (having both direction and mag-nitude) we must resolve the momenta of the electron and photon into x and y components. Conservation of momentum can then be applied as follows: hf/c = (hf '/c)cos \texttheta + mv cos \varphi (x component)(1) (hf '/c)sin \texttheta = mv sin \varphi (y component)(2) The law of conservation of energy can also be applied to the collision analysis because the collision is elastic. The total energy of the electron- photon system must then remain constant throughout the col-lision. Or hf + m_0c^2 = hf ' + mc^2(3) where m_0 is the rest mass of the electron. And m is the new relativistic mass [= m_0 / \surd{1 - (v^2/c^2)}] It incorporates the kinetic energy the electron has, after the col-lision. The angle \varphi can be eliminated by squaring (1) and (2) and adding. Then [(hf/c) - (hf '/c) cos \texttheta]^2 = m^2v^2 cos^2 \varphi [(h^2f '^2)/c^2] sin^2 \texttheta = m^2v^2 sin^2 \varphi h^2/c^2(f^2 + f '^2 - 2ff ' cos \texttheta) = m^2v^2(cos^2\varphi + sin^2\varphi) orh^2(f^2 + f '^2 - 2ff ' cos \texttheta) = m^2v^2c^2(4) forcos^2 \varphi + sin^2 \varphi = 1 Next the energy equation is squared to obtain [h(f - f ') + m_0c^2]^2 = m^2c^4(5) Consider now the quantity m^2c^4 - m^2v^2c^2 = c^2m^2(c^2 - v^2)(6) We know that m^2 = m2_0 / {1 - (v^2/c^2)} Thus m^2(c^2 - v^2) = m2_0c^2 and (6) simplifies to m^2 _0c^4. Therefore by subtracting (4) from (5) we obtain [h(f - f') + m_0c^2]^2 - h^2(f^2 + f'^2 - 2ff' cos \texttheta) = m^2c^2(c^2 - v^2) = m2_0c^4. Simplifying this equation gives the result 2hm_0c^2(f - f') - 2h^2ff'(1 - cos \texttheta) = 0 or 1 - cos \texttheta = [(m_0c^2)/h] [(f - f')/ff'](7) Using the relationship \lambdaf = c, equation (7) is reduced to 1 - cos \texttheta = (m_0c / h) (\lambda' - \lambda)(8) This process of scattering of photons is known as the Compton effect, and is a clear indication of an instance in which a wave motion be-haves like a particle. Notice that the change in wavelength depends only on the angle of scattering \texttheta and is independent of the wave-length itself. The quantity \lambda_0 = h / m_0c =0.0242\AA is known as the Compton wavelength.

Question:

10 ml of 0.200 M HC_2H_3O_2 is added gradually to 25 ml of 0.200 M NaOH. Calculate the pH of the initial 0.200 M NaOH solution and after each successive addition of 5 ml of acid. K_w = 10^-14.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0434.htm

Solution:

This is a titration problem. One is asked to find the pH of the 0.200 M NaOH at different concentration levels. First one wants to calculate the pH before any acid is added. To do this, consider the fact that the equilibrium constant for the autodissociation of water, K_w ,is defined as K_w = [H_3O^+] [OH^-]. Since, pH = - log [H_3O^+], one needs to calculate [H_3O^+], which can be determined from K_w,once [OH^-] is known. One knows [OH^-] from the fact that one has a 0.2 M NaOH solution. M = molarity = moles/liter. In solution, NaOH dissociates according to NaOH \rightleftarrows Na^+ + OH^-. Thus, [OH^-] = [NaOH]. Since [NaOH] = 0.2, [OH^-] = 0.2, which allows for substitution into the equilibrium expression, K_w = 1.0 × 10^-14 = [H_3O^+] [0.2], or [H_3O+] = [(1.0 × 10^-14) / 0.2] = 5.0 × 10^-14, which means pH = - log [5.0 × 10^-14] = 13.3. What is the pH when 5 ml of acid (HC_2H_3O_2) is added? To answer this, note that a neutralizing reaction occurs, (i.e. a reaction between an acid and base to produce a salt and water) when the acid is added to the NaOH solution. HC_2H_3O_2 + NaOH \rightarrow HOH + NaC_2H_3O_2 . It is necessary to find how much base is consumed, since this will reflect the remaining [OH^-]. From this, return to the expression for K_w and substitute this new value for [OH^-] to determine [H_3O^+] from which the pH can be calculated. To find the amount of NaOH that was consumed, use Molarity = moles/liter. If the acid is 0.200 M and one adds in 5 ml or 0.005 l (1000 ml =1 l), one has (0.2) (0.005) = 0.001 moles of acid. Amount of base present in moles equals: molarity × volume = (0.2) (0.025) = (0.005) moles. According to the reaction, they will react in equi-molar amounts. Thus, since only 0.001 moles of acid is present, only 0.001 moles of base react. This leaves 0.004 moles of base (NaOH), and as such 0.004 moles of OH^-. Since volume = base (l) + acid (l) = 0.03, M = 0.004 / 0.03 = 0.13 M. Recalling, K_w = 1.0 × 10^-14 = [H_3O^+] [OH^-] and that [OH^-] = 0.13, such that [H_3O^+] = (1.0 × 10^-14) / 0.13 = 7.5 × 10^-14. Thus, pH = - log [7.5 × 10^-14] = 13.13. If one adds another 5 ml of acid, the total acid added is 10 ml. The same procedure is followed, except that volume of acid = 10 ml or 0.01 l. Molarity of acid = 0.2, so that (0.2) (0.01) = 0.002 moles of acid are present. From the previous calculation, it can be seen that 0.005 - 0.002 = 0.003 moles of OH^- are now left in a volume of 0.025 l from base + 0.010 l from acid or 0.035 liters. Thus, its molarity is 0.003 / 0.035 = 0.086 M. K_w = 1.0 × 10^-14 = [H_3O^+] [OH^-] = [H_3O^+] [0.086]. Solving, [H_3O^+] = 1.17 × 10^-13.This means, pH = - log [H_3O^+] = - log [1.17 × 10^-13] = 12.9.

Question:

What is meant by the term "renal threshold"?"renal clearance"?

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0447.htm

Solution:

Although glucose is present in theglomerularfiltrate, there is normally little or no glucose in the urine, due to itsreabsorptionby the cells of the renal tubules. If the plasma level of glucose were increased drastically, the level of glucose in theglomerularfiltrate would likewise increase. In this situation, not all the glucose could be reabsorbed as the filtrate passed through the kidney tubules, and some glucose would appear in the urine. The concentration in the plasma of a sub-stance such as glucose at the point where it just begins to appear in the urine is referred to as the "renal threshold" of the substance. The renal threshold for glucose is about 150 mgs. of glucose per 100mls. of blood. When the concentration of glucose in the blood exceeds this level, glucose will begin to appear in the urine. There are comparable thresholds for many other substances. Physiologists have adopted the concept of "renal clearance" to express quantitatively the kidneys' ability to eliminate a given substance from the blood. The elimination of a substance is dependent on its concentration in the blood, the filtration rate at theglomerulus(or volume of blood filtered per minute), and the rate of secretion orreabsorptionof that substance by the renal tubules. The renal clearance relates the rate of appearance of a substance in the urine to its concen-tration in the blood, and is expressed as the volume of plasma cleared of that substance per minute. One can also think of renal clearance as the volume of plasma that would contain the amount of a substance excreted in the urine per minute. The renal clearance of a substance x is defined as C_X=[(mgs. of substance × secreted in urine per minute) / (mgs. of substance × per ml. of plasma)] Thus, if x is present in the plasma in a concentration of .02 mg/ml, but secreted at a rate of .2 mg/min, its clearance would be CX= (.2 mg x/min.) / (.02 mg x/ml plasma) = 10mlsplasma/min. In other words, 10mlsof plasma are completely cleared of substance x per minute, thus .2 mgs. of that substance appears in the urine per minute.

Question:

You are in a room with 29 other people. You are asked to write a program which computes the probability of at least two people having the samebirthdate.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0472.htm

Solution:

Denote the requested probability by Q. Then Q = 1 - P where P is the probability of no two identical birthdates. From elementary probability theory P = (365/365) × [(365-1)/365] ×...× [(365-29)/365] = (365 × 364 × 363 × ... × 336)/365^30 Here, the probability is 365/365 that one person will not have two identical birthdates. Given two people, the probability is 364/365 since the first person has 365 choices and the second only has 364. Continuing, we obtain P. The program appears as follows : 1\O LET P = 1 2\O FOR D = 365 TO 336 STEP -1 3\O LET P = P \textasteriskcentered D/365 4\O NEXT D 5\O LET Q = 1 - P 6\O PRINT "THE PROBABILITY OF TWO OR MORE" 7\O PRINT "IDENTICAL BIRTHDATES AMONG 30 PEOPLE IS"; Q 8\O END

Question:

The double-slit experiment can be simulated using two small blocks of wood oscillating up and down together in a pool of water as shown in the figure. Suppose that the blocks are 3 cm apart and the wavelength of the water waves is 1 cm. Locate the points 10 cm from one of the blocks where constructive interference occurs.

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/Users/wenhuchen/Documents/Crawler/Physics/D29-0894.htm

Solution:

The motion of the blocks produce circular waves which are emanating from two sources S_1 and S_2 separated by d = 3 cm. In order to observe the interference pattern that forms at 10 cm from one of the blocks, draw a circle of radius 10 cm around source S_2 as shown in the figure. Constructive interference occurs at points where the dif-ference in the distances to the two blocks is either zero or some whole number of wavelengths. Let A be a point on the circle at a distanceL_1 from S_1. All the points on the circle are L_2 = 10 cm from S_2. Constructive interference occurs at points for which the length \vert L_1 - L_2 \vert is an integer multiple of the wavelength \lambda = 1 cm. Therefore, we have constructive interference at points on the circle that are L_1 = 10 cm - 3 cm = 7 cm = 10 cm - 2 cm = 8 cm = 10 cm - 1 cm = 9 cm = 10cm = 10 cm + 1 cm = 11 cm = 10 cm + 2 cm = 12 cm = 10 cm + 3 cm = 13 cm from S_1.

Question:

Describe the structure of an insect wing. How are wings used in flight?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0490.htm

Solution:

The wings of an insect are evaginations or folds of the integument, and are composed of two sheets of cuticle. A vein runs through the wing at a point where the two cuticular membranes are thickened and separated, forming an effective supporting skeletal rod for the wing. The wings of the more primitive insects are netlike, but there has been a general tendency in the evolution of wings toward reduction of this netlike appearance. Each wing articulates with the edge of the dorsal cuticle, called the tergum, but its inner end rests on a dorsal pleural process (the pleura is the cuticle covering the side of the body), which acts as a fulcrum. The wing is thus somewhat analogous to an off-centered seesaw. Upward movement of the wings results indirectly from the contraction of vertical muscles within the thorax, depressing the tergum (see Figure A). Downward movement of the wings is produced either directly, by contraction of muscles attached to the wingbase, or indirectly, by the contraction of transverse horizontal muscles raising the tergum. Insects such as dragonflies and roaches exhi-bit direct contraction, while bees, wasps, and flies show indirect contraction. Downward movement can come about by both direct and indirect muscles in insects such as grasshoppers and beetles. The raising or lowering of the wings involves the alternate contraction of antago-nistic muscles. Up and down movement alone is not sufficient for flight; the wings must also move forward and backward. A complete cycle of a single wing beat describes an ellipse in grasshoppers, and a figure eight in bees and flies; the wings are held at different angles to provide both lift and forward thrust. The wing beat frequency varies from 4 beats per second in certain gnats. The fastest flying insects are the moths and horse flies, which can fly over 33 miles per hour. Gliding, an important form of flight in birds, occurs in only a few large insects. Insect flight muscles are very powerful. The fibrils in the muscle cells are relatively large and the mitochond-ria are huge. Insects are the only poikilothermic (cold-blooded) fliers; their low body temperature and corres-pondingly low metabolic rate impose limitations on mobility. On a cold day some butterflies are known to literally "warm up" before flight. They remain stationary on a tree trunk or some other location, and move their wings up and down until sufficient internal heat is generated to permit the stroke rate necessary for flight.

Question:

Using APL, simulate the game of craps.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0555.htm

Solution:

First of all, let us describe how the game of craps is played: A player rolls two dice. If the sum of the two faces on the first throw is 7 or 11, he wins the amount bet. If the sum turns up to be a 4, 5, 6, 8, 9 or 10, the player continues to roll until he obtains (1) a 7 in which case he loses, or (2) the sum he obtained on the first throw; in this case he wins. Our APL program for craps is: \nablaCRAPS [1]A \leftarrow 10 [2]AGAIN:A \leftarrow A, B \leftarrow +/?2\rho6 [3]\rightarrow( ( (1 =\rhoA)\wedge\simB \epsilon 7 11 2 3 12) \vee (1 \not =\rhoA) \sim \wedge\simB \epsilon (A [1] , 7) ) /AGAIN \sim \sim [4]\rightarrow( ( (1 =\rhoA) \wedge B \epsilon 7 11) \vee (1 \not =\rhoA) \wedge B =A [1] ) /WIN [5]`YOU LOSE WITH' ; A [6]\rightarrow0 [7]WIN:`YOU WIN WITH' ; A [8]\nabla In step [1] weintializethe vector A to be the null vec-tor. In [2] we find the sum of two random numbers. The random numbers are positive integers between 1 and 6. Wecatenateall sums in the vector A. Step [3] checks two things: (1) whether it was the first roll and if the roll automatically gave you a win (7 or 11) or a loss (2, 3, 12), (2) whether you rolled more than once, and if the last roll gave you a 7 (loss) or the same sum as the one obtained on the first throw (win). If you do not win or lose in step [3] you must roll again (go to step 2). In step 4: If you won, you are sent to step 7, where it is printed out that you win and vector A is also printed out. If you do not win, step [5] will indicate that you are a loser.

Question:

At a counting rate of 5 atoms per second, how long would it take to count the atoms in a spherical droplet of water 0.1 take to count the atoms in a spherical droplet of water 0.1 cm in diameter? cm in diameter?

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0919.htm

Solution:

We can calculate the number of atoms in the water droplet from the density of water and the gram molecular weight of a water molecule. The density of water is defined as its mass per unit volume, or \rho = m/v Hence,m =\rhov.(1) Now, the volume of water in the spherical droplet is the volume of a sphere of radius .05 cm. The volume of a sphere of radius r = 4/3 \pir^3. Volume of droplet = 4/3 \pi(0.05)^3 =5.24 × 10^-4 cm^3 The density of water is 1 gm per cubic centimeter, and from (1), the mass of water in the droplet is m = (1 gm/cm^3) (5.24 × 10^-4 cm^3) m = 5.24 × 10^-4 gm We may now find the number of molecules in this mass by dividing it by the mass of one molecule. A molecule of water contains two atoms of _1H^1 and one atom of _8o^16 and its molecular mass is therefore (2 × 1.0078amu) + 15.995amu= 18amu But 1amu1.66 × 10^-24gms. Therefore, a water molecule has a mass, in grams, of (18) (1.66 × 10^-24gms) = 29.88 × 10^-24gms The total number of molecules in 5.24 × 10^-24gmsis then n = [{5.24 × 10^-4 gm} / {2.99 × 10^-23 (gm/molec)]. = 1.75 × 10^19 molecules to sufficient accuracy. Since each molecule contains 3 atoms, the number of atoms in the water droplet is 3n = [(1.75 × 10^-19 molecules) (3 atoms/molecules)] 3n = 5.25 × 10^19 atoms At a rate of 5 per second the time taken to count these atoms would be [{5.25 × 10^19 atoms} / {5 (atoms/sec)}] = 1.05 × 10^19 sec There are approximately 3.16 x 10^7 sec in a year, so the time taken would be [{1.05 × 10^19 sec} / {3.16 × 10^7(sec/yr)}] = 3.3 × 10^11 years.

Question:

Given 0.207 nm for the S-S distance and 105\textdegree for the S-S-S bond angle, calculate the S-to-S distance across the S_8 ring.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0636.htm

Solution:

One is asked to find the distance from S_1 to S_2 as seen in figure A. This can be done by using trigo-nometry. Draw a line perpendicular to S_1S2 to A from S_0. One can solve for the length of S_2A with the knowledge that the angle S_2- S_0-A is equal to 1/2 of the S-S-S angle or 52.5\textdegree. This is true because the triangle S_0S_1S_2 is isosceles. sin 52.5o= (S_2A / S_2S_0) = 0.793 S_2A= 0.793 × S_2S_0 = (0.207) × (0.207 nm) = 0.164 nm. S_2S_1= 2 × S_2A = 2 × 0.164 nm = 0.328 nm.

Question:

Find the theoretical speed of sound in hydrogen at 0\textdegreeC. For a diatomic gas \Elzpbgam = 1.40, and for hydrogen M = 2.016 gm/mole.

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0825.htm

Solution:

Sound waves are longitudinal mechanical waves. That is, they are propagated in matter and the particles transmitting the wave oscillate in the di-rection of propagation of the wave. The speed of the wave is determined by the elastic and inertial pro-perties of the medium. For a gaseous medium, the elastic property depends on the undisturbed gas pressure p_0 and the inertial property onl_o, the density/mole of the gas. With \Elzpbgam a constant called the ratio of specific heats for the gas, the velocity is v = \surd[(\Elzpbgamp_o)/(p_o)] Since for one mole of gas, p_0 = RT/V = (RTp_0)/Mwe have v = \surd[(\ElzpbgamRT)/M] = \surd([1.40{8.317 joules/(mole K0)} (2730K)] / {2.016 × 10^-3 kg/mole}) = \surd[{1.40 × 8.317 × 273 joules} / {2.016 × 10^-3 kg}] = 1.25 × 10^3 m/sec

Question:

A capillary tube of internal radius 0.25 mm is dipped into water of surface tension 72 dynes \textbullet cm^-1. How high does the water rise in the tube? The capillary tube is gradually lowered into the water until only 1 cm is left above the surface. Explain what happens to the water in the tube.

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0422.htm

Solution:

Consider a capillary tube of radius r. The liquid in it makes contact with the tube along a line of length 2\pir. Let the liquid in the tube be a height y above the surface of the liquid in which it is dipped. The up-ward force, T, is defined as the product of the surface tension, \Upsilon, and the length perpendicular to the force, along which it acts. Then T = (\Upsilon) (2\pir) (cos \texttheta) where \texttheta is the contact angle. The downward force on the liquid in the tube is equal to its weight w. Then w equals the liquid's weight density, \rhog, multiplied by its volume \pir^2y. w = \rhog\pir^2y For the liquid in the tube to be in equilibrium, these forces must be equal. w = T \rhog\pir^2y = \Upsilon^2\pir cos \texttheta Theny =2\Upsilon cos \texttheta/ \rhogr. Since the liquid in this case is water, the angle of contact is 0\textdegree, and the density is 1 g\bulletm^-3. Hence y = (2 × 72 dynes \bullet cm^-1)/(1 g \textbullet cm^-1 × 980 cm \textbullet s^-2 × 0.025 cm) = 5.88 cm. As long as more than 5.88 cm of tube shows above the liquid surface, there is no problem. The liquid rises to that height. But, as the tube is lowered, a stage will be reached when less than 5.88 cm are above the surface. What cannot happen is that liquid pour out over the top. If it did, the liquid pouring over the edge could be used to drive a water wheel to provide energy; and the process would continue as liquid would always rise up the tube to take the place of that pouring from the end. In other words, a perpetual motion machine would be establish-ed, which is in direct contradiction with the principle of conservation of energy. What does happen is that the angle of contact at the top of the tube increases. Only the vertical component of the surface tension is used to balance the weight of the water column held up. As the height of the projecting tube gets smaller and smaller, the angle of contact gets larger and larger until, with y = 0, \texttheta = 90\textdegree, and the surface at the top of the tube is flat. In particular, when y = 1 cm, 1 cm = (2\Upsilon cos \texttheta)/(\rhogr) = 5.88 cos \texttheta cm. \thereforecos \texttheta = 1/5.88 = 0.17\therefore \texttheta = 80.2\textdegree.

Question:

Natural boron is made up of 20 per cent^10B and 80 per cent 11 B. What is the approximate atomic weight of natural boron on the scale of 16 units for ^16o (chemical scale)?

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Solution:

The scale of 16 units for ^16o means that the superscript represents the atomic weight of the atom. There-fore: atomic weight of ^10B = 10 atomic weight of ^11B = 11 average atomic weight of boron (natural) = [(20 × 10 + 80 × 11) / (100)] = 10.8 .

Question:

Write an APL program to evaluate the roots of the quadratic equation of the form ax^2 +bx+ c = 0.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0375.htm

Solution:

To do this, we use the quadratic formula: ROOT = [-B \pm \surd(B^2 - 4AC)] / [2A] The quantity given by B^2 - 4AC is named thediscriminant. If it is negative, then the roots are complex numbers. In that case the program will be terminated. (Complex numbers can be evaluated by a built-in library function, but we will omit this for simplicity sake.). If thediscriminantis positive, then we want to calculate the roots. The program is written below: \nablaQUADRATIC [1]DISCRIM \leftarrow (B\textasteriskcentered2) - 4 × A × C [2]\rightarrow (0 > DISCRIM) / 8 [3]'THE ROOTS ARE' [4]ROOT1 \leftarrow (-B-DISCRIM\textasteriskcentered0.5) \div 2 × A [5]ROOT2 \leftarrow (-B + DISCRIM\textasteriskcentered0.5) \div 2 × A [6]ROOT1, ROOT2 [7]\rightarrow0 [8]'THE ROOTS ARE COMPLEX' \nabla Line [2] may be translated as follows: "If zero is greater than the discriminant , then go to line [8]. If zero is less than or equal to the discriminant , proceed with the next line." This is called a branch statement. Line [7] is also a branch: after the roots have been cal-culated, the program goes to line [0], which actually means the end of the program. Literal text has been added to make clear the meaning of the values being computed. Now, for the polynomial 3x^2 - 19x - 14 = 0, we can give a sample output, with A = 3, B = -19, and C = -14. This polynomial has real roots. A \leftarrow 3 B \leftarrow -19 C \leftarrow -14 QUADRATIC The roots are -0.6666666667 +7.000000000.

Question:

A physicist, Mavis, walks at a speed of 1 m/s past a stationary physicist, Stan. They each have measuring sticks and clocks and they observe a bird hop from a tree branch to the lawn. Each measures the position and the time when this event occurred. Show in the special case of small velocities (v very much smaller than c) that the Lorentz equations may be replaced by the following equations: x' = x - vt y' = y t' = t where the primed and unprimed frames represent the frames of Mavis and Stan respectively.

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Solution:

When the velocities with which we deal approach the speed of light, we can no longer use classical mechanics, and must replace this theory by relativistic mechanics. The purpose of this problem is to show us how the Lorentz transformation (which is part of the relativis-tic mechanics) reduces to the Galilean transformation (which is part of classical mechanics) when the velocities we are concerned with are small when compared with c. The Lorentz transformation, relating the space and time coordinates of an event as observed in S' to the space and time coordinates of the same event as observed in S is x' = (x - vt) / \surd{1 -(v^2c^2)}(1) y' = y z' = z t' = {t - (xv/c^2)} / \surd{1 -(v^2c^2)} where v Is the relative velocity of S and S'. The event which we wish to locate in S and S' landing of the bird. Since S' moves relative to S at a velocity of v = 1 m/s, and the speed of light is 3 × 10^8 m/s, we find v/c = 1/(3 × 10^8) = 0.333 × 10^-8 \ll 1 Hence, we may neglect v^2 / c^2 in the equation of (1) because it is negligible when compared with 1. Therefore x' = (x - vt) / \surd1 = x - vt y' = y(2) z' = z t' = {t - (xv/c^2)} / \surd1 = {t - (xv / c^2)} Furthermore , v/c^2 = 1 / ( 9 × 10^16) = 0.111 × 10^-16 << 1 Now x would have to be enormous if xv/c^2 were to be comparable to t. Whence, we may neglect xv/c^2 in the last equation of (2) and t'= t

Question:

Define the term catalyst. Give an example of an inorganic catalyst and of an organic catalyst.

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Solution:

A catalyst is a substance which affects the speed of a chemical reaction without affecting its final equilibrium point. The addition of a catalyst enables the reaction to reach equilibrium at a faster rate than it otherwise would. A catalyst is not incorporated into the products of the reaction and hence is capable of being regenerated. To understand how a catalyst speeds up a chemical reaction, it is first necessary to define the activation energy of a reaction. There is usually an energy barrier to every chemical reaction which prevents the reaction from spontaneously occurring, even if the reaction is exergonic, that is, one with a negative ∆G (a decrease in free energy). This energy barrier is called the activation energy (Ea). One way to overcome the energy barrier, is through the addition of heat. This increases the internal energy content and therefore the frequency of collisions and reactions. A catalyst can also overcome the energy barrier by lowering the activation energy, and thereby increasing the number of molecules with a suffi-cient energy content to react and form a product. The catalyst acts by forming an unstable intermediate complex with the reactants. This complex is subsequently trans-formed into the product with the release of the catalyst. (C + R)\rightleftharpoonsC \textbullet R\rightleftharpoons(C + P) (catalyst + reactant)(complex)(catalyst + product) The catalyst is therefore free to complex with a second molecule of the reactant. The intermediate complex formed provides an alternative pathway of reaction with a corres-pondingly lower activation energy, thus enabling the reaction to proceed at a faster rate. Figure showing a higher activation energy for an uncata-lyzed reaction than for a catalyzed reaction, note that ∆G remains the same for both types of reactions. There are both inorganic catalysts and organic catalysts. Most inorganic catalysts are non-specific, i.e., they will catalyze various similar reactions. For example, platinum black will catalyze any reaction involving H_2, since it serves to weaken, the bond between the hydrogen atoms. Other inorganic catalysts include water, iron, nickel, and palladium. On the other hand, organic catalysts such as enzymes are highly selective as to the reactions they will catalyze. The specific molecules which an enzyme will bind to and react with are called substrates. There are many enzymes synthesized by living cells, all with specific functions and names. An example of an organic catalyst is sucrase, which is the enzyme that hydrolyzes sucrose into glucose and fructose. Other important enzymes include lysozyme, chymotrypsin, trypsin, and elastase.

Question:

Given that the surface tension of water is 72.62 (dyn / cm) at 20\textdegreeC, how high should water rise in a capillary that is 1.0 mm in diameter?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0272.htm

Solution:

The surface tension is related to the radius of the capillary tube by the equation: \Upsilon = (1/2) rh\rhog where \Upsilon is the surface tension, r is the radius of the capillary tube, h is the height the water rises, \rho is the density and g is the downward force of gravity. The density of H_2O is (1 g / cm^3), the force of gravity is 980 (cm / s^2) and the surface tension is given as 72.62 (dyn / cm). The radius is found by dividing the diameter of the tube by 2. (1 mm =10^-1 cm). Solving for h: 72.62 (dyn / cm) = 72.62 (g / sec^2) 72.62 (g/sec^2) = (1/2) (10^-1 / 2 cm) (h) (1 g / cm^3) (980 cm / sec^2) h = [(72.62 g / sec^2) (2)(2)] / [(10^-1 cm) (1 g / cm^3) (980 cm / sec^2)] h = 2.96 cm.

Question:

What are the distinguishing features of the several orders of insects?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0303.htm

Solution:

Insects are divided into different orders based primarily upon wing structure, type of mouthparts, and the type of metamorphosis they undergo, i.e., complete or incomplete. There are approximately twenty-five orders of in-sects. The class Insecta (Hexapoda) is divided into two subclasses. SubclassApterygota contains the wingless insects, which are believed to be the most primitive living insects. SubclassPterygotacontains most of the insect orders. These are winged insects, or if wingless, the loss of wings is secondary. Some of the better known orders will be further discussed. OrderThysanuracontains primitive wingless insects with chewing mouthparts and long tail-like appendages. Silverfish and bristletails are members of this order. Some species are commonly found in houses and eat books and clothing. OrderOdonatacontains dragonflies and damsel flies. The two pairs of long membranous wings are held permanent-ly at right angles to the body. They have chewing mouth-parts and large compound eyes. Immature forms of these insects (nymphs) are aquatic (fresh water). Other winged insects can fold their wings back over the body when they are not flying. OrderOrthopteracontains grasshoppers, crickets and cockroaches. The forewings are usually leather-like. They do not function in flying, but function as covers for the folded hind-wings. The chewing mouthparts are strong. Termites are social insects that belong to the orderIsoptera. Both winged and wingless varieties comprise the termite colony. OrderHemipteracontains the true bugs. They have piercing sucking mouthparts. The forewing has a distal membranous half and a basal, leathery, thick half. OrderAnopluracontains the sucking lice. These insects are wingless and have piercing-sucking mouthparts. The legs are adapted for attachment to the host. These lice are external parasites on birds and mammals- theheadlouseand crablouse are parasites on man. They are often vectors of disease, such as typhus. All of the orders just discussed have incomplete metamorphosis. Order Lepidoptera contains butterflies and moths. They undergo complete metamorphosis, as do the rest of the orders to be discussed. Lepidoptera have two large pairs of scale covered wings, and sucking mouthparts (in adults). OrderDipteracontains true flies, mosquitoes, gnats, and horseflies. They have one pair of flying wings, with thehindwingsmodified as balancing organs. Mouthparts are piercing-sucking, or licking. Adults are often disease vectors. OrderColeoptera, containing beetles and weevils, is the largest order. They have hard forewings which cover membranoushindwings. They have chewing mouth-parts and undergo complete metamorphosis. OrderSiphonapteracontains the fleas. They are small, wingless parasites. They have piercing-sucking mouthparts, and long legs adapted for jumping. Order Hymenoptera contains ants, wasps, bees, and sawflies. There are winged and wingless species. Wings, when present, are two membranous pairs which interlock in flight. Mouthparts are for chewing, chewing- sucking or chewing-lapping.

Question:

Zn, Zn^+2(.50m) \vert\vert Cu^2+(0.20m), Cu . The concentrations of the ions are given in parentheses. Find the E for this cell at 25\textdegreec. Assume the following standard oxidation potentials: Zn \rightarrow Zn^+2 + 2e^-E\textdegree = .763 Cu \rightarrow Cu^+2 + 2e^-E\textdegree = -.337 .

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Solution:

First of all, it is important to note that a Daniell cell is a typical galvanic cell, which means that spontaneous chemical reactions occur at each electrode of the cell to produce an electric current. To calculate E, the potential at other than standard conditions (25\textdegreec and 1 molar concentrations are the standard conditions), you can employ the Nernst equation. It is stated E = E\textdegree -(.0592 / n) log K, where E\textdegree = standard electrode potential, n = number of electrons gained or lost as shown in the overall reaction and K is the ratio of con-centration of ions on right side of equation to those on left side. The overall equation for this cell is given by Zn + Cu^+2 \rightarrow Zn^+2 + Cu . Thus, for this problem, K = [Zn^+2] / [Cu^+2] First, let us try to find the values of these parameters of the Nernst equation, so that E can be found. E\textdegree = E\textdegree_oxidation + E\textdegree_reduction . For the reaction to be spontaneous, E\textdegree must equal a positive value. Thus, you want to choose the oxidation half-reaction (loss of electrons) and reduction half-reaction (gain of electrons) carefully, so that their sum, E\textdegree, is a positive value. You are given the standard oxidation potentials, E\textdegree_ox , of all species in question. The reduction will be the reverse of this reaction, with a reduction potential, E\textdegree_red to negative of the E\textdegree_ox given. The electrode potential of the overall reaction is the sum of these two half-reactions.. For the E\textdegree (overall) to be positive, it follows that the oxidation is Zn \rightarrow Zn^+2 + 2e^-E\textdegree = .763 and the reduction is Cu^2+ + 2e^- \rightarrow CuE\textdegree = -(-.337) overall reaction Zn + Cu^2+ \rightarrow Zn^+2 +Cuwith E\textdegree = .763 + [-(-.337)] =1.1v. From the overall reaction, recall K= [Zn+^2] / (Cu^+2]. (The concentrations of solids are not included in a dissociation constant). You are told Zn^+2 = .5m and Cu^+2 = .2m, so that K = (.50/.20). In the overall reaction, 2 electrons are lost and gained, which means n = 2. If you substitute these values in the Nernst equation, you obtain E = 1.1 - (.-592 / 2) log (.50 / .20). Solving, E = 1.088v for this Daniell cell.

Question:

A list of numbers (1 through 9) is given. The object is to arrange the numbers in ascending order. A move is made by specifying how many numbers (starting from the left) are to be placed in reverse order. For example, suppose 214356789 is the starting posi-tion. Then "reverse 2" gives 124356789. "Reverse 5" gives 534126789. Write a Basic program to play Reverse with the computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0559.htm

Solution:

There are two approaches to the game - one algorithmic, and the other heuristic. The algorithmic solution guarantees a win in a certain number of moves (if there are N numbers, it takes 2N - 3 moves). The heuristic method takes advantage of the information at any moment to improve the strategy for minimizing the number of moves. The player decides which method to pursue. Here is the program for this game: 5PRINT "REVERSE - A GAME OF SKILL" 10DIM A (20) 20REMN = NUMBER OF NUMBERS 30N = 9 40REMMAKE A RANDOM LIST A (1) TO A (N) 50A (1) = INT ((N - 1) \textasteriskcenteredRND (1) + 2) 60FOR K = 2 TO N 70A (K) = INT (N\textasteriskcenteredRND (1) + 1) 80FOR J = 1 TO K - 1 90IF A (K) = A (J) THEN 70 100NEXT J 110NEXT K 120REM PRINT ORIGINAL LIST AND START GAME 130PRINT "THE LIST IS:" 140T = 0 150GOSUB 430 160PRINT "HOW MANY TO BE REVERSED?\textquotedblright 170INPUT R 180IF R = 0 THEN 370 190IF R < = N THEN 240 200REM THE COMPUTER CANNOT REVERSE MORE THAN 210REM N NUMBERS 220PRINT "TOO MANY. I CAN REVERSE AT MOST" ;N 230GOTO 160 240T = T + 1 250REM REVERSE R NUMBERS AND PRINT NEW LIST 260FOR K = 1 TO INT(R/2) 270Z = A (K) 280A (K) = A (R - K + 1) 290A (R - K + 1) = Z 300NEXT K 310GOSUB 430 320REM CHECK FOR A WIN 330FOR K = 1 TO N 340IF A(K) <> K THEN 160 350NEXT K 360PRINT "YOU WON IT IN"; T; "MOVES!" 370PRINT "TRY AGAIN (YES OR NO)"; 380INPUT A$ 390IF A$ = "YES" THEN 50 400PRINT "GOOD GAME, THANKS." 410GOTO 470 420REM SUBROUTINE TO PRINT LIST 430FOR K = 1 TO N 440PRINT A (K) 450NEXT K 460RETURN 470END

Question:

In an experiment, a block of beryllium is bombarded with \alpha- particles. A nearby block of paraffin, shielded from the \alpha- particles, is observed to be emitting protons and nitrogen nuclei in separate events. The ratio of the velocity of the proton to that of the nitrogen nucleus is measured to be 7.5. It is suspected that this is a result of a chargeless particle being emitted by the beryllium and absorbed by the paraffin. What is the mass of this particle?

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Solution:

Let the mass of the unknown particle be m and its velocity \upsilon. Paraffin contains many hydrogen atoms (for it is a hydrocarbon). Hence we assume that protons are emitted after a head-on collision with one of the hydrogen atoms, (initially at rest). The momentum of the unknown "particle-hydrogen" system, before the collision, must equal the total momentum of the system after the collision, by the law conservation of momentum. m\upsilon = m_p\upsilon_p' + m\upsilon'(1) where \upsilon_p' and \upsilon' are the new velocities of the emitted proton and of the unknown particle, respectively. By con-servation of energy, the sum of the kinetic energies of the particles before the collision must be equal to the total kinetic energy after the collision or (1/2) m\upsilon^2 = (1/2) (m_p\upsilon'_p)^2 + (1/2) m (1/2) m\upsilon^2 = (1/2) (m_p\upsilon'_p)^2 + (1/2) m \upsilon'^2(2) Equation (1) is rewritten as \upsilon - \upsilon' = [(m_p)/(m)] \upsilon' \upsilon - \upsilon' = [(m_p)/(m)] \upsilon' p (3) and (2) as \upsilon^2 - \upsilon'^2 = \upsilon^2 - \upsilon'^2 = [(m_p)/(m)] \upsilon'_p^2(4) Dividing (4) by (3) gives \upsilon + \upsilon' = \upsilon'_p(5) Then combining (3) and (5) we obtain \upsilon = [(m_p + m)/(2m)]\upsilon'_p(6) Similarly for the collision with the nitrogen atom we obtain \upsilon = [(m_N + m)/(2m)] \upsilon'_N(7) Thus from (6) and (7) [(\upsilon'_p)/(\upsilon ׳ N )] = (m_N +m)/(m_p + m) This ratio equals 7.5 so that 7.5 = (m_N + m)/(m_p + m) 7.5 m_p + 7.5 m = m_N + m 7.5 m_p - m_N = - 6.5 m (m_N - 7.5 m_p)/(6.5) = m Butm_N = 14.008 am\upsilon m_p = 1.008 am\upsilon Hence,m = [(6.458)/(6.5)] am\upsilon= .99 am\upsilon m = (m_N - 7.5 m_p)/(6.5) In other words the unknown particle has a mass almost identical to that of the proton. The particle is called a neutron, and actually has a mass larger than the proton by about 0.18 per cent.

Question:

40.0 kJ of energy is added to 1.00 gram of magnesium atoms in the vapor state. What is the composition of the final mixture? The first ionization potential of Mg is 735 kJ/mole and the second ionization potential is 1447 kJ/mole.

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Solution:

Ionization potential may be defined as the energy required to pull an electron away from an isolated atom. The second ionization potential is the amount of energy re-quired to pull off a second electron after the first has been removed. The composition of the final mixture is determined by calculating the number of electrons that will be removed from the magnesium ions. To do this one must determine the number of moles of Mg present in 1 g. From this one can determine the number of electrons that will be liberated by using the values for the first and second ionization po-tentials of Mg. The atomic weight of Mg is 24.3. Since moles = grams/atomic weight, there are in 1 gram of Mg, 1 / 24.30 or 4.11 × 10^-2 moles present. The first ionization potential of Mg is 735 kJ/mole. Therefore, 4.11 × 10-2 moles of Mg re-quires 4.11 × 10^-2 moles × 735 kJ/mole or 30.2 kJ to ionize all of the atoms once. 40 kJ was added to the system leaving 40 kJ - 30.2 kJ or 9.8 kJ to remove the second electron. If one has 9.8 kJ and 1447 kJ/mole is required to remove the second electrons, then [9.8 KJ / (1447 KJ/mole)] = 6.77 × 10^-3 moles of atoms can have their second electron removed. 4.11 × 10^-2 moles of Mg are present. [(6.77 × 10^-3) / (4.11 × 10^-2)] × 100 = 16.5%. This means that 16.5% of the atoms can have a second electron removed. Therefore, the composition of the mixture is: Mg^++ 16.5%, Mg^+ 100 - 16.5 or 83.5%.

Question:

What is meant by a hydrostatic skeleton? In which organisms is such a structure found?

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0487.htm

Solution:

Cnidarians (coelenterata) such as hydra, flatworms such as planaria , and annelids like the earth worms all move by the same basic principle of antagonistic muscles. They have no hardexo- or endo - skeleton to anchor the ends of their muscles. Instead the noncompressible fluid contents of the body cavity serve as a hydrostatic skeleton. Such animals typically have a set of circular muscles, the contraction of which decreases the diameter and increases the length of the animal, and a set of antagonistic longitudinal muscles, which, when contracted, decrease the length and increase the diameter of the animal. As an example, the leech is often used. The leech attaches its posterior end to the substrate by means of a sucker, extends its body forward by contraction of the circular muscles, attaches its front end by a sucker, and then detaches the posterior end and draws it forward by contraction of the longitudinal muscles. Note that circular and longitudinal muscle are found in man: the smooth muscle lining his hollow visceral organs, for example. By alternate contractions of each muscle layer, peristaltic waves propel the contents along the hollow organ. For instance, this occurs in the esopha-gus, intestines, oviducts,ureters, etc. The relatively soft internal tissues of solid-bodied animals like the leech can function as a hydrostatic skeleton only to some extent. More active wormlike animals, like the annelids, have evolved segmented bodies in which fluid is contained in a partitioned series of cavities. In addition to segmentation of the body cavity, there is a segmentation of the musculature; the fact that each segment of the body has its circular and longi-tudinal muscles makes possible effective use of the compartmented hydrostatic skeleton. The arrangement of the hyd-rostatic skeleton allows for peristaltic movements down the length of the organism. This aids the organism in burrowing. Some marine worms have additional diagonally arranged muscles that permit more complex movements- of the body and thepaddlelike parapodia that extend laterally from the body wall. Many marine worms live in tubes and the movements of theparapodiaare important not only in locomotion, but in moving currents of water laden with oxygen and nutrients through these tubes.

Question:

A piece of uniform wire is made up into two squares with a common side of length 4 in. A current enters the rectangular system at one of the corners and leaves at the diagonally opposite corner. Show that the current in the common side is one-fifth of the entering current. What length of wire connected between input and output terminals would have an equivalent resistive effect?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0686.htm

Solution:

Let each side of the double square have resistance R, and let the lettering of the diagram and the currents flowing be as shown in Figure A. Applying the Kirchoff current rule, \sumI = 0, to the junctions A, B and E in turn (with the convention that current entering a junction is positive, and current leaving a junction is negative), gives I_1 - I_2 - I_3 = 0,(1) I_2 - I_4 - I_5 = 0,(2) I_3 + I_4 - I_6 = 0,(3) Applying the Kirchoff voltage rule to the loops ABED and BCFE gives I_2 R + I_4 R - I_3 × 2R = 0,(4) I_5 × 2R - I_6 R - I_4 R = 0.(5) Eliminating I_5 and I_6 from Eqs. (2), (3) and (5), we obtain I_1 - I_2 - I_3 = 0(6) I_2 - 2I_3 + I_4 = 0(7) 2I_2 - I_3 - 4I_4 = 0(8) Eliminating I_2 from these three equations gives I_1 - 3I_3 + I_4 = 0,(9) and2I_1 - 3I_3 - 4I_4 = 0,(10) orI_4 = (1/5) I_1 .(11) Further, the potential drop from A to F by route ADEF is V_AF = I_3 × 2R + I_6 × R = R(2I_3 + I_3 + I_4) using equation (3). By (9) this becomes V_AF = R (I_1 + 2I_4) and using (11),V_AF = I_1 R [1 + (2/5)] = (7/5) RI_1 The equivalent effect is therefore obtained if a wire 7/5 times the length of any side of the square is connected between A and F, because it produces the same potential drop as the double square between these points (see fig. B).

Question:

Calculate the mean free path for oxygen at 25\textdegreeC at (a) 1atm pressure and (b) 10^-3torr.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0104.htm

Solution:

The mean free path is the average distance traversed by a molecule between collisions. In a second, a molecule will, on the average, traverse v meters and collide Z times (where Z = \surd2 \pi\sigma^2vn) . Thus, the mean free path is l = v/Z = 1/(\surd2 \pi\sigma^2n) Substituting n = P/kTfrom the ideal gas law, we have l =kT/(\surd2 \pi\sigma^2P) To solve this problem, we must 1) find n, the con-centration, 2) substitute the known values into the equation for the mean free path a) P = 1atm\sigma. = 3.61 × 10^-10 m T = 25\textdegreeC + 273 = 298\textdegreeKR = 0.082 liter-atm/0K-mole n = PN_A / RT = {[(1atm)[6.022 ×10^23(molecules/mole)] [10^3(leter/m^3)]} / {[0.082 (liter-atm/ 0K-mole)] (29 8 \textdegreeK)} n = 2.46 × 10^25 molecules/m^3 Thus,l = \surd2 \pi\sigma^2n = [(1.414) (3.14) (3.61×10^-10 m)^2 (2.46×10^25)]^-1 = 7.02 × 10^-8 m b) P = 10^-3torr= {10^-3torr}/{760 (torr/atm)} = 1.3157 × 10^-6 atm. T = 2980K\sigma = 3.61 × 10^-10 m Here, it is necessary to converttorrunits to units of atmospheres. n = {[(1.3157 × 10^-6atm) [6.022 ×10^23 (molecules/mole)] [10^3(leter/m^3)]} / {[0.082 (liter-atm/ 0K-mole)] (29 8 \textdegreeK)} = 3.24 × 10^19 molecules/m^3 Thus,l = 1/\surd2 \pi\sigma^2n = [(1.414) (3.14) (3.61 × 10^-10 m)^2 (3.24 × 1019/ m^3)]^-1 = 0.053 m = 5.3 cm.

Question:

For the indicator phenolphthalein (In^-),HInis colorless and In^- is red;K_dissis 10^-9. If one has a colorless solution of 50 ml phenolphthalein, how much 0.10 MNaOHsolution would one have to add to make itred: K_W = 10^-14 .

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Solution:

To answer this question, write out the dissocia-tion reaction of phenolphthalein in solution. After this, write out the equilibrium constant expression. This allows one to calculate the [H^+]. This informs one of the con-centration of OH^- required and allows the calculation of the amount of NaOHto be added. With this in mind, proceed as follows: An indicator, which changes color, undergoes a dissociation reaction like an acid. For phenolphthalein the reaction is HIn\rightarrow H^+ + In^-. It is given that In^- makes the solution red andHInmakes it colorless. The equilibrium constant expression, which measures the ratio of concentrations of products to re-actants, each raised to the power of its coefficient in the equation, is K_diss= 10^-9 = {[H^+][In^-]}/[HIn] The object is to make a red solution, from an originally colorless one. In general, a specific color of an indicator will show itself when the concentration of the corresponding species is 10 times as great as the concentration of the other species. Since one wants the solution to be red, and In^- = red color, [ln^-]/[HIn] = 10/1. Substituting this into the equilibrium constant expression, 10^-9 = [H^+] (10/1) . Solving for [H^+], [H^+] = 10^-10. From the K_W constant of water, it is known that [H^+][OH^-] = 10^-14 , where K_W is theautodissociationconstant of H_2O. If [H^+] = 10^-10, one can calculate [OH^-] and obtain [OH^-] = 10^-4 M. Thus, to have a red solution, one wants [OH^-] to be 10^-4 M. TheNaOHsolution being added is 0.1 M. (M =molarity= moles/liter.) The volume of the solution is 50 ml. or 0.05 l. (1000 ml = 1 l.) Thus, one needs 10^-4 M or 0.050 l ×10^-4 M = 5.0 ×10^-6 moles. Since volume in liters = number of moles/molarity, andmolarity= 0.1, liters ofNaOHto be added = = [(5.0 × 10^-6)/0.1] = 5.0 × 10^-5 liters.

Question:

When the wavelength of the incident light exceeds 6500 \AA, the emission of photoelectrons from a surface ceases. The surface is irradiated with light of wave-length 3900 \AA. What will be the maximum energy, in electron volts, of the electrons emitted from the surface?

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Solution:

Einstein's photoelectric equation relates the energy of the incident quanta to the maximum energy of the emitted electrons (= W) by the relation hf= h(c/\lambda) = W + W_0 = W + hf_0 =W + h (c/\lambda_0) , where W_0 is the work function of the surface and \lambda_0 the cut-off wavelength. Hence the maximum energy of the emitted photoelectrons in the problem is W =hc[(1/\lambda) - (1/\lambda_0)] = 6.6 × 10^-34 J \textbullet s × 3.0 × 10^8 m \textbullet s^-1 [{1/(3.9 × 10^-7 m)} - {1/(6.5 × 10^-7 m)}] = 19.8 × 10^-19 [(1/3.9) - (1/6.5)] J = [(19.8 × 10^-19 J) / (1.6 × 10^-19 J/eV)] × [(6.3 - 3.9) / (3.9 × 6.5)] = [(19.8 × 10^-19) / (1.6 × 10^-19)] × [(2.6) / (3.9 × 6.5)]eV= 1.27eV.

Question:

A bank of cells having a total emf of 12 V and negligible internal resistance is connected in series with two resistors. A voltmeter of resistance 5000 \Omega is connected across the resistors in turn, and measures 4 V and 6 V, respectively. What are the resistances of the two resistors?

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Solution:

The voltmeter is connected across R_1 as in diagram (a) and since (a) and since the resistance of the voltmeter is in parallel with R_1 the circuit is equivalent to that shown in diagram (b), where that shown in diagram (b), where (1/R) = (1/R_1 ) + (1/5000 \Omega) = [(5000 \Omega + R_1 )/(5000 \Omega R_1 )] R = [(5000R_1 \Omega)/(5000 \Omega + R_1 )](1) Now, the voltmeter measures a 4 V drop across R_1 . By Kirchoff's Voltage Law, the net voltage around the circuit shown in figure (b) is zero. Hence, traversing the circuit as shown \epsilon - IR - IR_2 = 0(2) But we know that the voltage drop across R_1 equals the voltage drop across R, since R_1 and the voltmeter are in parallel. Then IR = 4 V and, from (2) IR_2 = \epsilon - IR = 12 V - 4 V = 8 V Therefore,[(IR_2 )/(IR)] = (8/4) = 2 orR = (1/2) R_2 Using (1)R = [(5000 R_1 \Omega)/(5000 \Omega + R_1 )] = (R_2 /2) Similarly, from diagrams (c) and (d) , showing the second connection of the voltmeter and the equivalent circuit, we have R' = [(5000 R_2 \Omega)/(5000 \Omega + R_2 )]and6 V = I'R_1 = I'R' . R' = [(5000 R_2 \Omega)/(5000 \Omega + R_2 )] = R_1 . Hence, from the two equations obtained, we have (by cross multiplication) 10,000 R_1 \Omega = 5000 R_2 \Omega + R_1 R_2(3) and5000 R_2 \Omega = 5000 R_1 \Omega + R_1 R_2 .(4) Subtracting these equations, 10,000 R_1 \Omega - 5000 R_2 \Omega = 5000 R_2 \Omega - 5000 R_1 \Omega or 15,000 R_1 = 10,000 R_2orR_1 = (2/3) R_2 . Substituting back into equation (3), we get 10,000 [(2/3) R_2 ] \Omega = 5000 R_2 \Omega + [(2/3) R_2 ] (R_2 ) 10,000 \Omega = 7500 \Omega + R_2 R_2 = 2500 \Omega andR_1 = (2/3) R_2 = (2/3) (2500 \Omega) R_1 = 1667 \Omega

Question:

Two point charges + 2q and \rule{1em}{1pt}q are separated by a distance L. At what point on the line between their centers is the L. At what point on the line between their centers is the electric field zero? electric field zero?

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Solution:

Suppose that this point is a distance x from the + 2q charge, as shown in the figure. By definition, the electric field at a point P is E = K_Eq / r^2, where q is the charge producing the field, r is the distance from q to P, and K_E is a constant (K= 9 × 10^9N \bullet m^2/c^2). The field at x due to 2q is E_+ = K_E [(2q) / (x)^2] The electric field E_\rule{1em}{1pt} due to the negative charge is E_- = \rule{1em}{1pt}K_E [(q) / (x \rule{1em}{1pt} L)^2] The resultant electric field at the point P is then, by the principle of superposition, equal to the sum of the fields due to the individual charges 2q and \rule{1em}{1pt}q. Because we want E_net at P to be zero, we have E = 0 = E_+ + E_\rule{1em}{1pt} = K_E [(2q) / (x)^2]\rule{1em}{1pt}[(K_E q) / (x \rule{1em}{1pt} L)^2] orx^2 = 2(x \rule{1em}{1pt} L)^2 x2= 2x^2 \rule{1em}{1pt} 4xL + 2L^2 x^2 \rule{1em}{1pt} 4xL + 2xL2= 0 x =[{4L \pm \surd(16L^2 \rule{1em}{1pt} 8L^2 )} / 2] = [{4L \pm \surd(2.4)L^2 / 2] = 2 L \pm \surd2 L = 3.41 L or 0.588 L All we have done so far is to find two points along the axis where the fields due to +2q and \rule{1em}{1pt}q are equal in magnitude. We have yet to take account of the vector nature of the fields. At the interior point x = 0.588 L the vectors are in the same direction, so that the fields are added. However at x = 3.41 L the field vectors are in opposite di-rections, so that the resultant field is zero.

Question:

What are the consequences of intenseinterspecific competition?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0763.htm

Solution:

The more similar two niches are, the more likely it is that both specieswill utilize at least one common limited resource (food, shelter, nestingsites, etc) . They will therefore be competing for that limited resource. The competitive superiority of one of the rival species may be such thatthe other is driven to extinction. Two species may fluctuate in competitivesuperiority according to the habitats they share, with the result thatone is eliminated in some places and the other is eliminated in other places. One or both of the competing species may evolve in divergent directionsunder the strong selection pressure resulting from their intense competition. Natural selection would favor individuals with characteristics differingfrom those of the other species, because such characteristics wouldtend to minimize competition. In other words, the two species would evolvegreater differences in their niches. Whether any one or a combination of these effects will be the outcomein any given case of intense compe-tition is determined by many complexfactors. If a pair of species occupies identical ecological niches, oneof them will indeed become extinct. This generalization is usually referredto asGause'sprinciple, or the principle of competitive exclusion, whichstates that two different species cannot simultaneously occupy the sameniche in the same place. It is implicit in the definition of niche that no twospecies could ever occupy the same niche. To do so, they would have tobe identical in every respect and hence they would be one species, not two. Thus, the competitive exclusion principle helps us see why the differentspecies living together in a stable community occupy quite distinctniches. Only if organisms living together avoid competition can theysuccessfully coexist. A fit organism is one that avoids competition, andis free from the necessity to struggle with others in its own niche.

Question:

During its lifetime of about 10^10 years, a normal star radiates an energy of about1.0 × 10^52 ergs. What is the energy equivalent inkilowatthours(kwh) ?

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Solution:

This problem illustrates the conversion between watts (w) and ergs/sec (1 w = 10^7 ergs/sec). One kilowatt (kw) is equal to 10^3 watts. Convertingkwto erg/sec, we have 1kw= 10^3 w × 10^7 ergs/sec - w = 10^10 ergs/sec.Also , converting seconds to hours (1 hour = 3600 sec), we have 1kw= 10^10 ergs/sec = 10^10 ergs/sec × 3600 sec/hour = 3.6 × 10^13 ergs/hour. Therefore, 1kw= 3.6 × 10^13 ergs/hour or 1kwh= 3.6 × 10^13 ergs. The total energy radiated by the star is 1.0 × 10^52 ergs. Converting tokwhwe get (1.0 × 10^52 ergs) / (3.6 × 10^13 ergs/kwh) = 7.8 × 10^38kwh. Therefore the total energy radiated by the star is 7.8 × 10^38kwh.

Question:

In contrast to density dependent effects, there are also density independent effects which operate without reference to population size. Elaborate on the major density independent effects and their relationship to population size.

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Solution:

There is no scientific principle which states that populations must be controlled as a function of density. Density effects will always be working in crowded populations, but it is also possible for actual control of population size to be exercised before crowding occurs. Control agents that are not dependent on density can take the form of sudden events that are catastrophic to animal populations. Catastrophic events can solve the problem of overcrowding just as efficiently as a density dependent device. Hurricanes and volcanic explosions can destroy entire populations but these events are too scattered and local to be used as examples for general population controls. More applicable are the common catastrophes which we know as changes in the weather. Nearly all places on the earth suffer seasonal changes from summer to winter, from warm to cold, and from wet to dry. Each of these cyclic changes represents hazards to the animals of each area. Thus, growing populations are frequently cut back, making their normal lives a race to reproduce so quickly that there shall be at least some survivors following the next catastrophe. Unless something happens to prevent the catastrophe from occurring such a popu-lation may never grow large enough to suffer the effects of crowding. Density independent factors may play an important role in limiting some organisms, particularly those with very short life cycles characterized by a growth curve in which the environmental limiting factors do not become effective until after many generations (see fig. 1). A sudden strong density independent limitation then brings growth to an end before density dependent factors can become operative. The density independent effects are definite factors in moving population growth upward or downward, but they cannot hold the population size at a constant level. As a result, populations effected by density independent factors are under control but their numbers fluctuate within wide limits (see fig. 2) and can hardly be described as "in balance."

Question:

The ease of formation of carbonium ions is 3\textdegree>2\textdegree>1\textdegree>Ch_3+. Explain why.

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Solution:

A carbonium ion is a group of atoms containing a carbon atom with only six electrons around it. Thus the carbonium ion is positively charged. The carbonium ions are classified as primary (1\textdegree), secondary (2\textdegree), and tertiary (3\textdegree) after the carbon bearing the positive charge. For example, where R can be any alkyl group. To understand why the ease of formation is 3\textdegree>2\textdegree>1\textdegree>CH_3+, you need to understand some of the proper-ties of a charged system. According to the laws of electrostatics, the stability of a charged system is in-creased by the distribution of the charge. Alkyl groups, for example, tend to donate electrons, so that any positive charge is dispersed. Thus, the more alkyl groups donating electrons, the greater the distribution of positive charges in a carbonium ion, and, as such, the more stable it is. The greater the stability of a substance, the easier it is to form. In our example, the 3\textdegree carbonium ion has 3 alkyl groups, the 2\textdegree ion has 2 alkyl groups, and the 1\textdegree ion has one alkyl group. Note CH_3+ can be written which shows that no alkyl groups are present. Therefore, 3\textdegree>2\textdegree>1\textdegree>CH_3+ in ease of formation, since this sequence is also the sequence for the number of alkyls that distribute the positive charge, which results in in-creased stability.

Question:

A student hears the ringing of a bell at the end of the day and begins to pack his books to go home. Trace the path of the stimulus from the source (bell) to its appro-priate center in the brain.

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Solution:

When the bell is struck, it sets off vibrations in the air in the form of sound waves. These sound waves travel through the air and some of them are collected by the pinna. They are then channeled along the auditory canal of the outer ear towards the middle ear. At the inner end of the auditory canal, the sound waves impinge on the tympanic membrane and set the membrane vibrating. These vibrations are first transmitted to the malleus which is in direct contact with the tympanic membrane. Then the vibrations are relayed across the middle ear by the incus and stapes, which are so arranged that they decrease the amplitude but increase the force of vibrations. The stapes transmits the vibrations, via the oval window, to the perilymph in the vestibular canal. This occurs as the vibrating stapes strikes the membrane of the oval window, causing it to oscillate. Since fluids are incompressible, movement of the membrane causes displacement of the perilymph from the vestibular canal to the tympanic canal. (Recall that these two canals are connected at the apex of the cochlea.) The perilymph in the tympanic canal hits the membrane of the round window, resulting in movement of the membrane. Thus vibrations from the bones of the middle ear are transformed into oscillations of the perilymph in the cochlea. The pressure waves of the perilypmh set the thin membranes separating the three canals into vibration, most important of which is the basilar membrane in the cochlea canal, since on it lie the receptor cells for hearing. Vibrations of the basilar membrane cause the sensory hairs of the receptor cells to move up and down against the less movable tectorial membrane. This movement results in the deformation of the hairs, which in some unknown fashion initiates and sends nerve impulses along the sensory neurons, lying at the base of the receptor cells. Axons of the sensory neurons come together to form the auditory nerve which leads the impulses from the organ of Corti to the auditory center in the temporal lobe of the brain. There the impulses are interpreted as the ringing of the bell.

Question:

List the three primary nitrogenous excretory wastes; tell which animals excrete each, and why. What is uremia?

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Solution:

The three primary nitrogenous waste products elimi-nated from animals are ammonia, urea, and uric acid. Nitrogenous wastes of aquatic animals, both marine and freshwater, are usually excreted as ammonia. Ammonia, a toxic substance, is very soluble in water. A considerable amount of water is needed, but since there is no danger of water loss in aquatic animals, the excretion of ammonia presents no difficulty. Urea is a nitrogenous waste produced in the liver and excreted by the kidneys in mammals and amphibians. It is formed by combining ammonia with carbon dioxide, thus converting ammonia to a relatively non-toxic form. Urea is highly soluble, and requires water for its ex-cretion. However, because it is non-toxic, urea, unlike ammonia, does not require extremely large amounts of water for excretion. Urea may be retained in the body for some time before being excreted. Uric acid is also excreted by terrestrial animals,most notably,the egg-laying or oviparous animals such as birds,snakes,and lizards.Uric acid is a semi-solid,highly insoluble waste product.It is formed in the tissues and in insects is selectively absorbed by the malpighian tubule cells. In birds,it is absorbed by the kidney tubules. The rectum of uric acid-producing animals has powerful water-reabsorptive capacities; the uric acid thus leaves the rectum as a nearly dry powder or hard mass. In its excretion, nearly no water need be lost, and water is highly conserved. Those animals excreting uric acid lay eggs enclosed within a relatively impermeable shell.Excretion of toxic ammonia is out of the question,and urea,if excreted,would build up to harmful concentrations over the period of embryonic development in the egg. Uric acid,on the other hand,being almost solid,can be safely stored in the egg until development is complete. Uremia is a condition where there is a gradual accu-mulation of urea and other waste products in the blood. This condition can result from many different forms of kidney disease.The severity of uremia depends upon how well the impaired kidneys are able to preserve the con-stancy of the internal environment.

Question:

How can you prove that oxygen is given off by green plants during photosynthesis?

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Solution:

We can show that oxygen is one of the products of photosynthesis by performing a simple experiment. Elodea or some other photosynthetic aquatic plant is placed in a beaker of water and covered with an inverted glass funnel. A rubber tube is attached to the stem of the funnel. This tube is closed off with a pinch cock. The funnel is subsequently raised, creating a column of vacuum in its stem. (See accompanying diagram.) The plant is then exposed to light. The gas given off during the photosynthetic process bubbles upward through the water and fills the column of the stem. To identify the gas, a glowing splinter of wood is applied over the stem of the funnel. Upon removal of the pinch - cock, the glowing splinter bursts into flame. This means that the gas is oxygen, since it supports combustion.

Question:

Consider again the RCL circuit of the previous problem, or, analogously, an automobile suspension system discussed before. Suppose a voltage source is connected with the circuit [In the case of the automobile sus-pension system, suppose an external force is applied to the system. The road would be such a force.] At time t = 0, S is closed. If E(t) is the voltage at time t, it can be shown that the behavior of the circuit is governed by the following differential equation: Lq̇ ̇+ Rq̇ + (1/C)q(t) = E(t) . where q(t) is the charge on the capacitor, q̇ is the current at time t and q̇ ̇is the rate of change of current. [If E(t) is the external force on the suspension system at time t, then: Mẋ ̇ + Dẋ + Kx(t) = E(t)] If q_0(0) = i_0(0) = 0 [if the suspension system is at equilibrium, so that x_0(0) = i_0(0) = 0], write a digital computer program that models and simulates the behavior of the system from time t = 0 to t = t_f , for the following test inputs: (i) E_1(t) = h where h is constant (ii)E_2(t) =0if t < t1 Hif t1\leq t < t_2 0if t \geq t_2 E_2(t) is a pulse of height h and width (t_2 - t_1). [E_2(t) can be used to simulate a bump in the road.] (See Fig. 1) (iii) E_3(t) = A_1 sin wt where w and A_1 are constants. [E_3(t) could represent a hilly road.] (iv)E_4(t) =0if t < t1 Hif t1\leq t < t_1 + w 0if t1+ w \leq t < t1+ 2w and E_4(t + 2w) = E_4(t) E_4(t) is a pulse train.(See Fig. 2)

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Solution:

Using the variables associated with the automobile suspension system, we call upon the variable v(t) = ẋ again to transform the equation to a simultaneous system of first-order equations to be used in the Euler predictor-corrector program: x = v(t)(1) v̇ = [E(t) - Dẋ ̇- Kx(t)]/M(2) (See Fig. 3) Since E(t) varies with time, it is given a different symbol (trapezoid). Writing equation (2) as a statement function definition, using it at the correct time in the program, substituting the new initial values (T = XCOR = VCOR = 0) and changing I/O statements, complete the solution: (i)REAL K,M,N, T/0.0/, XCOR/0.0/,VCOR/0,0/ F(Y,Z) = (H - D{_\ast}Y - K{_\ast}Z)/M READ K, M, N, TFIN, ACCUR, D,H DT = (TFIN - T)/N Missed segment of the problem is similar to the ones in the previous problems. (ii)REAL K,M,N,T/0.0/, XCOR/0.0/, VCOR/0.0/ F(Y,Z) = (H - D\textasteriskcenteredY - K\textasteriskcenteredZ) /M G(R,S) = - (D\textasteriskcenteredR + K\textasteriskcenteredS)/M READ K,M,N,TF IN, CCUR, D,H,T1,T2 DT = (TF IN - T)/N 50X = XCOR V = VCOR A = G(V, X) IF (T.GE.T1.AND.T.LT.T2)A = F(V, X) 75PRINT T, X T = T + DT IF (T.GT.TFIN) STOP IF (T.LT.T1) GO TO 75 XPRED = X + V{_\ast}DT VPRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(V,XPRED) IF (T,GE.T2)VD0T = G(V,XPRED) XCOR = X + 0.5\textasteriskcentered(V + XD0T)\textasteriskcenteredDT : : : Missed segment of the problem is similar to the ones in the previous problems. (iii)REAL K,M,N, T/0.0/, XC0R/0.0/,VC0R/0,0/ F(TIME,Y,Z) = (A1{_\ast}SIN(OMEGA{_\ast}TIME) - D{_\ast}X - K{_\ast}Z)/M READ K,M,N,TFIN,ACCUR,D,A1,OMEGA. DT = (TFIN - T)/N 50X = XCOR V = VCOR A = F(T,V,X) PRINT T,X T = T + DT IF (T.GT.TFIN) STOP XPRED = X + V{_\ast}DT VPRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(T,V,XPRED) XCOR = X + 0.5\textasteriskcentered(V + XD0T)\textasteriskcenteredDT : : : Missed segment of the problem is similar to the ones in the previous problems. (iv)REAL K,M,N,T/0.0/,XCOR/0.0/,VCOR/0.0/ F(Y,Z) = (H - D{_\ast}Y - K{_\ast}Z)/M G(R,S) = -(D{_\ast}R + K{_\ast}S)/M READ K,M,N,TFIN,ACCUR,D,H,T1,W, DT = (TFIN - T)/N PER = T1 + 2W CPER = TIME OF COMPLETION OF FIRST PERIOD HPER = T1 + W CHPER = TIME OF COMPLETION OF FIRST HALF-PERIOD TIME = T 50X = XCOR V = VCOR A = G(V,X) IF (TIME.GE.T1.AND.TIME.LT.HPER)A = F(V,X) 75PRINT T,X T = T + DT IF (T.GT.TFIN) STOP TIME = T IF (T.GE.PER) TIME = T - 2\textasteriskcenteredW IF (TIME.LT.T1) GO TO 75 XPRED = X + V{_\ast}DT VRED = V + A{_\ast}DT 100XDOT = VPRED VDOT = F(V,XPRED) IF (TIME. GT. HPER)VDOT = G(V,XPRED) XCOR = X + 0.5{_\ast}(V + XD0T){_\ast}DT : : : Missed segment of the problem is similar to the ones in the previous problems.

Question:

It has become apparent that a large number of hormones actually have identical initial biochemical actions. This involves the activation of a system containing the enzyme adenyl cyclase, which is found in cell membranes throughout the body. This system has become known as the second-messenger or cyclic AMP system. Using this system; outline the sequence of events involved in producing glucose from glycogen via the action of epinephrine.

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Solution:

Adenyl cyclase is an enzyme found in membranes which, when activated, catalyzes the modification of ATP into cyclic AMP. It is the sole function of the hormone to interact with receptors on the cell membranes and activate the adenyl cyclase. The cyclic AMP is called the second messenger, and it serves to initiate a complex sequence of reactions, leading to the ultimate generation of the final active enzyme. For the case of epinephrine, the following sequence of events occurs: epinephrine, released from the adrenal medulla by stimulation of sympathetic nerves, activates adenyl cyclase on the surface of the liver cell (note that the hormone itself does not gain entry into the cell); adenyl cyclase in turn catalyzes the conversion of ATP into cyclic AMP; the cyclic AMP then stimulates conversion of the inactive enzyme in the cell into an active form; this active enzyme, in turn, catalyzes the critical reaction leading to the breakdown of glycogen. There may appear to be a small inconsistency in this adenylcyclase system. If the generation of cyclic AMP is the common biochemical action of many hormones, why don't all these hormones produce identical effects on different cell types of the body? The reason for this is that the adenyl cyclase systems in different cell types differ in their abilities to be activated by different hormones. There are qualitative differences in membrane receptor sites in different tissues. A second inconsistency which arises is how several hormones, all of which influence cyclic AMP, have different effects on the same cell. One possible explanation is that there are quali-tativelydifferent receptor sites in the cell that respond to different hormones. Within the cell there are separate "compartments" of cyclic AMP, each of which when stimulated gains access to a different intracellular site and catalyzes a reaction there. At least twelve hormones act through intracellular synthesis of cyclic AMP, the most notable exception being the steroid hormones. It is interesting to note that cyclic AMP also plays important roles in non-endocrine mechanisms, such as the control of antibody production and the regulation of vision.

Question:

How can one demonstrate that amino acids are used tobuild proteinsin cells, while the proteins already in the cell are usedin catabolism?

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Solution:

One way of following the path of the amino acids is by a process calledautoradiography. We can make amino acids radioactive by the incorporationof radioactive isotopes such as tritium (^3H) or carbon-14. We canthen inject theselabelledamino acids into cells in tissue culture, and notechanges in the radioactive level. This is done by exposing the cells to anauto-radiographicemulsion, which upon exposure to any radioactivity in thecells, produces small grains visible microscopically. If we givelabelled aminoacids to the cells in culture and we look for radioactivity at successiveintervals, we should initially see an increase in the number of grainsas thelabelledamino acids are used to build proteins. As time progresses, the amount of radioactive protein should remain dynamically stable: as protein is broken down, new protein is synthesized. If we cease providingthe cells withlabelledamino acids and give them unlabelled aminoacids instead, the amount of radioactive protein in the cells will decreaseand eventually disappear as radioactive protein is degraded and excreted. The disappearance of radioactive protein indicates that protein isbeingcatabolized. (Note: the new protein synthesized is not radioactive sinceunlabelled amino acids are involved.)

Question:

When a swimmer leaves cold water on a warm, breezy day, he experiences a cooling effect. Why?

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Solution:

The equilibrium vapor pressure, is defined as the pressure exerted by the gas of that substance in equilibrium with the liquid state of the same substance. When the swimmer leaves the cold water, he is, coated with water. As the water evaporates a layer of gaseous water molecules form around the swimmer's body. This vapor attains an equilibrium with the atmosphere which hinders further vaporization of the water. However, when the breeze comes, it blows away the vapor above the water, and thereby keeps the partial pressure of the vapor low above the skin. This, then, causes increased evapora-tion so that the partial pressure can be reestablished. This is true because the concentration and vapor pressure of the gaseous water, molecules must be below a certain point for evaporation to occur. Once thisponthas been reached no more liquid molecules of water can go into the gaseous state. For the water to evaporate, it must go from a liquid to a gaseous phase, a process which requires heat. The water removes heat from the swimmer's body. Thus, the swimmer feels a cooling effect.

Question:

Four thieves have just robbed a bank of all its $100 bills. They decide to hide out for the night in an old man's cabin. While the other three are asleep, one of them wakes up and decides to be greedy by dividing the money into four equal piles, hiding one pile for himself and leaving one $100 bill for the old man. Then he recombines the 3 remaining piles and goes back to sleep. Well, it seems that each of the other three thieves has a similar idea. In separate turns, each awakens, divides the remaining money into 4 equal piles, takes a pile for himself, and slips a left over bill to the old man. When they awaken in the morning, they again divide the considerably smaller pile into four equal piles, with a single bill left over for the old man. Write a FORTRAN program to determine the minimum number of bills in the original pile.

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Solution:

First, let us look at an algebraic interpretation of the night's festivities: T = 4N + 1first portion 3N = 4P + 1second portion 3P = 4Q + 1third portion 3Q = 4R + 1fourth portion 3R = 4S + 1final portion To begin, let S = 1 and evaluate the final expression (4S + 1) to see if it is a multiple of 3. If so, the value of R can be substituted in the next from the last equation. This upward climb is continued as long as integers that are multiples of 3 are being obtained. If this condition fails somewhere, S is increased to the next trial value, and the process is repeated until the top equation of the system is reached. To program this brain teaser, two arrays, K and L, are needed. The equations become L (1) = 4K (1) + 1 = 3K (2) L (2) = 4K (2) + 1 = 3K (3) L (3) = 4K (3) + 1 = 3K (4) L (4) = 4K (4) + 1 = 3K (5) T = 4K (5) + 1. The following flowchart will illustrate the program logic. The answer for T is 1021 $100 bills, or $102, 100 (quite a haul!) The components of the arrays are K:80107143191255 L:321429573765 INTERGER T, K, L DIMENSIONK (5), L (4) 5K (1) = 0 10I = 1 K (I) = K (I) + 1 40L (I) = 4\textasteriskcenteredK (I) + 1 K (I+1) = L (I)/3 IF (3\textasteriskcenteredK (I+1) - L (I)) 10, 50, 60 60PRINT "ERROR - TRY AGAIN" GO TO 5 50I = I + 1 IF (I - 5) 40, 90, 60 90J = 4\textasteriskcenteredK (5) + 1 PRINT 100, T, K, L 100FORMAT (I8 // 5I8 // 5I8) STOP END

Question:

Consider the manometer, illustrated below, first con-structed by Robert Boyle. When h = 40 mm, what is the pressure of the gas trapped in the volume, V_gas . The temperature is constant, and atmospheric pressure is P_atm = 1 atm.

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Solution:

We do not need to know any gas law to solve this problem. All we must realize is thatthe pressure exerted on the gas, P_total is equal to the sum of the pressure exerted by the mercury, P_Hg , and the pressure exerted by the air, P_atm . Since 1 mm Hg = 1 torr and 1 atm =760 torr, P_Hg = 40 mm Hg = 40 torr and P_atm = 1 atm =760 torr. Then P_total = P_Hg + P_atm = 40 torr + 760 torr = 800 torr,

Question:

Using the previous problem write the complete ENVIRON-MENT DIVISION. The SOURCE-COMPUTER is a Honeywell 200, and the system on which the program will be executed is an IBM System/360 Model 40 (IBM-360-40).

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Solution:

Use the system flowchart given in fig. 1 to describe the equipment for a program to randomly update an indexed sequential file. The program uses card input and print output in addition to the random access file. The indexed file requires 2 extra clauses, RECORD KEY and SYMBOLIC KEY if the file is used for random access. The RECORD KEY, C-NR in this program, is used by the computer to search the file for the specific record requested in a program. The SYMBOLIC KEY, SYM-KEY in this program, must be specified if the file is used for output. The variable associated with SYMBOLIC key must be defined in the WORKING STORAGE SECTION. The clause, ACCESS IS RANDOM, sets up the file so that records may be located randomlyby their key, C-NR in this program.

Question:

(a) 0.20 MHCl, (b) 0.10 MNaOH.

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Solution:

A pH scale has been devised to express the H_3O^+ concentration in solution. By definition, pH = - log [H_3O^+] or [H_3O^+] = 10^-pH It has been shown that water dissociates to H_3O^+ and OH^- ions to a small degree. H_2O + H_2O \rightarrow H_3O^+ + OH^-_ . Theequlibriumconstant is defined as K_W for this reaction and is expressed as [H_3O^+][OH^-]. The H_2O does not appear, since it is presumed to be a constant. From the dis-sociation equation, it can be seen that the concentration of H_3O^+ equals OH^-. By experimentation, K_W has been shown to equal 1.0 × 10^-14. This means that in water, therefore, H_3O^+ and OH^- each have a concentration of 1.0 × 10^-7 M. With this information in mind, one can now solve the problem. (a) The concentration ofHClis 0.20 M. SinceHClis a strong electrolyte, dissociation is complete. Therefore, the concentration of H_3O^+ is also 0.20 M = 2.0 × 10^-1 M. By definition, then pH = - log (2.0 × 10^-1) = 1 - 0.3 = 0.7. (b) The [OH^-] equals the concentration ofNaOH, since it is also a strong electrolyte. One wants the pH, there-fore, employ the expression for K_W. [H_3O^+] = (K_W) / [OH^-] = [(1 × 10^-14)/(0.10)] = 1.0 × 10^-13 M. Therefore, pH = - log (1.0 × 10^-13) = 13.

Question:

Given the consecutive reaction Ak_1\rightarrowBk_2\rightarrowC with k_1 = k_2. Draw a graph for the time variation of the con-centrations of A, B, and C.

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Solution:

To draw this graph, you need to determine the concentration with respect to time of each species in this reaction. This will necessitate writing rate law expressions for each of the species. The rate law states that rate is equal to a proportionality constant k times the concentration of one species or more than one species, each raised to a power. The power is equal to the number of moles reacting. You are told that Ak_1\rightarrowBk_2\rightarrowCwith k_1 = k_2. The rate at which A disappears can be written as {-d[A]} / dt= k_1[A]. A will disappear at a certain rate as given by the rate constant. How long it takes to dis-appear will reflect how much was initially present: the concentration. The rate at which C appears may be written as {-d[C]} / dt=k_2[B]. C forms from B at a certain rate as indicated by k_2. Thus, the amount of B initially present determines the time it take for C to appear. What about the concentration of B? It increases with time as A dis-appears, yet decreases as C appears. The net increase in B will be the differences in these processes. Thus, {-d[B]} / dt = k_1[A] - k_2[B]. If you start with pure A at t = 0, {d [B]} / dt is positive, since k_1[A] >> k_2[B]. After a time, how-ever, A is used up, so that {d [B]} / dt becomes negative. This means the concentration of B will peak sometime along in the reaction. At that instant {d[B]} / dt= 0, so that k_1[A] - k_2[B] = 0, which means [A]/[B] = [k_2]/[k_1] Thus, the behavior of the system depends on k_2/k_1. As k_1 grows small, B becomes less notable. But, you are told that k_2 = k_1, which indicates that k_2/k_1 = 1. [C] will rise as [B] falls. Thus, you have the follow-ing graph. If all reactions proceed to completion C will be the only species present at the conclusion of the process.

Question:

The outside curve on a highway forms an arc whose radius is 150 ft. If the roadbed is 30 ft. wide and its outer edge is 4 ft. higher than the inner edge, for what speed is it ideally banked?

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Solution:

We wish to relate the velocity of the car to \varphi, the banking angle. Note that the car is undergoing circular motion, hence its acceleration in the x-direction is a = v^2/R , where R is its distance from the center of the circle (see figure). Applying Newton's Second Law, F = ma, to the x circle (see figure). Applying Newton's Second Law, F = ma, to the x component of motion, component of motion, ma = N sin \varphi Buta = v^2/R and mv^2/R = N sin \varphi(1) The acceleration of the car in the y-direction is zero, since it remains on the road. Applying the Second Law to this component of motion, N cos \varphi = mg(2) Dividing (1) by (2), (mv^2/R)/mg = (N sin \varphi)/(N cos \varphi) = tan \varphi tan \varphi = v^2/Rg Hencev = \surd(Rg tan \varphi) Now, note that the width of the road bed is much smaller than the inner radius of the road. Hence, we may approxi-mate R as the inner radius. R \approx 150 ft v = \surd[(150 ft) (32 ft/s^2) tan \varphi] From the figure, sin \varphi = 4/30 cos^2 \varphi = 1 - sin^2 \varphi = (900/900) - (16/900) = 884/900 Hencecos \varphi= \surd884/30 andtan \varphi= (4/30)/(\surd884/30) = 4/\surd884 = .1345 Thereforev = \surd[(150 ft)(32 ft/s^2)(.1345)] v = \surd(645.6 ft^2/s^2) v = 25.41 ft/s

Question:

Clay contains 30 % AI_2 O_3, 55 % SiO_2, and 15 % H_2 O. What weight of limestone is required per ton of clay to carry out the following chemical change? 6CaCO_3 + AI_2O_3 + SiO_2\rightarrow3CaO\bulletAl_2 O_3 + 3CaO\bulletSiO_2 + 6CO_2. (limestone)

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Solution:

From the coefficients in the reaction equation, one sees that for every mole of AI_2 O_3 and SiO_2 present, moles of CaCO_3 (limestone) are required for the reaction to occur. As such, to determine how much limestone is necessary to react with 1 ton of clay, compute the number of moles of AI_2 O_3 and SiO_2 present. From this figure, the number of moles of CaCO_3 can be found. Because a mole is defined as grams divided by molecular weight (MW), the weight (grams) can be found once the molecular weight is calculated. It is given that 1 ton or 2000 1b of clay will be used. Because clay contains 30 % AI_2 O_3 and 55 % SiO_2 .30 × 2000 = 600 lbs of A1_2 O_3 and .55 × 2000 = 1100 lbs of SiO_2 are present in one ton of clay. To calculate the mole amounts, one must convert lbs to grams. This can be done by using the conversion factor 454 g/lb. Thus, 600 lbs × 454 g/lb = 2724 x 10^2 grams of Al_2 O_3 and 1100 × 454 g/lb = 4994 × 10^2 grams of SiO_2. The MW of Al_2 O_3 = 102 grams/mole. The MW of SiO_2 = 60 grams/mole.Therefore, moles of Al_2 O_3 = [(2724 × 10^2 g ) / (102 g/mole)] = 2.67 ×10^3 moles and moles of SiO_2 = [(4994 × 10^2 g) / (60 g/mole)] = 8.32 × 10^3 moles Notice, in one ton of clay, the mole amounts of SiO_2 and Al_2 O_3 are not equal. The mole amount of limestone re-quired will be six times the mole amount of Al_2 O_3, 2.67 × 10^3 , and not SiO_2. The reason stems from the fact that AI_2 O_3 is the limiting reagent. The amount of any reagent required (or product produced) depends only on the limiting reagent. SiO_2, with a mole of 8.3 × 10^3, is in excess; the amount of limestone required will not depend on it. Consequently, moles of limestone = 6(moles of AI_2 O_3) = 6 (2.67 × 10^3) = 1.60 × 10^4 moles. The molecular weight of CaCO_3 (limestone) = 100 g/mole. Grams of CaCO_3 required per ton of clay = (moles of CaCO_3) (MW of CaCO_3) = (1.60 × 10^4) (100) = 1.60 × 10^6.

Question:

Discuss the modes of asexual reproduction exhibited by Discuss the modes of asexual reproduction exhibited by higher plants.

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Solution:

A number of commercial plants-bananas, seedless grapes, navel oranges, and some varieties of potato, to name just a few-have lost the ability to produce functional seeds and must be propagated entirely by asexual means. Many cultivated trees and shrubs are asexually reproduced from the cuttings of stems, which will develop roots at their tips when placed in moist ground or in water containing a small amount ofindole-acetic acid (auxin). Many plants such as the strawberry develop long, horizontal stems calledstolonsor runners. These grow some distances along and above the ground in a single sea-son and may develop new erect plants at every other mode. Other plants spread by comparable, but underground, stems called rhizomes. Such weeds as witch grass and crab grass spread very fast by means of either runners or rhizomes. Swollen underground stems, or tubers, also serve as an asexual means of reproduction in plants such as the potato. New plants grow out at the buds of the tuber. The stems of raspberries, currants, and wild roses, and the branches of several kinds of trees maydroopto the ground. Adventitious roots and a new erect stem may develop at one of the nodes touching the ground. The stem or branch may lose its connection with the parent plant and grow into a new, independent plant.

Question:

What volume of a block of wood (density = 0.80 g / cm^3), which weighs 1.0 kg, will be above the water surface when . the block is afloat? (Note: A floating object displaces its own weight of water.)

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Solution:

Since a floating object displaces its own weight in water, 1 kg of water is displaced by this block of wood. One can find the volume of the block of wood above the water by solving for the volume of the block and subtracting the volume of 1 kg of water from it. One uses the density to solve for the volume. density = (Weight) / (Volume) = (g / cm^3) Therefore: volume = (Weight / density) Solving for the volume of the woods: 1 kg = 1000 g, Volume = 1000 g /(.80 g/cm^3) = 1.25 × 10^3 Solving for the volume of the water displaced: By definition the density of water is (1.0 g /cm^3). volume = (1000g) / [1.0(g / cm^3)] = 1.00 × 10^3cm^3 volume of wood above water = (volume of wood)- (volume of water) volume of wood above water = (1250 cm^3) - (1000 cm^3) = 250 cm^3

Question:

Determine ∆H\textdegree for the following reaction of burning ethyl alcohol in oxygen: C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l) ∆H\textdegree_f of C_2H_2OH(l) = \rule{1em}{1pt} 65.9 Kcal/mole ∆H\textdegree_f ofCO_2(g) = \rule{1em}{1pt} 94.1 Kcal/mole ∆H\textdegree_fofH_2O(l) = \rule{1em}{1pt} 68.3 Kcal/mole

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Solution:

The heat reaction (∆H\textdegree) may be found from the heats of formation (∆H\textdegree_f) by subtracting the sum of the heats of formation of all reactants from the sum of the heats of forma-tion of all products. The heat of formation of all pure elements is zero. When more than one mole of a compound is either reacted or formed, the heat of formation is multiplied by the stoichiometric coefficient for the specific compound in the equation. heat of reaction = [2 × ∆H\textdegree_f of CO_2 + 3 × ∆H\textdegree_f of H_2O] \rule{1em}{1pt} [∆H\textdegree_f of C_2H_2OH + 3 moles × ∆H\textdegree_f of O_2] heat of reaction = [2 moles × (\rule{1em}{1pt} 94.1 Kcal/mole) + 3 moles × (\rule{1em}{1pt} 68.3 Kcal/mole) - 1 mole × (\rule{1em}{1pt} 65.9 Kcal/mole + 3 moles × 0] = \rule{1em}{1pt} 327.2 Kcal.

Question:

Calculate the volume of oxygen necessary to burn completely 1000 cubic feet of butane gas according to the equation 2C_4 H1 0+ 13O_2 \rightarrow 8CO_2 + 10H_2 O

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Solution:

Since volumes are concerned, the procedure for solving this problem is to set up a ratio between the volumes present and the mole requirements. For this re-action, 2molesof butane react with 13 moles of O_2. When two gases are under the same temperature and pressure conditions, a mole of either gas will occupy the same volume. Therefore, since 13/2 times as many moles of O_2 are required, the volume of O_2 must be 13/2 times that of methane. Given the volume of methane is 1000 cu. ft., the volume of O_2 = (13/2) (1000) = 6500 cubic feet.

Question:

A uniform wooden beam, of length 20 ft and weight 200 lb, is lying on a horizontal floor. A carpenter raises one end of it until the beam is inclined at 30\textdegree to the horiz-ontal. He maintains it in this position by exerting a force at right angles to the beam while he waits for his mate to arrive to lift the other end. What is the magni-tude of the force he exerts?

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Solution:

Consult the diagram: AB is the beam and C its midpoint. The weight W acts through C, since the beam is uniform and the two other forces acting on the beam are the force F\ding{217} exerted by the carpenter at B at right angles to AB and a total force P\ding{217} exerted by the floor at A in an unspecified direction. Since the beam is acted on by three forces which maintain it in equilibrium, the lines of action of the three forces must be concurrent. Thus, the direction of force P\ding{217} is from A to the point D at which F\ding{217} and W\ding{217} meet. Since the board is not in motion, the second condition of equilibrium can be applied. Taking moments about point A, we have ^T\sum_A = P × 0 + F × AB - W × AE = 0 F = W(AE/AB) = W[(AC cos 30)/(AB)] = (1/2)W cos 30 = (\surd3/4) W. \therefore F = 86.6 lb.

Question:

In water vapor at 25\textdegreeC, what is the average speed of a water molecule in meters/sec? Recall that 1 Joule = 1Kg-m^2/sec^2. The Boltzmann constant, k = 1.3806 × 10^-23 J/deg.

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Solution:

Recall, that for all bodies in motion, kinetic energy (K.E.) = (1/2) MV^2, where M = mass of body andV= average velocity of the molecules. Thus, if K.E. and the mass are known, thenVcan be solved. To find K.E., note, that for gases average K.E. is proportional to temperature, This can be written as K.E. = 3/2kT, where k is theBoltzman constant 1.3806 × 10^-23 J/deg and T = temperature in Kelvin (Celsius plus 273\textdegree) . You are given the tem-perature of the system (25\textdegreeC or 298\textdegreeK). Thus, K.E. = 3/2(1.38 × 10^-23)(298) = 6.17 × 10^-21 J/molecule. Since K.E. = (1/2)MV^2, 6.17 × 10^-21 = (1/2)MV^2. If you knew the mass of one water molecule, you could substitute it into this expression and find V. The mass of the H_2O molecule can be found by using the molecular weight (M.W.) and Avogadro's number (6.02 × 10^23/mole). H_2O has an M.W. of 18 grams/mole. Thus, the mass of one molecule = (18 g/mole) / {(6.02 × 10^23 (molecule/mole)} = 2.99 × 10^-23 grams/molecule. = 2.99 × 10^-23 grams/molecule. Substituting and solving for V, V = \surd[2(K.E.)/ M] = \surd[2{6.17 × 10^-21( kg - m^2 / sec^2 - molecule)} / {(2.99 ×10^-23 g/molecule) (.001 kg/g)}] V = 642 m/sec. (Note: .001 kg/g is a conversion factor.)

Question:

Calculate the quantity of heat required to (a) convert a liter of H_2O at 30\textdegreeC to a liter of water at 60\textdegreeC, and (b) heat a 1 kg block of aluminum from 30\textdegree to 60\textdegreeC. Assume the specific heat of water and aluminum is, re-spectively, 1 cal/g\textdegreeC and .215 cal/g\textdegreeC.

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Solution:

When heat is added to a mass of unspecified substance, the temperature will rise. The quantitative relationship between the quantity of heat, expressed in kilocalories, and the rise in temperature - provided there is no phase change - is: quantity of heat = mass × specific heat × ∆T, where ∆T is the change in temperature and the specific heat equals the amount of heat required to raise the temperature of 1 g of any substance by 1\textdegreeC. (a) To find the mass of H_2O, remember that density = mass/volume , and the density of H_2O is one. Thus, mass = (1 liter H_2O) (1000 ml/liter) (1.00 g H_2O)/ml. H_2O). Therefore, quantity of heat= (mass) (specific heat) (∆T) mass= (1 l H_2O) (1000 ml/l) (1 g/ml) = 1000 g specific heat= 1 cal/g\textdegreeC ∆T= (60 - 30)\textdegreeC = 30\textdegreeC quantity of heat= (1000 g)(1 cal/g\textdegreeC)(30\textdegreeC) = 30,000 cal = 30 Kcal (b) A similar calculation can be used to determine the amount necessary to raise the temperature of the aluminum from 30 to 60\textdegreeC. mass= (1 kg Al)(1000 g/kg) = 1000 g specific heat= .215 cal/g\textdegreeC ∆T= (60 - 30)\textdegreeC = 30\textdegreeC quantity of heat= (1000 g)(.215 cal/g\textdegreeC)(30\textdegreeC) = 6450 cal = 6.45 Kcal.

Question:

Find the vapor pressure of CCl_4 (carbon tetrachloride) at 38\textdegreeC, if H^1 = 32,000 (J / mole) and at 23\textdegreeC, the vapor pressure is .132 atm.

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Solution:

To solve this problem, employ the Clausius- Clapeyron equation, which permits evaluation of vapor press-ure in terms of ∆H^1 (.the heat required to transform one mole of liquid to the ideal-gas state) , the original vapor press-ure and the two temperatures. The equation states: log (p_1 / p_2) = (∆H^1 / 19.15)[(1 / T_2) - (1 / T_1)], where p_1 and p_2 are vapor pressures and T_1 and T_2 are the temperatures in Kelvin. Let p_2 be the vapor pressure of CCl_4 at 38\textdegreeC. Substitute the known values for solution, log(.132 / p_2) = [(32,000) / (19.15)] [(1 / 311) - (1 / 296)]. solve for p_2, to obtain p_2 = .250 atm.

Question:

In pea plants, tall plants (T) are dominant to dwarf (t)t yellow color (Y) is dominant to green (y), and smooth seeds (s) are dominant to wrinkled seeds (s). What would be the phenotypes of the following matings? a) Tt Yy Ss×ttyyss b) Tt yy Ss×ttYySs

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Solution:

This is an example of a trihybrid cross, which means a cross involving three different traits. The method of solving follows the same principles as that for any other cross. a) Before we can do the cross, we must determine the gametes involved. We can obtain the total number of different possible gametes from each parent by using a simple mathematical rule. The total number of possible combinations is 2^n, where n is the number of heterozygous traits involved in the cross. For the first parent (Tt Yy Ss), n equals 3 and the total combinations is 2^3 or 8. The genotypes of these eight gametes can be de-termined by using a dichotimous branching system (or forkedline method). By following any set of lines from T (or t) to S (or s) one can determine all possible combinations of factors in the following fashion: Gametes Gametes Y S TYS Y S Tys T s TYs t s tYs y S TyS y S tyS S TYs s tys since the second parent is homozygous (ttyyss) there is only 1 possible gamete type, tys, because n = 0 and 2\textdegree = 1. in the cross: F_1 tys TYS TtYySs TYs TtYyss TyS TtyySs Tys Ttyyss tYS ttYySs tYs ttYyss tyS ttyySs tys ttyyss Phenotypically,TtYySsistall,yellow,smooth; TtYyssistall,yellow,wrinkled; TtyySsistall,green,smooth; Ttyyssistall,green,wrinkled; ttYySsisshort,yellow,smooth; ttYyssisshort,yellow,wrinkled; ttyySsisshort,green,smooth;and ttyyssisshort,green,wrinkled. b)There are two heterozygous traits in the parent with genotype TtyySs, so n equals 2^2. The number of different gametes is or 4. Determining their genotypes: Gametes T-y S TyS s Tys t-y S tyS s tys There are four different gametes from the second parent with genotype ttYySs (2^2 = 4). Gametes Y S tYS t s tYs y S tyS s tys In the cross: F^1 TyS tys Tys tys tYS TtYySS ttYySS TtYySs ttYySs tyS TtyySS ttyySS TtyySs ttyySs tYs TtYySs ttYySs TtYyss ttYyss tys TtyySs ttyySs Ttyyss ttyyss This cross could also be done using the forked line method. View the trihybrid cross as three separate monohybrid crosses. (Since genes segregate independently, we can legitimately do this). Then the genotypic pairs resulting from each cross can occur in any possible combinations with the gene pairs produced in any other cross. The ratios from each cross remain the same as they would if done separately. (Note that this method assumes that you can do the fairly simple monohybrid crosses by inspection, having had sufficient exposure and experience with them.) Summarizing our results: Phenotypes Genotypes Genotypic Frequency Phenotypic Frequency Tall, yellow, smoot TtYySS 1 3 TtYySs 2 1 Tall, yellow, wrinkled TtYyss 1 1 Tall, green, smooth TtyySS 1 3 TtyySs 2 Tall, green, wrinkled Ttyyss 1 1 Short, yellow, smooth ttYySS 1 ttYySs 2 3 Short, yellow, wrinkled ttYyss 1 1 Short, green, smooth ttyySS 1 3 ttyySs 2 Short, green, wrinkled Ttyyss 1 1

Question:

If a particle of mass 100 g initially had speed 1 × 10^2 cm/s, what would be its kinetic energy and velocity at the end of its 10-cm fall?

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Solution:

The initial kinetic energy is E_i = (1/2) mv^2_i = (1/2) (100 g) (10^4 cm^2/s^2) = 5 × 10^5 ergs During the fall, the particle moves as a result of the gravitational force F_G^\ding{217} acting on it. The work done by F_G^\ding{217} as the particle falls through the height h is, since F_G^\ding{217} is constant and in the same direction as the displacement, W = F_G × h = (100 g) (980 cm/s^2) (10 cm) = 9.8 × 10^5 ergs W is the change in kinetic energy as the particle moves in the direction of the gravitational force. Therefore, the final kinetic energy at the end of the fall is E_f = E_i + W \approx 15 × 10^5 ergs or(1/2) mv^2_f \approx 15 × 10^5 ergs v^2_f \approx (30 × 10^5 ergs)/(100 g) v^2_f \approx 3 × 10^4 cm^2/s^2 v_f \approx 1.73 × 10^2 cm/s This result agrees with what we would calculate from F^\ding{217} = Ma^\ding{217}, but note that we have not specified above the di-rection of the initial speed 1 x 102 cm/s. If it were in horizontal x direction (see the Fig.) it would remain constant and from E_f - E_i = W,we would have (1/2)m(v^2_xf + v^2_yf) - (1/2) m(v^2_xi + v^2_yi) = W But v_xf = v_xi and v_yi = 0, hence (1/2) m(v^2_xf + v^2_yf)^2 - (1/2) mv^2_xi = Wor(1/2)mv^2_vf = W and v_yf^ = \surd(2w/m) = \surd[(2)(9.8 × 10^5 erg)/100 g] v_yf= 139 cm/s. or if v_y^\ding{217} were initially downward in the negative y^\ding{217} direction, we could call upon the familiar relationships for falling bodies: y = y_i + v_yit + (1/2)gt^2 v_yf = v_yi + gt where x_i, v_yi are the initial y position and y velocity of the particle. Hence h = y - y_i = v_yit + (1/2)gt^2(1) v_yf = v_yi + gt(2) where y - y_i is the distance the particle falls through (h). To eliminate t in equation (1), solve (2) for t and substitute in (1). (v_yf - v_yi)/g = t h = v_yi [(v_yf - v_yi)/g] + g/2 [(v_yf - v_yi)/g]^2 orh = 2v_yi [(v_yf - v_yi)/2g] + [(v_yf - v_yi)/2g]^2

Question:

Two men lift the ends of a 20 ft beam weighing 200 lb onto their shoulders. Both men are of the same height so that the beam is carried horizontally, but one is much the stronger of the two and wishes to bear 50% more of the weight than his mate. How far from the end of the beam should he put his shoulder ?

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Solution:

The beam exerts downward forces on the shoulders of the 2 men. This total downward force is just the weight of the beam and, since the beam is uniform, the weight acts through it's center (see figure). The 2 men exert upward forces R_c\ding{217} and R_m\ding{217}, the former at distance x from one end and the latter at the other end. The beam is in equilibrium vertically since it experiences no motion in that direction. Therefore we can apply the first condition of equilibrium in the vertical direction. \sum F_y = R_m + R_c - 200 lb = 0 R_m + R_c = 200 lb.(1) But we were given in the problem that one man, (the one on the right end) carries 50% more of the weight than his mate. Therefore R_c = (150/100) R_m orR_m = (2/3) R_c(2) Substituting equation (2) in (1), we find (2/3) R_c + R_c = 200 lb (5/3) R_c = 200 lb R_c = 120 lb Note that we solved for R_c\ding{217} since it is this force's distance from end D that we wish to know. Since the beam is in equilibrium rotationally, the second condition of equilibrium can be applied. Taking moments about end A, the magnitude of the force R_m\ding{217} is not needed. We have T \sum_A = (R_m) (0) + Rc(20 ft - x) - (200 lb) (10 ft) = 0 20 ft - x - 2 = (2000 ft-lb)/R_c = (2000 ft-lb)/(120 lb) = (16)(2/3) ft. x = 20 ft - 16 (2/3) ft = 3 (1/3) ft

Question:

A chemist mixes together 100 ml of 0.5 M sodium acetate, 100 ml of 0.25 M hydrochloric acid (HCl) , and 100 ml of a 1.0 M salt solution. She dilutes this to 1000 ml. Deter-mine the concentrations of all the ions present,un-dissociated acetic acid, and the final pH of the solution, using K_a of acetic acid = 1.7 × 10^-5 andK_w= 1.0 × 10^-14.

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Solution:

To start, determine if any reaction took place in this mixture. If no reaction took place, then the molar concentrations will be numerically the same in 1 liter as the computed moles from the initial concentrations. How-ever, a reaction does take place. Namely, HOAc(acetic acid) + H_2O \rightleftarrows H_3O^+ +OAc^-. The H_3O^+ (fromHCl) reacts with the acetate ion (OAc^-) from the NaOAc(sodium acetate) to produce acetic acid and water. Start the analysis of the reaction by computing the number of moles involved. M =molarity= moles/liter. Thus, in 100 ml of 0.5 MNaOAc, there are 0.5 (100/1000) = 0.05 moles ofNaOAc. Similar calculations show that 0.025 moles ofHCland 0.1 moles ofNaCl(the salt solution) are present. The problem asks one to find the pH, the concentration of the ions present, and the concentration ofundissociatedacetic acid. This necessitates writing the equilibrium constant expression for this reaction, from which these concentrations can be calculated. Thus, write that K_a, the equilibrium constant, equals {[H_3O^+] [OAc^-]} / [HOAc]. Suppose the reaction went to completion. If it did, 0.25moleseach ofHOAcand excessOAc^- would have been produced. One starts with 0.5 moles and only 1 mole ofOAc^- can react per mole of H_3O^+. If there exist 0.25 moles H_3O^+,then, only 0.25molesof theOAc^- can react, leaving 0.25 moles in excess. If the reaction, however, stops x moles short of completion, x moles of H_3O^+ would remain and the excess ofOAc^- would increase from 0.25 to 0.25 + x. TheHOAcwould be reduced to (0.25 - x) moles. Since they are in 1000 ml or 1 liter, these become concentrations that can be substituted into K_a = {[H_3O^+] [OAc^-]}/[HOAc] to give K_a = {x (0.025 + x)}/(0.025 - x). It is given that K_a = 1.75 × 10^-5. Equating, {x (0.025 + x)}/(0.025 - x) = 1.75 × 10^-5. When the K_a for an acid is small, one knows that dissociation is minimal and therefore the concentration of H_3O^+ is very small compared to the concentration ofOAc^- andHOAc. This means that one can assume that their concentrations will not be appreciably changed by the dissociation. Thus, 0.025 - x \allequal 0.025 and 0.025 + x \allequal 0.025. The equation for the K_a can be rewritten as K_a = {x (0.025)} / (0.025) = 1.75 × 10^-5 {x (0.025)} / (0.025) = 1.75 × 10^-5 x = 1.75 × 10^-5. Therefore,[H_3O^+] = 1.75 × 10^-5 M. Since,pH = - log [H_3O^+], pH = - log [1.75 × 10^-5] = 4.76. [OAc^-] = 0.025 + x = 0.025 + 1.75 × 10^-5 \allequal 0.025. [HOAc] = 0.025 - x = 0.025 - 1.75 × 10^-5 \allequal 0.025. The other ions: [Na^+] = 0.150 M fromNaOAcandNaCl. [Cl^-] = 0.125 M fromHClandNaCl. Since [H_3O+] [OH^-] =K_w= 1.0 × 10^-14 and[H_3O+] = 1.7 × 10^-5, [OH^-] = (1.0 × 10^-14) / (1.7 × 10^-5) = 5.9 × 10^-10 M.

Question:

A 1-gram block rests on a frictionless surface and we measure the position of the block to a precision of 0.1 mm. What velocity have we imparted to the block by the act of measuring its position?

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Solution:

If we try to determine the position of a particle at a point x_0 along the x axis as shown in the figure we find that our measurement of the position x_0 cannot be made as precise as required (assuming that it is at rest). As we try to confine the particle to a more exact position, we become less certain about whether it really is at rest at that position. This is a consequence of quantum mechanics and is known as the Heisenberg uncertainty principle. If the uncer-tainty involved in the measurement of position is \Deltax, then the antic-ipated uncertainty in the momentum of the particle will be such that \Deltax \Deltap_x \approx h where h is Planck's constant. Since p_x = mv_x , we have \Deltax m\Deltav_x \approx h or \Deltav_x \approx (h/m \Deltax) = [(6.6 × 10^-27 erg-sec) / {(1 g) × (10^-2 cm)}] = 6.6 × 10^-25 cm/sec This velocity is so small that we must consider the block to be still "at rest". The implications of the uncertainty principle for macro-scopic objects are unimportant, but for microscopic objects they are crucial. That there is any uncertainty at all in the measurement arises from the fact that we can see the block only by virtue of the scatter-ing of photons from the block and this process imparts momentum to the block.

Question:

A copper strip 2.0cm wide and 1.0mm thick is placed in a magnetic field with B = 1.5T, as in Figure 1. If a current of 200 A is set up in the strip, what Hall potential differ-ence appears across the strip?

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Solution:

The Hall electric field is given by E_H = (jB) / (ne); But E_H = [(V_(x)y) / d], j = (i / A) = (i / dh), and n = (dN_0 / M), where h is the thickness of the strip. Combining these equations gives V_xy = [(iB) / neh] = [{(200A)(1.5T)} / {(8.4 × 10^28/m^3)(1.6 × 10^-19c)(1.0 × 10^-3m)}] = 2.2 × 10^-5v = 22\muV. These potential differences, though quite measurable, are not large.

Question:

(a)Use a truth table to show that the sum (disregarding the carry) of two binary digits x and y is given by the Boolean expression: (x\bullet \simy) + (\simx \bullet y)(Statement 1) (b)Write a Boolean expression for the carry. (c)Set up a circuit to find the sum and carry of two binary digits, using AND, OR and NOT gates.

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Solution:

Recall that the rule of addition of two binary digits is given in the truth table of Figure 1. First addend Second addend Sum digit Carry digit 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 Since x and y each could be either 0 or 1, there are 2^2 = 4 possible combinations of values of x and y. Evaluating the terms in parentheses of Statement 1, and then Statement 1 itself, the truth table of Figure 2 is created. x y \simx \simy x\bullet \simy \simx \bullet y (x\bullet \simy) + (\simx \bullet y) 0 0 1 1 0 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 0 0 We see that the entries in the last column of the truth table match the corresponding entries (that is, given the same values for x and y) in the sum-digit column. Therefore, the Boolean sum (that is, disregarding the carry) of x and y can be represented by the logical ex-pression given in Statement (1). Statement (1) is also called the nonequivalence function of x, y, written x (NOT =) y = (x\bullet \simy) + (\simx \bullet y). Note that x (NOT =) y = 1 if and only if x \equiv y = 0, that is, when x \not = y, which is why it is called the nonequivalence function. It is also known as the EXCLUSIVE - OR function. (b)Note that the carry digit is one if and only if both addends, x and y, are 1. But this is the definition of the AND function. Therefore, the carry digit can be represented by x \bullet y. (c)The circuit for the carry is simply the gate repre-sentation for x \bullet y (see Figure 3). Using the rule that oper-ands of the + operation are inputs to an OR gate, operands of the \bullet operation are inputs to an AND gate and the operand of the \sim operation is the input to a NOT gate, we can repre-sent the terms in parentheses as shown in Figure 4. Since each term in parentheses is an operand of the + operation, each of the above gate circuits is an input to an OR gate. Running the carry circuit in parallel to the sum circuit would complete the network. The network in Figure 5 is known as a half-adder. The half-adder has two inputs, corresponding to the two addends, and it will have two outputs, one giving the sum digit and the other giving the carry digit. It performs only half the task for full addition, since it provides neither for a previous carry nor for placing the carry bit to the next column. For complete addition two half-adders are required. Methods for addition may be classified as either parallel or serial. In parallel addition each column is added simultaneously by a separate full-adder. In serial addition each column to be added is transmitted one digit at a time, beginning with the digit in lowest position. The carry bit is retained to be combined with the addition results of the next column. The sum is given as a sequential output.

Question:

A block hangs at rest from the ceiling by a vertical cord. Find the forces acting on the block and on the cord.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0022.htm

Solution:

Part (b) of the figure is the free-body diagram for the body. The forces on it are its weight w_1\ding{217} and the upward force T_1\ding{217} exerted on it by the cord. If we take the x-axis horizontal and the y-axis vertical, there are no x-components of force, and the y-components are the forces w_1\ding{217} and T_1\ding{217}. Then, from the condition that \sumF_y = 0, we have \sumF_y = T_1 - W_1 = 0,T_1 = \sumF_y = T_1 - W_1 = 0,T_1 = W_1 from Newton's first law. In order that both forces have the same line of action, the center of gravity of the body must lie vertically below the point of attachment of the cord. Let us emphasize again that the forces W_1\ding{217} and T_1\ding{217} are not an action-reaction pair, although they are equal in magnitude, opposite in direction, and have the same line of action. The weight w_1\ding{217} is a force of attraction exerted on the body by the earth. Its reaction is an equal and opposite force of attraction exerted on the earth by the body. The reaction is one of the set of forces acting on the earth, and therefore it does not appear in the free-body diagram of the suspended block. The reaction to the force is T'_1\ding{217} an equal down-ward force, T'_1, exerted on the cord by the suspended body. T_1 = T'_1 (from Newton's third law). The force T'_1 is shown in part (c), which is the free-body diagram of the cord. The other forces on the cord are its own weight w_2\ding{217} and the upward force T_2\ding{217} exerted on its upper end by the ceiling. Since the cord is also in equilibrium, \sumF_y = T_2 - w_2 - T'_1 = 0 T_2 = w_2 + T'_1 .(1st law) The reaction to T_2\ding{217} is the downward force T'2\ding{217} in part (d), exerted on the ceiling by the cord. T_2 = T'_2(3rd law) As a numerical example, let the body weight 20 lb and the cord weigh 1 lb. Then T1= w_1 = 20 lb, T'_1 = T_1 = 20 lb, T_2= w2+ T'_2 = 1 lb + 20 lb = 21 lb, T'_2 = T_2 = 21 lb.

Question:

What will be the final concentration of I^- in a solution made by mixing 35.0 ml of 0.100 M NaHSO_3 with 10.0 ml of 0.100 M KIO_3? Assume volumes are additive.

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Solution:

The equation for this reaction is From equation 2 one notices that one mole of each H_2O, HSO^-_3, and I_2 are needed to form 2 moles of I^-. There are 5.0 ×10^-4 moles of H_2O, 1.0 ×10^-4 moles of HSO^-_3, and 5.0 ×10^-4 moles of I_2 present. Thus, the I_2 is the limiting reagent because there is the least of it present to react. Therefore, 2 × 5.0 ×10^-4 moles or 1.0 ×10^-3 moles of I^- formed. If the volumes are additive, the final volume present is 45 ml. Thus, there are 1.0 ×10^-3 moles of I^- present in the final volume of 45 ml of solution.

Question:

Draw a circuit realization of a D (delay) flip-flop, explain the operation of the device.

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Solution:

A circuit realization of a D flip-flop is shown in fig. 1. Note that the circuit is implemented by modi-fying the cross coupled NAND gates. It is important to remember that the clock input can be normally 1 or normally 0; however, in either case data is always clocked to the output Q on a 0 to 1 transition of the clock input C. Inputs Present State Next State D(t) C(t) Q(t) Q(t + E) 0 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 0 0NO 0 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 1 1change 0 0\rightarrow1 0 0Data is 0 0\rightarrow1 1 0clocked in 1 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 0 0No 1 0\rightarrow0, 1\rightarrow1, 1\rightarrow0 1 1change 1 0\rightarrow1 0 1Data is 1 0\rightarrow1 1 1clocked in Fig.2 The characteristics of the device are summarized by the truth table, shown in figure 2, and the state table shown in figure 3.

Question:

A bag of candies is emptied onto the pan of a spring scale which was originally reading zero. Each piece of candy weighs 1 oz. and drops from a height of 4 ft. at the rate of 6 per second. What is the scale reading at the end of 10 sec. if all collisions are perfectly inelastic?

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Solution:

Each piece of mass m_0 loses\DeltaE_p= m_0gh amount of potential energy while falling through a height h, g being the gravitational acceleration. This loss of poten-tial energy is compensated for by an increase\DeltaE_kin the kinetic energy since the total energy of each piece remains constant during the fall. \DeltaE_k=\DeltaE_p (1/2) m_0v^2 = m_0gh where v is the speed with which each piece strikes the pan. We have v^2 = 2gh = 2 × 32 ft/s^2 × 4 ft = 256 ft^2/s^2 v = 16 ft /s. Since the collision is perfectly inelastic, each time a piece hits the pan, it loses all its momentum and stays on the pan. The mass m striking the pan will exert a force F^\ding{217} on it as a result of the change in its momentum such that the impulse created by the impact equals the momentum change, Ft^\ding{217} =mv^\ding{217}, F^\ding{217} = v^\ding{217} (m/t). Here F is the force acting on the candies due to the pan, and t is the duration of the impact. But, the candy pieces are constantly striking the pan and there is a continuous flux of candies incident on the pan. Hence, if m/t = w/gtis the mass of candies striking the pan per second, F = (16 ft/s) × (6 pieces/s) × (1 oz/piece × {1/16 oz/1b} × {1/32 ft/s^2}) = 6/32 1b = 3 oz. The candies must provide an equal and opposite force F on the pan. After 10 seconds, the force acting on the scale is the continuous force F due to the constant impact plus the weight of the candies already in the pan. The scale reading = 3 oz. + (6 pieces/s) × (1 oz/piece) × 10s = 3 oz. + 60 oz. = 63 oz. = 3 1b. 15 oz.

Question:

A common rocket propellant is a mixture of one part liquid hydrazine (N_2H_4) as fuel and two parts liquid hydrogen peroxide (H_2O_2) as oxidizer. These reactants are hypergolic, that is, they ignite on contact. The reaction proceeds according to the equation N_2H_4(l) + 2H_2O_2 (l) \rightarrow N_2(g) + 4H_2O(g). If reactants are mixed at 25\textdegreeC, what is the heat of reaction?

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Solution:

We will make use of the following expression for the heat of reaction at 25\textdegreeC (298\textdegreeK), ∆H\textdegree_298: ∆H\textdegree_298 = sum of heats of formation of products - sum of heats of formation of reactants. Denoting the heat of formation of species s by ∆H\textdegree_f,s, this becomes ∆H\textdegree_298 = ∆H\textdegreef, N2 (g)+ 4∆H\textdegreef, H2O (g)- [∆H\textdegreef, N2H4 (l)+ 2∆H\textdegreef, H2O2 (l)] From a table of the standard heats of formation at 298\textdegreeK, we find ∆H\textdegreef, N2 (g)= 0 Kcal/mole, ∆H\textdegreef, H2O (g)= -57.8 Kcal/mole, ∆H\textdegreef, N2H4 (l)= 12 Kcal/mole, and ∆H\textdegreef, H2O2 (l)= - 46 Kcal/mole. Then ∆H\textdegree_298= ∆H\textdegreef, N2 (g)+ 4∆H\textdegree_f,H2O (g) - [∆H\textdegreef, N2H4 (l)+ 2∆H\textdegreef, H2O2 (l)] = 0 + 4(- 57.8) - (.12 + 2 (- 46)) = - 151 Kcal/mole.

Question:

K_dissof water for 2H_2O \rightleftarrows H_3O^+ + OH^- changes from 1.0 × 10^-14 at 25\textdegreeC to 9.62 × 10^-14 at 60\textdegreeC. Does the pH of water or its neutrality change when the temperature is increased from 25\textdegreeC to 60\textdegreeC?

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Solution:

To answer this question, it is necessary to find [H^+], since pH = - log [H^+]. To do this, write out the equilibrium dissociation expression for water, which equates the dissociation constant,K_diss, with the ratio of concentrations of products to reactants, each raised to the power of its coefficient in the chemical reaction. ThusK_diss= [H_3O^+][OH^-]. The concentration of water is omitted; it is assumed to be constant. For the chemical reaction one can see that OH^- and H_3O^+ are formed inequimolar amounts, which means [OH^-] = [H_3O^+]. Thus, whenK_diss= 1.0 × 10^-14, one has 1.0 × 10^-14= [H_3O^+][OH^-] = [H_3O^+]^2 = [OH^-]^2. Solving for either, [H_3O^+] = [OH^-] = 1.0 × 10^-7. Thus, pH = - log [H_3O^+] = 7. WhenK_diss = 9.62 × 10^-14 at 60\textdegreeC, 9.62 × 10^-14 = [H_3O^+][OH^-] = [H_3O^+]^2 = [OH^-]^2. Solving for either, [H_3O^+] = [OH^-] = 3.1 × 10^-7. Thus, pH = - log [H_3O^+] = 6.51. Thus, the pH does change. What about neutrality? Neutrality occurs when [H^+] = [OH^-]. In both cases, such a condition exists, so that the neutrality was not altered when the temperature was changed.

Question:

Consider a market model showing a positive relationship between number of new houses sold and washing machines sold. Assuming that the number of households increases exponentially and that washing machines break down, draw a system dynamics model for the above system.

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Solution:

Let H be the number of households, y - the num-ber of houses sold and x - the number of washing machines installed. Since the number of houses sold increases exponentially, the differential equation that gives the rate of change of houses sold over time is y ̇ = k (H - y) .(1) If the washing machines, once installed, never broke down, the rate of growth of washing machines would be ẋ = k_2 (y - x). But washing machines do break down. This means that x, the number of washing machines in use should fall, for every level of x. A new equation needs to be given for x. Assuming that the number of breakdowns is pro-portional to the level of x, we have ẋ = k_2 (y - x) - k_3x.(2) Next, the number of households increases exponentially. This is a more realistic assumption regarding the number of households than assuming that H is constant. The new equation for H is Ḣ = k_4H.(3) A systems diagram for the decay model would trace the interdependencies in the equation system (1) - (3).

Question:

Describe, using equations, what takes place when the following solutions are prepared: A.A solution containing equal amounts of 0.10M lead nitrate and 0.10M potassium chromate; B.A solution containing 1 mole of sodium chloride and 1 mole of potassium bromide.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0387.htm

Solution:

In such problems, two processes can occur: mixing of ions in solution and precipitation. A.Both lead nitrate (Pb(NO_3 )_2 ) and potassium chromate (K_2 CrO_4 ) are highlysolube, so that four ions are present in solution: Pb^2+ , K+ , NO-_3 , CrO^2-_4. There are four possible combinations of these ions:Pb(NO_3 )_2 , PbCrO_4 , KNO_3 , and K_2Cr_2 O_4 . Of these, only lead chromate, PbCrO_4 , is insoluble, so that the reaction Pb^2++ CrO_4^2-\rightarrow PbCrO_4 is driven to the right, and yields a precipitate, leaving K+ and NO-_3 ions in solution (along with trace amounts of Pb^2+ and CrO_4^2-, which exist in equilibrium with solid lead chromate). B.Both sodium chloride (NaCI) and potassium bromide (KBr) are highly soluble, giving rise to Na+ , K+ ,Cl-, and Br- ions in solu-tion. The four possible combinations of these ions areNaCI,NaBr,KCl, andKBr, all of which are highly soluble. Therefore no preci-pitate is formed, but only mixing of the ions according to the equation (Na+ ,Cl-) + (K+ , Br-) \rightleftarrows (Na+ , Br- ) + (K+ ,Cl-) .

Question:

Calculate the quantity of heat required to raise the temperature of one gallon of water (3.78 liters) from 10\textdegree to 80\textdegreeC.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0485.htm

Solution:

The unit of heat is called the calorie. It is defined as the quantity of heat necessary to raise the temperature of one gram of water one degree centigrade. In this problem, one is told that 3.78 liters of water are heated from 10\textdegreeC to 80\textdegreeC. If one subtracts the final tem-perature from the original temperature, one can find the number of degrees the temperature of the water is raised. number of \textdegreeC the temperature is raised = final temperature - original temperature no. of \textdegreeC the temperature is raised = 80\textdegreeC - 10\textdegreeC = 70\textdegreeC. One must now determine the number of grams in 3.78 liters of water. Because 1 milliliter of water weighs one gram, one must determine the number of milliliters in 3.78 liters of water. There are 1000 milliliters in 1 liter; therefore, liters can be converted to milliliters by multiplying the number of liters present by the factor 1000 ml/l liter. 3.78 liters × 1000 ml/l liter = 3780 ml Because number of ml = number of grams for water, there are 3780 g of water present in 3.78 liters. One now knows that 3780 g of water have been raised 70\textdegreeC in tem-perature. Remembering the definition for calorie, one can find the number of calories absorbed in this process by multiplying the number of grams of water by the number of degrees the temperature was raised by the factor 1 calorie/1g \rule{1em}{1pt} 1\textdegreeC, the specific heat of water. number of calories absorbed = 3780 g × 70\textdegreeC × 1 calorie/1g \rule{1em}{1pt} 1\textdegreeC = 265,000 calories There are 1000 calories in 1 kilocalorie, so that calories can be converted to kilocalories by multiplying the number of calories by the factor 1 Kcal/1000 cal. 265,000 cal × 1 Kcal/1000 cal = 265 Kcal. 265 Kcal are absorbed in this process.

Question:

Describe in words the function and the output of the following subroutine. SUBROUTINE GCD (I,J,K) IF (I - J) 1,4,2 1 L = J J = I I = L 2L = I-(I/J){_\ast}J IF(L) 3,4,3 3 I = J J = L GO TO 2 4 K = J RETURN END

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Solution:

The program is written to find the greatest common divisor K of two positive integers I and J. Following the first IF statement, the program pointer will go directly to 1 if the value I - J is negative, to 4 if the value is zero, and 2 if the value is positive. Statement number 1 and the following two statements interchange the values of I and J; therefore prior to statement 2, I is greater than or equal to J. Statement 2 seems useless at first glance. But do not forget that I,J,K and L represent integer variables. Therefore (I/J){_\ast}J may not be equal to I. For example: Let 1 = 4 and J = 3. I/J then will be equal to 1.25, but will be truncated by the computer into integer 1. Then, after multiplying it by J = 3 the computer will get 3 as the result, not 4 as it had originally. In the second IF statement using the same argument as above, the program pointer will go to statements 3 or 4. Statement 3 and the fol-lowing statement replace J with I and L with J. Then the computer goes back to the statement 2 and repeats the calculations. This goes on until I will be divided by J with no remainder left to truncate. In that case value of L will become zero, and the computer will go to statement 4, equate K to J and store this value under K as the desired divisor.

Question:

What is the heat capacity of uranium? Its atomic heat is 26 J/mole and its heat capacity is 238 g/mole.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0483.htm

Solution:

The atomic heat of a solid element at room tem-perature is defined as the amount of heat required to raise the temperature of one mole of an element by one degree Celsius. The heat capacity, or specific heat, is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. These two quantities can be related by the equation, atomic heat = atomic weight × heat capacity. Thus, to find the heat capacity of uranium, sub-stitute the given values of atomic heat and weight. heat capacity of uranium= 26 J/moleC\textdegree/238 g/mole =0.11 J/g\textdegreeC.

Question:

A sample of polymer contains 0.50 mole fraction with molecular weight 100,000 and 0.50 mole fraction with molecular weight 200,000. Calculate (a) the number average molecular weight, M_n and (b) the weight average molecular weight, M_w.

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Solution:

Not all particles (or molecules) in polymers have the same weight. Therefore, one seeks some average value and information regarding the distribution of the individual values. (a) The ordinary (un weighted) arithmetic mean is the number average; each item is counted once. The number average molecular weight is equal to the weight of the whole sample divided by the number of molecules in it. M_n= [\sum_i n_iM_i] / [\sum_i n_i] =\sum_i X_iM_i Here, n_i is the number of molecules of molecular weight M_i per gram of dry polymer and X_i is the mole fraction of each component. Solving for M_n: M_n = 0.5(100,000) + 0.5(200,000) = (50,000) + (100,000) = 150,000 (g / mole). (b) In the weight-average M_w, each item counts not as a single unit but in proportion to its weight. The molecular weight is multiplied by the weight (n_i M_i) of material of that molecular weight rather than by the number of molecules. The weight average molecular weight M_w is defined M_w = (\sumin_iM_i^2) / (\sumn_iM_i) Because(\sum_i n_iM_i) / (\sum_i n_i)= \sumiXiM_i; M_w = (\sum_iX_iM_i^2) / (\sum_i X_iM_i) Solving for M_w: M_w = [0.5 (100,000)^2 + 0.5(200,000)^2] / [0.5 (100,000) + 0.5 (200,000)] = [(1.0 × 10^5)^2 + (2.0 × 10^5)^2] / [(1.0 × 10^5) + (2.0 × 10^5)] = [( 5.0 × 10^10) / (3.0 × 10^5) = 1.67 × 10^5 g/mole.

Question:

Explain the use of the terms ortho, meta, and para in systematic nomenclature of benzene hydrocarbons.

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Solution:

In establishing a consistent systematic scheme for the naming of benzene hydrocarbons several problems arise. The prefixes ortho, meta, and para (abbreviated o, m, and p) are commonly used to designate the 1,2-, 1,3-, and 1,4-, relationship of substituents on the benzene ring. The problem arises when, for example, the compound named systematically as 1,2-dimethylbenzene is also known as ortho-xylene or ortho- methyltoluene or ortho-dimethylbenzene. Four isomeric compounds are possible with benzene hydrocarbons having the molecular formula C_8H_10. One is obtained by substituting an ethyl group for a hydrogen of benzene, whereas the other three are obtained by sub-stituting methyl groups for two hydrogens.

Question:

What pigments may be present in plant cells?What are the functions of these pigments?

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Solution:

Chlorophyll a occurs in all photosyntheticeucaryoticcells and is considered to be essential for photosynthesis of the type carried out by plants. It functions in the capture of light energy by either directly absorbing it or receiving it in the form of high energy electrons from the accessory pigments. These accessory pigments, such as chlorophyll b, are found in vascular plants, bryophytes, green algae andeuglenoid algae. Chlorophyll b differs from chlorophyll a structurally and in its absorption spectra. Chlorophyll b shares with chlorophyll a the ability to absorb light energy and produce in the molecule some sort of excited state. The excited chlorophyll b molecule transfers light energy via high energy electrons to a chlorophyll a molecule, which then proceeds to transform it into chemical energy. Since chlorophyll b absorbs light of wavelengths that are different from chlorophyll a, it extends the range of light that the plant can use for photosynthesis.Chorophyllc or d takes the place of chlorophyll b in some algae and plant - likeprotists. Thecarotenoidsare another class of accessory pig-ments. Carotenoids are red, orange, or yellow fat - soluble pigments found in almost all chloroplasts.Carotenoidsthat do not contain oxygen are called carotenes, and are deep orange in color; those that contain oxygen are called xanthophylls, and are yellowish. Like the chlorophylls, the carotenoids are bound to proteins within the lamellae of the chloroplast. In the green leaf, the color of thecarotenoidsis masked by the much more abundant chlorophylls. In some tissues, such as those of a ripe red tomato or the petals of a yellow flower, thecarotenoidspredominate. As accessory pigments, thecarotenoidsfunction in absorbing light not usable by the chlorophylls and in transferring the absorbed energy to chlorophyll a. Another pigment that may be found in plants is the light - sensitive, blue phytochrome. Phytochromes play a fundamental role in the circadian rhythms of plants; they allow the plant to detect whether it is in a light or dark environment.

Question:

Describe the hormonal interactions that control the development and functioning of the breasts.

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Solution:

Perhaps no other process so clearly demonstrates the intricate interplay of various hormonal control mechanisms as milk production. The breasts are composed of numerous ducts which branch through the breast tissue and converge at the nipples. These ducts terminate at their other end within the breast, in saclike structures (alveoli) typical of exocrine glands. It should be noted that an exocrine gland, by definition, secretes material into ducts leading to a specific compartment or surface. The alveoli are the actual milk secreting structures. Both the alveoli and the ducts are lined by contractile cells whose role will be eludicated later. Before puberty, the breasts are small with little internal structure. With the onset of puberty, estrogen and progesterone act upon the ductiletissue and alveoli to produce the basic architecture of the adult breast. In addition to these two hormones, normal breast development at puberty requires prolactin and growth hormone, both secreted by the anterior pituitary. During each menstrual cycle, the morphology of the breast changes in response to fluctuating blood concentra-tions of estrogen and progesterone. The greatest change in breast morphology occurs, however, during pregnancy. The most important hormone which promotes milk production is prolactin. But the onset of lactation requires more than just high prolactin levels. This is seen in that the level of prolactin is elevated and the breasts are enlarged as pregnancy progresses, and yet there is no secretion of milk. The question arises as to what occurs during delivery to permit the onset of lactation. There have been numerous experiments performed with different mammals, which has shown that if the fetus is removed during pregnancy without interfering with the placenta, lactation will not be induced; when the placenta is removed at any stage of the pregnancy, without removal of the fetus, lactation is induced. Present evidence indicates that this inhibitory effect of the placenta is due to its secretion of estrogen and progesterone, which in large concentration appears to inhibit miik release by some direct action on the breasts. Therefore, delivery removes the source of the two hormones, and thereby removes the inhibition to allow lactation to occur. The hypothalamus also influences lactation. The hypo-thalamus releases a Prolactin Inhibitory Factor. This chemi-cal is decreased during suckling; thus increased prolactin allows secretion of milk to occur. While it is still unclear as to what stimulates increased prolactin secretion during pregnancy, it is known what the major factor is in maintaining thesecretion during lactation. It is believed that receptors in the nipples, which are stimulated by suckling, have nervous inputs to the hypothalamus. These inputs ultimately cause prolactin to be released from the anterior pituitary. One final reflex is essential for nursing. The infant cannot suckle milk out of the alveoli where it is produced, but can only remove milk that is present in the ducts. Milk must first move out of the alveoli into the ducts, a process known as milk let-down. This is accomplished by contraction of the contractile cells surrounding the alveoli. The contraction of these cells is under the direct control of the posterior pituitary hormone oxytocin. This hormone is reflexively released by suckling just like prolactin.Oxytocin is also responsible for the contraction of the uterine muscles during delivery, and it is for this reason that many women experience uterine contractions during nursing. Another important neuroendocrine reflex triggered by suckling is the inhibition of follicle stimulating hormone and luteinizing hormone released by the pituitary. Low levels of FSH and LH subsequently prevent ovulation. This inhibition is relatively short-lived in many women, and approximately 50 per cent begin to ovulate despite continued nursing. The end product of all these processes is the milk which contains four major constituents: water, protein, fat, and the carbohydrate lactose. All of these components are very important for the proper development of the infant.

Question:

How does asexual reproduction take place in the bryophytes ?In the lower vascular plants?

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0188.htm

Solution:

The mosses and liverworts carry out asexual as well as sexual reproduction . The young gametophyte of the moss, theprotonema, is derived from a single spore but may give rise to many moss shoots simply by budding, a process of asexual reproduction. Some liver-worts, such as Marchantia , formgemmaecups on the upper surface of thethallus. Small disks of green tissue, calledgemmae, are produced within these cups. The maturegemmaeare broken off and splashed out by the rain and scattered in the vicinity of the parentthalluswhere they grow into new plants . Like the bryophytes, some lower vascular plants also propagate by vegetative reproduction. In some species of club mosses, small masses of tissue , called bulbils, are formed. These drop from the parent plant and grow directly into new youngsporophyteplants. The ferns reproduce asexually either by death and decay of the older portions of the rhizome and the subsequent separation of the younger growing ends, or by the for-mation of deciduous leaf-borne buds, which detach and grow into new plants .

Question:

You are given a box in which PCl_5(g), PCl_3(g), and Cl_2(g) are in equilibrium with each other at 546\textdegreeK . Assuming that the decomposi-tion of PCl_5 to PCl_3 and Cl_2 is endothermic, what effect would there be on the concentration of PCl_5 in the box if each of the follow-ing changes were made? (a) Add Cl_2 to the box, (b) Reduce the volume of the box, and (c) Raise the temperature of the system.

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Solution:

You are told that the following equilibrium exists in the box PCl_5\rightleftarrows PCl_3 + Cl_2 (all gases) and asked to see what happens to [PCI_5] when certain changes are made. This necessitates the use of Le Chatelier'sprinciple, which states that if a stress is applied to a system at equi-librium, then the system readjusts to reduce the stress. With this in mind, proceed as follows: (a) Here, you are adding Cl_2 to the box. This results in a stress, since one of the components in the equilibrium has its concentration increased. According to LeChatelier'sprinciple, the system will act to relieve this increased concentration of Cl_2 - the stress. It can do so, if the Cl_2 combines with PCl_3 to produce more PCl_5 . In this fashion, the stress is reduced, but the concentration of PCl_5 is increased. (b) When the volume of the box is reduced, the concentration of the species is increased, i.e., the molecules are crowded closer together. Thus, a stress is applied. The stress can only be relieved (LeChatelier's prin-ciple) if the molecules could be reduced in number. Notice, in our equi-librium expression you have 2 molecules, 1 each of PCl_3 and Cl_3 , pro-ducing 1 molecule of PCI_5 . In other words, the number of molecules is reduced if the equilibrium shifts to the left, so that more PCI_5 is pro-duced. This is exactly what happens. As such, the [PCl_5 ] increases . (c) You are told that the decomposition of PCl_5 is endothermic (absorb-ing heat). In other words, if must absorb heat from the surroundings to proceed. If you increase the temperature, more heat is available, and the decomposition proceeds more readily, which means [PCI_5 ] decreases. This fact can also be seen from the equilibrium constant of the reaction, K. This constant measures the ratio of products to reactants, each raised to the power of its coefficients in the chemical reaction. Now, when a re-action is endothermic, K is increased. For K to increase, the re-actant's concentration must decrease. Again, therefore, you see that [PCI_5 ] decreases .

Question:

What does the "pH" of a solution mean? Why is a liquid with a pH of 5 ten times as acidic as a liquid with a pH of 6?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0020.htm

Solution:

The pH (an abbreviation for "potential of hydrogen") of a solution is a measure of the hydrogen ion (H^+) concentration. Specifically, pH is defined as the negative log of the hydrogen ion concentration. A pH scale is used to quantify the relative acid or base strength. It is based upon the dissociation reaction of water: H_2 O \rightarrow H^+ + OH^-. The dissociation constant (K) of this reaction is 1.0 × 10^-14 and is defined as: K = {[H^+] [OH^-]} / {H_2O}where [H^+] and [OH^-] are the concentrations of hydrogen and hydroxide ions, respectively and [H_2O] is the concentration of water (which is equal to one). The pH of water can be cal-culated from its dissociation constant k: K = 1.0 × 10^-14 = {[H^+] [^-OH]} / {H_2O} = [H^+] [^-OH] Since one H^+ and one ^-OH are formed for every dissociated H_2 O molecule, [H^+] = [^-OH]. 1.0 × 10^-14 = [H^+]^2 ; [H^+] = 1.0 × 10^-7 pH = - log [H^+] = - log(1.0 × 10^-7) = 7 A pH of 7 is considered to be neutral since there are equal concentrations of hydrogen and hydroxide ions. The pH scale ranges from 0 to 14. Acidic compounds have a pH range of 0 to 7 and basic compounds have a range of 7 to 14.

Question:

In what ways do the Upper Paleolithic (Cro-Magnon) and the Neolithic cultures differ?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0756.htm

Solution:

The Neolithic (or New Stone Age) culture originated somewhere between Egypt and India about 10,000 years ago. It is characterized by implements bearing the marks of careful grinding and polishing and by the beginnings of agriculture and animal husbandry. The earliest animals to be domesticated after the dog (first by Cro-Magnon man) were the pig, sheep, goat, and cow. The horse was not do-mesticated until much later. Our human ancestors gradually changed from wandering hunters and food gatherers of the Cro-Magnon stage to more settled and efficient food producers, raising grain, making pottery and cloth and living in villages. At this time, more diversified tools were made, the hunting men stayed closer to the home and migrated less, people became more self-sufficient and made or obtained what they needed from nearby sources. They no longer lived in small breeding groups but expanded their communities. The increase in food supply enabled an increase in the size of the popu-lation. Neighboring populations interbred and the tendency toward genetic drift was thus greatly reduced. Neolithic people also invented the dugout canoe and the wheel. With the Neolithic age we enter the historical times, for the oldest cultures of Egypt and Mesopotamia were Neolithic. The use of metals began with copper and was followed with bronze. They were used for making tools, vessels and weapons beginning about 4000 B.C. The culture of Cro-Magnon man in comparison to the Neolithic culture may at first seem primitive. Actually, both peoples had about the same intelligence. But culture is an additive phenomenon i.e., one generation will usually build on what is inherited from the previous generations. Conse-quently, culture gets more and more complex and sophisticated. As we can see in the transition from Upper Paleolithic to Neolithic ages, people are living in larger social groups, language is progressing and is an expanding communication device for expressing ideas and experiences, both past and future. The interaction of language and a large social order speed up cultural advancements and the attainment of new knowledge.

Question:

At 273\textdegreeK and 1atm, 1 liter of N_2O_4 decomposes to NO_2 according to the equation N_2O_4\rightleftarrows 2NO_2. To what extent has decomposition proceeded when the original volume is 25 percent less than the existing volume?

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Solution:

Volume is proportional to the number of moles present when Pressure (P) and Temperature (T) are held constant. This can be seen from the equation of state, PV =nRTor V = (nRT/ P ) , where R = universal gas constant, n = moles, and V = volume. When P and T are constant, RT/P is a constant, and V varies directly with n. Thus, in this problem you can discuss volumes in terms of fractions decomposed. Let a be the volume fraction of the original N_2O_4 , that decomposes. Since moles are proportional to volumes and 2 moles of NO_2are produced for every mole of N_2O_4, this fraction, a, results in the production of 2a of N0_2 and 1\Elzbar a of N_2O_4. The total final volume fraction is 1\Elzbar a + 2a = (1 + a); that is, the final volume = (1 + a) times the original volume. The original volume is 1 liter of N_2O_4. The existing volume is (1 + a)(1 liter) = (1 + a) liter. But 1 = .75 (1 + a). Solving, a = .333 .Thus, you have 33% decomposition.

Question:

The pressure in a static water pipe in the basement of an apartment house is 42 lb \textbullet in^-2, but four floors up it is only 20 lb \textbullet in^-2. What is the height between the basement and the fourth floor?

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Solution:

If A is the cross sectional area of the pipe, then the pressure (p_1 = F1/A where is the force exerted) in the pipe at the basement must balance both the pressure P_2 [= F_2/A] in the pipe at the fourth floor and the weight of the water in the column of height h (see the figure). If \rho is the density of water, then by definition \rho = m/v where m is the mass of the water in the column of volume v = Ah. Then m = \rhoAh and the weight of the water is W = mg = \rhoghA. Thus the pressure due to the weight of the water is W/A = \rhogh. Hence P_1 = P_2 + \rhogh (P_1 - P_2)= (42 - 20) lb \bullet in^-2 = 22 lb/in^2 × 144 in^2/1 ft^2 = \rhogh = 1.94 slug \bullet ftr^-3 × 32 ft \bullet s^-2 × h \thereforeh = [(22 × 144)lb \bullet ft^-2]/[1.94 slug \bullet ft^-3 × 32 lb × ft^-3] = 99/1.94 ft = 51.03 ft.

Question:

A 100-kg man jumps into a swimming pool from a height of 5 m. It takes 0.4 sec for the water to reduce his veloci-ty to zero. What average force did the water exert on the man?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0318.htm

Solution:

The man's initial velocity (before jumping) is zero. Therefore, as he strikes the water, his velocity v is v^2 = v^2 _0 +2gh, which reduces to v^2 = 2gh v = \surd(2gh) = \surd[2 × (9.8 m/sec^2) x (5 m)] = 10 m/sec Therefore, the man's momentum on striking the water was p_1 =mv = (100 kg) × (10 m/sc) = 1000 kg-m/sec The final momentum was p_2 = 0, so that the average force was F =\Deltap/\Deltat= (p^2 - p^1)/\Deltat= (0 - 1000 kg-m/sec^2)/(0.4 sec) = 2500 N but this is the total force : friction and gravity: F=F_water+ mg = - 2500 - 980 = - 3480 N The negative sign means that the retarding force was directed opposite to the downward velocity of the man.

Question:

A proton (mass 1.67 x 10^-27kg) collides with a neutron (mass almost identical to the proton) to form a deuteron. What will be the velocity of the deuteron if it is formed from a proton moving with velocity 7.0 x 10^6m/sec to the left and a neutron moving with velocity 4.0 x 10^6m/sec to the right? (Figure A)

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Solution:

According to the conservation of momentum we write: m_dv_d = m_pu_p + m_nvn where m_d, m_p, and m_n are the masses of the deuteron, proton, and neutron respectively . Since the masses of the proton and neutron are almost identical: m_p = m_n = 1/2 md Inserting this into the momentum conservation equation (we adopt the convention that velocities to the right are positive): 2m_pv_d = m_pv_p + m_pv_n v_d = (v_p + v_n)/2 = (-7.0 × 10^6m/sec + 4.0 × 10^6m/sec)/2 = -1.5 ×10^6 m/sec(figure B). Thus the neutron moves to the left with speed 1.5 × 10^6 m/sec. In the actual collision a photon is produced and carries off some of the momentum, therefore the velocity calculated above is somewhat too large.

Question:

What are the main methods used for debugging a program and what are their advantages?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G25-0592.htm

Solution:

DEBUGGING is the procedure of eliminating bugs (errors) that almost every program has the first time it is run through the computer. These are some of the indications that the program does indeed have mistakes : answers . closer inspection reveals the answers to be wrong. The following is a list of some of the most commonly used debugging methods : assumes the program is fit enough to generate out-put. 4)Planning ahead is based upon occasionally including output statements into the program. Those output statements usually print intermediate results obtained up to that point. This allows the programmer to follow the execution of his program and check the inter-mediate results, accuracy of which is essential. 5) Running the program step by step: When the computer is in its MANUAL condition, pushing the SINGLE STEP key makes the computer perform one instruction. Each time this key is pushed, the computer will perform the instruction and change the lights on the control panel to indicate the address of the next Instruction. In this way, the entire program is performed step by step. This method can only be used when working directly with the central processing unit of the computer.

Question:

The composition of dry air by volume is 78.1% N_2, 20.9% O_2 and 1% other gases. Calculate the partial pressures, in atmospheres, in a tank of dry air compressed to 10.0 atmospheres.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E03-0084.htm

Solution:

A partial pressure is the individual pressure caused by one gas in a mixture of several gases. The total pressure,P_total, according to Dalton's Laws, is the sum of these individual partial pressure, p_1, p_2 and p_3. i.e. P_total= p_1 + p_2 + p_3. The term dry air indicates that no water vapor is present. The partial pressure is found by multiplying the percent of each gas in the volume by the total pressure. partial pressure = proportion by volume × total pressure P_(N)2 = .781 × 10atm= 7.81atm P_(O)2 = .209 × 10atm= 2.09atm P_other gasses = .010 × 10atm= .1 atm. The total pressure, as required, is 10 atm.

Question:

At what point between moon and earth do the gravitational fields of these two bodies cancel? The earth's mass is 5.98 × 10^24 kg, and the moon's is 7.35 × 10^22 kg. The distance between the centers of the earth and the moon is 3.85 × 10^7 m.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0141.htm

Solution:

Let the point where the gravitational fields cancel be at a distance R_1 from the earth's center, R_2 from the moon's center. The attraction of the earth at this point will equal that of the moon's: GM\mu/R_1^2 = Gm\mu/R_2^2 where G is the gravitational constant, M the mass of the earth, and m that of the moon, or M/R_1^2 = m/R_2^2. The term on the left side of the equation is called the earth's gravitational field, the term on the right is the moon's. The gravitational field at a point in space is the gravitational force experienced by a unit mass at that point. (It is similar to the electric field, which is the electrostatic force per unit charge experienced due to a particular charged body). Gravitational fields are vector fields and the resultant gravitational field due to two or more masses is calculated by adding the field vectors from all sources at every point in space. The only point in space where the gravitational fields of the earth and the moon cancel, must be collinear to the centers of both bodies since no two vectors cancel if they are not oppositely directed. From the equation above, we get [R_1/R_2]^2 = M/m R_1/R_2 = [M/m]^1/2 = [5.98 × 10^24 kg/7.35 × 10^22]^1/2 = 9. Since the point in question is collinear to both the earth and moon's centers, the distance between the centers must equal R_1 + R_2 (see diagram), thus: R_1/R_2 = 3.65 × 10^8 m R_1 = 3.85 × 10^8 m - R_2 R_1/R_2 = (3.85 × 10^8 m - R_2)/R_2 = {(3.85 × 10^8 m)/R_2} - 1 = 9 R-_2 = (3.85 × 10^7 m) R_1 = 9R_2 = 9(3.85 × 10^7 m) = 34.7 × 10^7 m.

Question:

If the vapor pressure of CCl_4 (carbon tetrachloride) is .132 atm at 23\textdegreeC and .526 atm at 58\textdegreeC, what is the ∆H' in this temperature range?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0268.htm

Solution:

Molecules in a vapor exert a pressure that is characteristic of its liquid state. The pressure exerted by this vapor, when in equilibrium with the liquid, is called the equilibrium vapor pressure. The equilibrium vapor pressure can be expressed quantitatively as log P = [(-∆H') / (19.15) T] + C where p = vapor pressure, T = temperature in degrees Kelvin, ∆H' = heat required to transform one mole of liquid to the ideal-gas state, and C = a constant that is characteristic of the liquid. C can be eliminated from the expression by denoting that at two temperature ranges, T_1 and T_2, you have; (log p_1) - (logp_2) = [{(-∆H') / (19.15) T_1} + {(∆H') / (19.15) T_2}] or log (p_1 / p_2) = [(∆H') / (19.15)] [(1 / T_2) - (1 / T_1)]. In this problem, you are asked to find ∆H'. Let p_1 = (.132 atm) at T_1 = 296\textdegreeK and P_2 = (.526 atm) at T_2 = 331\textdegreeK, then substitute these values in the above equation. Therefore, log (.132)/(.526) = (∆H' / 19.15) [ (1 /331) - (1 / 296)] Solving for ∆H', you obtain ∆H' = 32,000 (J / mole).

Question:

An electric field is set up by two point charges q_1 and q_2, of the same magnitude (12 x 10^\rule{1em}{1pt}9 coul; see diagram) but opposite sign, as shown in the figure. What is the electric intensity at points a, b, and c?

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Solution:

By definition, the magnitude of the electric field intensity, E^\ding{217}, is E = kq / r^2 where q is the charge causing the field, and r is the distance from q to the point at which we wish to cal-culate E^\ding{217}. (k = 9 × 10^9 N \textbullet m^2 / c^2). If the field is due to more than one charge, then the total field at a point is the vector sum of the fields due to each charge at that point. At every point, the intensity due to the positive charge is directed radially away from that charge, and the intensity due to the negative charge is radially inward toward the charge. At point a, the intensity due to each charge is directed toward the right. The resultant intensity E^\ding{217}_a is also toward the right and its magnitude is the arith-metic sum of the individual intensities. Hence, E_a = (kq_1) / (r_1^2) \rule{1em}{1pt}(kq_2) / (r_2 ^2) E_a = [9 × 10^9 (N \textbullet m^2 / c^2)][{(12 × 10^\rule{1em}{1pt}9c ) / (36 × 10^\rule{1em}{1pt}4 m^2)} \rule{1em}{1pt} {(\rule{1em}{1pt}12 ×10^\rule{1em}{1pt}9 c) / (16 ×10\rule{1em}{1pt}4m^2)}] E_a= 9.75 × 10^4 N / Ctoward the right. At point b, the intensity set up by q_1 is directed toward the left and that set up by q_2 is toward the right. The magnitude of the first is greater than that of the second because q_1 is closer to b than q_2. The resultant intensityE^\ding{217}_b is toward the left and its magnitude is the difference between the individual intensities. Therefore, E_b = \rule{1em}{1pt}[(kq_1) / (4cm)^2 ] \rule{1em}{1pt} [(kq_2) / (14cm)^2 ] Eb=[{\rule{1em}{1pt}9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C)^2}] [{(12 × 10^\rule{1em}{1pt}9c)/( 16 × 10^\rule{1em}{1pt}4m^2)} \rule{1em}{1pt} {(12 × 10^\rule{1em}{1pt}9c) / (196 × 10^\rule{1em}{1pt}4 m^2)}] E_b = \rule{1em}{1pt}6.2 × 10^4 N/C, which is toward the left. At point c, the individual intensities have the directions shown and the resultant intensity E^\ding{217}_c is their vector sum. Note, from the figure, that the vertical components of E^\ding{217}_1 and E^\ding{217}_2 will cancel, and their horizontal components will reinforce. Hence E_c = │E^\ding{217}_1│ cos 60\textdegree + │E^\ding{217}_2│cos 60\textdegree = [│E^\ding{217}_1│ / 2 ]+ [│E^\ding{217}_2│/ 2] = (1 / 2) [│{kq_1 / (10cm)^2}│] + (1 /2) [│{kq_2 / (10cm)^2}│] = (1 / 2)[│({9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C^2)} {(12 × 10^\rule{1em}{1pt}9c)}) / (100 × 10^\rule{1em}{1pt}4 m^2)│] +1 / 2[│({9 × 10^\rule{1em}{1pt}9(N \bullet m^2 / C^2)}{(\rule{1em}{1pt}12 × 10^\rule{1em}{1pt}9c)}) / (100 × 10^\rule{1em}{1pt}4 m^2)│] = 1.08 × 10^4 N / C towards the right.

Question:

Explain what is meant by Initialization or Initial Attributes. Explain how a variable is initialized. Also, explain how initialization is done, and what is stored in the computer memory by the following statement: DCL A FIXED(3,2) INITIAL(1\textbullet5), B FIXED(2,1) INIT (2\textbullet25) , C FLOAT(4) INIT(170), J INIT(10010B), P INIT(-11E2).

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Solution:

It is often necessary that a variable appearing in a program have a value before it is used in expressions in the program. This value is given to the variable by explicit-ly assigning an initial value or initial attribute to the variable. The assigning of an initial attribute is called Initialization. Initialization can be done by means of a statement as follows: A = 3 \textbullet 2; B= - 4E - 1; C = 2 In the above, A is given an initial value of 3\textbullet2. B is ini-tialized to a floating point value of -0\textbullet4, and the initial value of C is 2. Initialization can be done anywhere within the body of the program, but it must be done before the variable appears in an expression in the program. A good place to do the initialization of the variables in a program is immediately following the DECLARE statements. A better and shorter way is to initialize the variables simultaneously as they are declared, that is, do the initial-ization in the DECLARE statement itself. This is done as il-lustrated below: DCL A FIXED(3,2) INITIAL(1\textbullet5), B FIXED(2,1) INIT (2\textbullet25), C FLOAT(4) INIT(170), J INIT(10010B), P INIT( - 11E2) ; In the above, INIT is a short form for INITIAL. A is de-clared as FIXED(3,2) with an initial value of 1\textbullet5. Hence, the initial value stored in memory for A is 1\textbullet50. B is de-clared as FIXED(2,1) with an initial value of 2\textbullet25. Therefore, the value stored for B is as follows: 2\textbullet2. Notice the 5 is truncated because the declaration FIXED(2,1) specifies only one position to the right of the decimal. C is declared as FLOAT (4) with an initial value of 170. Therefore, the value stored for C is as follows: 0\textbullet1700E + 03. If a PUT DATA(C) statement were used in the program before the value of C was changed, the print out for C would be as follows: C = 1\textbullet700E + 02. For J, except for the initial value, no other attributes are declared. Hence, by Default Rules, J receives the following attributes: BIN REAL FIXED 15,0). The initial value of J is 10010B, which is stored in memory. A PUT DATA(J) statement would print out as below: J = 000000000010010B For P, by Default Rules, the attributes are as follows: DEC REAL FLOAT(6). Hence, the value of P stored in memory is as follows: -0\textbullet110000E + 04. A PUT DATA(P) statement will cause a print out as follows: P = - 1\textbullet10000E + 03

Question:

The air in a tank has a pressure of 640 mm of Hg at 23\textdegreeC. When placed in sunlight the temperature rose to 48\textdegreeC. What was the pressure in the tank?

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Solution:

The law of Gay-Lussac deals with the relation-ship existing between pressure and the absolute temperature (\textdegreeC + 273\textdegree) , for a given mass of gas at constant volume. The relationship is expressed in the law of Gay-Lussac: volume constant, the pressure exerted by a given mass of gas varies directly with the absolute temperature. That is: P \alpha T(volume and mass of gas constant). The variation that exists between pressure and tem-perature at different states can be expressed as (P_1 / T_1 ) T_1 ) = (P_2 /T_2 ) = (P_2 /T_2 ) where P_1 = pressure of original state, T_1 = absolute tem-perature of original state , P_2 = pressure of final state, and T_2 = absolute temperature of final state . Thus this problem is solved by substituting the given values into Gay-Lussac's Law. P_1 = 640 mm HgT_1 = 23\textdegreeC + 273\textdegree = 296\textdegreeK P_2= ?T_2 = 48\textdegreeC + 273\textdegree = 321\textdegreeK Substituting and solving, [(640 mm)/(296\textdegreeK)] = [(P_2 )/(321\textdegreeK)] P_2 = 640 mm × [(321\textdegreeK)/296\textdegreeK)] = 694 mm of Hg.

Question:

A very large conducting plate has a surface charge density of 10^-2 coul/m^2. What is the energy density of the field produced by this charge distribution?

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Solution:

The electric field due to the evenly charged sheet can be obtained by applying Gauss' Law at the sur-face of the sheet (see figure). By the definition of a conductor, the electric field at the surface of the con-ductor must be at right angles to it. (The y components of E^\ding{217} due to 2 symmetrically placed elements of the con-ductor's surface cancel, whereas the x components re-inforce. ) If we allow the Gaussian surface S to be very close to the surface of the conductor and neglect the fringing at the edges, we may use Gauss' Law, \epsilon_(0)s \intE^\ding{217} \bullet dS^\ding{217} = Q,or\epsilon_0AE = \deltaA, where A is the total surface area of the conductor. Therefore, E = \delta/\epsilon_0 where \delta is the surface charge density. The energy density of the field E is, by definition, u = \epsilon_0E^2/2 oru = (\epsilon_0/2) (\delta/\epsilon_0- )^2 = \delta^2/2\epsilon_0 But\epsilon_0 = 1/4\piK_E, where K_E = 9 × 10^9 [(N\bulletm^2)/c^2]. Henceu = (\delta^2/2) 4\piK_E = 2\piK_E \delta^2 In this problem u = 2\pi × [9 × 10^9 {(N\bulletm^2)/c2}] × 10^-4m (c^2/m^4) u = 56.52 × 10^5 (N/m^2) Because 1 Joule = 1N \bullet m u = 56.52 × 10^5 Joule/m^3.

Question:

Write a computer program in FORTRAN to find the correlation coefficient r of the two variables. What is r for the data below? X Y 30 .9 20 .8 10 .5 30 1.0 10 .8

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0490.htm

Solution:

Answering the second question first, the correlation coefficient r of the given data is r = ([n \sumxy - (\sumx) (\sumy)] / [\surd{n\sumx^2 - (\sumx)^2}\surd{n\sumy^2 - (\sumy)^2}])(1) = [5(86) - (100) (4.0)] / [\surd{5(2400) - (100)^2}\surd{5(3.34) - 4^2}] = .80 To understand what r is and what it measures, suppose that a regression line is drawn through the given data. Also, letybe drawn through the points. The regression line and the sample average line are predictors of how y varies when x varies. Note that (y -y) = (y - y˄) + (y˄ -y˄) (see Fig. 1) The explained variation of y is defined as \sum(y˄ -y)^2 . The unexplained variation is then \sum(y - y˄)^2 . The coefficient of variation, r^2 is defined as r^2 = (Explained Variation of y) / (Total Variation of y) or, r^2 = {\sum(y˄ -y)^2} / {\sum(y -y)^2}(2) The correlation coefficient r is the square root of equation (2). The computational formula for r, (1), follows from (2). Thus, r measures how much of the variation in y is explained by the regression line, in this case - 80%. This means that 20% of the variation is due to chance or other factors The program looks as follows: CPROGRAM TO FIND CORRELATION COEFFICIENT DIMENSION KF(2\O), A(1\O\O), S(1\O\O), R(1\O\O,1\O\O) 1READ (5,3) NV, NS, KF 3FORMAT (2I5)/2\OA4) DO 5 I = 1,NV A (I) = 0 DO 5 J = 1,NV 5R(I,J) = 0 DO 15 J = 1,NS READ (5,KF) (S(J), J= 1,NV) DO 15 J = 1,NV A(J) = A(J) + S(J) DO 15 K = J,NV 15R(J,K) = R(J,K) + S(J) \textasteriskcentered S(K) T = NS DO 2\O I = 1,NV A (I) = A(I)/T 20S(I) = SQRT(R(I,I)/T - A(I) \textasteriskcentered\textasteriskcentered 2) DO 25 I = 1,NV DO 23 J = I,NV IF (S(I) \textasteriskcentered S(J) .EQ. 0.0) GO TO 23 R(J, I) = (R(I,J)/T - A(I)\textasteriskcentered A(J))/S(I)\textasteriskcenteredS(J)) 23R(I,J) = R(J,I) 25R(I,I) = 1.0 WRITE (6,3\O) (A(I), I = 1,NV) 3\OFORMAT (8H1 × MEANS., (/1 OF 10.4)) WRITE (6,31)(S(I), I = 1,NV) 31FORMAT (8H SIGMAS., (1 1 OF 10.4)) DO 4\O K = 1,NV, 1\O L = K + 9 IF (NV - L) 32, 33, 32 32L = NV 33WRITE (6,35) K,L 35FORMAT (14H1R MATRIX. COL, I3, 3H TO I3) DO 4\O I = 1,NV 4\OWRITE (6,45) I, (R(I, J) , J = K,L) 45FORMAT (I5, 10F 10.4) STOP END

Question:

A chemist dissolvesPbSin water . TheK_spis found to be 1 × 10-^28 . What is themolarityof each ion in solution ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0395.htm

Solution:

The key to answering this question is the determination of the equilibrium equation. SincePbSis a salt, it dissociates into ions in solution. Upon inspection of the oxidation states of the periodic table, the equilibrium will be found to be PbS(s) \rightleftarrows Pb^2+ (Aq) + S^2- (Aq) . The presence of the equilibrium derives from the fact that a saturated salt solution's ions will exist in equilibrium with any additional salt. This additional salt is called the solid phase.K_spmeasures this equilibrium and can be expressed as K_sp= [Pb^2+] [S^2-] . This, equals 1 × 10-^28 . Therefore, you have 1 × 10-^28 = [Pb^2+] [S^2-]. The question asks for the concentrations represented on the right side. From the equilibrium equation, you see that whatever the concentration of S^2-, it will be equal to Pb^2+ . Therefore, you can represent both by x . As such, you now have 1 × 10-^28 = x \textbullet x 1 × 10-^28 = x^2 . Solving for x you obtain 1 × 10-^14 M. This means, therefore, that the concentration of each ion in solution is 1 × 10-^14 M .

Question:

A block of steel 2 ft. square and 1/4in. thick is to be compressed .01 in. in length by application of a force F to faces A_1 and A_2. If Young's modulus for steel is 29 × 10^6 lb. in.^2, find F.

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/Users/wenhuchen/Documents/Crawler/Physics/D08-0351.htm

Solution:

Young's modulus is defined as Y = (STRESS)/(STRAIN)(a) where STRESS = (NORMAL FORCE)/(AREA) = F/(24 × 1/4) = F/6 and STRAIN = (CHANGE IN LENGTH)/(ORIGINAL LENGTH) = .01/24 = 1/2400 Substituting our values in (a), 29 × 10^6 = (f/6)/(1/2400) Thus F = 72,500 lb.

Question:

What is the output of the following program? main ( ) { intx = 5; inty = 5; printf("%d%d\textbackslashn", x++, x); printf("%d%d\textbackslashn", ++y, y); }

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0424.htm

Solution:

Here it is important to distinguish between the two styles of increment operations. 1) x++: Use variable x first and then increment afterwards,eg. Let z = x++; means z = x; x = x + 1; 2) ++y: Increment the variable y first and then use it afterwards.eg. Let z = ++y; means y = y + 1; z = y; Therefore the output is as follows: 56/\textasteriskcentered 1stprintf\textasteriskcentered/ 65/\textasteriskcentered 2ndprintf\textasteriskcentered/

Question:

(a) Compute the maximum height attained by a projectile launched with velocity of magnitude v_o directed at an angle \texttheta_0 to the horizontal, using the principle of con-servation of energy. (b) What is the magnitude of the projectile's velocity u^\ding{217}, when it has reached half its greatest height?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0238.htm

Solution:

(a) At the start of the motion, the projectile possesses kinetic energy of amount (1/2) mv^2_0. At its greatest height, h, it possesses kinetic energy due to the com-ponent of its velocity in the x-direction (v_O cos \texttheta) only, (1/2) m(v_0 cos \texttheta_0)^2 and also potential energy due to its increased height, mgh. By the principle of conserva-tion of energy, (1/2)mv^2_0 = (1/2)mv^2_0 cos^2 \texttheta_0 + mgh \therefore h = [v^2_0(1 - cos^2 \texttheta_0)] / (2g) = (v^2_0 sin^2 \texttheta_0) / (2g) (b) By the principle of conservation of energy, the sum of the projectile's potential and kinetic energies at half its greatest height must equal its initial kinetic energy at the time of launching. (The potential energy at the point of launching is also zero.) At half the greatest height, the potential energy possessed by the projectile is mg(h/2) = mg [(v^2_0 sin^2 \texttheta_0) / (4g)] = (1/4) mv^2_0 sin^2 \texttheta_0. and its kinetic energy is (1/2)mu^2. Thus: (1/2) mv^2_0 = (1/2) mu^2 + (1/4) mv^2_0 sin^2 \texttheta_0 u = \surd[v^2_0 - (1/2) v^2_0 sin^2 \texttheta_0]

Question:

Explain how peristalsis moves food through the digestive tract .

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0425.htm

Solution:

In each region of the digestive tract, rhythmic waves of constriction move food down the tract. This form of contractile activity is called peristalsis, and involves involuntary smooth muscles. There are two layers of smooth muscle throughout most of the digestive tract. Circular muscles run around the circumference of the tract while longitudinal muscles traverse its length. Once a food bolus is moved into the lower esophagus, circular muscles in the esophageal wall just behind the bolus contract, squeezing and pushing the food downward. At the same time, longitudinal muscles in the esophageal wall in front of the bolus relax to facilitate movement of the food. As the bolus moves, the muscles it passes also contract, so that a wave of contraction follows the bolus and constantly pushes it forward. This wave of constriction al-ternates with a wave of relaxation. Swallowing initiates peristalsis and once started, the waves of contraction cannot be stopped voluntarily. Like other involuntary responses, peristaltic waves are controlled by the autonomic nervous system. When a peristaltic wave reaches a sphincter, the sphincter opens slightly and a small amount of food is forced through. Immediately afterwards, the sphincter closes to prevent the food from moving back. In the stomach, the waves of peristalsis increase in speed and intensity as they approach the pyloric end. As this happens, the pyloric sphincter of the stomach opens slightly. Some chyme escapes into the duodenum but most of it is forced back into the stomach (see figure). This allows the food to be more efficiently digested. There is little peristalsis in the intestine, and more of a slower oscillating con-traction. This is why most of the 12-24 hours that food requires for complete digestion is spent in the intestine.

Question:

Although different simulation studies require variations in procedure, certain basic steps are usually followed. State the stepsof the general procedure.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0569.htm

Solution:

1)Definition of the problem. 2)Planning the study (considering the exogenous factors that willaffect the system; deciding if such factors will be included in the study; etc.) 3)Formulation of a mathematical model. Simplifying assumptions aboutthe system components and their behavior are usually made at this stage . 4)Gathering of relevant data. 5)Construction of a computer program for the model. 6)Validation of the model. 7)Design of experiments. 8)Execution of simulation run and analysis of results. In common use, there are two different types of simulation, continuous anddiscrete. Continuous simulation deals with variables (such as velocity, current, work, etc.)thattake on real values and change continuously with time. These changes are defined as being smooth andnonabrupt. In discretesimulation, the model deals with random events that occur at isolatedtimes. For example, the arrivals of motorists at a gas station couldbe considered random, and the sim-ulation model might have to determinethe average waiting time per customer under various organizationsof attendants and gas pumps. This process of using random variablesis called a stochastic process. There exists a problem-oriented programming language called GPSS (General Purpose System Simulation) which can be used for solving the simulationproblems.

Question:

Differentiate between the monocots and dicots.

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0204.htm

Solution:

The two classes of angiosperms, the monocotyledons and dicotyledons, differ in eight respects. First, the monocot embryo has one cotyledon (seed leaf) while the dicot has two. The cotyledon of the monocots functions generally in food absorption while that of the dicots can function both in absorption and storage. Second, the endosperm typically persists in the mature seed of the monocots, while it is usually absent from the mature seed of the dicots. Third, the leaves of the monocots have parallel veins and smooth edges; dicot leaves have net-like veins and lobed or indented edges. Fourth, cambium or meristematic cork and vascular tissue is usually present in the stems of dicots but absent in the monocots. Fifth, the vascular bundles of xylem and phloem are scattered throughout the stem of monocots. In the dicots they occur either as a single solid mass in the center of the stem or as a ring between the cortex and pith. Sixth, the flowering parts of monocots - petals, sepals, stamens, and pistils - exist in threes or multi-ples of three , whereas dicot floral parts usually occur in fours, fives or multiples thereof. Seventh, the monocots are mostly herbaceous plants while many dicots are woody plants. Finally, the roots of monocots are typically fibrous and adventitious (outgrowths of the stem) whereas the root system of dicots usually consist of one or more primary tap roots and numerous secondary roots.

Question:

A certain boat can move at a speed of 10 mi/hr in still water. The helmsman steers straight across a river in which the current is 4 mi/hr. What is the velocity of the boat?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0017.htm

Solution:

The boat has a speed of v_b = 10 mi/hr per-pendicular to the river due to the power of the boat. The current gives it a speed of v_c = 4 mi/hr in the direction of flow of the river. The boat's resultant velocity (having both magnitude and direction) can be found through vector addition. v\ding{217} = v\ding{217}_b + v\ding{217}_c The magnitude of the velocity which is the speed of the boat is found using the Pythagorean theorem (see figure). v = \surd(v^2_b + v^2_c ) = \surd[(10)^2 + (4)^2] = \surd116 = 10.8 mi/hr. The angle \texttheta, which determines the direction of the velocity is, \texttheta = tan^-1 (v_c/v_b) = tan^-1 (4/10) = 22\textdegree

Question:

In a cloud chamber, a proton crosses at right angles a uniform magnetic field of 1.0 weber/m^2 . From the photograph of the proton's path the radius of curvature is measured as 0.22 m. Calculate the speed of the proton.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0720.htm

Solution:

The centripetal force acting on the proton causes it to curve. This force is due to the magnetic deflecting force Bqv where q, the charge on the proton, is 1.6 × 10^-19 coulombs. The force can also be represented in terms of the radius of curvature and the velocity of the proton as (mv^2 /r), with the mass of the proton, m, equaling 1.67 × 10^-27 kg. Setting the two expressions equal and solving for the velocity, Bqv = (mv^2 /r) v = (Bqr/m) = [{(1.0 weber/m^2 ) (1.60 × 10^-19 coul) (0.22 m)} / (1.67 × 10^-27 kg)] = 2.1 × 10^7 (m/sec). (Since this speed is less than a tenth the speed of light, we are justified in neglecting any relativistic change in the mass of the proton.)

Question:

Two substances A and B have the same 0.132-atm vapor press-ure at 15\textdegreeC, but A has a heat of vaporization of 25,000 (J / mole), whereas B, 35,000 (J / mole). What will be the respective normal boiling points?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0264.htm

Solution:

The vapor pressure, heat of vaporization and absolute temperature are related in the Clausius-Clapeyron equation. This equation is written: ln (p_1 / p_2) = (∆H / R) [(1 / T_2) / (1 / T_2)] where p_1 is the initial pressure, p_2 is the final pressure, ∆H is the heat of vaporization when T_2 is the boiling point, R is the gas constant, (8.314 J / mole \textdegreeK), T_1 is the initial temperature and T_2 is the final temperature. To solve for the boiling point of A, note that at the boiling point of a liquid the vapor pressure is equal to the atmospheric pressure. Atmospheric pressure is 1 atm. P_1 = 0.132 atmT_1 = 15\textdegreeC + 273 = 288\textdegreeK p_2 = 1 atmT_2 = ? ∆H = 25,000 (J / mole)R = 8.314 (J /mole \textdegreeK) In (0.132 / 1) = [(25,000 J/mole) / (18.314 J/mole \textdegreeK)] {(1 / T_2)-(1/288\textdegreeK)}] -2.025 = (3.00 × 10^3\textdegreeK) (1 / T_2) - (3.47 × 10^-3 /\textdegreeK) -6.75 × 10^-4 / \textdegreeK = (1 / T) - (3.47 × 10^-3 /\textdegreeK) 2.795 × 10^-3 / \textdegreeK = 1/T 357.78 = T The boiling point of A in \textdegreeC is (357.78 - 273) = 84.78\textdegreeC Solving for the boiling point of B: P_1 = 0.132 atmT_1 = 15\textdegreeC + 273 = 288\textdegreeK p_2 = 1.0 atmT_2 = ? ∆H = 35,000 (J / mole) R = 8.314 (J / mole \textdegreeK) In (0.132 / 1.0) = [(35,000 J/mole)] / (8.314 J/mole \textdegreeK)] [(1/T_2) - (1/288\textdegreeK)] -2.025 = 4.21 × 103 \textdegreeK (1 / T_2) - 3.47 × 10^-3 / \textdegreeK) -4.81 × 10^-4 / \textdegreeK = (1 / T_2) - 3.47 × 10-3/ \textdegreeK 2.989 × 10^-3 / \textdegreeK = (1 / T_2) 334.56\textdegreeK = T_2 The boiling point of B is equal to 334.56\textdegreeK or 334.56 -273 = 61.56\textdegreeC.

Question:

Distinguish between the two types of proteins produced by ribosomes.

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0630.htm

Solution:

Ribosomes are the cellular organelles that provide the sites for protein synthesis. The two types of proteins that are synthesized on the ribosomesare structural proteins and functional proteins. Structural proteins are actualstructualcomponents of the cell. They constitute part of the cell protoplasm, and as such, are generally insoluble. They are an important component of cell membranes. Some structural proteins, such as collagen, or keratin are specialized for strength and support. Structural proteins help to form the cell, and the organism of which the cell is a part, therefore, struc-tural proteins contribute to both cell and body growth. Functional proteins are responsible for the control of cell activity, such as hormone production or nutrientmetablism. Some of these proteins are known as enzymes their function is to catalyze chemical reactions in the cell. Others function as carrier molecules. Still others function as regulatory molecules which control the synthesis of other proteins. Most functional proteins are soluble in the cell, since their action depends on their interaction with various molecules throughout the cell. Some are soluble in cell membranes. The distinction between structural and functional proteins, then, is based on what they do in the cell.The former are involved in the structure of the cell or organism, whereas the latter deal with its functions.

Question:

Explain the difference between truncation and rounding. How can one operation be made to perform the function of the other?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0015.htm

Solution:

Consider the real number 5.6158 whose integer part is 5 and fractional part is .6158. This number can be converted into a whole number (integer) in one of two ways. First, the fractional part can be chopped off to give 5, and this is called truncation. Second, the integer whose value is closest to 5.6158 is 6, and this process is known as rounding. Suppose the computer is capable of truncation only. How can we accomplish rounding? Just add .5 to every number to be rounded before the truncation. The truncated increased value will be equal to rounded initial value. For example: 5.6158 ROUNDED= 6 5.6158 + .5000 = 6.1158 TRUNCATED = 6 Suppose the computer is capable of rounding only, and we need to truncate. In that case, simply subtract .5 from every number to be truncated before the rounding. The rounded de-creased value will be equal to truncated initial value. For example: 5.6158 TRUNCATED = 5 5.6158 - .5000 = 5.1158 ROUNDED = 5

Question:

Standing waves are produced by the superposition of two waves of the form y_1= 15 sin (3\pit \rule{1em}{1pt} 5x)(1) y_2 = 15 sin (3\pit + 5x)(2) Find the amplitude of motion at x = 21.

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0804.htm

Solution:

We note that the two waves are of the form y = A sin(wt \pmkx) where y is the displacement of the wave at position x and time t. A is the amplitude of the wave (i.e., the maximum dis-placement) , w is the angular frequency (= 2\pif) and k is the angular wavelength (= 2\pi/\lambda) of the wave. Both waves have the same characteristics (i.e., the same amplitude, frequency and wavelength). They differ only in that they travel in opposite directions. The negative sign in wave (1) indicates that it is travelling to the right. Wave (2) is travelling to the left. The resultant wave produced by the superposition of these two waves can be found as follows We use the relationships sin(\alpha + \beta) = sin \alphacos\beta +cos\alpha sin \beta sin(\alpha \rule{1em}{1pt} \beta) = sin \alphacos\beta \rule{1em}{1pt}cos\alpha sin \beta Therefore sin (\alpha + \beta) + sin (\alpha \rule{1em}{1pt} \beta) = 2 sin \alphacos\beta(3) Thus, comparing (3) with (2) and (1), y = y_1 + y_2 = 2 (15) sin (3\pi)tcos(5) x = 30 sin 3\pitcos5x . This wave pattern is called a standing wave (as opposed to a travelling wave). The wave remains in one location; or alternatively, the energy associated with the wave is not transferred from one location to another. With x = 21, 5x = 105 radians = 33.4\pi radians. Nowcos33.4\pi =cos(.4\pi + 33\pi) =cos.4\pi =cos72\textdegree = 0.309. Thus, x = 21, y = 30 sin 3\pitcos5x y = (30) (.309) sin 3\pit y = 9.27 sin 3\pit. The amplitude of this wave is the maximum value of y. This maximum value is 9.27.

Question:

From what you have learned of the processes of heredity andevolution, how do you think new species arise with respectto mutation? Which is more important, the accu-mulationof small mutations or a few mutations with largephenotypic effects? Give the reasons for your answer.

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Solution:

Most adaptive mutations produce barely distinguishable changes in thestructure and function of the organism in which they occur. On the otherhand, mutations with the potential to produce large phenotypic effects, necessary to elicit major changes in one jump, are practically alwayslethal or at least detrimental to survival. Major mutations are deleteriousbecause they constitute a great disturbance to the delicately balancedgenetic systems in which they arise. Therefore, major genotypic changesthrough mutation are not likely to be important in evolution. In contrast, the accumulation of small mutations over the course of manygenerations has significant evolutionary value. This has been supportedby fossil evidence, laboratory experiments and field studies. Evolutionary changes appear to arise as a result of many small changes thathave accumulated over time, ultimately producing a change in the compositionof a given population's gene pool. An example of this can be seenin the evolution of the orchid. The modern orchid flower, in both shapeand color, resembles the female wasp. Male wasps, thinking it is a female, are attracted to the flower and stimulated to attempt copulation. In doingso, they become coated with the sticky pollen grains of the flower andcarry them along to the next orchid. Any mutation that gave the orchid anysimilarity, however slight, to the female wasp would increase its chancesof being pollinated. Thus, such mutations would give those orchidsbearing them a selective advantage over orchids without them. The increased rate of pollination of the mutated forms, would promote the propagationof the mutant gene. It is easy to see how, over time, such mutationswould tend to be propagated and accumulate to the point where theorchid flower is today.

Question:

A scientifically minded Romeo has found a method of sending secret messages to a beautiful Juliet who is immured in the top floor of a castle 50 ft from the ground. Romeo places two light metal rods (too light to use for climbing up)against her windowsill, and between the rods he mounts a wire 10 cm long, to which is attached the message and a magnet so placed that the wire is permanently in a magnetic field of strength 0.049 Wb \textbullet m^-2, at right angles to the plane of the rods. When he passes a current of 10 A up one rod, through the connecting wire and back down the other rod, the message, wire, and magnet travel at uniform speed up the rods. The moving assembly weighs 0.25 kg. Neglecting friction, calculate what the length of the rods must be.

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Solution:

From figure A, we see that the magnetic field must be at right angles to the plane of the rods and acting downward. The magnetic field vector is per-pendicular to both the direction of current and to the force. The right hand rule determines that the direc-tion is downward. The magnitude of the force experienced by the wire and attachments is \vertF^\ding{217}\vert = lIB sin \varphi where \varphi is the angle between the direction of I and B \textbullet \varphi = 90\textdegree, and F = lIB = 0.1 m × 10 A × 0.049 Wb \bullet m^-2 = 0.49 N . Considering fig. B, we see that the forces acting on the wire and attachments are three in number: the weight acting vertically downward, the force P acting up the plane of the rods, and N, the normal reaction of the rods on the wire acting at right angles to the plane of the rods. Since the assembly moves up the rods at uniform speed, F = mg sin \texttheta . \thereforesin \texttheta = [(0.049 N) / (0.25 kg × 9.8 m \textbullet s^-2 )] = 0.02 . From the diagram, h/L = sin \texttheta = 0.02 . \thereforeL = (h/0.02) = (50 ft/0.02) = 2500 ft.

Question:

The "law of cooling" states that the temperature of a body, warmer than the surrounding substance, decreases at a rate proportional to the difference in temperature between the body and the surrounding substance. If the difference in temperature at time t is D(T), the constant of proportionality is r (the so-called "rate of cooling") and D(0) = D_0, use the modified Euler's method to simulate this system in FORTRAN, from time t = 0 to t = t_f . Compare the approxi-mations with the exact values.

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Solution:

If Ḋ is the rate of change of temperature difference, a process diagram looks as shown in Fig. 1. The corresponding equation is: Ḋ = -r \textasteriskcentered D(T)(1) The minus sign Is included because the temperature difference decreases at the rate r. The "goal" of this system is to reach a state of zero temperature difference (i.e., equilibrium). The exact solution of equation (1) is: D(T) = D_0e^-rt Note that t \ding{217} \infty \Rightarrow D(T) \ding{217} 0 as expected. The program looks as follows: DATA. T/0.0/ COMMON R READ, N, TFIN, ACC UR, R, DO PRINT,T,DO REALN = N DT = TFIN/REALN D = DO DO 10 I = 1,N T = T + DT CALL MEULER(T,D,ACCUR,DT) EXACT = DO \textasteriskcentered EXP(-R\textasteriskcenteredT) ERROR = ABS((EXACT-D)/EXACT) \textasteriskcentered100. PRINT, T, D, EXACT, ERROR 10CONTINUE STOP END FUNCTION G(W) COMMON R G = R\textasteriskcenteredW RETURN END

Question:

What is the adaptive importance of display in aggressive behavior?

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Solution:

It is necessary to distinguish aggressive behavior between members of the same species from violent predatory behavioral patterns, which are usually directed at members of different species. Aggressive behavior is usually exhibited between same-species males in defending their territory or establishing their status in a social order. The critical factor in aggressive behavior is that most of the fighting consists of display. Aggressivedisplay between fishes. Male pike blennies (Chaenopsis ocellata ) defend territories on the sea floor with highly ritualized exchanges that include lowering of the gill case, raising the fins, gaping of the jaws, and changes of color. Winner is determined by size and aggressiveness. Display consists of ritualized, highly-exaggerated mo-vements or sounds which convey the attack motivation of the contestants. In attempting to appear as formi-dable as possible, the animal often changes the shape of certain body parts in an attempt to make itself appear as large as possible. The adaptive significance of display is that the two contestants are rarely seriously hurt. Aggression is a type of bluff. The contestant's goal is to repel his opponent to a point outside of his territorial boundary or to convince his rival that a certain female is "taken." Aggressive behavior is therefore highly ritualized, with each movement conveying a certain message. It seldom results in bodily harm, even if physical contact is made. The animals can tell without serious fighting which one would be the probable winner if they were to actually fight. It would be biologically deleterious if during every aggressive contest between rivals, the loser were to be killed. Display behavior allows the "would-be loser" to escape and establish a territory or acquire a mate elsewhere. Survival and reproduction are therefore preserved.

Question:

Prove that the virtual image observed in a plane mirror is the same distance behind the mirror as the object is in front of the mirror.

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Solution:

As shown in the figure, let OA be the ray of light that strikes normal to the reflecting surface, while OB represents the ray that strikes the mirror at point B. The law of reflection states that the angle of incidence i equals the angle of reflection r, i = r. Rays DB and OA are extended back through the mirror to form triangle AIB, where point I is the apparent position of the image. Angle r must equal a; therefore, we obtain a = r = i. Angles b and c are equal since they are both right angles. Therefore, we have shown that the two triangles OAB and IAB are congruent, that is, coincide perfectly when super-imposed because they share a common side, AB. We conclude that OA = IA and that the virtual image appears as far behind the mirror as the object is in front of the mirror.

Question:

Discuss the role of pheromones in chemical communication.

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/Users/wenhuchen/Documents/Crawler/Biology/F31-0795.htm

Solution:

In general, communication between organisms can be defined as actionon the part of one individual that alters the behavior of another individual. Communication need not necessarily be visual or auditory, nor mustit always be confined to the present. If it is chemical, it obviously has theability to span time. Specific chemicals involved in social communicationare called pheromones, by analogy with hormones. Pheromones are small quantities of specific chemical substances released byan organism which act upon another organism at a distance from their pointof release and in very specific ways. Pheromones can be classified in two groups: those that possess releasereffects, which elicit immediate behavioral responses, and those thathave primer effects, which work by altering the physiology and subsequentbehavior of the recipient. Releaser pheromones are used by a variety of animals for different purposes, such as the attraction of mates, individual recognition, or trail or territorialmarking. The females of many moth species secrete pheromonesto attract males over large areas. The area within which the substanceis dense enough to be detected by a male moth is called the activespace. When a sexually active male flies into the active space, it fliesupwind until it reaches the pheromone-emitting female. This particular pheromoneis very potent since only one molecule of it is needed to stimulatea receptor on the male's antennae. The sugarglider ,asquirrel- likemarsupial, uses its own glandular secretions to mark its territorial boundaries. Male sugar gliders also mark their mates with pheromone, for recognitionand in order to warn other males that their mate is claimed. Ants use pheromones to lay down a trail from a recent food source to their nest. The pheromone, which is laid from the tip of their stinger, evaporates toform an active space. This enables other ants to rapidly find the food source. Primer pheromones change the physiology of the recipient by activatingcertain endocrine glands to secrete hormones. These hormones subsequentlycause changes in the behavioral patterns of the recipient. For example, social insects such as ants, bees and termites use primer pheromonesin caste determination .In termites the king and queen secreteprimer pheromones which are ingested by workers, preventing themfrom developing reproductive capabilities. In mammals, the estrous cycleof female mice can be initiated by the odor of a male mouse. In addition, the odor of an unfamiliar male mouse can block the pregnancy of afemale that has been newly impregnated by another male. This enables thestrange mouse to mate with the female. However, if the olfactory bulbs ofthe female are removed, the pregnancy block will not occur. This signifiesthat pheromones, unlike hormones, have an external rather than internalsource.

Question:

Steroid hormones do not act through the adenyl cyclase system. What is the mechanism of steroid hormone action?

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Solution:

Steroid hormones serve to increase the synthesis of proteins in target cells. The mechanism by which steroid hormones do this has been called the mobile-receptor model. The first step in this mechanism is the entry of the hormone into the cytoplasm. Steroid hormones are lipid soluble and can readily cross cell membranes. Once within the cell, the hormone binds to a soluble hormone-specific cytoplasmicprotein, known as the receptor. The hormone-receptor complex then moves into the nucleus and combines with specific proteins associated with DNA in the chromosome. The molecular interaction triggers off the specific RNA synthesis which results in increased protein synthesis by the cell. The thyroid hormones (amino acid derivatives) also act via direct diffusion into target cells. The distinction to be made is that thyroid hormones enter the nucleus and bind to a nuclear receptor. Diagram showing mobile-receptor model of asteriod hormone.

Question:

Give a list and briefly define the primitive (built in) functions of SNOBOL IV language.

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Solution:

ANY -matches any character appearing in its argument. APPLY -creates and executes a function call. ARBNO -takes a value of pattern structure that matches zero or more consecutive occurrences of strings matched by its argument. ARG -takes a value of the indicated argument from programmer- defined function. ARRAY -creates an array of variables. BACKSPACE - backspaces the indicated record. BREAK -matches a string up to indicated character. CLEAR -sets the values of all natural variables to the null string. CODE -converts a string of characters into object code. COLLECT -returns as value the amount of storage available after regeneration. CONVERT - converts a given object into given data type. COPY -produces a copy of an array, DATA -defines a new data type. DATATYPE - identifies the type of given data. DATE -gives the date of program execution. DEFINE -defines a programmer defined function. DETACH -removes any input and output association from the argument. DIFFER -succeeds if two given arguments are not identical. DUMP -lists all natural variables and their values in the program if argument \not = 0. DUPL -repeats the first argument the number of times specified by second argument. ENDFILE -closes the data set, indicated by argument. EQ-predicate "EQUAL". EVAL -evaluates an unevaluated argument. FIELD -gives function a value equal to the name of a given field of the programmer defined data type, specified in the argument. GE-predicate "GREATER THAN, or EQUAL TO" GT-predicate "GREATER THAN" IDENT -succeeds if two given arguments are identical. INPUT -specifies a programmer - defined input. INTEGER -succeeds if the value of argument is an integer. ITEM -assigns values to the specified members of the specified arrays. LE -predicate "LESS THAN, or EQUAL TO". LEN -returns a pattern structure that matches any string of the specified length. LGT -succeeds if the first argument is lexically greater than the second. LOCAL -gives function a value equal to the name of a given local variable from a specified defined function. LT -predicate "LESS THAN". NE -predicate "NOT EQUAL". NOTANY -matches any character not appearing in the argument. OPSYN -provides synonyms for existing functions, or operators. OUTPUT -specifies a programmer formatted output of a pro-grammer defined data. POS -succeeds if the cursor's position is just to the right of the one, specified by the integer argument. PROTOTYPE - returns a value of the dimension or range of an array. REMDR -returns an integer that is the remainder of dividing the first argument by the second. REPLACE -replaces character, given in the argument, by another in the specified string. REWIND -repositions the data set associated with the given number to the first file. RPOS -succeeds if the cursor's position is just to the left of the one, specified by the integer argument. RTAB -matches all the characters of a string up to, but not including the one specified in the argument, from the right end of a subject string. SIZE -returns an integer equal to the length of a given string. SPAN -matches a run of characters given in the argument. STOPTR -cancels a single trace association. TAB -matches all the characters of a string up through the one, specified in the argument. TABLE -creates a table of variables. TIME -gives the elapsed time of program execution (in MSEC.) TRACE -traces a specified data segment. TRIM -removes trailing blanks from a string in the argument. VALUE -assigns a given value to a specified variable.

Question:

(a) What is the acceleration of gravity on the moon's surface, g_m (as compared to the acceleration of gravity on the earth surface,g_e) if the diameter of the moon is one-quarter that of the earth's and its density is two-thirds that of the earth? (b) An astronaut, in his space suit and fully equipped, can jump 2 ft vertically on earth using maximum effort. How high can he jump on the moon?

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Solution:

(a) On the earth's surfaces: G is the Universal Gravitational Constant, M_e is the earth's mass and R_e is the radius of the earth. The volume of the earth is v_e= (4/3)\piR3_e If\rho_eis the earth's density, then, by definition of density M_e =\rho_ev_e= (4/3)\piR^3_e\rho_e Hence, by Newton's Law of Universal Gravitation g_e=GM_e/R2_e = (4\pi/3)[(GR^3_e\rhoe)/R2_e] = (4/3)\pi\rho_eGR_e(1) On the surface of the moon, we have a similar relation for the acceleration of gravity g_m =GM_m/R2_m = (4/3)\pi\rho_mGR_m(2) Therefore, upon division of equation (1) by equation (2), g_e/g_m =\rho_eR_e/\rho_mR_m= 3/2×4/1 = 6; that is, g_m = (1/6)g_e. Here, use was made of the given ratios\rho_e/\rho_mand D_e/D_m (D_e and D_m are the diameters of the earth and the moon respectively). (b) When he jumps at maximum effort, the astronaut can launch himself upward with an initial velocity v_0. On earth, whenv_final= 0, then, using the kinematic relation for constant acceleration, v2_final = v2_0 + 2as 0 = v2_0 - 2g_e×2 ft where the negative sign forg_eindicates that it is in the downward direction. \therefore v2_0 = 2g_e×2 ft = 12 g_m×2 ft. Similarly, on the moon, when he starts with the same initial velocity, he jumps the distance y vertically, where 0 = v^2_0 - 2g_m × y. \therefore y = v^2_0/2g_m = (24g_m/2g_m)ft = 12 ft.

Question:

Suppose a fisherman was interested in trying to extract the maximum number of fish from a pond or lake. Why would it not be to his economic advantage to reduce the fish populations by excessive harvesting?

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Solution:

The ideal economic strategy for the fisher- man to follow would be to catch only enough fish so as to keep the population at a level of optimal yield. If the population is allowed to reach the maximum that the environment can support, or if it is exploited to the point where the organisms become scarce, the yield will decline. This phenomenon exists because the greatest amount of increase in population size does not occur either when the population is very low or when the population has reached the carrying capacity of the environment. The point of greatest growth and hence the maximum replacement capacity of a population occurs at a point midway on the exponential part of an S-shaped growth curve, that is, half way between the baseline and the maximum carrying capacity (see figure). This point corresponds to the point of steepest rise on the S- shaped curve. At this point, the greatest number of new individuals are added to the population in a given amount of time. This maximum rate of population growth is referred to as the optimal yield. Fish populations, as well as other animal populations that man harvests, can be obtained with much less effort at a point of optimal yield than populations reduced by overfishing. Moreover, the greater the exploitation of populations, the more the population will come to consist of younger and smaller individuals, which are usually the least valuable commercially. The proportion of large fish caught is higher when the population is at a point of optimal yield. Consequently, as the fish populations are reduced, more expensive and sophisticated ships and trawling techniques are required to produce the same amount of yield.

Question:

Explain the various features of DO statements used for iter-ation .

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Solution:

By means of condition testing, one can execute only a single statement. There is a DO statement which al-lows a group of statements to be executed on condition test-ing. This is called a THEN DO statement. It is illustrated by the following example. IF A>0THEN DO ; SUM = SUM + X ; A = A + 1 ; B = SUM/Y ; PUT LIST(SUM,B) ; END ; Observe that four statements are executed when the condition tested is true as compared to only one statement that would be executed without the use of the DO statement. This DO statement allows a grouping of statements. The DO statement must have as its last statement an END statement. The DO and END statement pair do not influence the execution of the PL/I machine but has an editing function in the program. The pair identifies the loop more explicitly. Besides the simple DO group, there is also an iterative DO group. The format of the iterative DO statement is as follows: DO CONTROL VARIABLE=INITIAL VALUE TO LIMIT VALUE BY INCRE- MENT VALUE. Control variable is a variable name whose value will change according to control specifications. The variable is fol-lowed by an assignment symbol (=), which in turn is followed by a control specification. The control specification may consist of up to three elements as follows: a) Initial value, which is the value assigned to the control variable initially. The initial quantity must always be specified in the DO statement. Also, this initial quan-tity must be the first item to be specified on the right hand side of the (=) sign in the DO statement. b) The TO clause describes the limiting value to be used by the loop control variable. Once the value of the control variable exceeds the limiting value, no statement within the loop is executed and the next statement immedi-ately outside the loop is executed. c) The BY clause describes the increment value to be used in the loop control mechanism. Every time the loop is executed the value of the control variable is incremented by the increment value. ' The new value of the control variable is tested against the limiting value and the loop is exe-cuted if the control variable's value is less than or equal to the limiting value. For example, consider the following: DO I = -5 TO 5 BY 2; STATEMENT 1; STATEMENT 2; \textbullet \textbullet \textbullet STATEMENT N; END; STATEMENT X; is the control variable, its initial value is -5. As -5 is less than the limiting value of 5, statements 1 to n are executed, then the value of I is incremented by the incre-ment value 2. Hence, the value of the control variable is -3. Still it is less than 5, and again the loop is executed. Once again I is incremented by 2, tested against the limiting value of 5, and so on. When 1 = 5, still I is not. greater than the limit value of 5 and therefore the loop is executed once again. Now 1 = 7 which is greater than 5 and exit from the loop to statement X is performed. With regard to DO loops, the points to remember are as follows: a) If the increment value is positive, the limiting value must be greater than the initial value. b) If the increment value is negative, the limiting value must be less than the initial value. c) The BY and TO clauses are optional and may be omitted. i) When the BY clause is omitted: If no BY clause is found, but the TO clause is present then, by default the PL/I machine assumes BY 1 (increment by one). That is, PL/I assumes that the follow-ing statements DO I = 1 TO 70 BY 1,and, DO I = 1 TO 70; are completely equivalent. ii) When the TO clause is omitted: If the DO statement has the form given below: DO X = 2 BY 3, it can be seen that the TO clause is omitted. But there are no default rules to supply a TO clause in PL/I. Instead, the PL/I machine proceeds without testing the limiting value when the TO clause is omitted. Hence, a mechanism must be provided within the body of the loop to exit out of the loop, otherwise infinite looping will result. e.g.LOOP:DO I = 1 BY 1; SUM = SUM + I; IF SUM>10 THEN GOTO LABEL; END LOOP; LABEL:PUT LIST(I,SUM); END; In the above, exit from the loop is achieved when the value of SUM exceeds 10. If the TO clause and the BY clause are both omitted, for instance, DO J = 3. The DO statement will be executed only once as a result of the above. The While Clause In the Specification of a DO Group: The flexibility and power of the DO statement is fur-ther enhanced by using a WHILE clause in the control speci-fication of the DO statement. The format is as follows: DO WHILE (CONDITION) STATEMENT 1 STATEMENT 2 \textbullet \textbullet \textbullet STATEMENT N END In the above, processing of the DO group is terminated as soon as the CONDITION specified in the WHILE clause is not satisfied anymore. There are various formats and features of DO statements used for iteration. The DO statements are extensively used for iteration purposes as compared to GOTO statements. For a DO statement, the following segmented flowchart symbol is used:

Question:

The kangaroo rat never drinks water, yet it survives in some of the hottest and driest regions of the world. Explain how it obtains and conserves water.

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Solution:

Animals which live in the desert have limited means of obtaining water. For these animals, water retention and conservation are very crucial. The kangaroo rat can survive without ever drinking water, and in this respect is fairly unique. This animal is found in theSoutwestern deserts of the United States. It lives on dry seeds and obtains all of its water from carbohydrate metabolism. The excretory organs of the kangaroo rat are modified to conserve as much water as possible. Its kidneys are remarkably efficient in concentrating the urine, allowing very little water loss. As was noted earlier, a concentration gradient is achieved in the interstitial medium surrounding the loop ofHenleand the collecting duct in the mammalian kidney. The longer the loop, the greater the gradient can become, and therefore, the greater is the amount of water that can be reabsorbed. In the kangaroo rat, the loops are extremely long and thus a large amount of water is retained by the body and very little excreted.

Question:

A man holds a ball of weight w = 1/4 lb at rest in his hand. He then throws the ball vertically upward. In this process, his hand moves up 2 ft and the ball leaves his hand with an upward velocity of 48 ft/sec. Find the force P\ding{217}with which the \ding{217} man pushes on the ball.

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Solution:

During the action of throwing the ball, it experiences a net force (see figure) equal to the force of the hand, P, less the gravitational force of the earth on the ball, or mg. Hence, F_net = P - mg By Newton's Second Law, F = ma, the ball's acceleration, a, is given by a = (F_net/m) = (P - mg)/m This acceleration is delivered to the ball over a distance s = 2 ft, and accelerates the ball from rest (v_0 = 0) to v_f = 48 ft/sec. Hence, by our kinematics equations (a = constant if P is constant) v2_f - v2_0 = 2as v2_f = 2[(P - mg)/m]s = (2ps/m) - (2mgs/m) v2_f + 2gs = (2Ps/m) P = m/2s(v2_f + 2gs) P = (mv2_f/2s) + mg To calculate the mass of the ball from its weight, w, note that w = mg andm = w/g ThereforeP = (wv2_f/2gs) + w = w [(v2_f/2gs) + 1] P = (1/4 lb)[{(48 f/s^2)} / {(2)(32 f/s^2)(2 f)} + 1] P = (1/4 lb)(18 + 1) = 4.75 lb.

Question:

What is the energy of a photon of blue light whose frequency is7 × 101 4Hz and the energy of a photon for FM electromagneticradiation if the frequency is 1 × 10^8 Hz?

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Solution:

We shall use an approximate value for h of 6.6 × 10^-34 J s. The energyof a light quantum(photon) is given by E =hf wheref is the frequency of the light. Therefore, we have bluelight E =hf= (6.6 × 10^-34 Js) (7 × 10^14 Hz) = 4.6 × 10^-19 J FM waves E =hf= (6.6 × 10^-34 Js) (1 × 10^8 Hz) = 6.6 × 10^-26 J Notice that the higher the frequency of the electro-magnetic radiation, the greater the energy of the photon.

Question:

A wire of length 1 m is moving at a speed of 2 m \bullet s^-1 perpendicular to a magnetic field of induction of 0.5 Wb\bulletm^-2. What is the potential' difference induced between the ends of the wire? The ends are joined by a circuit of total resistance R = 6 \Omega. At what rate is work being done to keep the wire moving at constant velocity?

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Solution:

Since the direction of motion and the magnetic induction are at right angles, the induced emf will have magnitude given by Faraday's law \epsilon = [(d\varphi)/(dt )](1) where the flux, \varphi = \int B^\ding{217} \bullet dA^\ding{217}, is evaluated over the area enclosed by the circuit, B^\ding{217} is the constant magnetic induction passing through the loop of area A. The angle between the direction of the vectors B^\ding{217} and dA^\ding{217} is 0\textdegree. Thus \varphi = BA (for A = \int dA). However, A = lx, where l is the length of the wire, and x is the position of the wire within the magnetic field (see the figure). Hence, (1) becomes \epsilon = Bl (dx/dt) = Bl v where v is the velocity of the wire.Hence, \epsilon = vBl, = 2 m \bullet s^-1 × 0.5 Wb \bullet m^-2 × 1 m = 1 V. The power dissipated in the external circuit is P = (\epsilon^2 /R) = (1^2 V^2)/(6\Omega) =(1/6) W, and this must equal the rate at which work is being done to keep the wire moving at constant velocity. If the closed system is the loop, magnetic field, and the agent moving the loop, then the energy given up by the agent is used to heat the resistance. Only an energy exchange occurred, and not a change in energy, as required by the principle of conservation of energy. This can be checked by considering that a current I = (\epsilon/R) = (1/6)A will flow in the wire in the direction shown in the diagram for only in this direction will the current create a flux which will oppose the flux of the magnet-ic field in accordance with Faraday's Law. Using the right hand rule, if the fingers of the right hand curl in the direction of the current about the loop, then the thumb will point in the direction of the magnetic field created by this current. In this case, the created B points out of the page, opposite to the direction of the given field. When a current flows in a wire in a magnetic field, the wire experiences a force of magnitude F = IlB, which will be in this case in the direction shown. The work done per second to keep the wire traveling at uniform speed v is F (x/t) = Fv = IlvB = (1/6)A × 1 m × 2 m \bullet s^-1 × 0.5 Wb \bullet m^-2 = (1/6) W, in agreement with the result obtained above.

Question:

A piece of ice floats in a vessel filled with water. Will the water level change when the ice melts, if the final temperature of the water remains 0\textdegree C?

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Solution:

The temperature of the system remains constant, therefore we do not expect any thermal expansion of the water. Now, suppose that somehow we were able to take the ice out of water while preserving the geometry of the water as shown in figures A and B. The volume V_c of the cavity that is going to result is the volume of the ice which was submerged in water. In order to keep the level of the water the same, the pressure we must exert on the walls of this cavity must be equal to that exerted by the weight of the ice. Therefore, we can fill the cavity with water whose weight is equal to that of ice in order to ac-complish this. But, we already stated that this amount of water will have exactly the same volume as the submerged part of the ice, i.e., the volume of the cavity (Fig. C). As a result, we see that the level remains the same if the cavity is filled up by water. When ice melts, the situation is equivalent to what has been described so far since ice becomes water upon melting and effectively fills up the volume vacated by the ice in water.

Question:

Explain why tetraethyl lead is added to gasoline.

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Solution:

To explain the function of this additive in gasoline, consider how an automobile uses gasoline. The composition of gasoline varies according to the time of year, the refinery at which it was processed and the climate. Basically, gasoline is composed of hydrocarbon isomers containing between seven and ten carbon atoms. In cold climates, hydrocarbons such as butanes are used, which have a lower boiling point, so that the volatility of the gasoline is increased. The increased volatility allows the engine to start more easily. In warmer climates, higher-boiling components are used so that the gasoline's volatility is decreased; this minimizes such problems as vapor lock. The octane rating is a measure of the manner in which gasoline burns in the engine. The fuel should be consumed at the end of the piston cycle and the gasoline should burn such that there is steady pressure on the piston. If these conditions are not met as closely as possible, the automobile engine will knock or ping. This can eventually damage some engine parts. The quantitative measure of a fuel's ability to knock resulted in the formation of octane rating. By definition, an octane rating of 100 is the highest premium gasoline, while an octane of zero is the lowest. The octane rating of a part-icular gasoline can be raised by blending in compounds of relatively high octane rating. The only problem is that such compounds are extremely expensive. The discovery of additives solved the problem of improving octane rating. Tetraethyl lead (as pictured above) is one of these additives that can improve the octane rating without using the expensive high-octane hydrocarbons.

Question:

What are the evolutionary advances shown in the anatomy of the proboscis worms? How do these animals compare with the round worms?

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Solution:

The proboscis worms (or nemerteans) are a relatively small group of animals. Almost all are marine forms, excepting a few which inhabit fresh water or damp soil. They have long, narrow bodies, either cylin-drical or flattened, and range in length from 5 cm. to 20 meters. Their most remarkable organ is the proboscis, a long, hollow, muscular tube which they evert from the anterior end of the body to capture food. A mucus secreted by the proboscis helps the worm in catching and retaining the prey. There are three important evolutionary advances achieved by the proboscis worms. First, these animals have a complete digestive tract, with a mouth at one end for ingesting food, an anus at the other for egesting feces, and an esophagus and intestine in between. This is in contrast to the coelenterates and planarians, in which food enters and wastes leave by the same opening. As in the flatworms, water and metabolic wastes are eli-minated from the proboscis' body by flame cells. The second advance exhibited by the proboscis worms is the separation of digestive and circulatory functions. These organisms have a primitive circulatory system consisting of only three muscular blood vessels which extend the length of the body and are connected by transverse vessels. Hemoglobin, the oxygen-carrying protein present in higher animals, is found in the red blood cells of this rudimen-tary circulatory system. In the absence of a heart, the blood is circulated through the vessels by the movements of the body and the contractions of the muscular blood vessels. Capillaries have not developed in these animals. The third advance in the anatomy of the proboscis worms is seen in the structure of the nervous system. There is a primitive "brain" at the anterior end of the body, consisting of two groups of nerve cells, called ganglia, connected by a ring of nerves. Two nerve cords run poste-riorly from the brain. Compared to the proboscis worms, the round worms (or nematodes) are a far more numerous and ecologically diver-sified group than the proboscis worms. There are about 8000 species of round worms, all with a similar basic body plan. They have elongate, cylindrical, threadlike bodies which are pointed at both ends. Their many habitats include the sea, fresh water, the soil, and other animals or plants which they parasitize. This last characteristic is in sharp contrast to the proboscis worms, none of which is parasitic and none is of economic importance. Common parasitic nematodes are the hookworm, trichina worm, ascaris worm, filaria worm and guinea worm, which all utilize man as the host. Because of their parasitic exis-tence, these worms are covered with a protective cuticle and have only longitudinal muscles for simple bending movements. Contrary to the proboscis worms, which have cilia all over the epithelium and the lining of the diges-tive tract, none of the nematodes has any cilia at all. However, like the proboscis worms, the round worms have evolved a complete digestive system, a separate circulatory system, and a nervous system composed of a "brain" and nerve cords.

Question:

Given the ground-state oxygen atom, tabulate each of the electrons by its quantum number.

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Solution:

Begin by writing the electronic configuration of the ground-state oxygen atom. An electron is in its ground state when it is in its lowest energy level, this, is the normal state of the electrons. The atomic number of oxygen is 8. Thus, the oxygen atom has 8 electrons. The electronic configuration of the oxygen atom is 1s^2 2s^2 2p^4. The quantum numbers are N, l, M, and s. N is the principal quantum number which, in the electronic con-figuration, is the integer in front of thesubshells(s and p) and corresponds to the energy level occupied by the electron. The other quantum numbers can be found once N is determined, l is equal to 0, 1, 2, . . . N - 1, depending upon N. M is equal to + l to - l and S is equal to +1/2 or - 1/2. l is the orbital quantum number which denotes thesubshelland angular shape of electron dis-tribution. M is the magnetic quantum number and s denotes the spin of electron. All four are needed to describe each electron. To tabulate the electrons according to their quantum numbers proceed as follows: the electron conf-iguration is 1s^2 s^2 2p^4. Take the first shell, or N = 1, 1s^2. The superscript indicates that 2 electrons are present in this shell. Thus, 2 sets of quantum numbers are needed. For both electrons, N = 1. Because l = 0 . . . n - 1, l = 0 for both electrons in the s subshell . M + l to - l, thus, M = 0 for both also, The four quantum numbers for each electron must be different, Therefore, one electron has a spin of + 1/2 and one has a spin of - 1/2. The sets of quantum numbers for these two electrons are 1, 0, 0, + 1/2 and 1, 0, 0, - 1/2. For the second shell, N = 2 There are twosubshells, s and p. Consider the ssub-shellfirst. In the 2s^2 orbital, N = 2 and there are two electrons to be described. For both electrons l = 0, because both are in the ssubshell, M = 0, because M = + l to - l. For one electron s = +1/2, for the other s = -1/2. The sets of quantum numbers for these two electrons are 2, 0, 0, + 1/2 and 2,0, 0, -1/2, in the 2p^4 orbital there are 4 electrons to be described. For an, N = 2,and l = 1 (l is always equal to 1 for a psubshell), M = + l to - l. Thus, one electron has M = + 1, the second electron has M = 0, and the third electron has M = - 1. The fourth electron will have an M = - 1, 0, or + 1. Whatever it is, it will differ in spin from the electron with the same M. The spins of the others can be + 1/2 or - 1/2. The sets of the quantum numbers for the first three electrons in a 2p orbital can be written (2, 1, -1, + 1/2) , (2, 1, 0. + 1/2) , and (2, 1, 1, + 1/2) , The fourth might be written (2, 1, - 1, - 1/2), (2, 1, 0 - 1/2), or(2, 1, 1, - 1/2).

Question:

Complete the following nuclear equations. (a) _7N^14 + _2He^4\rightarrow_8O^17 + ..... (b) _4Be^9 + _2He^4\rightarrow_6C^12 + ..... (c) _15P^90\rightarrow_14Si^30 + ..... (d) _1H^3\rightarrow_2He^3+ .....

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0715.htm

Solution:

The rules for balancing nuclear equations are: (1) the superscript assigned to each particle is equal to its mass number and the subscript is equal to its atomic number or nuclear charge; (2) a free proton is the nucleus of a hydrogen atom, and is therefore written as _1H^1; (3) a free neutron has no charge and is therefore assigned zero atomic number. Its mass number is one and its notation is _0n^1; (4) an electron, \beta^-, has zero mass and its atomic number is - 1, hence the notation _-1e^0; (5) a positron has zero mass and its atomic number is + 1, hence the notation _+1e^0; (6) an alpha particle (\alpha-particle) is a helium nucleus, and is represented by _2He4or \alpha; (7) Gamma radiation (\gamma) is a form of light, and has no mass and no charge; (8) in a balanced equation, the sum of the sub-scripts must be the same on both sides of the equation; the sum of the superscripts must also be the same on both sides of the equation. In equation (a), _7N^11 + _2He^4\rightarrow_8O^17 + ..., the sum of the subscripts on the left is (7 + 2) = 9. The subscript of one of the products is 8, thus, the other product must have a subscript or net charge of 1. The sum of the super-scripts on the left is (14 + 4)= 18. The superscript of one of the products is 17, thus the other product on the right must have a superscript or mass number of 1. The particle with a + 1 nuclear charge and a mass number of 1 is the proton, _1H^1. In equation (b) , _4Be^9 + _2He^4\rightarrow_6C^12 + ..., the nuclear charge of the second product particle (that is, its subscript) is (4+2) -6=0. The mass number of the particle (its superscript) is (9 + 4) - 12 = 1. Thus, the particle must be the neutron, _0n^1. In equation (c),is _15P^30 _14Si^30 + ..., the nuclear charge of the second particle is 15 - 14 = + 1. Its mass number is 30 - 30 = 0. Thus, the particle must be the positron, _+1e^0. In equation (d), _1H^3\rightarrow_2He^3 + ..., the nuclear charge of the second product is 1 - 2 = - 1. Its mass number is 3 - 3 = 0. Thus, the particle must be a \beta^-1 or an electron, _-1e^0.

Question:

Examine the following COBOL program and describe in words the function of each section.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0263.htm

Solution:

We have the first three divisions (IDENTIFICATION DIVISION, ENVIRONMENT DIVISION, and DATA DIVISION) of the program LIST- REPORT given. The IDENTIFICATION DIVISION gives the program name, who wrote it and when it was written, and may include remarks about the program. The ENVIRONMENT DIVISION consists of two sections. In the CONFIGURATION SECTION the compiling (source) and executing (object) computers are specified. In this program both compilation and execution of the program is done on an IBM-370. The INPUT-OUTPUT section specifies the interface between the program files and the actual external devices. In this program the actual file asso-ciated with the identifier TAPE-FILE is the UT-S-TAP01 tape reader. Similarly, CARD- FILE is associated with the UR-S-CAR01 card reader and PRINT-FILE is associated with the UR-S-PRINT01 printer. The DATA DIVISION is also composed of two sections. The File Section describes the structure of the files used in the program. Every file selected in the INPUT- OUTPUT SECTION of the Environment Division must have an FD (File description) entry in this section. Every FD contains: 1) a file-name 2) a LABEL RECORDS clause. The label record can be either STANDARD or OMITTED. OMITTED is used when there is no label record, STANDARD is used when the label is a standard system label. 3) A description of the record associated with the file-name. As an example of an FD, the file TAPE-FILE will be used from our program. In this file the label records are standard. The statement BLOCK CONTAINS 10 RECORDS. means that the tapereader UT-S-TAP01 sendsdata to the computer in the form of Data Blocks, each block containing 10 records of the type described in TAPE-RECORD. The WORKING STORAGE SECTION describes the independent or group variables which will be used for data storage and manipulation in the PROCEDURE SECTION of a program.

Question:

Consider the simple regression model. When there is a large number of observations, storage in single subscripted arrays can create problems. Devise a flowchart and write the program for performing the linear regression without using arrays.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G19-0485.htm

Solution:

Let (x_i,y_i) be data points. The simple regression problem is to find numbers\alpha˄, \beta˄ such that \alpha˄ ^n\sum_i=1 [y_i - (\alpha˄+ \beta˄x_i)]^2 \alpha˄ is minimized. The solution is given by the formulae \beta˄^ = {[\sum(x -x) (y -y)] / [\sum(x -x)^2]}\alpha˄=y- \beta˄x \alpha˄ \alpha˄ where x= (1/n) ^n\sum_1 x_i,y= (1/n) ^n\sum_1 y_i. One method of performing linear regression on a computer is to read and store the data (x_i,y_i) in arrays. The sums of squares and means are then found by operating with these arrays. However, the regression formulae can be rewritten so that \sum(x-x)^2, \sum(y-y)^2 , \sum(x-x) (y-y) can be expressed in terms of ^n\sum_1 x_i , ^n\sum_1 y_i , ^n\sum_1 y^2_iand^n\sum_1 x^2_i . These latter sums can be computed directly from the observations without storing them in arrays. To clarify, consider \sum(x-x)^2 : ^n\sum_i=1 (x_i -x)^2 = ^n\sum_1 x^2_i - 2x^n\sum_1 x_i + ^n\sum_1x^2(1) But 2x^n\sum_1 x_i = 2xnx(^n\sum_1 x_i = nx= 2nx^2 . Also, ^n\sum_1x^2 =x^2 + x^2 +...+ x^2= nx^2 \downarrow n times Thus, (1) is equal to ^n\sum_1 (x_i -x)^2 = ^n\sum_1 x^2_i - nx^2.(2) To compute (2), all we need is ^n\sum_1 x^2_i andx. No array is necessary for either of these since^n\sum_1 x^2_i can be computed directly by squaring each observation, as it is read in, and summing. Similarly,x= (1/n) \sumx_i can be computed directly. From (1) and (2) above, \sum (y_i -y)^2 = ^n\sum_1 y^2_i - ny^2.(3) To simplify \sum(x-x) (y-y) consider the following: ^n\sum_1 (x_i -x) (y_i -y) = x_1 y_1 - x_1y-xy_1 +xy + x_2 y_2 - x_2y_2 -xy_2 +xy+...+ x_n y_n - x_ny-xy_n +xy = ^n\sum_1 x_i y_i -y^n\sum_1 x_i -x^n\sum_1 y_i + nxy(4) But -y^n\sum_1 x_i = -ynx= -x^n\sum_1 y_i . Thus (4) equals ^n\sum_1 (x_i -x) (y_i -y) = ^n\sum_1 x_1 y_1 - nxy(5) The program looks as follows: READ, N SX = 0 SY = 0 SXX = 0 SYY = 0 SXY = 0 J = 1 1IF (J.GT.N) GO TO 2 READ, X,Y SX = SX + X SY = SY + Y SXX = SXX + X\textasteriskcenteredX SYY = SYY + Y\textasteriskcenteredY SXY = SXY + X\textasteriskcenteredY J = J + 1 GO TO 1 2XM = SX/N YM = SY/N SXX = SXX - N\textasteriskcenteredXM\textasteriskcentered\textasteriskcentered2 SXY = SXY - N\textasteriskcenteredXM\textasteriskcenteredYM SYY = SYY - N\textasteriskcenteredYM\textasteriskcentered\textasteriskcentered2 B = SXY/SXX A = YM - B\textasteriskcentered XM SD = SYY - B\textasteriskcenteredSXY PRINT, A,B,SD STOP END

Question:

Calculate the voltage (E) of a cell with E\textdegree = 1.1 volts, If the copper half-cell is at standard conditions but the zinc ion concentra-tion is only .001 molar. Temperature is 25\textdegreec . The overall reaction is Zn + Cu^+2 \rightarrow Cu + Zn^+2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0597.htm

Solution:

To solve this problem requires a relation between the cell potential (voltage), E, and the standard cell potential, E\textdegree . The equation used is the Nernst equation, which states E = E\textdegree -(RT / nF) In(Q), where n is the number of mole of electrons transferred, R is the constant 8.314 joules per degree, F is 96,500 coulombs, and Q is the equilibrium constant expression. In other words, Q = [product] / [reactant] You are told the concentration of Zn^+2 is .001. = [Zn^+2] / [Cu^+2 ]= .001 / 1 Standard conditions indicate that the concentration of the substance is 1M and at a temperature of 25\textdegree c . Thus [Cu^+2] = 1 and there-fore, from the reaction equation, it can be seen that 2 electrons are transferred. Thus, n = 2. Substituting these values into the Nernst equation, one obtains E = 1.1 - {[8.314(298)] / [2(96,500)]} log (.001 / 1) = 1.1 - [.0295(-3)] = 1.19 volts.

Question:

Why must mosses and liverworts (phylumBryophyta) always live in close association with the water?

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0178.htm

Solution:

The bryophytes are generally considered to be primitive land plants because they are relatively ill-adapted to the terrestrial environment compared to the higher land plants. Mosses, for instance, are found frequently on stream banks or moist roadsides. Liverworts, lacking a cuticle , are not as well protected againstdessicationas are the mosses, and are even more restricted in their distribution - the majority of them grow in moist, shady localities, and some are even true aquatic plants. The fact that mosses and liverworts always live in close association with water can be explained by their anatomical structures and reproductive mechanism. They have rhizoids which are simple filaments of cells or cellular projections performing the function of water absorption. The rhizoids are, however, not efficient absorbers and in relatively dry areas cannot withdraw adequate materials from the ground. In addition, mosses and liverworts lack vascular tissues to conduct water, minerals, and organic substances to different parts of the plant. Therefore transport of materials in the mosses and liverworts depends to a great extent on simple diffusion and active transport through the leaves and plant axis. Furthermore, mosses and liverworts are small plants with a high surface to volume ratio. This means that evaporation can cause a rapid dehydration of inner as well as outer tissues. These disadvantages can be avoided in a moist habitat where evaporation by the atmosphere is slower and sufficient water is available through diffusion to compensate for water lost through the surface. A moist habitat also favors the reproductive process of the bryophytes . The gametophyte plant produces flagellated sperm which can swim to the egg only when water is present. In order to reproduce successfully , mosses and liverworts must grow in close proximity to water.

Question:

(1) Cadmium and Hydrogen (2) Silverand Hydrogen (3) Cadmium and silver, using the following data: Reaction E\textdegree volts Cd \rightarrow Cd+^2 + 2e^- +.403 H_2 \rightarrow 2H^+ + 2e^- 0.00 Ag \rightarrow Ag^+ +e^- -.799

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0584.htm

Solution:

You are asked to calculate the standard cell potential (∆E\textdegree) for each of the given pairs. To do this, you must realize that in such cells, you have 2 half-reactions. Namely, an oxidation re-action (loss of electrons) and a reduction reaction ( gain of elec-trons). The sum of these half- reactions yields the overall reaction and the ∆E\textdegree of the whole cell. Thus, to find the ∆E\textdegree for each of these pairs, you need to know the E\textdegree of the half- reactions. There is one other important fact to be kept in mind. In cells, if the reaction is to proceed spontaneously, ∆E\textdegree must have a positive value. This means, therefore, that you must choose the half- reactions such that their sum always gives a positive ∆E\textdegree You are told the E\textdegree of the oxidation half-reaction for each element in each pair. The reduction half-reaction is the reverse of the oxidation reaction for each element, with a change in sign of E\textdegree . For example, if you have A \rightarrow A^+1 + e^- with an E\textdegree = B for oxidation, the reduction is A^+1 + e^-\rightarrow A with a E\textdegree = -B. With this in mind, the procedure is as follows: (1) Cadmium (Cd) and hydrogen (H_2). The reaction for this cell must be the sum of the oxidation and reduction such that the ∆E\textdegree is positive. This can only occur if the anode (oxidation) has the higher oxidation potential. Thus, you calculate ∆E\textdegree as + (.403) - (0.000) = + .4303v . Similarly, for (2)and (3), a positive ∆E\textdegree can only be obtained with the anode having the higher oxidation potential. Thus, for (2), E\textdegree =(0.000) - (-.799) = +( .799v.) For (3), E\textdegree = (.403) - (-.799) = +1.202v.

Question:

What is the period of a small oscillation of an ideal pendulum of length l, if it oscillates in a truck moving in a horizontal direction with acceleration a? (see figure a).

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0372.htm

Solution:

Let the equilibrium position be given by the angle \alpha. In this position, the net force on m is along the horizontal axis and equals ma. The angle \alpha is determined by the equations T sin \alpha = ma(1) T cos \alpha = mg(2) where g is the gravitational acceleration. When the pendu-lum is displaced by a small amount \texttheta, it will perform a simple harmonic motion around the equilibrium position. The force on m along the horizontal x-axis is T sin(\texttheta + \alpha), as shown in Fig. b. Newton's second law for the motion along this axis is m[d^2x/dt^2] = -T sin (\texttheta + \alpha)(3) where x is the distance from the vertical. For small \texttheta, we can expand sin(\texttheta + \alpha) as sin(\texttheta + \alpha) \approx sin \alpha + \texttheta cos \alpha. Substituting in (3) m[d^2x/dt^2] \approx -T sin \alpha\Elzbar \textthetaT cos \alpha.(4) Combining (1), (2) and (4), we get m[d^2x/dt^2] \approx -ma - \textthetamg d^2x/dt^2 \approx -a - \textthetag.(5) \texttheta and l are related geometrically as x = l sin (\texttheta + \alpha) = l sin \alpha + \textthetal cos \alpha therefore d^2x/dt^2 \approx l cos \alpha (d^2\texttheta/dt^2) Substituting in (5), l cos \alpha (d^2\texttheta/dt^2) \approx -\alpha - \textthetag d^2\texttheta/dt^2 \approx - [g/(l cos \alpha)][\texttheta + (a/g)](6) If we make the following substitution in (6) \Phi = \texttheta + (a/g) we get d^2\Phi/dt^2 = - [g/(l cos \alpha)]\Phi. This is the differential equation for a simple harmonic motion. Therefore its solution is \Phi = A sin(\omegat + B) where A and B are the constants of integration and \omega = \surd[g/(l cos \alpha)]. The expression for \texttheta is \texttheta = A sin(\omegat + B) - (a/g). The boundary conditions are \texttheta = 0 at t = 0 and \texttheta = \texttheta_max at t = \tau/2 (where \tau is the period). Determine the constants A, B: \cyrchar\cyrii) t = 0: A sin B = a/g \cyrchar\cyrii\cyrchar\cyrii) t = \tau/2: A sin[(\omega\tau)/2 + B] - a/g = \texttheta_max . The period of motion is \tau = 2\pi/\omega = 2\pi\surd[(l cos \alpha)/g].

Question:

What is eutrophication? What problems are associated with it? Why would a reduction of phosphate ions from waterways alleviate eutrophication in lakes and streams?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0856.htm

Solution:

The enrichment of water with nutrients is a naturally occurring biological process called eutrophication. It is caused by excess nutrients in water that enable algae to grow to great abundance. The problem of massive algae growth occurs when it dies. As it dies, it sinks to the bottom of lakes and rivers and begins to decay. This decay-ing process consumes oxygen thereby deleting the oxygen content of the water. Fish and other aquatic life dependent on oxygen suffocate. Therefore, the decaying of excessive amounts of algae can lead to the death of marine life. This is one major problem associated with eutrophication. A second problem also results from the depletion of oxygen. Remember that anaerobic bacteria, i.e. bacteria that can live without oxygen, still function effectively under the conditions described above. These bacteria feed on decaying algae. As a result of their metabolic process, they produce compounds such as hydrogen sulfide, which cause the putrid odor associated with decaying organic matter. The water becomes foul smelling and foul tasting. A reduction of phosphate would alleviate eutrophica-tion because it is one of the major nutrients of algae. A smaller amount of phosphate in the water would limit the growth of algae.

Question:

Minimize the given minterm function f(A,B,C,D) via the Karnaugh map. f (A., B, C, D) = \summ(0,1,3,8,9,11,13,14)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0056.htm

Solution:

We are given the function f(A,B,C,D) = \summ(0,1,3,8,9,11,13,14) which consists of minterms to be minimized. The first step in obtaining the solution is to plot the function on the K-map. In figure # 1 a K-map for four variables is given. These maps are standard for each problem, however they change size according to the number of variables in the function. To plot the given function into the K-map, 1's are put into the boxes which number matches with the given minterm variable. The final K-map with the variables plotted looks as shown in figure 2. Secondly, the number of adjacencies for each minterm is counted. The number of adjacencies for each minterm is shown in the lower right-hand corner of the minterm block of figure 3. Now the map is ready for the simplification process. This is done by choosing first the minterm m_14 which has no adjacencies, as in figure 4. This minterm must be taken as a group itself; hence the first term in the minimized function is ABCD since minterm box #14 is not covered by variable D. Next, minterms with one adjacency are examined; m_13 the only one. Consequently, m_13 has one way of being grouped and that is with m_9 .This is illustrated in figure 5. Note that minterms m_13 and m_9 have the following variables: m_13 = ABCD , m_9 = ABCD , therefore, B (in m_13), andB(in m_9) may be cancelled according to Boolean algebra theorems. Continuing the solution; four minterms with two adjacencies are taken into account. These are m_0,m_3, m_8 and m_11 . One of these is picked to group at random. If m_3 is picked (it is adjacent to m_1, and m_11), since m_9) is also available (any minterm may be used as many times as it is needed), a large group of four minterms may be formed. Note that m_1 =A B CD m_3 =A BC D m_11= ABC D m_9 = AB CD in these four minterms A, and -Aterms also C, andCterms cancel each other, using Boolean theorems a\bulleta= 0 , therefore the minimized form of minterm m_1,m_3,m_11, and m_9 isBD. The function now becomes f(A,B,C,D) = ABCD+ ACD +BD. Finally there are still two minterms which have not been accounted for. These two, m_0 and m_8, can be grouped with m_1 and m_9 respectively, to form a last group of four minterms as shown in figure 6. This last group adds the final product termBCto_the function f(A,B,C,D), and hence: f(A,B,C,D) = ACBD+ ACD +BD +BC.

Question:

Find Antilog2.3625.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0968.htm

Solution:

By definition, b = Antilog a, Is equivalent to log b = a. Let N = Antilog 2.3625. Therefore, log N = 2.3625. The characteristic is 2. Hence, the number that corresponds to the mantissa 0.3625 will be multiplied by 10^2 or 100. In a table of four-place common logarithms, the mantissas 0.3617 and 0.3636 are those given that are closest to 0.3625. Therefore, since the mantissa 0.3625 does not appear in the table, the number that corresponds to this mantissa will be found through inter-polation. The following proportion is now established: d / 0.1 = .0008 / .0019 Cross multiplying, .0019d =(.01) (.0008) d = .01 [(.0008) / (.0019)] = 1 × 10^\rule{1em}{1pt}2[(8 × 10^\rule{1em}{1pt}4) / (1.9 × 10^\rule{1em}{1pt}3)] = (8 × 10^\rule{1em}{1pt}2+(\rule{1em}{1pt}4)) / (1.9 × 10^\rule{1em}{1pt}3) = (8 × 10^\rule{1em}{1pt}6) / (1.9 × 10^\rule{1em}{1pt}3) = (8/1.9) × (10^\rule{1em}{1pt}6 /10^\rule{1em}{1pt}3) = 4.2 × 10^\rule{1em}{1pt}6\rule{1em}{1pt}(\rule{1em}{1pt}3) = 4.2 × 10^\rule{1em}{1pt}3 d = .0042 Hence, x = 2.30 + 0.0042 = 2.3043 \approx 2.304. Therefore, Antilog 2.3625 = N= 2.304 ×10^2 = 2.304 ×100 = 230.4

Question:

An observer moves past a meter stick at a velocity that is one-half the velocity of light. What length does he measure for the meter stick?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0931.htm

Solution:

Einstein stated two postulates which form the basis for the theory of special relativity. They are From these postulates, it is found that the length of an objectviewed by an observer travelling with velocity v with respect to the object is less in the direction of motion than the length seen by a person stationary with respect to the object. The apparent shortened length l ' of the meter stick as it moves with respect to the observer (Lorentz contraction) is given by l ' = [ l \surd{1 -(v/c )^2} =l \surd( 1 - \beta^2) where l is the length seen by a person stationary with respect to the object and \beta is v/c. For this problem, \beta = v/c = 0.5c/c = 0.5 and the length l ' measured by the observer is l ' = { l ' \surd(1 - \beta^2)} = (10cm) × \surd{1 - (0.5)2} = (100cm) × {\surd0.75} = 86.6 cm

Question:

A converging lens with a focal length of 3 m forms an image of an object placed 9 m from it. Find the position of the image and the magnification.

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0875.htm

Solution:

The simple lens equation for the converging lens of this problem is 1/f= 1/p + 1/q where f, p and q are respectively the focal length of the lens and the distances of the object and the image from the lens. The image is real and inverted (see the figure). Substituting the given values in the above equation, we get 1/q = 1/f - 1/p = 1/3m - 1/9m = (2/9)m-1and q = (9/2)m = 4.5m Since the value of q is positive, the image occurs on the right side of the right side of lens. The magnification M is M = q/p = 4.5m/9m = 0.5 so the image is one-half as high as the object.

Question:

One problem concerning space flight is the removal of carbon dioxide (CO_2, molecular weight = 44.0 g / mole) emitted by the human body (about 924 g of CO_2 per day). One solution is to react CO_2 with sodium hydroxide (NaOH, molecular weight = 40.0 g / mole) to form water and sodium carbonate (Na_2CO_3, molecular weight = 94 g / mole) according to the followingreation: 2NaOH + CO_2 \ding{217} NaCO_3- + H_2O. How muchNaOHmust be carried on board a space capsule to remove the CO_2 produced by an astronaut on a 10 day flight?

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Solution:

Two moles ofNaOHare required to remove one mole of CO_2, as indicated by the coefficients in the reaction. The number of moles of CO2 produced is calculated from the mass emitted by the astronaut during the flight. From this, the number of moles ofNaOHrequired for the removal of he CO_2 is calculated, and this number of moles is then con-verted to grams. At the rate of 924 g per day, in 10 days the astronaut will produce 924 g / day × 10 days = 9240 g of CO_2. Dividing this mass by the molecular weight of CO_2, it is calculated that the astronaut will produce a total of [9240 g / (44.0 g / mole)] = 210 moles of CO_2. The number of moles ofNaOHrequired to remove 210 moles of CO_2 is 2 × 210 moles = 420 moles. Multiplying this by the molecular weight, the minimum amount ofNaOHthat must be carried on board is 420 moles × 40 g / mole = 16,800 g = 16.8 kg.

Question:

Illustrate transfer of control in a PL/I program, usingGOTO statements.

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Solution:

The PL/I machine has an implicitly defined sequential execution. The execution of statements is normally a sequential process, however, a specialstatement, the GOTO statement can alter the processing sequenceof the PL/I ma-chine. This GOTO statement discontinues the normalsequen-tial execution and processing resumes at a point that is in-dicatedby a label. Examples of well formed GOTO statements are given below: 1) SUM:A= B + C; \textbullet \textbullet \textbullet \textbullet GOTO SUM; 2)GOTO NEXT; \textbullet \textbullet \textbullet \textbullet NEXT:X=Y\textasteriskcenteredZ; 3) JUMP:GOTO TERMINATE; \textbullet \textbullet \textbullet \textbullet TERMINATE:END; The following are a few examples of not well-formed GOTO statements. a)JOE:PROCEDURE OPTIONS (MAIN); \textbullet \textbullet \textbullet \textbullet GOTO JOE; The above is wrong due to the fact that the control can only be transferred withinthe body of the main program. In other words, GOTO cannot transfercontrol to the starting point of a procedure as above. b) GOTO END_1 \textbullet \textbullet \textbullet \textbullet END END_1 The above is wrong because there is no label END_1 anywhere in the bodyof the program. END_1 is used as a part of an-other statement, but END_1 is not a label of any statement in the program body.

Question:

What would be the simplest and safest reliable method of distinguishing between the following pairs of materials? (a) Water and rubbing alcohol. (b) Gasoline and kerosene, (c) Baking soda and baking powder.

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Solution:

(a) Pour to a depth of about one-half inch of each compound into two separate test tubes. Cover each of the test tubes with your thumb and shake them vigorously. There will be an outward pressure on your thumb from the contents of one of the test tubes. This tube contains the rubbing alcohol. This pressure is called vapor pressure. It is proportional to the ability of a liquid to vaporize or evaporate. Vapor pressure is a measure of the physical property called volatility. Rubbing alcohol is more volatile than water and will have a greater vapor pressure and lower boiling point. (b) Gasoline and kerosene are both products of the distillation of crude oil. Distillation is a process used to separate mixtures of liquids. When a liquid is distilled it is boiled in a round bottomed container at the end of a glass enclosed tube. The various compounds in the mixture burn at their own distinctive boiling points. Thus, first the lowest boiling substance will boil and evaporate, then as the mixture becomes hotter the next lowest and so on until no compounds are left in the container. As each com-pound boils its vapors enter the tube or column where they are condensed and eventually collected. A different vessel is used to collect each fraction with a distinct boiling point and hence the mixture becomes separated into its various components. The more times a fraction is distilled the more pure a sample made. This is the best method for separating gasoline and kerosene. Gasoline burns over a range from 40\textdegree to 200\textdegreeC and kerosene from 175\textdegree to 275\textdegreeC. They can be easily separated by distillation. The accompany-ing figure shows a fractional distillation of crude oil. (c) The difference between baking soda and baking powder is that baking soda is a compound and baking powder is a mixture. A compound is a homogeneous substance composed of two or more elements, the proportions of which are fixed and invariable by weight. A mixture is a substance made of two or more elements or compounds in a nonhomogeneous combination. Under a magnifying glass or microscope one will not be able to distinguish the components of a compound whereas one can readily identify the components of a mix-ture. Baking powder is a mixture of baking soda, an acid substance and starch, while baking soda is made up of a single compound, sodium hydrogen carbonate (NaHCO_3).

Question:

Five resistors, each of resistance 10 \Omega, are connected to form a letter H, a 2-V cell of internal resistance 1.86 \Omega being connected across the upper ends and an ammeter of resistance 5 \Omega across the lower ends. What current passes through the ammeter?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0681.htm

Solution:

Because the 5 \Omega resistance of the ammeter and the 10 \Omega resistance of the lower branch of the circuit in figure (A) are in series, figure (A) is equivalent to figure (B). This follows because the equivalent resistance of n resistors in series is equal to the sum of their individual resistances. The resistances of 10 \Omega and 25 \Omega are in parallel. Hence the equivalent resistance is R, where (1/R) = (1/10 \Omega) + (1/25 \Omega) = [(35 \Omega)/(250 \Omega^2)] \thereforeR = (50/7) \Omega = 7(1/7) \Omega = 7.14 \Omega. The circuit is therefore equivalent to the one shown in diagram (C). It is now possible to find the current I_0 in the battery circuit, for, by Ohm's Law I_0 = (\epsilon/R) = [(2 V) / {(10 + 10 + 7.14 + 1.86) \Omega}] = (2/29) A.(1) This current splits up into currents I_1 and I_2 through the lower parts of the circuit, as shown in diagrams (A) and (B). Using figure (B), note that branches CD and EF are in parallel. Therefore, the voltage drops across CD and EF are equal. Using Ohm's Law V_CD = V_EF or(10 \Omega)I_2 = (25 \Omega)I_1 (I_1 /I_2 ) = (R_2 /R_1 ) = [(10 \Omega)/(25 \Omega)] = (2/5) Since no charge can accumulate in the circuit, then I_1 + I_2 = I_0. Therefore, I_1 + (25/10) I_1 = I_0 I_1 = (10/35) I_0 = (10/35) × (2/29) A = 0.0197 A, where we have used (1). This is the current flowing through the ammeter.

Question:

Two small loudspeakers A and B vibrate in phase at 800 cps, and are set apart by 10 ft. The speakers emit spherical waves which obey the inverse square law with respect to intensity. The intensity varia-tion (for distances away from the speakers greater than a few inches) is I_A = 60/R^2_A I_B = 30/R^2_BThe units are arbitrary; R_A is the distance from speaker A, R_B is the distance from speaker B. Find the point of minimum intensity along the line joining the speakers. Find the first three points of zero intensity that are nearest to the speakers.

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Solution:

Assume that points of minimum and zero intensity occur only at points of destructive inter-ference. Destructive interference of two waves occurs when the waves are 180\textdegree out of phase. The resultant amplitude at the point of interference will be the difference of the amplitude of the two waves at the point. The speed of sound is 1200 ft/sec. Consider point C (in the figure) such that R_A =CAR_B =CB Wave cancellation occurs when the two waves are 180\textdegree out of phase. Since the two waves originate with a 0\textdegree phase difference, one wave must undergo a 180\textdegree phase shift (relative to the second wave) for destruct-ive interference to occur. In our case, the phase change is effected by letting the two waves move through different path lengths before they meet and interfere. Destructive interference will then occur when this path difference is an odd number of half wavelengths. This will insure a phase shift of 180\textdegree. Therefore, │R_A - R_B│ = (2n - 1) \lambda/2n = 1,2,3,... Now, if \lambda and f are the wavelength and frequency of the sound wave, and c is its velocity, \lambdaf = c \lambda = c/f = (1200 ft/see)/(800 sec^-1) = 1.5 ft We are also given thatAB= 10 ft = R_A + R_B, when C is on the line joining AB ; thus │R_A - R_B│ = │R_A -(10 - R_A)│ = │2R_A - 10│ and cancellation will occur for 2R_A - 10 = (2n - 1) \lambda/2n = 1, 2, 3, 4, ... = (1) 1.5/2 = 0.75if n = 1 = (3) 1.5/2 = 2.25if n = 2 = (5) 1.5/2 = 3.75if n = 3 = (7) 1.5/2 = 5.25if n = 4 and so forth. The resultant intensity at the point of destructive interference is The resultant intensity at the point of destructive interference is I_C = │I_A - I_B│ = │[60/R^2_A] - [30/R^2_B]│ This follows from the very nature of destructive inter-ference; the intensities of the interfering waves tend to try to cancel one another.

Question:

A recipe for artificial sea water is NaCl, 26.518 g/kg; MgCl_2, 2.447 g/kg; MgSO_4, 3.305 g/kg; CaCl_2, 1.141 g/kg; KCl, .725 g/kg; NaHCO_3, .202 g/kg; NaBr, .083 g/kg. All of these salts are dissolved in water to make a total weight of 1000 g. What are the chlorinity and salinity of the solution?

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Solution:

The chlorinity of a solution is defined as the percent of Cl^- ions in the solution by weight. The salinity is a measure of the salt content of seawater. To solve this problem, the number of moles of Cl^- must be determined. Once this value is known the weight of Cl^- can be determined and thus the percentage of Cl^- can be found. The molecular weights of the salts containing Cl^- are 58.5 g/mole for NaCl, 95.3 g/mole for MgCl_2, 111 g/mole for CaCl_2, and 74.5 g/mole for KCl. Since there are 1000 g in 1 kg of sea water, the weights of the salts are 265.18 g for NaCl, 24.47 g for MgCl_2, 11.41 g for CaCl_2, and 7.25 g for KCl. Thus, moles of Cl^- from NaCl= 26.518 g / (58.5 g/mole)= .453 moles moles of Cl^- from MgCl_2 = [2(2.447 g)] / [95.3 g/mole = .0514 moles moles of Cl^- from CaCl_2 = [2(1.141 g)] / [111 g/mole]= .0206 moles moles of Cl^- fromKCl= (.725 g) / (74.5 g/mole)= .00973 moles Total number of moles of Cl^- = .535 moles The total weight of Cl^- in the sea water is (.535 moles) (35.5 g/mole) = 19.0 g. Thus the chlorinity of the sea water is [(19.0 g Cl^-) / (1000 g sea water)] × 100 = 1.9% chlorinity. [(19.0 g Cl^-) / (1000 g sea water)] × 100 = 1.9% chlorinity. This value of chlorinity is average for sea water. To find the salinity of the sea water use the formula: % salinity = 1.805 (% Cl) + 0.03 % = (1.805) (1.9%) + 0.03 % = 3.433 %. = (1.805) (1.9%) + 0.03 % = 3.433 %.

Question:

Plant tissues that are neither considered surface tissues nor vascular tissues, are referred to as fundamental tissue. Describe the various types of fundamental tissues.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0055.htm

Solution:

Fundamental tissues make up most of the plant body. Examples include the soft parts of the leaf, the pith (central core of stems and roots), the cortex (outer area of stems and roots), and the soft parts of flowers and fruits. The chief functions of the fundamental tissues are the production and storage of food. Some fundamental tissues may also function in physical support of the plant body. There are four types of fundamental tissues, each composed of a single type of cell. One type is parenchyma tissue, which occurs in roots, stems and leaves. Parenchyma consists of small cells with a thin cell wall and a thin layer of cytoplasm surrounding a large vacuole. The cells are loosely packed, resulting in abundant spaces in the tissue for gas and nutrient exchange. Most of the chloroplasts of leaves are found in these cells. A second type is collenchyma tissue. It contains cells that are generally more elongate and their cell walls are irregularly thickened compared to the cell walls of parenchyma. The corners of their cell walls are thickened to provide the plant with support. These tissues occur just beneath the epidermis of stems and leaf stalks. Sclerenchyma tissue, a third type, also functions in support. Most mature cells of this tissue are dead. They have very thick cells walls to provide support and mechanical strength. The cell wall is usually impregnated with an additional tough substance called lignin. The cell wall may be so thick that the internal space of the cell is nearly obliterated. Sclerenchyma is found in many stems and roots. Sclerenchyma cells are divided into two cate-gories, fibers and sclereids. Fibers are elongate cells with tapered ends. Fiber cells are tough and strong, but flexible. Sclereids are more irregularly shaped cells. Stone cells are rounded sclereids. They- are found in the hard shells of nuts and in seeds. Endodermis, the fourth type of fundamental tissue, occurs in a layer surrounding the vascular core of roots. Endodermal cells have cell walls thickened with lignin and suberin which are chemical substances that make the cells waterproof. The endodermal layer is one cell thick, and the cells are compactly arranged. This cell layer is called the Casparian strip. The Casparian strip regulates the entry of water into the vascular tissues of roots.

Question:

Using methyl orange as an indicator, a solution of unknown pH was matched to a Bjerrum wedge at a point where the acid wedge was 40 percent of the total thickness of the combined wedges. This point is indicated by the arrow in the diagram. Determine the pH of the solution.

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Solution:

From the Bjerrum wedge, the ratio of the concentration of methyl orange indicator in basic solution ([Ind^-]) to the concentration in acid solution ([HInd]) will be determined. This ratio will then be used to find the pH of the unknown solution. A Bjerrum wedge is a glass box divided into two wedge-shaped compartments by a glass plate placed diagonal-ly in the box (see diagram, above). In one compartment is placed an acidic solution of the chosen indicator (HInd) and into the other a basic solution of the indicator (Ind^-). The concentration of indicator is the same in both wedges. A view from the top will produce a continuum of color varying from that of pure HInd on one side to that of pure Ind^- on the other. A solution of indicator of the same concentration as the Bjerrum wedge is prepared using a solution of unknown pH. The resulting solution is then placed in a glass box having a thickness equal to that of the Bjerrum wedge. The boxes are placed side- by-side and the color of the unknown solution is matched to a color on the Bjerrum wedge. At the point where the two colors match, let the acidic wedge be x percent of the total thickness of the box. Then the basic wedge is 100 - x percent of the total thickness. The ratio [lnd^-]/[HInd] is then [Ind^-]/[HInd] = [(100 - x)/x] . In this problem, the acidic wedge is 40 percent the total thickness, hence [Ind^-]/[Hind] = [(100 - 40)/40] = (60/40) . To make use of this ratio, we need an expression relating the pH to [lnd^-]/[HInd]. Consider the dissociation of acidic indicator: HInd + H_2O \rightleftarrows H_3O^+ + Ind^-. The equilibrium constant, K_a , for this reaction is K_a = [H_3O^+] × ([Ind^-] / [HInd]) . Taking the logarithm of both sides and multiplying by negative one gives - log K_a = - log [H_3O^+] - log ([Ind^-]/[Hind]) . - log K_a = pK_aand- log [H_3O^+] = pH. Hence, pK_a = pH - log ([Ind^-]/[Hind]) , or,pH = pK_a + log ([Ind^-]/[Hind]) . For methyl orange, pK_a = 3.7. Using the value of [lnd^-]/[HInd] obtained above, we get pH = 3.7 + log [60/40] = 3.9 as the pH of our unknown solution.

Question:

What advantages and disadvantages do the fungi have in comparison withchlorphyll-bearing plants in terms of survival?

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/Users/wenhuchen/Documents/Crawler/Biology/F06-0166.htm

Solution:

The chlorophyll-bearing plants require no organic food source. Their sole needs are light, carbon dioxide, water, and inorganic minerals. Plants are able to grow in any environment where these needs are met. The ocean is a habitat very favorable to green plants (e.g. the green algae), but not to fungi. Since fungi do not photosynthesize, they are able to grow in the dark. Green plants cannot do this. Fungi do not require light penetration and they can survive with very thick, tough cell walls. The cell walls of fungi sometimes contain cellulose, but chitin is usually their most im-portant constituent. Chitin, a polysaccharide' of acetyl glucosamine, is also the principal constituent of the exoskeleton of arthropods-insects, lobsters, shrimp, and crabs. The strong cell wall of fungi permits them to grow in environments where no plants or other organisms can grow. Certain fungi are extremely resistant toplasmolysis(cell shrinkage in hypertonic medium), and can grow in concentrated salt solutions and sugar solutions. Certain fungi are able to withstand high con-centrations of toxic substances. An example is the ethanol-producing yeasts, which grow at extremely high concentrations of ethanol. Other organisms are killed at these ethanol concentrations. Fungi are less adept at surviving than green plants in that fungi require an organic food source, and are only able to grow where large quantities of foodstuffs are present. Fungi cannot, capture their food, they must obtain nutrition by growing directly on or within their food source. For these reasons, they are not commonly found in ocean habitats.

Question:

A luminous object and a screen are placed at a fixed distance D apart. Show that if a converging lens of focal length f, where f < D/4, is inserted between them it will produce a real image of the object on the screen for two positions separated by a distanced ={ \surd(D - 4f)}, and that the ratio of the two image sizes for these two positions of the lens is(D - d)^2 / (D + d)^2.

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Solution:

In both cases, 1/s + 1/s' = 1/f where s is the object distance, and s' is the image distance. ss' / (s + s') = for ss'/D = f. Sinces + s' = D (see figure) ss' = fD(1) ands + s' = D(2) letd = s - s' . d^2 = (s - s')2= (s -s')^2 - 4ss' = D2- 4fD. Therefore d = \surdD(D - 4f). Also, from Eqs. (1) and (2), s + (fD/s) = Dor s^2 - Ds + fD = 0. s= [D\pm \surd(D2- 4fD)] / 2 = (D \pm d) / 2 This is a general formula for s. It has 2 roots, one of which must be s_1, and the other of which must be s2.(see figure ). s_1 =(1/2)(D-d)and S_2 = (1/2)(D + d) From (2) , s'_1= D - s'1= (1/2)(D-d)ands'2= D - s_2 = (1/2)(D - d). Therefore S_2 - S_1 = d, and the two positions of the lens are separated by the distance d = \surd[D(D - 4f)]. Also, the magnifications for the 2 positions are m_1 = -(s'1/ s_1) = y-1/ yand m2= -(s'_2 / s_2) = y2/ y. \thereforey_2 / y = m_2 / m1= (s_1 s'_2)/(s'_1 s_2) ={1/4(D - d)^2} / {1/4D + d)^2} =(D - d)^2 / (D + d)^2 . Note that this method only works iff < D/4; for otherwise d^2 is negative and d is thus imaginary.

Question:

A 40-lb stone is carried up a ramp, along a path making a 30\textdegree angle to the horizontal, to the top of a building 100 ft high. How much work is done? (Neglect friction.)

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0259.htm

Solution:

Work is defined as the component of the force in the direction of the displacement multiplied by the displace-ment, for constant forces. In mathematical terms, W = F^\ding{217} \textbullet S^\ding{217} = FS cos \texttheta where \texttheta is the angle between the force and the displacement. We may compute the length of the ramp because from the figure (part a), sin 30\textdegree = (1/2) = 100/(length ot ramp) Therefore, length of ramp = 200 ft. Since the stone is being carried up the ramp the force is upwards and we see that the angle \texttheta is 60\textdegree (see figure (b)). Hence the work is W = (40 lb)(200 ft)cos 60\textdegree = (40)(200)(1/2) ft-lbs = 4000 ft-lbs. As a check we can use the fact that this work done must equal the change in the potential energy of the stone. ∆PE = (weight) (∆ height) = (40 lb) (100 ft) = 4000 ft-lbs.

Question:

Besides temperature, what other physical conditions must be taken into account for the growth of bacteria?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0125.htm

Solution:

Although all organisms require small amounts of carbon dioxide, most require different levels of oxygen. Bacteria are divided into four groups according to their need for gaseous oxygen. Aerobic bacteria can only grow in the presence of atmospheric oxygen .Shigelladysenteriaeare pathogenic bacteria (causingdysentery) which require the presence of oxygen. Anaerobic bacteria grow in the absence of oxygen. Obligate anaerobes grow only in environments lacking O_2.Clostridiumtetaniare able to grow in a deep puncture wound since air does not reach them. These bacteria produce a toxin which causes the painful symptoms of tetanus (a neuromuscular disease). Facultative anaerobic bacteria can grow in either the presence or absence of oxygen.Staphylococcus,a genera commonly causing food poisoning , is a facultative anaerobe. Microaerophilic bacteria grow only in the presence of minute quantities of oxygen.Propionibacterium, a genus of bacteria used in the production of Swiss cheese, is a microaerophile. The growth of bacteria is also dependent on the acidity or alkalinity of the medium. For most bacteria, the optimum pH for growth lies between 6.5and7.5, although the pH range for growth extends from pH 4 to pH 9. Some exceptions do exist, such as thesulfer-oxidizing bacteria:Thioba- cillus thiooxidansgrow well at pH 1. Often the pH of the medium will change as a result of the accumulation of metabolic products. The resulting acidity or alkalinity may inhibit further growth of the organism or may actually kill the organism. This phenomenon can be prevented by addition of a buffer to the original medium. Buffers are compounds which act to resist changes inpH.During the industrial production of lactic acid from whey by Lactobacillusbulgaricus, lime, Ca(OH)_2, is periodically added to neutralize the acid. Otherwise, the accumulation of acid would retard fermentation.

Question:

Give a brief history of the six generations of computers. What do you think the next generation has in store for us?

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Solution:

The first generation, starting with the ENIAC machine that filled an entire room, is based on "valve technology," where vacuum tubes were used as the basic electronic units to regulate the flow of electricity in much the same way as the flow of water in pipes. The memory was based on a primitive mercury delay line. ENIAC and UNIVAC I (Universal Automatic Computer) were the forerunners of this generation. The second generation was based on "transistor" technology. With transistors replacing the vacuum tubes, these computers were less expensive , more reliable, more compact, and more pow-erful. With this generation , we also begin to see the "symbolic" processing of letters, that is to say text, as well as numbers. The third generation was based on "integrated circuits," also loosely referred to as "chips" that were theby productof space technology . This technology was introduced by the Depart-ment of Defense and the fledgling aerospace program. The fore-runners of this generation were IBM 360 and 370 mainframe com-puters, Honeywell 200 series mainframes, and Digital Equipment Corporation (DEC) PDP-8 and PDP-11 series. This generation of computers normally used ferrous magnetic core memories. The fourth generation was based on LSI (Large Scale Inte-gration) and VLSI (Very Large Scale Integration) technology. This technology used very small area to pack hundreds of tran-sistors and circuitry. This technology for processor (e.g. CPU: Central Processing Unit) design was coupled with solid-state semiconductor memories (generally using CMOS manufacturing tech-nology) to yield low cost, compact systems with impressive ex-ecution speeds and primary memory capacities. The fifth generation, a term coined by the Japanese, is not as well defined , except that it heavily depends upon "symbolic computing" where a computer is a symbol-processing engine rather than just a number- cruncher . It is characterized by heavy use of parallelism and knowledge- based system software technologies. The sixth generation, with the advent of "neural networks" and neurocomputing , will be computers that are capable of func-tioning very much like biological brains: Very good in practical tasks that require approximations such as scene analysis and cognition, speech synthesis, etc., but not toogoodat number crunching operations such as multiplying two 18-digit numbers in nanoseconds. We can venture to state the seventh generation as "molecu-lar computing " where the architecture of the computer will uti-lize a large number of molecules that are capable of self-organization and intelligent behavior . In other words, these comput-ers will be "chemical" in nature rather than "electronic." It is too early to give any details regarding this generation ! So, a plausible answer to the second part of the question is that the sixth generation will be neural computing machines and the seventh molecular computing systems.

Question:

One method for determining the molecular weight of large, biologically important molecules, is by measuring the density by standard procedures and determining the average volume occupied by a single molecule by X-ray crystallographic analysis. If a biochemist measures the density of a sample of deoxyribonucleic acid (DNA) as 1.1 g/cm^3 and X-ray analysis of the same sample estimates the volume of a single DNA molecule as 0.91 × 10^-15 cm^3, what is the molecular weight of this type of DNA?

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Solution:

The number density (molecules/cm^3) is the reciprocal of the volume of a single molecule and is re-lated to the density (g/cm^3) by the following formula: number density = 1 / (volume per molecule) = (density / molecular weight) × Avogadro's number The validity of this formula is readily seen by considering the dimensions of the quantities involved. Dividing the density (g/cm^3) by the molecular weight (g/mole) gives a quantity with units of (g/cm^3) / (g/mole) = mole/cm^3 (molar density) . Multiplying this number by Avogadro's number (molecules/mole) gives a quantity with dimensions of (mole/cm^3 ) × (molecules/mole) = molecules/cm^3 (number density). Solving for the molecular weight in the above ex-pression we obtain molecular weight = density × Avogadro's number × volume per molecule = 1.1 g/cm^3 × 6 × 10^23 molecules/mole ×0.91 × 10^-15 cm^3/molecule = 6 × 10^8 g/mole.

Question:

Write a PL / I program to initialize, search, insert and delete items in a one-way list.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0157.htm

Solution:

To simulate a one-way list in PL / I we shall use two dimensional array. The first column stores the Data and second column the pointers to next OWL element. ONE-WAY LIST / \textasteriskcentered OWL: TWO DIMENSIONAL ARRAY \textasteriskcentered / / \textasteriskcentered 1st ENTRY OP OWL IS DATA, 2nd ENTRY IS THE POINTER \textasteriskcentered / / \textasteriskcentered TO THE NEXT OWL ELEMENT, HEAD IS PNTR \textasteriskcentered / / \textasteriskcentered TO THE HEAD OP LIST \textasteriskcentered / /\textasteriskcentered INITIALIZE \textasteriskcentered/ DCL OWL (10, 2) FIXED (5); DCL HEAD FIXED (5) INIT (\cyrchar\CYRF); / \textasteriskcentered INITIAL CHECK AND FIRST INSERT \textasteriskcentered / IF HEAD = \cyrchar\CYRF THEN DO; / \textasteriskcentered OWL IS EMPTY \textasteriskcentered / HEAD = J; / \textasteriskcentered J CAN HAVE A VALUE FROM 1 TO 10 \textasteriskcentered / OWL (J, 1) = X; / \textasteriskcentered INSERT X IN OWL (J, 1) \textasteriskcentered / OWL (3, 2) = \cyrchar\CYRF; / \textasteriskcentered REARRANGE THE PNTR \textasteriskcentered / / \textasteriskcentered SEARCH AND INSERT \textasteriskcentered / / \textasteriskcentered SEARCH OWL ELEMENT KEY AGAINST THE INSETING ELEMENT, INSERT \textasteriskcentered / / \textasteriskcentered AFTER KEY REARRANGE THE POINTERS \textasteriskcentered / NEXT = HEAD; DO WHILE OWL (NEXT, 1)\rceil = KEY; / \textasteriskcentered SEARCH FOR KEY \textasteriskcentered / NEXT = OWL (NEXT, 2); / \textasteriskcentered GOES ON MATCHING DOWN \textasteriskcentered / IF NEXT = \cyrchar\CYRFTHEN [go to error routine]; / \textasteriskcenteredTHE LIST, \cyrchar\CYRF PTR INDICATES LAST ELEMENT \textasteriskcentered / END; OWL [Insert, 2] = OWL [NEXT, 2]; / \textasteriskcentered PTR OF KEY ELEMENT IS TO BE ASSIGNED TO INSERT \textasteriskcentered / OWL (NEXT, 2) = INSERT / \textasteriskcentered NOW OWL (NEXT, 2) POINTS TO INSERT \textasteriskcentered / / \textasteriskcentered SEARCH AND DELETE \textasteriskcentered / Initially NEXT points to the top element and there is no previous pointer. But as the search proceeds NEXT points out the next element and PREV points to the element having pointer NEXT. We want to delete B. The goal is to make PREV point to C and if the match is successful for the 1^st element A we want to change the value of the head. NEXT = HEAD; PREV = 0 DO WHILE (OWL (NEXT, 1)\rceil = KEY) ;/ \textasteriskcentered SEARCH \textasteriskcentered / PREV = NEXT NEXT = OWL (N, 2)/ \textasteriskcentered LOOPING TO SEARCH NEXT ELEMENT \textasteriskcentered/ IF NEXT = \cyrchar\CYRF THEN [go to error routine];/ \textasteriskcentered NO MATCH \textasteriskcentered / END; IF PREV=0 THEN HEAD = OWL (NEXT,2) / \textasteriskcentered MATCH SUCCEEDS AT FIRST ELEMENT \textasteriskcentered / ELSE OWL (PREV, 2) = OWL (NEXT, 2); / \textasteriskcentered MATCH SUCCEEDS IN THE LIST END;AND REARRANGE POINTERS \textasteriskcentered /

Question:

A cylinder rests on a horizontal rotating disc, as shown in the figures. Find at what angular velocity, \omega, the cylinder falls off the disc, if the distance between the axes of the disc and cylinder is R, and the coefficient of friction \mu = D/h , where D is the diameter of the cylinder and h is its height.

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Solution:

The centripetal force that keeps the cylinder at rest on the disc is the frictional force f. According to Newton's third law of motion, the cylinder reacts with an equal and opposite force, F, which sometimes is referred to as the centrifugal force, F = M \omega^2R where M is the mass of the cylinder. The cylinder can fall off either by slipping away (Fig. A) or by tilt-ing about point P (Fig. B), depending on whichever takes place first. The critical angular speed, \omega_1, for slipping occurs when F equals f; F = f M \omega^2_1 R = \mugM where g is the gravitational acceleration. Hence \omega_1 = \surd(\mug/R). F tries to rotate the cylinder about P, but the weight W opposes it. The rotation becomes possible, when the torque created by F is large enough to take over the opposing torque caused by W; F (h/2) = W(D/2) F = W(D/h) M \omega^2_2 R = Mg (D/h) giving\omega_2 = \surd(D/hR) Since we are given that \mu > D/h , we see that \omega_1 > \omega_2 and the cylinder falls off by rolling over at \omega = \omega_2 .

Question:

The following reaction is performed at 298\textdegreeK. 2NO (g) + O_2 (g) \rightleftarrows 2NO_2 (g). The standard free energy of formation of NO (g) is 86.6 kJ/mole at 29 8\textdegreeK; find the standard free energy of formation of NO_2 (g) at 298\textdegreeK.k_p= 1.6 × 101 2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0555.htm

Solution:

Free energy, ∆G\textdegree, is a measure of the useful work that a system can perform. There is a quantitative relation between standard free energy and the equilibrium constant at a constant temperatures: ∆G\textdegree = - RT Ink_p, where R = the universal gas constant, which equals 8.31 J/mole deg, T = temperature in Kelvin (Celsius plus 273\textdegree) andk_p= the equilibrium constant for the reaction. You have all the information to find the ∆G\textdegree for the en-tire reaction. Therefore, to find ∆G\textdegree_NO2, you (1) find ∆G\textdegree for the reaction. (2) Use the fact that ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 to find ∆G\textdegree NO2 . ∆G\textdegree = - RT Ink_p = - (8.31 J mol^-1 deg^-1) (298) In 1.6 × 101 2 = - (8.31 J mol^-1 deg^-1) (298) (2.303) log (1.6 × 101 2) = - 69.6 kJ/mole. ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 The ∆G\textdegree for the entire reaction must just be the ∆G\textdegree of the product minus the ∆G\textdegree's of the reactants. ∆G\textdegree _react = 2∆G\textdegree_NO2 - 2∆G\textdegree_NO - ∆G\textdegree_O2 You have calculated ∆G\textdegree overall and are given ∆G\textdegree_NO. ∆G\textdegree_O2 is zero, since the ∆G\textdegree of any element in its standard state is taken as zero. To find ∆G\textdegree_NO2, therefore, substitute in these values and solve for it. Thus, ∆G\textdegree_NO2 = (1/2) [∆G\textdegree + 2∆G\textdegree_NO] = (1/2) (- 69.6 + 2 × 86.6) = 51.8 kJ/mole.

Question:

Assuming complete ionization, calculate (a) the pH of 0.0001 NHCl, .(b) thepOHof 0.0001 N KOH.

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Solution:

(a) pH is defined as the negative log of the hydrogen ion concentration. pH = - log [H^+] The normality of an acid is defined as the number of equivalents of H^+ per liter of solution. The ionization ofHClcan be written HCl\leftrightarrows H^+ +Cl^- This means that there is one H^+ for everyHCl, and that the concentration of H^+ equals the concentration of completely ionizedHCl. [H^+] = [HCl] We are told that [HCl] = 0.0001 N = 1 × 10^-4 N. Therefore, [H^+] = 1 × 10^-4 N. We can now solve forpH.Note: In this problem, normality =molarity (concentration), since equi-valent weight = M.W. pH = - log [H^+] pH = - log (1 × 10^-4) = 4. The pH of this solution is 4. (b) ThepOHis defined as the negative log of the OH^- ion concentration. pOH= - log [OH^-]. The ionization of KOH can be stated KOH \leftrightarrows K^+ + OH^- Therefore, one OH^- is formed for every KOH, and when KOH is completely ionized, their concentrations are equal. [KOH] = [OH^-] We are told that [KOH] = 0.0001 N, thus [OH^-] = 0.0001 N (again, normality =molarity.) Solving forpOH: pOH= - log [OH^-] = - log (0.0001) = - log (1 × 10^-4) = 4. ThepOHof this solution is 4.

Question:

What must be the length of a simple pendulum that will have a period of 1 sec at the surface of the Earth?

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Solution:

In the simple pendulum figure, the torque about point 0 is \cyrchar\cyrt = - mglsin \texttheta, wherel is the length of the string, and the negative sign is introduced to show that the, torque acts to decrease \texttheta. For \texttheta less than 30\textdegree, \texttheta \cong sin \texttheta is a good approxima-tion, thus: \cyrchar\cyrt = -mg l\texttheta We note that this equation resembles the standard form for simple harmonic oscillators: F = -kx where mglis analogous to k, and \texttheta to x. We find the expression for the period T of a simple pendulum by analogy with the equation for that of the harmonic oscillator: T = 2\pi\surd(m/k) thus: T = 2\pi\surd(I/mgl) = 2\pi\surd(ml^2/mgl) = 2\pi\surd(l/g) where I = ml^2 is the moment of inertia of the mass m about point 0. From the above equation: l = T^2g/4\pi^2 l = [(1 sec)^2 (9.8 m/sec^2)] / [4(3.14)^2] = 0.248 m

Question:

Electrons with energies of 1 MeV are injected into the vacuum tube of a betatron of radius 1 m, through the center of which is a magnetic flux which changes at the rate of 100 Wb \textbullet s^-1 . The electrons make 250,000 re-volutions before being ejected from the betatron. What is their final energy?

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Solution:

A betatron (see Fig. A) basically consists of an evacuated tube in the shape of a circular ring with a magnetic field passing through the center of the ring. The magnetic field is slowly increased. This induces an increasing tangential electric field around the circumference of the ring by Faraday's Law. An electron placed in the ring will be accelerated around the ring by this electric field. The tube is evacuated to pre-vent collisions between the accelerating electrons and air particles. The force due to the changing magnetic flux acting on the electron is found using Faraday's law and the definition of emf \epsilon. \epsilon \equiv^-\int_C E^\ding{217} \textbullet dl^\ding{217} = - (d\varphi/dt) \equiv - \varphi(1) The line integral is taken about the contour C defined by the circumference of the circular ring. E^\ding{217} is the electric field vector induced by the changing magnetic field and dl^\ding{217} is an element of arc length (see figure B). \varphi is the rate of change of magnetic flux through the contour. The magnitude of E^\ding{217} is constant at all points on the contour. E^\ding{217} is also parallel to dl^\ding{217} at any point on the contour. Thus (1) becomes E \int_C dl = E(2\pir) = \varphi(2) where r is the radius of the circular contour within the evacuated tube. But by definition of electric field, E is equal to the force per unit charge, or E = (F/q) The work, then, done on the electron (of charge e) by the induced electric field in moving the electron once around the ring is the product of the force act-ing on it times the distance it moves. Or W = Fs = eE(2\pir) = e\varphi Here, we used equation (2). This is equal to the energy gained by the electron in one revolution. Making use of the electronvolt-joule conversion factor, W = e\varphi Joules × [(1 eV)/(1.6 × 10-1 9J)] = (1.6 × 10-1 9\varphi)J × [(1 eV)/(1.6 × 10^-19 J)] = \varphi eV The total energy the electrons acquire while in the betatron is thus the energy gained per revolution times the number of revolutions performed ( = 250, 000). \thereforeW = 100 × 250, 000 eV = 25 MeV. The final energy is thus E = 1 MeV + W = 26 MeV.

Question:

Briefly explain thesignificance of the ASSEMBLER language.

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Solution:

Any high level-language, such as BASIC, FORTRAN, COBOL, and others, is a very convenient and relatively simple set of instructions that a programmeruses to record a program needed to be executed. However, theselanguages are not readable by a computer, and have to be translatedinto an intermediate language, which, in turn, is changed into binarycode before the actual execution pro-cess starts. This intermediate language is called the ASSEMBLER language. It is not a mandatory rule to use a high level-language for writing a program. A program can be written in assemb-ler language. This saves a computersome execution time by eliminating the process of translation of thehigh-level in-structions into assembly instructions. It adds, however, the procedureof checking the input assembly program for syntax errors. But theloss of time from this check is minimal compared to the gain of time fromeliminating the translation step. Being the next step from high-level language coding to the actual executionprocess, the assembler language is called a low-level programminglanguage.

Question:

Calculate the value of the universal gravitational constant, G, assuming that the earth's radius is 6.4 × 10^6 m the earth has an average density of 5.5 × 10^3 kg/m^3. The volume V of a sphere is V = (4/3)\pir^3 , where r is the radius of the sphere.

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Solution:

Begin by computing the volume of the earth: V = (4/3)\pir^3 = (4/3\pi)(6.4 × 10^6m)^3 = 1.1 × 10^21m^3. Since Density = Mass/Volume, we have Mass = Density × volume or, the total mass of the earth, m_2 = (1.1 × 10^21m^3)(5.5 × 10^3 kg/m^3) = 6.0 × 10^24 kg. If an object of mass m_1 is placed on the Earth's sur-face, the gravitational force of attraction (W) between it, and the Earth (i.e., its weight) is found by use of Newton's second law. W = m_1g where g is the acceleration of the object due to the gravi-tational force (i.e., acceleration of gravity). According to the Universal Law of Gravitation, however, the gravitational force of attraction between m_1 and m_2 is W = [Gm_1m_2/d_2] where d is the radius of the Earth. Then [Gm_1m_2/d^2] =m_1g After rearranging, G = (gd^2/m_2) = [{(9.8 m/s^2)(6.4 × 10^6 m)^2}/(6.0 × 10^24 kg)] = 6.7 × 10^-11 Nm^2/kg^2 Since the value of the earth's radius is more accurately known today, Newton's estimate of G differed from this calculated value.

Question:

A 5 -kg block slides down a frictionless plane inclined at an angle of 30\textdegree with the horizontal as shown in figure (a). Calculate the amount of work W done by the force of gravity for a displacement of 8 m along the plane.

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Solution:

We will solve this problem first by the dynamics method and then by the energy method. The formula for calculating work is: W = F^\ding{217} \textbullet s^\ding{217} = F s cos \texttheta where \texttheta is the angle between the force F^\ding{217} and the dis-placement s^\ding{217} of the mass in question. We see from figure (a) that the angle between F^\ding{217} and s^\ding{217} is 60\textdegree: W = Fs cos 60\textdegree = (1/2) mgs = (1/2)(5 kg) (9.8 m/sec^2) (8 m) = 196 kg-m^2/sec^2 = 196 Joules Another way to solve this problem is to calculate the difference in gravitational potential energy that the block goes through as it slides 8 m down the incline. We know that this equals the amount of work that gravity does on the block. As the block slides 8 m down the incline it falls through a vertical height ∆h (see figure (b)) : ∆h = 8(sin 30\textdegree)m = 4m The gravitational potential energy difference that the block experiences is: W = ∆E_p = mgh_2 - mgh1= mg(h_2 - h_1) = mg∆h = (5 kg) (9.8 m/sec^2) (4 m) = 196 Joules

Question:

Describe the phenomena of the critical angle in optics. What is the critical angle for a glass-air interface, if the index of refrac-tion of glass with respect to air is 1.33?

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Solution:

Consider two media,1 and 2,such that the index of refrac-tion of 1 with respect to 2 is less than unity, as shown in Fig. A. That is, medium 1 is "denser", and the angle of refraction will be greater than the angle incidence. In general part of the incident ray is reflected and part re-fracted. As the angle of incidence is increased, the angle of refraction will increase until r = 90\textdegree. At the critical angle ,i_c (see Fig. B), we have (sin i_c/sin 90\textdegree) = sin i_c = n_21 At the critical angle, and for values of i greater than i_c, refraction cannot occur and all the energy of the incident beam appears in the reflected beam. This phenomenon is called total internal reflec-tion. The index of refraction of air with respect to glass is (1/1.33) = 0.75. The critical angle for a glass-air interference is therefore sin i_c = 0.75 or i_c = 48.6\textdegree Thus, for angles of incidence \geq 48,6\textdegree, total internal reflection will occur for a glass-air combination similar to the one shown in the figure.

Question:

At 395 K, chlorine and carbon monoxide react to form phosgene, COCI_2 , according to the equation CO (g) + Cl_2 (g) \rightleftarrows COCI_2 (g). The equilibrium partial pressures are PCI_2 = 0.128atm, P_CO = 0.116atm, and P(COCl)2= 0.334atm. Determine the equilibrium constantK_pfor the dissociation of phosgene and the degree of dissociation at 395 K under a pressure of 1 atm.

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Solution:

After obtaining an expression forK_pfor the dissociation of phosgene, the degree of dissociation under 1atmof total pressure will be obtained by combiningK_pwith Dalton's law of partial pressures. The dissociation of phosgene may be written as COCI_2 (g) \rightleftarrows CO (g) + Cl_2 (g) . By definition,K_pis the product of the partial pressures of the products divided by the product of the partial pressure of the reactants. Hence, K = { P_CO P(Cl)2} / {P(COCl)2} = {(0.116atm) (0.128atm)} / {.334atm} = 0.0444 atm. K_pfor the dissociation of phosgene is thus 0.0444 atm. Let \alpha denote the fraction of the original number of moles of phosgene that decomposed. Then, 1 - \alpha is the fraction of the original number of moles of phosgene re-maining. From thestoichiometryof the dissociation re-action, one mole of CO and one mole of Cl_2 are formed for every mole of phosgene that decomposes. Thus, \alpha moles of phosgene decomposes to \alpha moles of CO and \alpha moles of CI_2. From Dalton's law of partial pressures, P_(C0Cl)2 = X_(C0Cl)2 P_T P_CO = X_CO P_TandP(Cl)2= X(Cl)2P_T where X(COCl)2is the mole fraction of phosgene, X_CO is the mole fraction of CO, X(Cl)2is the mole fraction of CI_2 , and P_T is the total pressure. Mole fraction may be defined as the number of moles of that particular substance divided by the total number of moles present. Since the total number of moles present after \alpha moles of phosgene decomposes is (1 - \alpha)(from remaining phosgene) + \alpha (from CO) + \alpha (from Cl_2 ) = 1 + \alpha we have P(COCl)2= X(COCl)2P_T= [(1 -\alpha) / (1 + \alpha)] P_T P_CO = X_CO P_T= [\alpha / (1 + \alpha)] P_T and P(Cl)2= X(Cl)2P_T = [\alpha / (1 + \alpha)] P_T Substituting these into the expression forK_p, we obtain K_p= {P_CO P(Cl)2} / {P(COCl)2} = {([\alpha / (1 + \alpha)] P_T ) ( [\alpha / (1 + \alpha)] P_T )} / ( [ (1 -\alpha) / (1 + \alpha) ] P_T ) = {\alpha^2 / [(1 + \alpha) / (1 - \alpha) ]} P_T = { \alpha^2 / (1 -\alpha^2 ) } P_T , Now, the total pressure is P_T = 1atmandK_phas been determined as K_p= 0.0444 atm. Hence , K_p= {\alpha^2 / (1 -\alpha^2)} P_T , 0.0444atm= {\alpha^2 / (1 -\alpha^2)} × 1atm, 0.0444-0.0444 \alpha^2 = \alpha^2 , or\alpha = (0.0444) / (1.0444)^1/2 = 0.206. The degree of dissociation of phosgene is equal to the fraction, \alpha, of original moles that have dissociated. Hence, the degree of dissociation of phosgene at 395 K under a pressure of 1atmis 0.206.

Question:

Examine the following program and explain the usage of Subprograms (i.e., Subroutines) COMPLEX X1, X2 COMM0N/QUADR/X1, X2, A,B,C DATA INPUT /60/ 7STOP END SUBROUTINE ROOTS COMPLEX X1, X2, W COMMON/QUADR/X1, X2, C1, C2, C3 W = C2 {_\ast}{_\ast} 2 - 4.0 {_\ast} C1 {_\ast} C3 W = W {_\ast}{_\ast} 0.5 X1 = (-C2 + W) / (2.0 {_\ast} C1) X2 = (-C2 - W) / (2.0 {_\ast} C1) RETURN END

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Solution:

This is a complete program which consists of a MAIN PART and a SUBPROGRAM. The MAIN part reads in data for the coefficients of a quadratic equation and then calls upon a subprogram to find its roots. The following points are significant in examining the program. common block named QUADR. QUADR - X1, X2, C1, C2, and C3. Getting into a more detailed explana-tion of Subprograms: Every subroutine program begins with a statement of the form SUBROUTINE NAME. There should be at least one blank space between the word SUBROUTINE and its name. The name of the subroutine must not appear anywhere else within the subprogram. Every subprogram has END as its last statement, and must contain at least one RETURN statement in order to continue the execution of the MAIN program from the next statement after the one that called this subroutine. A subroutine is entered from another program unit (main program or a different program) by means of a CALL statement. The general form of the CALL statement is: CALL name of the subroutine A subroutine may assign values to one or more of its arguments in order to "return" results to the main part of the program. Furthermore, there is the possibility of storing results in COMMON AREA so that values obtained in a subprogram could be available to other program units.

Question:

Write a routine that implements shell sorting. Assume there is an array of integers and size of array is known in the main program. The input parameters to the routine are a) array. b) size of array.

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Solution:

The shell sorting algorithm 1) Compares elements that are far apart and then compares adjacent elements. 2) It uses another: routine, values _ swaped ( ) where the values sent are interchanged. 3) The logic used is that number of elements in an array is divided by 2, then element {0] and element [no. of ele-ments/2] are compared. The difference between [no. of elements/2 - 0] is a constant for the first pass and is referred in the program as difference. shell _ sort (array, number _ of _ elements) int array [ ]; int number _ of _ elements; { int i; int difference = number _ of _ elements/2; int swap _ occurred; /\textasteriskcenteredTRUE if a swap occurred \textasteriskcentered/ do do { swap _ occurred =0; for (i = 0; i < number _ of _ elements - difference; i++) if (array [i] > array [i + difference]). { values _ swaped (&(array[i]), &(array [i + difference])); swap _ occurred = 1; } } while (swap _ occurred); while ((difference/=2) > 0); /\textasteriskcenteredmeaning (difference = difference/2) >0 \textasteriskcentered/ } values _ swaped (value1, value2) int \textasteriskcenteredvalue1 = \textasteriskcenteredvalue2; { int temp; /\textasteriskcenteredstore temporary values \textasteriskcentered/ temp = \textasteriskcenteredvalue1; \textasteriskcenteredvalue1 = \textasteriskcenteredvalue2; \textasteriskcenteredvalue2 = temp; } } C REFERENCE GUIDE C REFERENCE GUIDE Arithmetic Operators + addition. - subtraction. \textasteriskcentered multiplication. / division. % modulus. Operators unique to C ++ increment. -- decrement. Assignment operators operator = eg. a\textasteriskcentered = 5; a = a \textasteriskcentered 5; a% = 5; a = a % 5; Bitwise Operators &bitwise AND. !bitwise OR. ^bitwise exclusive - or. <>shift right. \simone's complement. Break Statement: The break statement is used within loops and the switch statement. Within A loop the break statement causes the execu-tion of the loop to cease and execution of the program to con-tinue at the statement that follows the loop. Format is break ; Global Variables: In C the use of global variables is done usingexternstatement. It is done as follows type variable _ name; /\textasteriskcenteredoutside of main \textasteriskcentered/ main ( ) { extern type variable _ name; /\textasteriskcenteredextern tells the compiler that the \textasteriskcentered/ /\textasteriskcenteredvariable is a global one\textasteriskcentered/ Relational Operators: greater than <=less than or equal to >=greater than or equal to Relational operators unique to C ==equal to !=not equal to Ternary Operator : ?:conditional operator. It is a conditional operator eg. conditional - expression: expression 1? expression 2: expression 3 Conditional expressions group right to left. They operate as follows a) If expression 1 (i.e. expression to be evaluated) is TRUE, expression 2 will be performed. If it is FALSE expression 3 will be performed. Logical AND Operator: logical - AND - expression expression 1 && expression 2 Logical OR Operator: logical-OR-expression expression 1 \vert\vert expression 2. Type Specifiers: type - Specifiers are char short int long unsigned float double stifuct -or-union specifier typedef name.

Question:

The plates of a parallel plate capacitor are 5 mm apart and 2m^2 in area. The plates are in vacuum. A potential dif-ference of 10,000 volts is applied across the capacitor. Compute (a) the capacitance, (b) the charge on each plate, and (c) the electric intensity.

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Solution:

(a)C_0 = \epsilon_0(A / d) = 8.85 × 10^\rule{1em}{1pt}12 × [2 / (5 × 10^\rule{1em}{1pt}3)] = 3.54 × 10^\rule{1em}{1pt}9 farad = 3.54 × 10^\rule{1em}{1pt}3 \muf = 3540 \mu\muf. (b)The charge on the capacitor is q = CV_ab = 3.54 × 10^\rule{1em}{1pt}9 × 10^4 = 3.54 × 10^\rule{1em}{1pt}5 coulomb. (c)The electric intensity is E = (1 / \epsilon_0) \sigma = (1 / \epsilon_0) (q / A) = (36\pi × 10^9) × (1.77 × 10^\rule{1em}{1pt}5) = 20 ×10^5 volts/meter. The electric intensity may also be computed from the poten-tial gradient. E = V_ab / d = 10^4 / (5 × 10^\rule{1em}{1pt}3) = 20 × 10^5 volts / meter.

Question:

A laboratory technician has a solution of 0.1 M NaC_2H_3O_2. Find its pH and the percent hydrolysis. AssumeK_w= 1.0 × 10^-4 andK_diss= 1.8 × 10^-5.

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Solution:

NaC_2H_3O_2 is a salt that exists as ions Na+ and C_2H_3O_2^- in water. Since water is a weak acid and weak base, which dissociates to H^+ and OH^- ions, it can react with ions from weak acids and bases. Such a process, termed hydrolysis, occurs in this problem. Because the problem asks for pH, one must determine the concentrations of the H_3O^+ or OH^- in this problem. This means that the hydrolysis reaction must be written and an equilibrium constant set up. The net hydrolysis reaction may be written as . C_2H_3O_2^- + H_2O \rightleftarrows HC_2H_3O_2 + OH^-. TheK_hydfor this reaction is {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] =K_w/K_diss. However,K_hydis defined asK_w/K_diss_' whereK_wis the dissociation constant for water andK_dissis the dissociation constant for the acid formed. One can equate the two to obtain {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] =K_w/K_diss. But the value ofK_wandK_dissis known. Therefore, {[OH^-] [HC_2H_3O_2]} / [C_2H_3O_2^-] = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.6 × 10^-10. If one can find [OH^-], one can determine [H_3O^+] from [H_3O^+] [OH^-] =K_w= 1.0 × 10^-14. Therefore, let x = moles/liter of OH^-. From the hydrolysis reaction, one can see that for each mole of OH^- generated, a mole of HC_2H_3O_2 is also generated. Therefore, let x = [HC_2H_3O_2] also. If one starts with 0.1 M of C_2H_3O_2^-, and x moles/liter reacts to form OH^- (and C_2H_3O_2^-), then, at equilibrium, one has 0.1 - x moles/liter of C_2H_3O_2^- left. Substituting these values into the previous expression: (x \bullet x) / (0.1 - x) = 5.6 × 10^-6. Solving for x, x = 7.5 × 10^-4 M. (assuming in this calculation that x is small, therefore approximating 0.1- x as 0.1.) To find [H^+], solve [H_3O^+] = K_W / [OH^-] = (1.0 × 10^-14)/(7.5 × 10^-4) = 1.33 × 10^-11. Since pH = - log [H_3O^+] pH = - log [1.33 x 10^-11] = 10.88. The percent hydrolysis of this reaction is the number of moles of hydrolyzed acetate ion, divided by the number of moles available for hydrolysis, times 100. % Hydrolysis = {(Moles C_2H_3O_2^- hydrolyzed) / (Moles C_2H_3O_2^- available) × 100 = [(7.5 × 10^-4) / 0.1] × 100 = 0.75%.

Question:

Distinguish between crystalline and amorphous solid substances, using some specific examples. To what extent is the distinction useful?

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Solution:

Crystalline substances can be generally thought of as species composed of structural units with specific geometric patterns. The accompanying drawing of sodium chloride would be an example of such a pattern. The important point is that there exists a regularity in the arrangement of structural units. Structures with regularity generally show a sharp and characteristic melting point, which is the case with crystalline substances. Amorphous substances, however, tend to be shapeless and without definite order. That is, you have a randomness. For example, glassy or glasslike materials such as Plexi-glas and silicate glasses. In substances with a general lack of order, the melting points vary over a range or temperature interval. For amorphous substances, this is exactly what you find. It would, however, beincorrectto state categorically that amorphous substances are without ANY order. For they do tend to have short range order even though they do contain long-range randomness.

Question:

Under standard temperature and pressure conditions, compare the relative rates at which inert gases,Ar, He, and Kr diffuse through a common orifice.

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Solution:

This problem involves the application of Graham's law of diffusion. It states that the relative rates at which gases will diffuse will be inversely proportional to the square roots of their respective densities or molecular weights. That is, rate \alpha 1/ \surdmass. Thus, to compare the rates of diffusion ofAr, He, and Kr, look up their weights in the Periodic Table of Elements and substitute this value into 1/ \surdM \alpha\mu_m, where M = mass of that element and\mu_m= rate. Therefore,\mu_Ar: \mu_He :\mu_Kr = 1/ \surdM_Ar: 1/ \surdM_He: 1/ \surdM_Kr = 1/ \surd39.95 : 1/ \surd4.003 : 1/ \surd83.80 = .1582 : .4998 : .1092.

Question:

Two different molecules, A and B, were reacted to give products according to the equation A + B \rightarrow products. The rate of reaction was measured for fixed concentrations of A and B and the following data was obtained: Experiment [A] A (moles/liter) [B] B (moles/liter) Rate (moles^2/liter^2 - sec) 1 1.0 1.0 0.05 2 1.0 2.0 0.10 3 3.0 1.0 0.45 Write an expression for the rate of reaction.

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Solution:

The rate of reaction is equal to some rate constant, k, multiplied by the concentrations of A and B raised to appropriate powers. That is, rate = k [A]^m [B]^n where the exponents m and n are to be determined. Comparing experiment 2 with experiment 1, we see that when the concentration of B is doubled from 1.0 M to 2.0 M, the concentration of A being held constant at 1.0 M, the rate of reaction doubles from 0.05 to 0.10 moles^2/liter^2 - sec, Hence, the rate of reaction is directly proportional to the concentration of B, and the exponent of [B] in the rate expression is n = 1. If we were to triple [B], holding [A] constant, the rate would triple, and so on. Comparing experiment 3 with experiment 1, we see that when the concentration of A is tripled from 1.0 M to 3.0 M, the concentration of B being held constant at 1.0 M, the rate of reaction is multiplied by 9 (9 × 0.05 = 0.45 moles^2/liter^2 - sec). Since the factor by which the rate is multiplied (9) is the square of the factor by which the concentration of A is multiplied (3), the exponent of [A] in the rate expression must be m = z. Thus if [A] is doubled, [B] being held constant, the rate is multiplied by 4. Substituting the exponents of [A] and [B] determined above into the rate expression gives rate = k [A]^2 [B].

Question:

What is meant by translocation? What theories have been advanced to explain translocation in plants? Discuss the value and weakness of each.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0260.htm

Solution:

Translocation is the movement of nutrients from the leaves where theyare synthesized to other parts of the plant body where they are neededfor a wide variety of metabolic activities. Translocation takes place throughthe sieve tubes of the phloem. Experimental data indicates that the high rate at which translocation occurs cannot be attributed to simple diffusion. Three theories have been offered to explain the mechanism of translocation. Each of them has its own value and weakness, but the first one presented (pressure-flow theory ) holds the most support. The pressure-flow theory, which had been widely accepted in the past , proposes that the nutrient sap moves as a result of differences in turgor pressure. Sieve tubes of the phloem in the leaf contain a high concentration of sugars, which results in high osmotic pressure into the cells. This osmotic pressure causes an influx of water, and build up ofturgor pressure against the walls of the sieve tube cells. In the roots however, sugars are constantly being removed. Here, a lower concentration of solute is present resulting in a lowering of the osmotic pressure. There is consequently a lowerturgorpressure exerted against the sieve tube walls of the root as compared to the leaf. This difference inturgorpressure along the different regions at the phloem is believed to bring about the mass flow of nutrients from a region of high turgor pressure - such as the leaves (where photosynthe-sis produces osmotically -active substances like glucose) -toregions of lowerturgor pressure - such as the stem and roots. The fluid containing the nutrients is pushed by adjacent cells, along the gradient of decreasingturgor, from the leaves to the roots. This theory predicts that the sap should be under pressure as it moves down the phloem, and this is experimentally verified. But there are problems with this hypothesis. Under some conditions, sugar is clearly transported from cells of lesserturgorto cells of greaterturgor. In addition , this theory fails to explain how two substances can flow along the phloem in different directions at the same time, a situation observed to occur by some investigators. Another theory proposes thatcyclosis, the streaming movement evident in many plant cells, is responsible for translocation. According to this theory, materials pass into one end of a sieve tube through the sieve plate and are picked up by the cytoplasm which streams up one side of the cell and down the other. At the other end of the sieve tube, the material passes across the sieve plate to the next adjacent sieve tube by diffusion or active transport. This theory is able to account for the simultaneous flowing of nutrient saps in different direc-tions. However, it is attacked by some investigators on the basis thatcyclosishas not been observed in mature sieve cells. A third theory proposes that adjacent sieve-tube cells are connected bycytoplasmictubules, in which sugars and other substances pass from cell to cell. The movement of these substances is powered by the ATP from the mitochondria-like particles that are believed to lie within these connecting tubules. This theory is, however, weak since it is supported by few experimental findings and much of its content is based on speculation.

Question:

Find the tension in the cable shown in Figure A. Neglect the weight of the wooden boom.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0025.htm

Solution:

Take the directions of the tensions in the cable and the boom to be as shown in the force diagram(fig. B). We assume at this point, that the given directions are correct. However, the forces may turn out to point in the opposite direction. If this is the case, our solutions for the tensions will be negative. We can thus correct ourselves at the end of the problem. The first condition of equilibrium yields \sumF_x = T_2 cos 60\textdegree - T_1 cos 30\textdegree = 0(1) \sumF_y = T_2 sin 60\textdegree - T_1 sin 30\textdegree - 2000 = 0(2) We wish to find T_1, the tension in the cable. Solving for T_2 in terms of T_1 in equation (1) gives T_2 = (T_1 cos 30\textdegree)/(cos 60\textdegree) Substituting this in equation (2), [(T_1 cos 30\textdegree)/(cos 60\textdegree)] sin 60\textdegree - T_1 sin 30\textdegree = 2000 Solving for T_1: T_1(cos 30\textdegree tan 60\textdegree - sin 30\textdegree) = 2000 T_1 = 2000/(cos 30\textdegree tan 60\textdegree - sin 30\textdegree) = 2000/ [(0.8660)(1.7321) - (0.5000)] = 2000/(1.5 - 0.5) = 2000 lb Since our answer is positive, the force acts in the direction assumed in the beginning.

Question:

Of the following two pairs, which member will more likely deviate from ideal gas behavior? (1) N_2 versus CO, (2) CH_4 = versus C_2 H_6 .

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Solution:

Characteristics of ideal gases include: (1)Gases are composed of molecules such that the actual volume of the molecules is negligible compared with the empty space between them. (2) There are no attractive forces between molecules. (3) Molecules do not lose net kinetic energy in collisions. (4) The average kinetic energy is directly proportional to the absolute tem-perature . From these assumptions came the ideal equation of state: PV = nRT, where P = pressure, V = volume, n = moles, R = universal gas constant, and T = temperature. No gas is ideal; they don't obey these assumptions absolutely. To reflect these limitations the ideal gas law can be written as: [P + (n^2 a/V^2 )] (V - nb) = nRT. The term V = nb can be thought of as representing the free volume minus the volume occupied by the gas molecules themselves. The magnitude of (b) is pro-portional to a gas molecule's size. Thus, the greater the size of a gas molecule, the greater the nb is, and, the greater the deviation, since the volume is assumed to equal exactly V in the ideal Gas Law. V - nb is smaller than V. The greater nb, the smaller it becomes. The term n/V, a concentration term, when squared, gives probability of collisions, (a) gives a measure of the cohesive force between molecules. Real molecules do have an attraction for each other. The greater the value of (a), the larger n^2 a/V^2 becomes, and the larger (P + n^2 a/V^2) gets. The ideal gas law assumes a value of ex-actly p. Thus, as (a) increases, the more deviation. To determine which gas deviates the most, look at the size of the gas molecules and their dipole moment. The dipole moment is an indication of unlike charges separated by a given amount of distance. Unlike charges on separate molecules will be attracted to each other. Thus, a higher dipole moment indicates greater cohesive attraction among molecules and greater deviation. Proceed as follows: (1)N_2 versus CO. The CO gas has a net dipole moment, while N_2 does not. This is because O is more electro-magnetic than C. In N_2 both of the atoms are the same and thus their electronegativities are equal and no net dipole moment exists. Thus, CO deviates to a greater extent. (2)CH_4 versus C_2 H_6. C_2 H_6 is a much larger molecule and occupies more volume. Therefore, it is more likely to deviate from ideal gas behavior.

Question:

A student sets up a Young's experiment using a sodium vapor lamp as a light source and placing his slits 1 m from a screen. He is unsure of the slit separation and, standing beside the slits, he varies the separation and finds the interference pattern vanishes if the slits are too far apart. If the angular resolution of his eye is 1 minute of arc, how far apart are the slits when he just cannot see the interference fringes ?

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Solution:

The position of the mth bright fringe of the fringe system from the center O is given by y_m = mR\lambda/d , where \lambda is the wavelength used (see figure (A)). Similarly, for the ( m + 1) th fringe, y_m+1 = [{(m + 1)R\lambda} / d] Thus, the fringe separation is ym+1- ym= R\lambda/d , a quantity independent of mand thus constant throughout the fringe system. The angle subtended by this fringe separation at the student's eye is (see figure (B)) \texttheta = [(R\lambda/d)/R] = \lambda/d rad. Here, we have approximated the arc shown in figure (B) (dotted line) by the straight line distanceAB. But the student cannot see the fringes as distinct unless the angle neighboring fringes subtend at his eye \geq1 minute of arc. But 1 minute of arc is 1/60\textdegree = \pi/60 × 180 rad. Hence, \texttheta_min = \lambda/d_max = [\pi/(180 × 60)] rad \therefored_max = [(180 × 60 × 5.89 × 10^-5) / (\pi rad)] cm = 2.025 mm.

Question:

A worker hangs a uniform bar of mass 12 kg horizontally from the roof of a laboratory by means of three steel wires each 1 mm in diameter. Two of the wires are 200 cm long and one, by an oversight, 200.05 cm long. The long wire is fastened to the middle of the bar, the others to the two ends. By how much is each wire stretched, and how much of the weight does each wire carry? Young's modulus for steel = 2.0 × 10^12 dynes \textbullet cm^-2.

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Solution:

If the bar is hanging horizontally (see the figure), two of the wires will be extended ∆1 and one ∆1 - 0.05 cm. Now the formula for Young's modulus is Y = stress/strain = (F_n/A)/(∆1/1_0), where A is the cross sectional area of the steel wire, 1_0 is the length of the wire with no stress acting on it, and F_n is the (stretching) force the bar exerts on the wire (equal in magnitude to the. force the wire exerts on the bar). Thus, F_n = (AY∆1)/(1_0) . Hence two wires exert upward forces on the bar of magni-tude F_n, and one wire exerts a force of magnitude F'_n = [AY(∆1 - 0.05 cm)]/[1_0 + 0.05 cm] Since the ratio 0.05/1_0 = .00025 is so small, we may ignore 0.05 in comparison with 1_0 in the denominator of the expression for F'_n. Then because thebar is in equilibrium, we obtain bar is in equilibrium, we obtain W = 2F_n + F'_n = (AY/1_0)(2∆1 + ∆1 - 0.05 cm). Therefore ∆1 = (1/3)[(1_0w)/AY) + 0.05 cm] However A = \piR^2 = \piD^2/4, where R and D are the radius and diameter, respectively, of the wire. We are given D = 1 mm = 1 mm × (1cm/10 mm) = 0.1 cm. Hence ∆1 = (1/3)[{(200 cm × 12 kg × 981 × 10^3 dynes)/(\pi/4 × 10^-2 cm^2 × 2 × 10^12 dynes \textbullet cm^-2)} + 0.05 cm] = (1/3)(0.15 + 0.05)cm = 0.0667 cm = 0.667 mm. Thus two of the wires are stretched by 0.667 mm and the other by (0.667 - 0.05)mm = 0.167mm. Also F_n/F'_n = [∆1/(∆1 - 0.05 cm)] = (0.667 mm)/(0.167 mm) = 4; 12 kg = W = 2F_n + F'_n = 9F'_n. \thereforeF'_n = 1(1/3) kgandF_n = 5(1/3) kg. \thereforeF'_n = 1(1/3) kgandF_n = 5(1/3) kg.

Question:

The energy of the doubly charged a-particles of mass 6.64 × 10^-27 kg emitted from The is 6.048MeV. What is their velocity and what magnetic field applied per-pendicular to a collimated beam of such particles would bend the beam into a circle of radius 40 cm?

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Solution:

The energy of the \alpha-particles is 6.048MeV = 6.048MeV× 1.602 × 10^-13 J/ 1MeV= 9.689 × 10^-13 J. The energy of the particle refers to its kinetic energy or KE = 1/2 mv^2 = 9. 689 × 10^-13 J, orv = [\surd{(19.378 × 10^-13 J)/(6.64 × 10^-27 kg)}] = 1.709 × 10^7 m \bullet s^-1 . In the magnetic field the magnetic force supplies the centripetal force necessary to keep the particles moving in a circle. This centripetal force, F, due to the magnetic field has magnitude F =qvBsin \texttheta where is the angle between v^\ding{217} and B^\ding{217}. Since we are given that this angle is 90\textdegree, then sin \texttheta = 1. The centripetal acceleration of a particle with velocity v about a circle of radius R is a = (v^2 /R). By Newton's second law, F = (mv^2 /R) =qvBor v = (qBR/m). Therefore B = (mv/qR). The \alpha-particles carry twice the electronic charge (for they contain two positively charged protons and two neutral neutrons). \therefore b = [(6.64 × 10^-27 kg × 1.709 × 10^7 m \bullet s^-1 )/(2 × 1.602 × 10^-19 C × 0.40 m)] = 0.885Wb\textbullet m^-2 .

Question:

A capacitor with capacitance C and initial charge q_0 and an inductor with inductance L are connected in series. At time t = t_0 , the switch Sis closed. If q(t) is the charge on the capacitor at time t it can be shown that the behavior of the circuit is governed by the following differential equation: Lq̇ ̇+ (1/C)q(t) = 0 where q̇ ̇= d^3q/dt^3 . If i_0 is the initial q̇ ̇ current, construct a model of this circuit and devise a digital program which simulates the behavior of the circuit from time t = t0to t = t_f .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0305.htm

Solution:

Following the steps of the general procedure outlined in the two previous problems: 1.If the given differential equation can be solved, and samples of the values q assumes from t = t0to t = t_f, recorded, we will have a model which effectively simulates the given system, i.e., the circuit. 2.The assumption that L and C are linear and time-invariant has already been included in the model, through the derivation of the given equation. Similarly, the assumption that all the capacitance of the circuit is considered "lumped" at C and all inductance "lumped" at L has already been made. 3.Circuit resistance is neglected for simplicity. 4.The given differential equation is a dynamic mathematical model of the system. 5.The input data is t_0, t_f, q(t_0) = q_0, i(t_0) = i_0, L and C. 6.The equation given here and the differential equation of the previous two problems appear to be similar. In fact, using the equivalences: q \rightarrow x, i = q̇ \rightarrow v = ẋ,q̇ ̇\rightarrow a = v̇ = ẋ ̇, L \rightarrow M, (1/C) \rightarrow K q̇ ̇ the differential equation of harmonic motion can be constructed from the equation of this problem. Hence, the programmer needs only to substitute his input data as follows: USE THE VALUE(1/C) AS K USE THE VALUEL AS M USE THE VALUEq_0 AS XCOR USE THE VALUEi_0 AS VCOR

Question:

Explain what is meant by reduction of vectors. Also explain the rule for APL arithmetic. Evaluate the expression: 6 + ((5 + 2 × 11 - 2) - 3 + 2) \div 2 × 3.

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Solution:

Reduction of a vector, in APL, is equivalent to placing a dyadic operator between each element of the vec-tor and evaluating the resulting expression. Examples of dyadic operators are plus +, minus -, multiply ×, divide \div, exponentiation \textasteriskcentered, etc. The symbol for reduction is a right-tilting slash, e.g.,/. The use of the symbol is illustrated below for a vector V. +/V, -/V, ×/V, \div/V, \textasteriskcentered/V, etc. If V is defined by V \leftarrow 1 2 3 then each of the above re-ductions is equivalent to the following, respectively. +/V is equivalent to 1 + 2 + 3 -/V is equivalent to 1 - 2 - 3 ×/V is equivalent to 1 × 2 × 3 \div/V is equivalent to 1 \div 2 \div 3 \textasteriskcentered/V is equivalent to 1 \textasteriskcentered 2 \textasteriskcentered 3 The rule for APL arithmetic is a right to left rule. That is, the arithmetic operations are performed on a string of operators from right to left. Items enclosed within parentheses are evaluated first, and the whole parenthesized content acts as a single opera-tor. The given expression evaluates as follows: Expression = 6 + ((5 + 2 × 11 - 2) - 3+ 2) \div 2 × 3 = 6 + ((5 + 2 × 9) - 3+ 2) \div 2 × 3 = 6 + ((5 + 18) - 3+ 2) \div 2 × 3 = 6 + ((23) - 3+ 2) \div 2 × 3 = 6 + (23 - 3+ 2) \div 2 × 3 = 6 + (23 - 5) \div 2 × 3 = 6 + (18) \div 2 × 3 = 6 + 18 \div 2 × 3 = 6 + 18 \div 6 = 6 + 3 = 9Answer Note that the innermost pair of brackets is evaluated out first using the right to left rule inside the brackets. Then the outer pair of brackets is evaluated using the right to left rule inside the brackets. Then, after all the brackets have been eliminated, the whole expression is evaluated using the right to left rule. Also note that the effect of a reduction1; operation performed on the vector V is to reduce it to a scalar quantity, which is the evaluated value of the resulting expression. Now, a vector has a dimension of 1. And, a scalar has a dimension of 0. Thus, the reduction operation has reduced the dimension of the variable V from 1 to 0. In general, it is found that for an n-dimensional ar-ray, a reduction operation reduces the dimension of the ar-ray by 1, from n to n-1. Finally, note that the arithmetical operations are not ordered in a hierarchy unlike FORTRAN, where multiplications and divisions are performed before additions and subtrac-tions .

Question:

A rope C helps to support a uniform 200-lb beam, 20 ft long, one end of which is hinged at the wall and the other end of which supports a 1.0-ton load. The rope makes an angle of 30\textdegree with the beam, which is horizont-al. (a) Determine the tension in the rope, (b) Find the force F\ding{217} at the hinge .

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Solution:

Since all the known forces act on the 20-ft beam, let us consider it as the object in equilibrium. In addition to the 200- and 2000-lb forces straight down, there are the pull of the rope on the beam and the force F which the hinge exerts on the beam at the wall. Let us not make the mistake of assuming that the force at the hinge is straight up or straight along the beam. A little thought will convince us that the hinge must be pushing both up and out on the beam. The exact direction of this force, as well as its magnitude, is un-known. The second condition for equilibrium is an ex-cellent tool to employ in such a situation for if we use an axis through the point 0 as the axis about which to take torques, the unknown force at the hinge has zero moment arm and therefore, causes zero torque. The remark-able result is that we can determine the tension T in the rope without knowing either the magnitude or the direction of the force at 0. (a) The torques about an axis through O are (200 lb) (10 ft)= 2000 lb-ft counterclockwise (2000 lb)(20 ft)= 40,000 lb-ft counterclockwise The moment arm of T is OP = (20 ft)(sin 30\textdegree) = 10 ft. Thus we have T (20 ft) sin 30\textdegree= T(10 ft)clockwise Since the beam is in equilibrium, the torque about any point of the beam is zero. We can then say -T (10 ft) + 2000 lb-ft + 40,000 lb-ft + F(0 ft) = 0 orT = 4200 lb = 2.1 tons The trick just used in removing the unknown force from the problem by taking torques about the hinge as an axis is a standard device in statics. The student should always be on the lookout for the opportunity to sidestep (temporarily) a troublesome unknown force by selecting an axis of torques that lies on the line of action of the unknown force he wishes to avoid. (b) Using the first condition for equilibrium, \sumF_x = 0 = T cos 30\textdegree - F cos \texttheta \sumF_y = 0 = T sin 30\textdegree + F sin \texttheta + W - 20001b The above two equations can be solved simultaneously, since there are two unknowns. Substituting numerical values, F cos \texttheta= T cos 30\textdegree = (4200 lb)(0.866) = 3640 lb F sin \texttheta=-W + 2000 lb - T sin 30\textdegree = 200 lb + 2000 lb - (4200 lb)(0.500) = 100 lb Dividing the first equation by the second, (F cos \texttheta)/(F sin \texttheta)= cot \texttheta = 3640/100 = 36.4 \texttheta\approx 1.5\textdegree Since \texttheta is almost zero, we have from the first equation F cos 1.5\textdegree \approx F cos 0\textdegree = F = 3640 lb.

Question:

During run of the Millikan oil - drop experiment, the following charges were observed on five different oil droplets: 1.44 × 10^-1 8, 2.56 × 10-1 8, 4.80 × 10-1 9, and 9.60 × 10-1 9 coulomb. Does this support the charge of an electron as being 1.60 × 10-1 9coulomb?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0667.htm

Solution:

This experiment, illustrated schematically in the above diagram, was first performed in 1909 by Robert A. Millikan. Oil droplets are sprayed between a top plate having a positive charge and a bottom plate having a negative charge. A beam of X-rays is simultaneously di-rected at this spray. The X-rays will knock electrons out of molecules in the air and these electrons will be picked up by the oil droplets, which will acquire a charge that is an integral multiple of the electronic charge. This occurs because each oil droplet will pick up an integral number of electrons. By adjusting the charge on the plates, the upward pull on the electrons due to attraction by the positive plate on the top and repulsion by the negative plate on the bottom can be balanced against the downward pull of gravity. The forces are balanced when the oil drop is stationary, as viewed through a telescope. By equating the gravitational and the electrostatic forces one can determine the charge on an oil droplet. It is given that the following charges were observed on oil droplets: 1.44 × 10^-18, 2.56 × 10^-18, 4.80 × 10^-19, and 9.60 × 10^-19 coulomb. If the elementary electronic charge is 1.60 × 10^-19 coulomb, each of these observed charges must be an integral multiple of the elementary charge. Hence, each of these charges must be exactly divisible by 1.60 × 10^-19 coulombs. [(1.44 × 10^-18 coulomb)/(1.60 × 10^-19 coulomb)] = 9.00, [(2.56 × 10^-18 coulomb)/(1.60 × 10^-19 coulomb)] = 16.00, [(4.80 × 10^-19 coulomb)/(1.60 × 10^-19 coulomb)] = 3.00and [(9.60 × 10^-19 coulomb)/(1.60 × 10^-19 coulomb)] = 6.00. Each of the observed charges is an integral multiple of 1.60 × 10^-19 coulomb, supporting the contention that 1.60 × 10^-19 coulomb is the charge of an electron.

Question:

The atomic radius of hydrogen is 0.037 nm, Compare this figure with the length of the first Bohr radius. Explain any differences.

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Solution:

The first Bohr radius, r, is equal to (N^2/Z) a_0 where a_0 is .05292 nm, N is an integer, and Z the atomic number. For hydrogen, Z = 1. For the first Bohr radius N = 1. Thus r = [(1)^2/(1)] a_0 = .05292 nm. Thus, the atomic radius of hydrogen (.037 nm) is consider-ably smaller than the first Bohr radius. This can be ex-plained by the fact that the atomic radius is calculated from the bond length of a H_2 molecule. There is consider-able orbital overlap in bonded atoms, which means the bond length is shorter than the sum of the radii of lone atoms.

Question:

Viruses do not have any ribosomes which are the structures needed for protein synthesis. How are they able to syn-thesize their protein coats during their replication? In your explanation, describe the life cycle of a typical virus.

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Solution:

Viruses consist of a nucleic acid core (either DNA or RNA single or double stranded) surrounded by a protective protein coat, which may be further enveloped. There are no mitochondria, ribosomes, or any of the other organelles normally found in eucaryotic cells. How does the virus acquire the energy and the structural machinery (ribosomes) for making proteins? Since viruses have no independent metabolic activity, they are incapable of reproduction by fission, budding, or other simple means of reproduction. Instead, they replicate by inserting their nucleic acid into a functional host cell. The host either cell provides both the energy and structural machinery necessary for the production of new viral components. After these components are assembled, they are released from the host cell and a new cycle of replication begins. The T_4, bacteriophage (a virus which attacks bacteria) may be employed to demonstrate viral replication. T_4 is one of the bacterial viruses which multiplies inE.colicells. The general structure of a bacteriophage, or simply phage, is shown in Figure 1. The replication process begins when the phage par-ticle collides with a sensitive bacterium and the phage tail fibers specifically attach to the bacterial cell wall. Host cells have specific receptor sites to which the phage attaches. An enzyme in the phage tail then breaks down a small portion of the cell wall, creating a small hole in the bacterium. The spikes are believed to help anchor the phage to the bacterium. The tail sheath then contracts, driving the core of the tail into the cell (see Figure IB). The viral DNA is then injected into the bacterium through the enzymatically weakened cell wall. The protein coat is left outside the bacterium. (Most animal viruses are completely engulfed by the cell through a process known as pinocytosis, so that no viral substance remains outside the cell.) The free viral chromosome (DNA molecule) serves as a template to direct the synthesis of new viral components. The DNA serves as a template for its own replication and for the viral-specific RNA necessary for the synthesis of its specific proteins. The viral-specific mRNA molecules attach to the host ribosomes, where viral protein synthesis takes place. Sometimes viral infection does not affect normal cell metabolism, but more often, most of the host cell's metabolism is directed toward the synthesis of new molecules needed for new viral particles. In some cases, all DNA and RNA synthesis on the host chromosome is inhibited and pre- existing bacterial RNA templates (mRNA) are broken down, so that all protein synthesis occurs only on new viral RNA templates.E.coli ribosomes may become modified, so that they work better with T_4, mRNA than with their own mRNA. This causes more viral molecular synthesis and less bacteri-al molecular synthesis. During infection, therefore, both host nucleic acid synthesis and protein synthesis are usual-ly suppressed. Once all the viral products are synthesized (there are over 40 of them), they interact in a definite sequence or morphogenetic pathway to produce the mature viruses. The complete infective phages that are produced must now be released from the bacterium, which has a rigid cell wall. To ensure release, many phages have a gene which codes for lysozyme, an enzyme which degrades bacterial cell walls. Lysozyme begins to be synthesized when the coat proteins appear, so that its accumulation causes rupture of the bacterial cell wall when the phages are mature and ready for release. Usually several hundred viral particles are released from each bacterial cell. These new viral particles then infect otherE.coli, replicating themselves and lysing the bacteria. The life cycle of a T_2 phage is summarized in Figure 3.

Question:

A particle which is performing simple harmonic motion passes through two points 20 cm apart with the same velocity, taking 1 s to get from one point to the other. It takes a further 2 s to pass through the second point in the opposite direction. What are the period and amplitude of the motion?

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0381.htm

Solution:

From the equations of simple harmonic motion, the velocity at any displacementx has a value v = \omega\surd(A^2 - x^2), where \omega is the angular frequency of the motion and A its amplitude. Thus v can be the same at two points only if the displacements of these points are \pm x. If the origin of the time scale is taken at the mean position of the motion, as shown in the rotor diagram, and if the second point is reached t_0 thereafter, the displacement at the first point is - x = A sin \omega(t_0 - 1 s)(1) at the second point + x = A sin \omegat_0,(2) and at the second point on the return journey, + x = A sin \omega(t_0 + 2 s)(3) Hence, using equation (1) and (2) sin \omegat_0 = x/A = - sin \omega(t_0 - 1 s) Since - sin \omega(t_0 - 1 s) = sin [-\omega (t_0 - 1 s)] = sin \omegat_0 t_0 = -(t_0 - 1 s) or t_0 = 1/2 s. Also, using equations (2) and (3), sin \omega(t_0 + 2 s) = x/A = sin \omegat_0. Since, in general, sin \texttheta_1 = sin \texttheta_2 implies \texttheta_2 = \pi - \texttheta_1 then, in this particular case \omega(t_0 + 2 s) = \pi - \omegat_0 or \pi/\omega = 2t_0 + 2 s = 3 s \therefore \omega = (\pi/3) rad\bullets^-1. Also sin \omega(t_0 - 1 s) = sin[\pi/3 rad\bullets^-1×-1/2 s] = - sin \pi/6 = -1/2. But sin \omega(t_0 - 1 s) = - x/A = -(0.1/A)m. \therefore 1/2 = (0.1/A)morA = 0.2 m.

Question:

A chemist reacts metal "B" with sulfur and obtains a compound of metal and sulfur. Assuming metal "B" weighed 2.435 g (MW = 121.75 amu), and the compound weighs 3.397g, what is the simplest or empirical formula of the compound? The atomic weight of sulfur is 32.06 g/mole.

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Solution:

To calculate the simplest formula, one must know the mole ratio of the elements that reacted. A mole is defined as the weight in grams of a substance divided by its atomic weight. The number of moles of "B" = (2.435g) / (121.75 g/mole) = .0200 moles of "B" . The number of grams of sulfur that reacted must be 3.397 - 2.435 = .962 g. The increase in weight can only be derived from the addition of sulfur. The number of moles of sulfur is (0.962 g) / (32.06g/mole) = .0300 moles of sulfur. The ratio of "B" to sulfur moles is .0200 B /.0300 S = 2:3. Therefore, the simplest formula must be B_2S_3 .

Question:

How does nutrient procurement by green plants differfrom thatof animals, fungi, andprotista?

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/Users/wenhuchen/Documents/Crawler/Biology/F17-0421.htm

Solution:

All green plants areautotrophs, subsisting exclusively on inorganic substancestaken from their en-vironment. The necessary molecules, being smalland soluble, are able to pass through cell membranes. Autotrophic organismstherefore do not need to digest their nutrients before taking theminto their cells. Green plants are photosynthetic, taking in carbon dioxideand water from their environment, and using the energy from sunlightto synthesize carbohydrates from them. 93% of the dry weight of plant was originally carbon dioxide in the air. Plants also need to synthesize proteins. Protein synthesis requires otherelements, such as nitrogen and sulfur. When rain water dissolves thesenutrients in the soil, the solution is taken up by the root system and transportedto various parts of the plant via the vascular system. Minerals requiredby plants are also transported up the plant body via the root and vascularsystems. Some plants supplement their protein diet by trapping insects(the Venus flytrap is an example). Many fungi such as mushroom, cannot produce their own carbohydratesfrom the air and water. They areheterotrophs: they must obtainthe carbohydrates they require from the surroundings. This is usuallysupplied by dead plant matter, but plant matter is mainly cellulose andother polysaccharides which must be first digested. Specialized secretorycells in these organisms release the enzymes for digestion into thesurrounding organic matter. The products of this extracellular digestion arethen directly absorbed through the cells surfaces or some-times by specializedabsorbing structures. Almost all animals are dependent on plant matter, or other animals thatfeed on plant matter, for food. They too areheterotrophs. Without plants, there would be virtually noheterotrophson earth. Like the mushroom, most animals digest the majority of their foodextracellularly. The simple products are then taken into the cells where they are further degradedby intracellular enzymes. Digestion is unimportant in green plantsbut fungi and animals have di-gestive process of the extracellular type. Protistaareautotrophsorheterotrophs. Protozoa are heterotrophic, nonphotosyntheticprotista. Protozoa includes theparemeciumand amoeba, in which digestion is intracellu-lar.

Question:

Write a program that performs matrix multiplication on a) array matrix [2] [4] and array vector [4], b) Result of multiplication is to be stored in array result [ ]. c) Print the result of multiplication i.e. print each row of result [ ]. d) Assume column vector contains {12 3 4} and matrix contains values [0 1 2 3] and [1 2 3 4].

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0429.htm

Solution:

# include < stdio. h > # define NUM _ C0L 4 /\textasteriskcenteredno. of columns\textasteriskcentered/ # define NUM _ R0WS 2 /\textasteriskcenteredno. of rows\textasteriskcentered/ main ( ) { int matrix [NUM _ R0WS][NUM _ COL]; /\textasteriskcenteredcreates a 2 × 4.array of type integer\textasteriskcentered/ int vector [NUM _ COL]; int result [NUM _ ROWS]; /\textasteriskcenteredTo store the result of multiplication \textasteriskcentered/ int i, j; /\textasteriskcenteredsubscripts or counters \textasteriskcentered/ /\textasteriskcentered1. Initialize the matrix, vector and result\textasteriskcentered/ for (i = 0; i < NUM _ ROWS; i++) /\textasteriskcenteredSet result to zero\textasteriskcentered/ result [i] = 0; for (i = 0; i < NUM _ C0L; i++) /\textasteriskcenteredset vector to 1 2 3 and 4 \textasteriskcentered/ vector [i] = i + 1; for (i = 0; i < NUM _ R0WS; i++) for ( j = 0; j < NUM _ COL; j++) /\textasteriskcenteredset matrix to 0 1 2 3\textasteriskcentered/ /\textasteriskcentered1 2 3 4\textasteriskcentered/ matrix [i] [ j] = i + j; /\textasteriskcentered2. Perform matrix multiplication \textasteriskcentered/ for (i = 0; i < NUM_R0WS; i++) for ( j = 0; j < NUM_COL; j++) result [i] = result [i] + matrix [i] [ j] \textasteriskcentered vector [ j]; printf ("The result of matrix multiplication = "); /\textasteriskcenteredPrint each row of result [ ]\textasteriskcentered/ for ( j = 0; j < NUM _ ROWS; j++) printf ("%d", result [ j]); /\textasteriskcenteredput newline character\textasteriskcentered/ put char ('\textbackslashn'); }

Question:

You want to plate out 50g of copper from an aqueous solution of CuSO_4. Assuming 100% efficiency, how many coulombs are required?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0571.htm

Solution:

In this question, you are dealing with the phenomenon of electrolysis. When an electric current is applied to a solution containing ions, the ions will either be reduced or oxidized to their electronically neutral state. To answer this question, you must realize that Cu^2+ ions exist in solution. To plate out copper, 2 electrons must be added to obtain the copper atom, Cu^0. Since the Cu^2+ must gain electrons, it must be reduced. The amount of electricity that produces a specific amount of reduction (or oxidation) is related by q = nF (Faraday's Law). Where q = the quantity of electricity in coulombs, n = number of equivalents oxidized or reduced and F = Faradays. The number of equivalents equals the weight of material oxidized or reduced (m) divided by the gram- equivalent weight of the material (M_eq) i.e., N = (M) / (M_eq). A faraday = 96,490 coulombs or one mole of electrons. Since copper ion requires two electrons for reduction, the gram- equivalent weight is one half of the atomic weight or 31.77g-equiv. You have, therefore, q = [(50) / (31.77)] (96,490) = 1.52 × 10^5 coul. required.

Question:

Find an expression for the hydrolysis constant of the bicarbonate ion HCO_3^-,K_h(HCO_3^-), using only the first dissociation constant for H_2CO_3, K_(a)1 (H_2CO_3), and the water constant, K_W .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0362.htm

Solution:

The solution to this problem is a direct application of the definitions ofK_h, K_(a)1 , and K_W . The removal of one proton may be represented by K_a1, H_2CO_3 \rightleftarrows H^+ + HCO_3^- Hence, the first dissociation constant is K_(a)1 (H_2CO_3) = {[H^+][HCO_3^-]}/[H_2CO_3] The water constant is given by K_W = [H^+][OH^-]. Hydrolysis of HCO_3^- proceeds according to the reaction HCO_3^- + H_2O \rightleftarrows H_2CO_3 + OH^-. We are trying to determine an expression for the hydrolysis constantK_h(HCO_3^-) in terms of K_(a)1 (H_2CO_3) and K_W . By definition, K_h(HCO_3^-) = {[H_2CO_3][OH^-]} / [HCO_3^-] . Hence,K_h(HCO_3^-) = {[H_2CO_3][OH^-]} / [HCO_3^-] = ({[H_2CO_3][OH^-]} / [HCO_3^-]) × [H^+]/H^+] = ([H_2CO_3] / {[H^+] [HCO_3^-]}) × [H^+] [OH^-] = [(1)/{K_(a)1 (H_2CO_3)}] × K_W, or,K_h(HCO_3^-) = [(K_W)/{K_(a)1 (H_2CO_3)}] .

Question:

In man, two abnormal conditions, cataracts in the eye and excessive fragility of the bones, seem to depend on separate dominant genes located on different chromosomes. A man with cataracts and normal bones, whose father had normal eyes, married a woman free from cataracts but with fragile bones. Her father had normal bones. Their daughter marries a man with normal eyes and bones. If these two people have a child, what is the probability that it will have both cataracts and fragile bones?

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0691.htm

Solution:

First, let us determine the possible genotypes of the individuals involved. Let C represent the dominant allele for cataracts, c the recessive allele for normal eyes, B the dominant allele for fragile bones, and b the recessive allele for normal bones. The man has cataracts, but he must be heterozygous for that trait (Cc) since his father had normal eyes (cc). Since the man has normal bones, his genotype for that trait is bb. The woman whom he married has normal eyes (cc). Since she has fragile bones, and her father had normal bones (bb), she must be heterozygous for that trait (Bb). The daughter has one of four possible genotypes: CcBb, Ccbb, ccBb, and ccbb. Her husband, who is normal for both traits involved, is ccbb. In order for their child to express both abnormalities, its genotype must be CcBb (since the daughter's husband can transmit only recessive alleles). This means that the daughter must donate both dominant alleles. Looking at the cross outlined above, we see that there is only one genotype possible in which the daughter carries both dominant alleles, namely CcBb. The probability of this occurring is 1/4 (one out of four possible genotypes). Given this, there is a 1/2 probability that the daughter will donate the B, rather than the b allele, to her child, and a 1/2 probability that she will donate the C, rather than the c allele. The probability that she will both possess and subsequently donate both dominant alleles is equal to the product of the separate probabilities of each of the three events, namely 1/4 × 1/2 ×1/2 or 1/16 probability that a child having both cataracts and fragile bones will be produced.

Question:

The gypsy moth produces a natural attractant, C_18H_34O_3. If a female moth is trapped behind a cellophane screen containing a pinhole and the carbon dioxide she produces diffuses through the pinhole at the rate of 1millimicromole per 90 seconds, what quantity of attractant will diffuse through the orifice in the same amount of time (90 seconds)?

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Solution:

The rate of escape of gases through an orifice is inversely proportional to the square root of the density of the gas. This principle is known as Graham's Law. If two gases are compared at the same temperature and pressure (as in this problem), then u_1 / u_2 = \surd(d_2 / d_1 ) where u is the velocity (or the rate at which the gas diffuses out of the orifice) and d is the density of the particular gas. To solve this problem, this equation is modified somewhat to avoid using densities in the expression (since the densities are unknown). However, the density is defined as mass per unit volume, and since the volume per mole of both gases, C_18H_34O_3 and CO_2 , are the same, Graham's Law becomes u_1 / u_2 = \surd(d_2 / d_1 ) = \surd[(m_2 / V) / (m_1 / V)] = \surd(m_2 / m_1 ), where V is volume and m_1 and m_2 are the masses of the gases. Another modification is necessary before solving this problem: the mass of any substance is its molecular weight times the number of moles of that substance. However, assume that 1 mole is involved of both substances and thus molecular weights of the gases (which can be determined) is substituted. Graham's Law then becomes u_1 / u_2 = \surd(M_2 / M_1 ) u_1 for CO_2 = 10-^9 mole/90 sec = 1.1 × 10-^11 moles/sec. Note: 1milli micromole = 10-^9 mole. M_1 for CO_2 = 44 g/mole M_2 for C_18H_34O_3 = 298 g/mole. Thus, solve for u_2. (1.1 × 10-^11moles/sec) / u_2 = \surd[(298 g/mole) / (44 g/mole)] = 2.6 u_2 = 4.2 × 10-^12 moles/sec. 1picomole= 10-^12 moles. u_2 = 4.2 × 10-^12 moles/sec = 4.2picomolesper sec u_1 = 1.1 × 10-^11moles/sec = 11.0picomolesper sec. The CO_2 diffuses 1millimicromole within 90 sec ([ 0.011millimicromole/sec] [90 sec] = 1millimicromole) and C_18H_34O_3 diffuses (0.0042millimicromoles/sec) (90 sec) = 0.38millimicromoles within the same time as the CO_2.

Question:

The deviation of the second-order diffracted image formed by an optical grating having 5000 lines/cm is 32\textdegree. Calculate the wavelength of the light used.

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/Users/wenhuchen/Documents/Crawler/Physics/D30-0906.htm

Solution:

An optical grating is a transparent piece of glass or plastic on which there are closely spaced parallel scratches. These scratches behave like opaque barriers and the spaces between slits like narrow slits. Each adjacent pair of slits produces a double-slit diffraction pattern so that this problem can be analyzed using double-slit interference. Point P_n in the diagram receives light from both slits. The path lengths from slits S_1 and S_2 to P_n differby an amount very close to ∆L. If the two waves are in phase as they leave slits Si and S2, then they will also be in phase at P_n if ∆L is an integral multiple of the wavelength of the light. That is, if ∆L = n\lambda then rein-forcement occurs at P_n and a bright band occurs there. If b is small compared to L, then the angle \texttheta in the small right triangle is nearly equal to \texttheta in the larger right triangle. Therefore, we can say sin \texttheta_N = N\lambda/b orb sin \texttheta_N = N\lambda where \texttheta_N is the deviation of the Nth order diffracted image and b is the distance between slits. b = 1/5000 cm = 0.00020 cm \lambda = [(b sin \texttheta_N)] / [N] = [(0.00020 cm × 0.53) / (2)] = 0.000053 cm = 5300 A

Question:

Write a program in FORTRAN that prints a calendar for any month of the year from March 1600 to year N.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0544.htm

Solution:

To grasp a thorough theoretical knowledge of the procedure behind this problem, one needs to know something about congruence and modulo arithmetic. Here, we will try to minimize the theoretical discussion and simply state some pertinent theorems. We will look at the Gregorian calendar, first implemented on October 15, 1582. Let N be any year later than 1600. Then, between the year 1600 exclusive, and the year N inclusive, there are [N/4] - 400 years divisible by 4 To prove this let y be any year such that 1600 < y \leq N, Year y is divisible by 4 only if y = 4x, where x is an integer. Thus, the number of years divisible by 4 is the same as the number of integers x satisfying 1600 < 4x \leq N, or 400 < x \leq N/4. There are clearly [N/4] - 400 of such integers. [N/400] - 4 years divisible by 400 [N/100] - 16 century-beginning years [N/100] - [N/400] - 12 century-beginning years that are not leap years [N/4] - [N/100] + [N/400] - 388 leap years [A/4] + [B/4] + 24A - 388 leap years, where A and B are integers governed uniquely by the restrictions N = 100A + B and 0 \leq B < 100 (Brackets are used to express the greater integer part of the division.) These relations tell us that leap years, containing 366 days, are those divisible by 4, and that century-beginning years are leap years only if divisible by 400. Thus, 1600, 2000, 2400,... are century-beginning leap years. Since February is a troublesome month, especially during leap years, we rearrange our calendar so that March is the first month of the year N. January and February of year N become the eleventh and twelfth months of year N - 1. Then, April through the following February become the second through twelfth months of year N. This places February at the end of the year, a manipulation which will make the program simpler to follow. We must also consider the days of the week. Starting with Sunday, we number the days 0,1,2,3,4,5,6, ending with Saturday. The theorem which gives the weekday number of theKthday of month M of year N (known as Zeller's congruence) is K + [2.6M - 0.2] + B + [B/4] + [A/4] + 2A(mod 7). To understand the derivation of this formula, consider the following: Since March 1, 1600 fell on a Wednesday, its weekday number is W = 3. Since 365 \equiv 1 (mod 7) (365 - 1 = 364 is divisible by 7), March 1, 1601 has weekday number 4. March 1, 1602 has number 5 and March 1, 1603 has number 6. But March 1, 1604 has number 1, not 0 because it is a leap year. In general, the weekday number W_N of March 1 of year N is: W_N = W + t + e (mod 7), where W = 3(1) where t is the total number of years and e is the number of leap years between 1600 exclusive and N inclusive. Now, t = N - 1600, N = 100A + B, and, e = [A/4] + [B/4] + 24A - 388. Thus, equation (1) is: W_N = 3 + 100A + B - 1600 + [B/4] + [A/4] + 24A - 388 (mod 7) or, after simplification, W_N = 3 + B + [B/4] + [A/4] - 2A (mod 7).(2) Equation (2) gives the weekday number of March 1 for any N. Since April 1 occurs 31 days after March 1, and 31 \equiv 3 (mod 7), its week-day number is 3 more than the weekday number of March 1, i.e., 6. Continuing for the other months we obtain the sequence of constants W: 3,6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32 \uparrow\uparrow MarApr .... Reverend Zeller noticed that the arithmetic function 1 + [2.6m - 0.2] assumes exactly this sequence of integers for m = 1,2,...,12. Hence, the weekday number of the first day of any month, m, of the Nth year is given by W_N \equiv 1 + [2.6m - 0.2] + B + [B/4] + [A/4] - 2A (mod 7)(3) Finally, if, instead of the first day, we consider thekthday, Zeller's congruence is obtained; W_N \equiv k + [2.6m - 0.2] + B + [B/4] + [A/4] - 2A (mod 7). Take for example November 1, 2060. K is 1, M is 9 (according to our renumbering scheme), B is 60, and A is 20 (since N = 2060, and N = 100A + B). Plugging in the values, we get 1 + 23 + 60 + 15 + 5 - 40 = 64. To complete the problem, we find that 64 \equiv 1 (mod 7) since 64 - 1 is divisible by 7. Hence, 1 corresponds to Monday, and the answer is obtained. We present the program forthwith, high-lighting control sections with comments. CCALENDAR PROGRAM INTEGER BLANK, DATE, YEAR, RMTHS CTHE ELEMENTS OF DATE, ALONG WITH THE CCHARACTER BLANK, ARE USED TO PRINT CALENDAR CIF FIRST DAY OF MONTH IS NOT SUNDAY. DIMENSION RMTHS (12), DATE (6) DATA RMTHS/'JAN ', 'FEB ', 'MAR ', 'APR, 'MAY', 'JUN', 1'JUL ', 'AUG ', 'SEP ', 'OCT', 'NOV ', 'DEC' / DATA BLANK, DATE/' ', '1', '2', '3', '4', '5', '6'/ CINPUT MONTH (JAN = 1, FEB = 2,..., DEC = 12) AND YEAR 10READ (5,100) MONTH, YEAR 100FORMAT (I2, I4) CDO WHILE YEAR GREATER THAN ZERO IF (YEAR.LE.O) GO TO 99 CNOW, MONTHS ARE RENUMBERED SUCH THAT CM = 1 FOR MAR, M = 2 FOR APR, AMD SO ON UNTIL CM = 10 FOR DEC. FOR JAN AND FEB, M = 11 AND M = 12 COF PRECEDING YEAR. 20NEWYR = YEAR IF (MONTH - 2) 25, 30, 35 CJANUARY SECTION 25M = 11 NEWYR = NEWYR - 1 CNDIM EQUALS NUMBER OF DAYS IN MONTH NDIM = 31 GO TO 40 CFEBRUARY SECTION 30M = 12 NEWYR = NEWYR - 1 CDETERMINE WHETHER NYEAR IS A LEAP YEAR IF (YEAR.NE. (YEAR/4) \textasteriskcentered4) GO TO 33 IF (YEAR.NE. (YEAR/100)\textasteriskcentered100) GO TO 34 IF (YEAR.EQ. (YEAR/400) \textasteriskcentered400) GO TO 34 CIF YEAR IS NOT A LEAP YEAR 33NDIM = 28 GO TO 40 CIF YEAR IS A LEAP YEAR 34NDIM = 29 GO TO 40 CNOW COMPUTE MARCH THROUGH DECEMBER M = MONTH - 2 CCASE STRUCTURE TO DETERMINE DAYS IN MONTH GO TO (36, 37, 36, 37, 36, 36, 37, 36, 37, 36), M 36NDIM = 31 GO TO 40 37NDIM = 30 CNOW CALCULATE WEEKDAY NUMBER FOR THE CFIRST DAY OF THE MONTH 40NC = NEWYR/100 ND = NEWYR - 100\textasteriskcenteredNC NW = 1 + (26\textasteriskcenteredM-2)/10 + ND + ND/4 + NC/4 - 2\textasteriskcenteredNC CDO LOOP REDUCES WEEKDAY NUMBER MODULO 7 CTO ONE OF THE VALUES 0, 1, 2,...,6. DO 41 I = 1,7 NWKDAY =I - 1 NDIFF = NW - NWKDAY IF (NDIFF.EQ.(NDIFF/7)\textasteriskcentered7) GO TO 45 41CONTINUE CWRITE HEADINGS 45WRITE (6, 101) RMTHS (MONTH), YEAR 101FORMAT (1HO, 10X, A4, I4) WRITE (6, 102) 102FORMAT (1HO, 27HSUN MON TUE WED THU FRI SAT/) IF (NWKDAY. EQ.O) GO TO 52 CTHIS SECTION PRINTS THE FIRST WEEK OF THE CMONTH IF SUNDAY IS NOT THE FIRST DAY LIM = 7 - NWKDAY WRITE (6, 103) (BLANK, I = 1, NWKDAY), (DATE(I), I = 1, LIM) 103FORMAT (7(3X,A1)) LIM = LIM + 1 GO TO 53 52LIM = 1 CFINISH PRINTING CALENDAR 53WRITE (6, 104) (I, I = LIM,NDIM) 104FORMAT (7I4) GO TO 10 CEND DO-WHILE 99STOP END

Question:

A chemist has an 18% solution and a 45% solution of a disinfectant. How many ounces of each should be used to make 12 ounces of a 36% solution?

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Solution:

Let x = Number of ounces from the 18% solution And y = Number of ounces from the 45% solution (1) x + y = 12 (2) .18x + .45y = .36(12) = 4.32 Note that .18 of the first solution is pure disinfectant and that .45 of the second solution is pure disinfectant. When the proper quantities are drawn from each mixture the result is 12 gallons of mixture which is .36 pure disinfectant, i.e., the resulting mixture contains 4.32 ounces of pure disinfectant. When the equations are solved, it is found that x = 4 and y = 8.

Question:

The faint light sometimes seen over Marshland at night, the "will-o'-the-wisp", is believed to come about as a result of the burning of a compound of phosphorus (P) and hydrogen (H) . What is the formula of this compound?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0613.htm

Solution:

To find the formula of this compound, it is necessary to determine the valence of the elements from which it is composed. The valence of an element is the number of electrons that are involved in chemical bonding. To find the valence of phosphorus and hydrogen, consider their atomic number and electronic configuration. Hydrogen: Atomic number= 1. Electronic configuration= 1s1 Phosphorus: Atomic number= 15. Electronic configuration= 1s^22s^22p^63s^23p^3. The outer electrons are 1s^1 for hydrogen, and 3p^3 for phosphorus. It takes one additional electron to fill hydrogen's s orbital; its valence is one. It takes 3 more electrons to fill phosphorus' p orbital; its valence is three. Elements react with the purpose of filling all their orbitals with the maximum number of electrons by either a transfer of electrons or by sharing electrons. It would take three hydrogen atoms to complete phosphorus' outer orbital. In turn, each electron of phosphorus would serve to complete the outer orbital of each hydrogen atom. This can be pictured in an electron-dot formula as shown above. In this figure, the X's represent the outer electrons of hydrogen and dots represent the electrons in the outer shell of phosphorus. The formula of this compound is, thus, PH_3.

Question:

A radio telescope sited on the edge of a cliff over-looking the sea operates on a wavelength of 100 m. A radio star rises above the horizon and is tracked by the telescope. The first minimum of the received signal occurs when the star is 30\textdegree above the horizontal. Explain why the minimum occurs and determine the height of the cliff, assuming that radio wave suffer a phase change of \pi on reflection at a water surface. (See figure.)

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Solution:

The radio telescope is receiving signals direct from the star (ray XBC) and also by reflection from the water surface, (YAC) which is acting like a plane mirror. In this case the source of radiation is far enough away to be considered at infinity and the rays are descending on the earth in a parallel beam. When the path difference between rays YAC and XBC (see figure) is such that they reach the telescope exactly out of phase, destructive interference occurs and the signal at C will be a minimum. Consider the diagram. If AB is drawn perpendicular to XC, it will also be perpendicular to YA, and AB is a plane wave front of the incoming beam. Thus the disturbances at A and B have the same phase at all times. A phase change of it occurs at A and this is equivalent to an increase in path of \lambda/2. Thus the path difference between the two rays is \delta = AC + (\lambda/2) - BC. But by the laws of reflection \angleYAZ = \angleCAO = 30\textdegree, and therefore \angleBAC = 180\textdegree - 90\textdegree - 30\textdegree - 30\textdegree = 30\textdegree. \delta = AC + \lambda/2 - AC sin 30\textdegree = \lambda/2+ AC(1 - sin 30\textdegree) Since sin 30\textdegree = Y/AC, AC = Y/sin 30\textdegree and \delta = \lambda/2 + [y/(sin 30\textdegree)] (1- sin 30\textdegree) = \lambda/2 + y. In order for minima to occur, the path difference \delta must equal an odd number of half wavelengths, or\delta = (2n + 1)\lambda/2n = 0, 1, 2, ... Then(2n + 1) \lambda/2 = \lambda/2 + y. If n = 0, \delta = \lambda /2, y = 0, and this corresponds to the center of the interference pattern being a minimum. Since this is the first minimum the telescope receives, C must be at the first minimum above 0 for the situation shown in the diagram. (Where it a minimum other than the first, a previous minimum must have occurred for an angle lower than 30\textdegree. But we are told that this doesn't happen.) Therefore, n = 1 and \delta must be 3\lambda/2. Hence,3\lambda/2 = \lambda/2 + y and y = \lambda = 100 m.

Question:

What electric field intensity is just sufficient to bal-ance the earth's gravitational attraction on an electron? If this electric field were to be produced by a second electron how far away must it be put?

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Solution:

The gravitational force on the electron is mg and the force due to the field E is eE. Thus if they are to balance (see the figure) eE = mg E = mg / e = {(9.1 × 10^\rule{1em}{1pt}31 kg)(9.8 m/s^2)} / (\rule{1em}{1pt}1.6 × 10^\rule{1em}{1pt}19 c) = \rule{1em}{1pt}5.57 × 10^\rule{1em}{1pt}11 N/coul If this field is supplied by another electron a distance R away (see figure) then by definition, the electric field due to this electron at R is E = ke / R^2 where k = 9 × 10^9 (N \bullet m2/ c^2), and e is the signed charge of an electron, Hence, solving for R, R^2 = ke / E. Using the previous calculation of E, R^2 = [{9 × 10^9 (N \bullet m^2 / c^2)} (\rule{1em}{1pt}1.6 × 10^\rule{1em}{1pt}19 c) ] / (\rule{1em}{1pt}5.57 × 10^\rule{1em}{1pt}11 N / c) R^2 = 25.8 m^2 R = 5.1m

Question:

A grain of sand has a mass of 10^-5 gram. Through what heightmust it fall from rest in order to have a kinetic energy equalto the energy of a quantum of visible light with a frequencyof 5 × 10^14 sec^-1 ?

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Solution:

The energy of the quantum is given by the product of its frequency, v, and Planck's constant h. \epsilon =hv = 6.625 × 10^-27 × 5 × 10^14 = 3.31 × 10^-12erg When the grain of sand falls through a height H from rest, its final kinetic energy, (1/2)mv^2 , is equal to its decrease in gravitational potential energy, mgH. (1/2)mv^2= mg H Therefore mg H =hv H = (hv/mg) = [(3.31 × 10^-12)/(10^-5 × 980)] = 3.38 × 10^-10 cm The grain of sand would have to fall through a height of only 3.38 × 10^-10 cm.The diameter of an atom is about 10^-8 cm.

Question:

Show that the angular momentum of a particle about a point 0 will be equal to zero if any one of the following conditions applies? (a) P^\ding{217} = 0 (b) the particle is at point 0 (c) P^\ding{217} is parallel (or anti-parallel) to r^\ding{217}

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1068.htm

Solution:

In general, for a particle of mass m and velocity v^\ding{217} = (v1i˄ + v_2 \^{\j} + v_3 k ̂) located a distance r from point 0, r^\ding{217} = (r_1 i˄ + r_2 \^{\j} + r_3 k ̂); angular momentum L^\ding{217} = r^\ding{217} × mv^\ding{217}. In scalar form, this equation becomes L = r m v sin \texttheta where \texttheta is the angle between r^\ding{217} and v^\ding{217}. a) P\ding{217}= m v^\ding{217}. If P^\ding{217} = 0 then m v must be 0. L = r (0) sin \texttheta L = 0 b) If the particle is located at point 0 then r^\ding{217} = 0 and r = 0. (0) (m v) sin \texttheta = 0 L = 0 c) If P^\ding{217} is parallel or anti-parallel to r^\ding{217} then v^\ding{217} is parallel or anti- parallel to r^\ding{217} since P^\ding{217} has the same direction as v^\ding{217} because m is a scalar. If v^\ding{217} is parallel or anti-parallel to r^\ding{217} then \texttheta = 0\textdegree or 180\textdegree. In either case, sin \texttheta = 0. r m v (0) = 0 L = 0

Question:

The diffusion coefficient for serum globulin at 20\textdegree in a dilute aqueous salt solution is 4.0 × 10^-11 m^2 s^-1. If the molecules are assumed to be spherical, calculate their molecular weight. Given \etaH_2O = 0.001005 Pascal sec at 20\textdegree andѵ= 0.75 (cm^3/g) for the protein.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0820.htm

Solution:

For spherical molecules, the molecular weight M is related to the diffusion coefficient D by the follow-ing equality D = [(RT) / (N_A6\pi\eta)][(4 \pi N_A) / (3Mѵ)]^1/3 where R is the gas constant in joules (8.314 J/mole \textdegreeK), N_A is Avogradro's Number (6.02 × 10^23), \eta is the viscosity of the solvent, andѵthe partial specific volume of the protein. Solving for M; D = 4.0 × 10^-11 (m^2/s) = [(8.314 J/mole \textdegreeK) (293\textdegreeK)] / [(6.02 × 10^23/mole) 6\pi (.001005 Js/m^2)] × [4\pi(6.02 × 10^23 mole)] / [3 × m × 0.75 × 10^-6 m^3/s]^1/3 4.0 × 10^-11 m^2/s = (2.14 × 10^-19 m^2/s) × (3.36 × 10^30/M)^1/3 1.87 × 10^8 = (3.36 × 10^30/M)^1/3 6.53 × 10^24 = 3.36 × 10^30/M M = 5.15 × 10^5 g/mole.

Question:

Spring-Heel Jack was a legendary English criminal who was never captured because of his ability to jump over high walls and other obstacles which his pursuers were unable to scale. It is believed that he had a powerful spring attached to each shoe for this purpose. Assuming that he weighed 150 lb and that his springs were com-pressed by 1 in. when he stood on them, by how much did he need to keep his springs compressed on one of his operations in order to be ready to clear a 10-ft wall in the event of an emergency?

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Solution:

The figure shows an idealized drawing of Spring-Heel Jack. When Jack stands up, the springs are compressed a distance x. When in equilibrium, the net force on Jack is zero. Hence, 2kx= mg ork= mg/2x where m is his mass and k is the spring constant. Thus k= [150 lb]/[(2)(1 in)] = (75 lb)/(1/2 ft) = 900 lb/ft. If Jack wishes to clear a height h while remaining erect, the potential energy stored in the springs must have been sufficient to raise his 150-lb weight through a vertical distance h. But if x was the compression of each spring, then by conservation of energy 1/2 kx^2 + 1/2 kx^2= mgh x^2= mgh/k \thereforex^2= (150 × 10 ft \bullet lb)/(900 lb \bullet ft^-1) = 20/12 ft^2. x= \surd(1.67) ft = 1.29 ft.

Question:

Two electrochemical cells were fitted with inert platinum electrodes and connected in series. The first cell contained a solution of a metal nitrate; the second cell, metal sulfate. After operation of the cell for a prescribed period of time, it was observed that 675 mg of the metal had been deposited in the first cell and 76.3 ml of hydrogen gas (25\textdegree and 1 atm) evolved in the second. Determine the equivalent weight of the metal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0564.htm

Solution:

Equivalent weight may be defined as that amount of substance which one mole of electrons will reduce or oxidize. Since the electrodes are connected in series, the same current passes through both cells. Thus, equal numbers of equivalents of the metal (m) and hydrogen must be liberated. The cathode reactions for the two cells are: first cell: M^n+ + ne^- \rightarrow M(s) ; second cell: 2H^+ + 2e^- \rightarrow H_2(g), where n, is the number of electrons needed to form the M(s) from the metal ion. One can solve for the number of moles of H_2 formed by use of the Ideal Gas Law. n = (PV/ RT) where n is the number of moles, R is the gas constant, 0.082 . liter-atm / mole\textdegreeK, T is the absolute temperature, V is the volume and P is the pressure. One is given the temperature in \textdegreeC, to convert to \textdegreeK add 273 to the temperature in \textdegreeC . T = 25\textdegree + 273 = 298\textdegree K . Now, solving for n: n = [(1 atm)(.0763 liters)] / [(0.082 liter-atm / mole\textdegreeK) (298\textdegreeK)] = 3.12 × 10^-3 moles. From the half equations, one can see that there are two equivalents for each mole of H_2 formed. Thus, there must be 2 × (3.12 × 10^-3) moles of equivalents of the metal present. Hence, equivalents of M = 2 × 3.12 × 10-^3 = 6.24 × 10^-3 equiv. One is given that 675mg or .675 g of metal are formed, thus, in .675 g of the metal there are 6.24 × 10^-3 equiv . The weight of one equivalent can be found by dividing .675 g by 6.24 × 10^-3 equiv . equiv. wt. =(.675g) / (6.24 × 10^-3 equiv.) = 108.2 g / equiv.

Question:

Explain the importance and structure of the endoplasmic reticulumin the cell.

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Solution:

The endoplasmic reticulum is responsible for transporting certain moleculesto specific areas within the cytoplasm. Lipids and proteins are mainlytransported and distributed by this system. The endoplasmic reticulumis more than a passive channel for intracellular transport. It containsa variety of enzymes playing important roles in metabolic processes. The structure of the endoplasmic reticulum is a complex of membranesthat traverses the cytoplasm. The membranes form interconnectingchannels that take the form of flattened sacs and tubes. When the endoplasmic reticulum hasribosomesattached to its surface, werefer to it as rough endoplasmic reticulum and when there are no ribosomesattached, it is called smooth endoplasmic reticulum. The rough endoplasmicreticulum functions in transport of cellular products; the role ofthe smooth endoplasmic reticulum is less well known, but is believed to beinvolved in lipid synthesis (thus the predominance of smooth end reticuluminhepatocytesof the liver). In most cells, the endoplasmic reticulum is continuous and interconnectedat some points with the nuclear membrane and sometimes, with the plasma membrane. This may indicate a pathway by whichmaterials synthesized in the nucleus are transported to the cytoplasm. In cells actively engaged in protein synthesis and secretion (suchasacinarcells of the pancreas), rough endoplasmic reticulum is abundant. By a well regulated and organized process, protein or polypep-tidechains are synthesized on theribsosomes. These products arethen transported by the endoplasmic reticulum to other sites of the cell wherethey are needed. If they aresecre-toryproducts, they have to be packagedfor release. They are carried by the endoplasmic reticulum to theGolgi appar-atus, another organelle system. Some terminal portions of theendoplasmic reticulum containing protein molecules bud off from the membranesof the reticulum complex, and move to the Golgi apparatus in theform of membrane-bounded vesicles. In the Golgi apparatus, the proteinmolecules are concentrated, chemically modified, and packaged sothat they can be released to the outside byexocytosis. This process is necessarybecause some proteins may be digestive enzymes which may degradethe cytoplasmanslysethe cell if direct contact is made.

Question:

How would one expect the bond strength of NO to compare with that of O_2?

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Solution:

The molecular orbital is the sum of the atomic orbitals. For example, when two atomic orbitals are combined, two molecular orbitals are formed. One orbital is called a bonding orbital and the other is an antibonding orbital. The bonding orbital is at a lower energy level than the antibonding orbital. If possible, electrons seek out the bonding orbital rather than the antibonding orbital. The orbitals in the production of H_2 from H atoms can be visualized as shown in the accompanying figure. The greater the number of antibonding orbitals in a molecule, the weaker the bond. Bond order is a quantity that indicates the strength of bonding orbitals. It is defined as half the number of bonding electrons minus half the number of antibonding electrons. Thus, the higher the bond order, the stronger the chemical bond. To compare the bond strength of NO with O_2, compare their bond orders. To do this, consider the total number of valence electrons in each element. Valence electrons are the outer electrons, which participate in bonding. For NO, the total number of valence electrons is 3 + 4 = 7. There exist 3 bonding p orbitals, which accommodate 6 electrons. The 1 unpaired electron must be in an anti-bonding orbital. This means that the bond order of NO is (1/2) (6) - (1/2) (1) = 3 - 0.5 = 2.5. In O_2 there is a total of 8 valence electrons. This means 2 electrons must be in antibonding orbitals, since the 3 bonding orbitals can accommodate only 6 electrons. The bond order of O_2 = (1/2) (6) - (1/2) (2) = 2. The bond order of NO is higher, which means, its chemical bond is stronger than that of O_2.

Question:

An experimenter crossed a stock of corn homozygous for the linked recessive genes colorless (c), shrunken (s) and waxy (w) to a stock homozygous for the dominant alleles of these genes (C, S, W) and then backcrossed the F_1 plants to the homozygous recessive stock. The progeny were as follows: Phenotype (all are -c-s-w) Number 1 CSW 17,959 2 csw 17,699 3 Csw 2,009 4 cSW 2,024 5 CSw 4,455 6 csW 4,654 7 CsW 288 8 cSw 312 49,400 Draw a linkage map for the three genes.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0679.htm

Solution:

In order to draw a linkage map for the three genes, we must determine the distances between the genes in map units and the order of the genes on the chromo-some. Let us first determine the parental type. We know that since the F_1 plants resulted from a cross between a homozygous dominant (CC, SS, WW) and a homozygous recessive (cc, ss, ww) parent, they are heterozygous (CcSsWw). They were crossed with homozygous recessive plants (ccssww) Therefore the parental types are CcSsWs ( 1 ) and ccssww ( 2 ). The other six phenotypes are recombinants. To determine the distances between the genes, let us determine the RF for two genes at a time. Consider a cross-over anywhere between c and S. The number or recombinants (from F_2) = 2009 3 + 2024 4 + 288 7 + 312 8 = 4633. RF = (4633) / (49400) = 0.094 An RF of 0.094 gives a separation of 9.4 units between C and S. Consider a cross-over between S and W. The number of recombinants = 4455 5 + 4654 6 + 288 7 + 312 8 = 9709. RF = (9709) / (49400) = 0.196 That is, S and W are 19.6 units apart. Consider a cross-over between C and Wx. The number of recombinants = 2009 3 + 2024 4 + 4455 5 + 4654 6 = 13142. RF = (13142) / (49400) = 0.266 That is, C and W are 26.6 map units apart. To construct a linkage map, we choose randomly C and Wx as our reference points. S is 19.6 units away from w. The only way s can be 9.4 units away from C is for s to lie between C and W. On a genetic map, we have: However, the sum of the map distances between C' and S, and between S and W = 9.4 + 19.6 or 29.0 unit. This sum is greater than the distance between C and W (26.6 units). How do we explain this discrepancy? The explanation lies in the fact that more than one cross-over can occur between two genes, expecially if they are far apart. In this case, there is a cross-over between C and S and between S and W making two cross-overs between C and Wx. Diagrammatically, this can be rep- resented as follows: The result of two cross-overs between two genes is called a double recombinant. Since single cross-overs are quite infrequent, the frequency of double recombinants is even lower. The double recombinants are usually identified by their lowest frequency in the F_2. From F_2, the number of double recombinants = 288 7 312 8 or 600. Percentage of = [(number of double cross-over)/(total number of progeny) × 100 = [(600)/(49400)] × 100 = 1.2% The cause of the discrepancy between the map distances can be traced to the fact that when we calculate the RF between C and S and between S and W, we include the number of double recombinants into the number of recombinants each time. (Refer to previous calculations). However, when we determine the RF between C and W, we neglect the double cross-overs because they do not affect the parental types CW and cw. To compensate for the double cross-overs, we have to add to the map distance between C and W twice the perc.entage of double cross- overs. This is because each double cross-over is the result of two separate events, a cross between c and S and a cross between S and W. In essence, four more crosses have occured between c and w that had not been considered before, having a total frequency of: (288)/(49400) (between c and s)(288)/(49400)(between s and w) (312)/(+49400) (between c and s)(312)/(+49400) (between c and w) or 2 × 1. 2. Thus the distance between c and w is 26.6 + (2 × 1.2) or 29.0 units. After the correction is made, everything falls in place. Note that the recombination frequency between each pair of genes is increased by the frequency of the double cross-overs.

Question:

Convert the following sequential machine which uses T type flip-flops into one using S-C type flip-flops only.

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Solution:

Step I. First, write out all the outputs and inputs of all the gates and flip flops in terms of x and the flip-flop outputs y_1, y'_1, y_2, y'_2 as follows: Hence, we find that T_1 = [(xy_1)' \textbullet y'_2]' (from the figure). \therefore T_1 = (xy_1) + y_2 (using De Morgan's laws). Also, T_2 = (xy_2)' = x' + y'_2 (by De Morgan's laws) and,z = (y'_1 y_2)' = y_1 + y'_2 (by DeMorgan's laws). Step II, Now, we plot each equation T_1, and T_2 into a Karnaugh Map for y_1y_2 and x, as follows: Now, in order to convert a T flip-flop into an S-C flip-flop, we make the following observations: i) The excitation equations of the T and S-C type flip-flops, for y_1\rightarroware: \rightarrow y_1\rightarrow= T_1 \textbullet y'_1 + T'_1 \bullet y_1 y_1\rightarrow= T_1 \textbullet y'_1 + T'_1 \bullet y_1 \rightarrow and, and, y_1\rightarrow= S_1 \textbullet y'_1 + C' \textbullet y y_1\rightarrow= S_1 \textbullet y'_1 + C' \textbullet y \rightarrow 1 ii) On putting y_1 = 0 (and hence y_1' = 1), in both the above equations, we get: y_1\rightarrow= T_1 \bullet 1 + T'_1 \bullet 0 = T_1 \rightarrow and, y_1\rightarrow= S_1 \bullet 1 + C'_1 \bullet 0 = S_1. \rightarrow Hence we see that in our Karnaugh Map for T_1, the first two rows (viz., for y_1y_2 = 00 and 01, i.e., for which y_1 = 0), correspond to T_1 or S_1. iii) Similarly, on putting y_1 = 1 (and hence y'_1 = 0) in the above equations, we get: y_1\rightarrow= T_1 \bullet 0 + T'_1 \bullet 1 = T'_1, \rightarrow and, y_1\rightarrow= S_1 \bullet 0 + C'_1 \bullet 1 = C'_1. \rightarrow \therefore (y_1\rightarrow)' = T_1 , \rightarrow and, (y_1\rightarrow)' = C_1 (complementing both sides in the above equation). \rightarrow Hence, we see that the last two rows of our Karnaugh Map for T_1, viz., y_1y_2 = 11 and 10 are complements of y_1\rightarrow, and correspond to the \rightarrow remainder of T_1 in a T flip-flop, or to C_1 in an S-C flip-flop. Step III. In view of the observations i), ii), and iii), we can now split up T_1 into its components for S_1 and C_1. The fig. (3a) for T_1 leads to fig. (3b) for y_1\rightarrowby complementing the entries \rightarrow For y_1y_2 = 11 and 10 from fig. (3a), as explained in observation iii) before. Fig. (3b) is now split up into two parts corresponding to y_1 = 0 and y_1 = 1 in fig. (3c) and fig. (3d), respectively. Note that fig. (3d) uses the complements of the entries in fig. (3b). We could have split up fig. (3a) directly into the figs. (3c) and (3d) without worrying about the fig. (3b) for y_1\rightarrow, and the complements. But this \rightarrow intervening step was used in the example only to illustrate the reasoning in-volved. Step IV. Now complete the partial Karnaugh Maps of figs. (3c) and (3d) by making use of any special properties of the type of flip-flop used. For example, for changing over to J-K flip-flops, all the remaining entries would be don't care conditions, "d". But, an SC type flip-flop, must satisfy the condition S\bulletC = 0. Hence, fill up the empty blanks such that the product of corresponding blanks in the Maps for S_1 and C_1 = 0. This is done as follows, in fig. (4): Where a zero already exists in one of the partial Karnaugh Maps, a 'd' can be entered in the corresponding square in the other Karnaugh Map. This is because the product of S_1 = 0 and C_1 = 0 or 1 will always be 0. For example, note that a 0 already exists in the Karnaugh Map for S_1 in the square for y_1y_2 x equal to 00 0. Hence, a 'd' can be entered in the corresponding square for y_1y_2 x equal to 00 0 in the Map for C_1, because S_1 \textbullet C_1 will equal 0. Similarly, a 'd' is entered in the square for y_1y_2 x equal to 00 1 in the Map for C_1 because there is a 0 in the square for y_1y_2 x equal to 00 1 in the map for S_1. But, the squares for y_1y_2 x equal to 01 0 and 01 1 are entered with zeroes in the Map for because there are 1's in the corresponding squares in the Map for S_1. Similarly, fill up the lower half of the Map for S_1 with d's or 0's according to whether there are 0's or 1's in the corresponding squares in the Map for C_1. Step IV. We now write out equations for S_1 and C_1 S_1 = y'_1y_2(from fig. (4a).) C_1 = y'_2 + y_1 x . (from fig. (4b)) We repeat the Steps III to V for T_2 to get S_2 and C_2. \therefore S_2 = y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5d).) (from fig. (5d).) and C2= y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5e).) C2= y'_1x' + y'_1y'_2 = y'_1 \bullet (x' + y'_2)(from fig. (5e).) (from fig. (5e).) Step VI. Step VI. We now draw a circuit diagram with two SC type flip-flops, and draw necessary input gates to satisfy the inputs S_1, C_1, S_2, C_2 of the flip flops obtained above.

Question:

A piece of iron of mass, M = 20g is placed in liquid air until thermal equilibrium is achieved. When it is quickly taken out and placed in water at 0\textdegreeC, a coating of ice of mass, M = 5.22g forms on it. The mean specific heat,c_iof iron over the range of temperature of the experiment is 0.095 cal/g -\textdegreeC and the heat of fusion L, of water is 80 cal/g. What is the temperature of the liquid air?

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Solution:

The iron is initially at the same temperature T as the liquid air. When placed in water, it takes heat from the water until its temperature reaches 0\textdegreeC. The amount of heat Q lost by the water can be found from the amount of ice formed as a result of this transfer of heat. Q = ML = (5.22gr)(80cal/gr) = 417.6 cal. The heat acquired by the iron as its temperature is raised by 0\textdegreeC - T\textdegreeC = - T\textdegreeC, must be equal to the heat Q lost by the water in turning to ice. Then, if M is the mass of the iron, Q = -C_iMT 417.6 cal = -(0.095 cal/gr-0C) (20gr) T giving T = -[417.6 / (20 × 0.095)] 0C = -220\textdegreeC

Question:

What is meant by the term tonus, or tone?

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Solution:

The term tonus refers to the state of sustained partial contraction present in skeletal muscles as long as the nerves to the muscle are intact. Unlike skeletal muscle, cardiac and smooth muscle exhibit tonus even after their nerves are cut. Tonus is a mild state of tetanus. It is present at all times and involves only a small fraction of the fibers of a muscle at any one time. It is believed that the individual fibers contract in turn, working in relays, so that each fiber has a chance to recover completely while other fibers are contracting before it is called upon to contract again. A muscle under slight tension can react more rapidly and contract more strongly than one that is completely relaxed, because of changes in the elastic component in the latter.

Question:

Compare the types of reactions in which the energy of ATP is made available for biosynthetic processes.

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Solution:

ATP has two high energy phosphate bonds and can transfer energy to another molecule in a variety of ways. The most common way is the transfer of the terminal phosphate group to the energy-requiring molecule while liberating ADP (adenosine diphosphate). This is known as a phosphorylation reaction, and is usually catalyzed by enzymes called kinases. The hydrolysis of ATP, in which the terminal phosphate group is transferred to water, releases energy (∆G0^' = -7 kilocalories/mole). By coupling the energy-yielding hydrolysis of ATP to the reaction of a phosphate group with a given compound, one can "capture" part of the energy of hydrolysis in the phosphorylated compound, as seen below: The phosphorylated compound is now "activated," that is, it is of a higher energy content than before, and as such, is able to undergo reactions that were previously unfeasible. A second type of reaction involves the transfer of AMP with the liberation of the last two phosphate groups as pyrophosphate (PPi). The energy released when ATP is split in this manner (-10 kcal/mole) is substantially greater than when a single phosphate group is cleaved off. A compound activated by AMP transfer can thus be of a significantly higher energy than its unactivated form. An example of this type of activation is the activation of an amino acid. Here, AMP is transferred to the COO- group of the amino acid. The activated amino acid is then transferred to the terminus of a transfer RNA molecule for use in protein synthesis, forming a new high energy bond to the tRNA and releasing AMP. Usually, the phyrophosphate group initially generated is quickly cleaved to form two orthophosphate groups: PPi + H2O \textemdash\textemdash> Pi + Pi∆G0^' = - 4.6 kcal/mole Note that the reaction releases a good deal of energy. Although this may Note that the reaction releases a good deal of energy. Although this may appear wasteful, the reaction is highly useful in that it provides an exergonic "push" to the initial reaction, assuring that it will indeed occur to completeness. A third type of reaction utilizing the energy of ATP involves the transfer of the last two pyrophosphate groups and the liberation of AMP. Again, this results in a highly activated compound (R\simP\simP) . The \sim \sim intermediate compounds in the synthesis of cholesterol and its derivatives are activated in this manner. The final major type of reaction involves the transfer of the adenosine group with the liberation of pyrophosphate and inorganic phosphate. This reaction is important in the activation of the amino acid methionine so that its methyl (CH3) group can be transferred to other compounds.

Question:

Write a program to read input variables length, width, and height and compute the area and the volume using those variables. At the end of the program print values found.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0261.htm

Solution:

To write this program we need four independent data items: INPUT-VARIABLE (which is nine numeric charac-ters long), VOLUME (which is nine numeric characters long), AREA (which is six numeric characters long), and PERIMETER (which is four numeric characters long). We also need one group item subdivided into three elementary data items: DIMENSIONS subdivided into LENGTH, WIDTH, and HEIGHT. Each of these elementary data items is three numeric characters long, with the decimal point at the rightmost end. Note the character types used in COBOL language are shown in figure 1. Data Type Picture Char. Allowable Characters Alphanumeric Alphabetic Numeric Numeric X A 9 V Letters, digits, spaces, special characters Letters and spaces only Digits 0-9 only Implied decimal point Arithmetic Operations: The formats for arithmetic operations are very straight-forward, and the four standard ADD, SUBTRACT, MULTIPLY, and DIVIDE statements are very similar. Either numeric data-names or numeric literals can be specified in these four statements. ADD statement format: SUBTRACT statement format: MULTIPLY statement format: DIVIDE statement format: COMPUTESTATEMENT: This statement has several uses. It can be used in much the same way as .the other arithmetic statements we have seen so far. Compute statement format Relational Operator Symbol GREATER THAN NOT GREATER THAN LESS THAN NOT LESS THAN EQUAL TO NOT EQUAL TO > NOT> < NOT< = NOT= Fig. 2 format for GO TO statement: GO TO statement specifies a paragraph name to which control is transferred, format for IF statement: the IF statement allows execution of statement - 1 if the condition is true. Relation Condition Operators are shown in fig. 2 Data values can be moved around in the Procedure section. One way to do this is to use the MOVE statement. This statement causes the data to be copied from one data location to another. format for MOVE statement: Using the data, information, and the statements given we write the program as shown in fig. 3.

Question:

A 5000-cm^3 container holds 4.90 gm of a gas when the pressure is 75.0 cm Hg and the temperature is 50\textdegree C. What will be the pressure if 6.00 gm of this gas is confined in a 2000-cm^3 container at 0\textdegree C?

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Solution:

From the ideal gas law, (P_1V_1)/(m_1T_1) = (P_2V_2)/(m_2T_2) Note we can use masses instead of number of moles since they are Note we can use masses instead of number of moles since they are proportional. (P_1 × 2000 cm^3)/(6.00 gm × 273\textdegree K) = (75.0 cm Hg × 5000 cm^3)/(4.90 gm × 323\textdegree K) P_1 = 194 cm Hg

Question:

Solve 6x^2 - 7x - 20 = 0.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0973.htm

Solution:

6x^2 - 7x - 20 = 0 is not factorable. Therefore, find the roots of the quadratic equation ax^2 + bx + c using: x = [{\rule{1em}{1pt}b \pm \surd(b^2 - 4ac)} / {2a}], where a = 6, b = \rule{1em}{1pt}7, c = \rule{1em}{1pt}20. x = [{7 \pm \surd(49 - 4(6) (\rule{1em}{1pt}20)} / {12}], x = [{7 \pm \surd(49 - 4(6) (\rule{1em}{1pt}20)} / {12}], x = [{7 \pm \surd529} / {12}] = [{7 \pm 23} / {12}]. Therefore,x_1 = (7 + 23) / 12 = 30/12 = 5/2 x_2 = (7 - 23) / 12 = \rule{1em}{1pt} (16/12) = \rule{1em}{1pt}(4/3).

Question:

Calculate the potential energy of the charge distribution shown in the diagram.

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Solution:

The potential energy of any two charges q_i and qjis U_ij = K_E [(q_iq_j) / R_ij ] where Rijis the distancebetween the two charges, ad K_E = 9 × 10^9 (N \bullet m^2 / c^2). The potential energy of the charge distribution is the sum of the poten-tial energies of every possible pair of charges within the distribution. Hence, U = U_12 + U_13 + U_23 = K_E[{(q)(2q) / (L)}+ {(\rule{1em}{1pt}q)(\rule{1em}{1pt}q) / (d)}+ {(+2q)(\rule{1em}{1pt}q) / (L)}] Since d^2 =L^2 + L^2 we have d = \surd2L and U = K_E[{(\rule{1em}{1pt}2q^2)/(L)} + {(q^2)/(\surd2L)} \rule{1em}{1pt} {(2q^2)/(L)}] = \rule{1em}{1pt}3.29 K_E(q^2 / L) joules If q = e and L = 1A\textdegree = 10^\rule{1em}{1pt}8 m, then U = \rule{1em}{1pt}3.29 × 9 × 109(nt \rule{1em}{1pt} m^2 / coul^2) × (1.6 × 10^\rule{1em}{1pt}19 coul)^2 / 10\rule{1em}{1pt}8 m = -7.58 × 10^-20 J The negative sign Indicates that work would be required to disassemble the charge distribution (i.e., work oust be done against attractive electrical forces).

Question:

For the reaction 2NO(g) + H_2 (g) \rightarrow N_2O(g) + H_2O(g) at 1100\textdegreeK, data, as shown in the following table, were obtained. Find the rate law and the numerical value of the specific rate constant. Initial Pressure of NO,atm Initial Pressure of H_2,atm Initial rate of pressure decrease,atm/min 0.150 0.400 0.020 0.075 0.400 0.005 0.150 0.200 0.010

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0444.htm

Solution:

The rate of a homogeneous reaction, i.e. one which occurs in only one phase, depends on the concentration of reactants in that phase. This reaction is a homogeneous reaction in the gaseous phase. It can be investigated kinetically by following the change in pressure of the gaseous mixture as the reaction proceeds. The pressure drops because 3 moles of gaseous reactants are converted to two moles of gaseous products. Since the reactants are being used up during the course of the reaction, their concentrations and their rate if reaction are constantly changing. The concentrations and rates listed are those at the very beginning of the reaction, when little change has occurred. From the data given, one can see that when the initial pressure of NO is halved with the pressure of H2 remaining constant, the initial rate is quartered . When the pressure of H_2 is halved and the NO remains constant, the initial rate is halved . When the pressure of H_2 is doubled and NO remains constant, the rate doubles. We conclude that the rate is proportional to [H_2]. When the H_2 pressure is kept constant and the NO concentration is doubled, the rate is quadrupled, this means that the rate is proportional to [NO]^2. The equation for the rate can thus be written Rate = k [H_2][NO]2 where k is the rate constant. Using the values in the first trial of the table one can solve for k. 0.020atm/min = k (0.400atm) (0.150atm) ^2 (0.020atm/min) / {(0.400atm)(0.150atm)^2} = k k = 2.22 atm^-2 min^-1.

Question:

Suppose you learned that "shmoos" may have long, oval or round bodies and that matings of shmoos resulted in the following: a) long × oval gave 52 long :48 oval; b) long × round gave 99 oval;and c) oval × oval gave 24 long :53 oval : 27 round. What hypothesis about the inheritance of shmoo shape would be consistent with these results? Assume that shmoos are diploid. (Note that the shmoo is a hypotheti-cal genetic organism.)

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Solution:

In this problem we have to decide what type of inheritance is used to determine a shmoo's body shape. From the data given in (b), we observe that the off-spring of that cross possess an intermediate characteristic rather than a parental one; that is, oval instead of long or round. (Note that an oval, geometrically, is somewhere in between round and long.) This would immediately suggest the concept of codominance. Codominance is defined as the expression in the heterozygote of neither the dominant nor recessive phenotype, but something intermediate between the two. Let us test our hypothesis. Let A be the gene for long body in shmoos, and B be the gene for round body in shmoos. According to our hypothesis of codominance, AB would then be the genotype of an oval- bodied shmoo. Outlining the given crosses in the usual way: The ratio of long to oval offspring is 1:1, which agrees with the data in (a). All the offspring in F_1 are oval-shaped; this is in agreement with the data in (b). the ratio predicted in F_1 tallies with the data in (c). Thus, body shape in shmoos is determined by means of the genes for long body and short body, and these two genes are codfaninant alleles.

Question:

Trace the path of an anesthetic, such as ether, from the ether cone over the nose to the cells in the brain.

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Solution:

The ether enters the body through the external nares, or nostrils, and would then enter the nasal cavities. The sense organs for olfaction are located here, and the individual would at this point smell the ether. The air containing the ether would be warmed, moistened and fil-tered here before passing via the internal nares into the pharynx. The pharynx is where the paths of the digestive and respiratory systems cross. From there the ether would pass through the glottis into the larynx, the trachea, the bronchi, the bronchioles and ultimately would reach the alveoli. At the alveoli, the ether will diffuse into the blood capillaries surrounding the alveoli, due to the con-centration gradient of ether that exists across the alveolar membrane. Diffusion is possible because of the extreme thinness and moistness of the membranes separating the pulmonary air from the blood in the capillaries. Electron micrographs have shown that these membranes, the alveolar epithelium and the endothelium of the capillary, are each one-cell thick. Once within the capillary, the ether will follow the circulatory routes to the heart. It is here that the blood containing the ether will be pumped to the brain, where the action of the gas will take effect.

Question:

The muscle fibers innervated by a motor unit will contract if a stimulus of sufficient magnitude is propagated or it will not contract at all. This is known as the all-or- none law. If this is true, how can muscle force be graded? For example, we use the same muscles to lift a one ounce object that we use to lift a 20 lb. object.

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Solution:

The total tension that a muscle can develop depends upon two factors: 1) the number of muscle fibers in the muscle bundle that are contracting at any given time and 2) the amount of tension developed by each contracting fiber. The number of fibers in a muscle that are contractiong at any given time depends upon the number of motor neurons to the muscle that are being stimulated. Recall that each motor neuron innervates several muscle fibers, forming a motor unit. The number of motor units that are activated is determined by the activity of the synaptic inputs to the motor neurons in the brain and spinal cord. With proper stimulation by the brain and spinal cord, more motor units may be activated at any one time, thus increasing the number of contracting fibers, and therefore, the strength of the muscle bundle. The process of increasing the number of active motor neurons and thus the number of active motor units is known as recruitment. The number of muscle fibers associated with a single motor neuron varies considerably in different types of muscle. In muscles of the hand, for example, which are able to produce very delicate movements, the size of the individual motor units is small. In the muscles of the back and legs, each motor unit contains hundreds of muscle fibers. The smaller the size of the motor units, the more precisely the tension of the muscle can be controlled by the recruitment of additional motor units. In addition to the variability of the number of active motor units, the tension produced by individual fibers can be varied. This can be accomplished by increasing the frequency of action potentials, resulting in stronger contractions, such as in summation or tetanus.

Question:

A chemist possessesKCl, PH_3, GeCl_4, H_2S, andCsF. Which of these compounds do not contain a covalent bond?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0620.htm

Solution:

A covalent bond is defined as one in which electrons are shared. The stability of covalent bonds in molecules depends on the difference in electronegativity values of the two atoms which make up the molecule. Electronegativity refers to the tendency of an atom to attract shared electrons in a chemical bond. If theelectronegativitydifference of two elements is greater than 1.7, an ionic bond is formed; if it is less than 1.7, a covalent bond is formed. To solve this problem, consult a table ofelectro-negativityvalues, and compute theelectronegativitydifference of the atoms in each of the given compounds. Proceed as follows: Electronegativity value Difference KCl K = 0.8Cl= 3.0 2.2 PH_3 P = 2.1P = 2.1 0 GeCl_4 Ge = 1.8Cl=3.0 1.2 H_2S H = 2.1S = 2.5 0.4 CsF Cs = 0.7F = 4.0 3.3 Thus, onlyKClandCsFexceed 1.7. They possess ionic bonds. The remainder of the molecules possess covalent bonds.

Question:

A horizontal capillary tube closed at one end contains a column of air imprisoned by means of a small volume of water. At 7\textdegreeC and a barometric pressure of 76.0 cm of mercury, the length of the air column is 15.0 cm. What is the length at 17\textdegreeC if the saturation pressures of water vapor at 7\textdegreeC and 17\textdegreeC are 0.75 cm and 1.42 cm of mercury, respectively?

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Solution:

Since the tube is horizontal and the pressure at the open end of the water column is always atmospheric, the pressure at the closed end of the water column is also always atmospheric. The pressure in the moist air is made up of the partial pressures of air and of water vapor. When the equilibrium between the liquid and gas phases of a liquid is reached in a closed volume, the pressure of the vapor acting on the liquid equals the saturated vapor pressure. The eva-poration process effectively stops once this pressure is attained by the vapor. The air and water vapor act on the liquid surface independently, therefore the pressure on the inner surface of the water is the sum of the pressures p and p_v, due to the air and vapor in the tube, respectively (as shown in the figure). p_a = P + p_v. Hence, for the two cases, the air pressure inside is p = (76.00 - 0.75) cm Hg at 7\textdegreeC, andp = (76.00 - 1.42)cm Hg at 17\textdegreeC. Applying the gas law to the air alone, since the air and water vapor exert effects independent of one another, pV/T = p'V pV/T = p'V '/T where p, V, T are the pressure, volume and temperature (in \textdegreeK) of the air. or(75.25 cm × 15 cm × A)/(280 \textdegreeK) = (74.58 cm × yA)/(290\textdegreeK), where A is the cross-sectional area of the tube and y is the length of the column at the temperature of 17 \textdegreeC. Hence y = [(290 × 75.25 × 15)/(280 × 74.58)] cm = 15.68 cm.

Question:

What is implied by the theory ofuniformitarianism?

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Solution:

The concept ofuniformitariansism, first developed by James Hutton in1785, states that the geologic forces at work in the past were the same asthose of the present. After carefully studying the erosion of valleys by riversand the formation of sedimentary deposits at the river mouths, Hutton concluded that the processes of erosion, sedimentation, disruption andup-lift, over long periods of time, could account for the formation of the fossil-bearingrock strata which had interested many scientists. Since the geologic factors operating on earth remain uniform throughoutthe ages, the rock strata bearing fossils of varying life forms reflectthe gradual changes in the structures and living habits of organisms throughtime. The theory ofuniformitarianismtherefore implies that animalsand plants continually undergo a process of organic evolution; theirremains being fossilized in the earth's crust in layers by gradual depositionof sediment.Uniformitarianismalso implies that the earthcan notbe only a few thousand years in age, as was widely accepted to be truetwo centuries ago. It must be much older, at least old enough for the processof organic evolution and the geological evolution of the earth's crust, both very slow, and gradual processes, to have occurred.

Question:

What is the structure of a sponge? How do sponges obtain food and water?

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Solution:

The most primitive metazoans belong to the phylum Porifera and are commonly called the sponges. Sponges may be radially symmetrical, but more commonly they are asymmetrical animals, consisting of loose aggregations of cells which are poorly arranged into tissues. The surface of a sponge is perforated by small openings, or in current pores, which open into the interior cavity, the atrium or spongocoel. The atrium opens to the outside by a large opening at the top of the tube--shaped sponge, called the osculum. There is a constant stream of water passing through the incurrent pores into the atrium, and out through the osculum. The body wall is relatively simple. The outer surface is covered by flattened polygonal cells called pinacocytes. The pores are guarded by cells called porocytes, which are modified pinacocytes, and are shaped like tubes extending from the outside of the sponge to the spongocoel. The pore, or ostium, can be regulated by contractions in the outer end of the porocyte. The mesohyl, a layer consisting of a gelatinous protein matrix, contains skeletal material and amoeboid cells, and lies directly beneath the pinacoderm. The skeletal material may be calcareous or siliceous spicules, or protein spongin fibers, or a combination of the last two. The composition, size and shape of these spicules form the basis for the classification of species of sponges. Though the spicules are located in the mesohyl, they frequently project through the pinacoderm. The skeleton, whether composed of spicules or spongin fibers, is secreted by amoebocytes called sclerocytes. Amoedboid cells in the mesohyl include the archaeocytes. These calls are capable of forming other types of cells that are needed by the sponge. The inner side of the mesohyl, lining the atrium, is a layer of cells called choanocytes, or collar cells. The characteristic shape of a choanocyte is ovoid, with a flagellum surrounded by a collar. The choanocytes are responsible for the movement of water through the sppnge and for obtaining food. The current is produced by the beating of the flagella of the choanocytes. The flagella are oriented toward the spongocoel, and beat in a spiral manner from their base to their tip. As a result of this, water is sucked into the spongocoel through the incurrent pores. Sponges feed on very fine particulate matter. Only particles smaller than a certain size can pass through the pores. Food particles are brought inside the spongocoel by water currents. Larger food particles are phagocytized by amoebocytes lining the inhalant chamber. Particles of bacterial size or below are probably removed and engulfed by the choanocytes. It is thought that the amoebocytes act as storage centers for food reserves. Egested wastes leave the body in water currents. There are three basic structural types of sponges. The simplest, called the asconoid type, has been described earlier when basic sponge structure was discussed. The syconoid and leuconoid types are more complex, having thicker mesohyls and a greater number of water channels, but their structures can be seen to be just complexes of asconoid sponges attached to one another.

Question:

Suppose we have a list consisting of people's names, sexes, eye colors, heights, weights, and telephone numbers. Sup pose in addition that we wish to look up the telephone num-bers of three particular people, namely Robert J. Coyle, Mary Holmes, and Victor E. Lee. Write a computer program in PL / I to accomplish this.

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Solution:

We use mnemonic notation for the alphanumeric data in the problem by using identifiers like NAME, SEX, and COLOR. We abbreviate phone numbers as NUMBER, and we desig-nate the height of the person by H and the weight by W. The program proceeds as follows. We start by declaring our four character strings and then transmitting the data to the program. Notice that in the GET LIST statement, we transmit to the program all 6 variables, even though we want only the name and phone number of that person. In the next three statements in the program, we make tests to determine if the name is Coyle, Holmes, or Lee. Note that the names must be put in quotation marks. If any of the tests yield TRUE, we go to PRINT where the desired message is printed. If any of the tests fail, we drop through to the next step. / \textasteriskcentered PHONE NUMBERS \textasteriskcentered / DECLARE NAME CHARACTER (15), SEX CHARACTER (7), COLOR CHARACTER (6), NUMBER CHARACTER (9); START: GET LIST (NAME, SEX, COLOR, H, W, NUMBER); IF NAME = 'ROBT J. COYLE' THEN GO TO PRINT; IF NAME = 'MARY HOLMES' THEN GO TO PRINT; IF NAME = 'VICTOR E. LEE' THEN GO TO PRINT; GO TO START; PRINT: PUT EDIT ('PHONE NO. OF', NAME, 'IS', NUMBER) (A (20), A (15), A (3), A (9)); PUT SKIP (2); GO TO START;

Question:

How does a worker bee communicate to other bees the distance and direction of food from the hive?

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Solution:

After one bee has discovered a food source, many bees soon appear. The first bee must somehow communicate to the others the location of the new food source. It has been determined that when a worker bee discovers food, it returns to the hive, feeds several other bees, and performs a dance on the surface of the honeycomb. If the food source is near the hive, the bee performs a "round" dance. This consists of a re-peating pattern of circling, first to the right and then to the left: The round dance excites the hive mates in the dancer's vicinity and they soon leave the hive in search of the food. This dance does not indicate direction or exact distance. It simply indicates that the food source is a short distance away. With increasing distance of the source from the hive, the round dance is gradually transformed into a "waggle" dance, which gives information about distance and direction. In the waggle dance, the bee repeatedly runs in a figure- eight pattern, wagging its abdomen from side to side as it traverses the "waist\textquotedblright of the figure eight: The number of turns per unit time is inversely proportional to the distance of the food source. The faster the bee runs (or turns), the closer the food source. For honey-bees, an estimate of distance is based on the amount of energy expended traveling the round trip. When conditions exist such as headwinds, uphill flights, clipped wings, or added weight to the bees, excessive distances will be reported. The waggle dance also indicates direction. The location of the food relative to the sun is indicated by the direction of the straight portion of the waggle dance. For instance, if the bee runs straight up the vertical comb, the food lies in the direction of the sun; if it runs down, the food is found in the opposite direction from the sun. A straight run at an angle indicates that the food source is to be found at that angle to the sun. Therefore a run 30\textdegree to the left of vertical indicates a food source 30\textdegree to the left of the sun. (See figure 3) If the bee should perform the waggle dance on a horizontal surface outside the hive, it simply dances in the direction of the food source relative to the sun. Waggle dance of bees: B and C show the same dance, performed inside and out-side the hive respectively, for locating the same food source in A.

Question:

In the design of a modern steel bridge, provisions must obviously be made for expansion. How much does this amount to in the case of a bridge two miles long which is subjected to temperatures ranging from - 40\textdegreeF to + 110\textdegreeF, assuming an average expansion coefficient of .000012/\textdegreeC?

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Solution:

By definition of the coefficient of linear expansion \alpha =\Deltal/ (l_0 \DeltaT) where\Deltal/l_0 is the fractional change in length of an object due to a temperature change \DeltaT. In our case \Deltal = \alpha l_0 \DeltaT \Deltal = (.000012 \textdegreeC^-1)(2 miles)[110\textdegreeF - (- 40\textdegreeF)] \Deltal = (.000012 \textdegreeC^-1)(2 miles) (150\textdegreeF) Since 150\textdegreeF = (5/9) \textbullet 150\textdegreeC = (750\textdegreeC/9) \Deltal = (1.2 × 10 ^-5) (2 miles) (750/9) \Deltal = .002 miles

Question:

Array A contains all odd integers from 1 to 100 (e.g. 1,3,5,7,.......99) in random order, and is declared as follows : TYPEARRAYTYPE = ARRAY [1..50] of integer; VARA :ARRAYTYPE; Write a short procedure that will accept A and return an array with the same elements in sorted, increasing order.

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Solution:

The key here is to recognize the relationship be-tween the indices and the elements of the array. Thus, given an element e, we can compute its correct place in the array - let's call it X, from the following formula: X: = ediv 2 + 1; This is, then, the procedure: PROCEDURE SORT (A: ARRAYTYPE; VAR B:ARRAYTYPE ); VARi, X:integer ; BEGIN FORi: =1 to 50 DO BEGIN X: = A [i] div 2 + 1; B[x]: = A[i] END END; Let's simulate this procedure for an Array of 5 elements con-taining odd integers from 1 to 10 in random order: A 7 9 3 5 1 Wheni= 1,x= A[1] div 2 + 1 = 7 div 2 + 1 = 4. B[4]: = A [1] = 7 The array B now looks: 7 at step 2 ,i= 2, x = A[2] div 2 + 1 = 5 B[ 5]: = 9.Array B now looks: 7 9 At step 3,i= 3, x = 3 div 2 + 1 = 2 B[ 2] : = 3 3 7 9 Fori=4, x=5 div 2+1 = 3 B[ 3] : 5 3 5 7 9 Finally , fori= 5, x = 1 div 2 + 1 = 1 B[ 1] : = 1.Now array B is in sorted increasing order: 1 3 5 7 9 The procedure works exactly like this for larger numbers.

Question:

The pilot of an airplane flying on a straight course knows from his instruments that his airspeed is 300 mph. He also knows that a 60-mph gale is blowing at an angle of 60\textdegree to his course. How can he calculate his velocity relative to the ground?

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Solution:

Relative to an observer on the ground, the airplane has two velocities, one of 300 mph relative to the air and the other of 60 mph at an angle of 60\textdegree to the course, due to the fact that it is carried along by the moving air mass. To obtain the resultant velocity, it is therefore necessary to add the two components by vector addition. In the diagram, A\ding{217} represents the, velocity of the air-craft relative to the air, and B\ding{217} the velocity of the air relative to the ground. When they are added in the normal manner of vector addition, C\ding{217} is their resultant. The magnitude of C\ding{217} is given by the trigonometric formula known as the law of cosines (see figure). C^2 = A^2 + B^2 - 2AB cos \texttheta. But A = 300 mph, B = 60 mph, and \texttheta = (180\textdegree - 60\textdegree) = 120\textdegree. Therefore C^2 = (300 mph)^2 + (60 mph)^2 - 2 × 300 mph × 60 mph (- 1/2) = 111,600 (mph)^2; \thereforeC= 334 mph. Also, from the addition formula for vectors, we have sin \alpha = (B/C) sin \texttheta = (60 mph)/(334 mph) × (\surd3/2) = 0.156. \therefore\alpha = 9\textdegree.

Question:

(a) ^3\surd\rule{1em}{1pt}511(b) ^4\surd(81/16)(c) ^3\surd\rule{1em}{1pt}16 \div ^3\surd\rule{1em}{1pt}2.

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Solution:

(a) By the law of radicals which states that ^n\surdab = ^n\surda ^n\surdb where a and b are any two numbers, ^3\surd\rule{1em}{1pt}512 = ^3\surd8(\rule{1em}{1pt}64) = ^3\surd8 ^3\surd\rule{1em}{1pt}64. Therefore, ^3\surd\rule{1em}{1pt}512 = ^3\surd8 ^3\surd\rule{1em}{1pt}64 = (2) (\rule{1em}{1pt}4) = \rule{1em}{1pt}8. The last result is true because (2)^3 = 8 and (\rule{1em}{1pt}4) = \rule{1em}{1pt}64. (b) By another law of radicals which states that ^n\surd(a/b) = (^n\surda) / (^n\surdb) where a and b are any two numbers, ^4\surd(81/16) = (^4\surd81) / (^4\surd16). Therefore, ^4\surd(81/16) = (^4\surd81) / (^4\surd16) = (3/2). The last result is true because (3)^4 = 81 and (2)^4 = 16. (c) By the law of radicals stated in example (b) ^3\surd\rule{1em}{1pt}16 \div ^3\surd\rule{1em}{1pt}2 = (^3\surd\rule{1em}{1pt}16) / (^3\surd\rule{1em}{1pt}2) = ^3[\surd(\rule{1em}{1pt}16) / (\rule{1em}{1pt}2)] = ^3\surd8 = 2. The last result is true because (2)^3 = 8.

Question:

Silver, atomic weight 107.9 and density 10.5 gm/cm^3, has one free electron per atom. Calculate the Fermi energy of the electrons.

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Solution:

The Fermi energy is given by \epsilon_F = (h^2/8m) [3N/\piV]^2/3(1) where V is the molar volume of silver, and N is the number of free electrons in 1 mole of silver. Sincedensity = (mass/volume) density= [{molar mass (M)}/{molar volume (V)}] Hence, V = (M/density) But M is the mass of an element in atomic mass units. Therefore V = [(107.9 amu)/(10.5 g/cm^3) = [(107.9 g/mole)/(10.5 g/cm^3)] = 10.26 cm^3/mole = 1.026 × 10^-5 m^3/mole. In order to find N, note that each silver atom con-tributes one electron, (that is, its free or valence electron). Since there are Avogadro's number, (6.02 × 10^23) of particles in a mole of a substance, the mole of silver will contain 6.02 × 10^23 free electrons. These electrons are distributed into energy states accord-ing to Fermi-Diracstatistics (in other words, each energy state can contain only one particle). The energy levels are filled by placing the electrons first into the lowest energy levels, and then progressively filling the higher levels. The highest energy level into which the last electron is placed is called the Fermi energy. Substitut-ing the values of N and V into (1) and using the given data, \epsilonF= [{(6.63 × 10^-34 J\textbullets)^2} / {(8)(9.11 × 10^-31kg)] [(3 × 6.03 × 10^23elec/mole)/ (\pi×1.026×10^-5 m^3/mole)]^2/3 = 8.85 × 10^-19 Joules Since1 eV = 1.6 × 10^-19 J \epsilon_F = [(8.85 × 10^-19 J)/(1.6 × 10^-19 J/eV)] = 5.54 eV We could interpret this as a maximum kinetic energy of the electrons. Then the maximum velocity would be (1/2) mv^2 = 8.85 × 10^-19 J v^2 = [(2 × 8.85 × 10^-19 J)/(9.11 × 10^-31 kg)] = 1.94 × 10^12 m^2/sec^2 v = 1.4 × 10^6 m/sec = 0.005 c where c = speed of light.

Question:

(a) 2-hexanone; (b) 2-methylbutanal ; (c) O-methylbenzaldehyde; (d) methyl phenyl ketone.

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Solution:

The aldehydes are derivatives of hydrocarbons whose molecules have a double bond to oxygen in place of two hydrogens at the end of a chain. Those derived from alkane hydrocarbons have the general formula RCHO, where R is the hydrogen for the first member of the series and an alkyl group for higher homologs. The longest chain containing the -CHO group is considered the parent compound and is named by changing the -e ending of the corresponding alkane to -al. The ketones are derivatives of hydrocarbons whose molecules have a double bond to oxygen in place of two hydrogens at a position other than the end of the carbon chain. Those derived from alkane hydrocarbons have the general formula RR'C = 0, where R and R' are alkyl groups. The longest chain containing the carbonyl group is considered the parent compound and is named by changing the -e ending of the corresponding alkane to -one. Where it is necessary, the locations of the carbonyl carbon and attached groups are indicated by numbers. Thus, compound (a) is a ketone whose longest chain has 6 carbons to which oxygen is attached on carbon number 2. The formula is Compound (b) is an aldehyde whose longest chain has 4 carbons to which a methyl group is attached to carbon 2 and whose end carbon atom is -CHO. The formula is Compound (c) is an aldehyde derived from benzene by replacing a hydrogen with thegroup. This compound also has a methyl group attached to the ortho position on the benzaldehyde (C_6H_5CHO) ring. The structure can be written as shown in Figure A. Compound (d) is an aromatic ketone (any compound that contains benzene derivatives is called aromatic), derived from phenols. Phenols are hydroxy derivatives of aromatic compounds whose general formula is ArOH (Ar is an aromatic group. Phenyl is C_6H_5- and is attached to In this case. This formula can be written asor as shown in the figure B.

Question:

For a T type (toggle) clocked, master-slave flip-flop, give a) the block diagram,b) the timing diagram, c) the Kar-naugh Map,d) the excitation equation, e) the state table,andf) the state diagram.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G04-0080.htm

Solution:

a) The block diagram is shown in figure 1. The slave gets a clock pulse which is a complement of the master's clock pulse. The input to the slave is taken from the x-output of the master. (It could be taken from the x'-output, too.) b) The timing diagram is shown in Figure 2. Note that the transitions take place in x at the leading edge of the clock pulse, while transitions in Y take place at the trailing edge of the clock pulse. This is becauseof the Inverter which is incorporated inside the master of the Inverter which is incorporated inside the master slave flip-flop. c)The Karnaugh Map of the T flip-flop is shown in figure3. The Karnaugh Map of the T flip-flop is shown in figure d) The excitation equation of the T-flip-flop is obtained from the Karnaugh Map as follows: y\rightarrow= Ty' + T'y. \rightarrow e) The State Table, shown in figure 4, is obtained from the excitation equation: Ty y\rightarrow \rightarrow REMARKS 00 01 0 1 No Change 10 11 1 0 FLIPS f) The state diagram, drawn from the Karnaugh Map is shown in fig. 5.

Question:

Howarethe type of muscle tissues differentiated?

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0052.htm

Solution:

The cells of muscle tissue have great capacity for contraction. Muscles are able to perform work by the summed contractions of their individualcells. The individual muscle cells are usually elongate, cylindricalor spindle-shaped cells that are bound together into sheets or bundlesby connective tissue. Three principal types of muscle tissue are found in vertebrates. Skeletal or striated muscle is responsible for most voluntary movements. Smooth muscle is involved in most involuntary movements of internal organs, such as the stomach. Cardiac muscle is the tissue of which much ofthe heart wall is composed; it is also involuntary. Most skeletal muscle, as the name implies, is attached to the bones ofthe body, and its contraction is responsible for the movements of parts ofthe skeleton. Skeletal muscle contraction is also involved in other activitiesof the body, such as the voluntary release of urine and feces. Thus the movements produced by skeletal muscle are primarily involved withinteractions between the body and the external environment. Skeletal muscle reveals a striated appearance, and is therefore alsoreferred to as striated muscle. These striations are actually due to the regulararrangement of thick and thinmyofilamentsin individual muscle fibercells. Skeletal muscle is an exception to the common observation thateach cell contains only one nucleus: each skeletal fiber cell is multinucleated: it has many nuclei. Skeletal muscle can contractcery rapidlybut cannot remain contracted; the fibers must relax before the next contrac-tioncan occur. Smooth muscle can be classified as visceral smooth muscle. It is foundin the walls of hollow visceral or-gans, such as the uterus, urinary bladder, bronchioles and much of the gastrointestinal tract. Vascular smoothmuscle refers to that smooth muscle in the walls of blood vessels. Unlike skeletal muscle cells that are cylindrical, multinucleate, and striated, smooth muscle cells are spin-dle-shaped,uninucleate, and lack striations. This lack of striations accounts for its smooth appearance. They havea slower speed of contraction, but can remain contracted for a longer periodof time than the striated muscle cells. Cardiac muscle has properties similar to those of both skeletal and smoothmuscles. Like skeletal muscle cells, cardiac muscle cells are striated. Like smooth muscle fibers, cardiac muscle fibers are uninucleatedand are designed for endured contraction rather than speedy orstrong contraction. Cardiac and smooth muscles are not voluntarily controlledbut have spontaneous activities, and are regulated by the autonomicnervous system. The skeletal muscle, being voluntary, is controlledby the somatic nervous system. Comparision of the types of muscle tissue Skeletal Smooth Cardiac Location Attached to skeleton Walls of visceral organs. Walls of blood vessels Wails of heart Shape of fiber Elongate, cylindrical, blunt ends. elongate , spindle shaped, pointed ends. Elongate, cylindrical, fibers branch and fuse. Number of nuclei per fiber Many One Many Position of nuclei Peripheral Central Central Cross striations Present Absent Present Speed of contractions Most rapid Slowest Intermediate Ability to remain contracted Least Greatest Intermediate Type of control Voluntary Involuntary Involuntary

Question:

A hoist raises a load of 330 pounds a distance of 15 feet in 5 seconds. At what rate in horsepower is the hoist working?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0293.htm

Solution:

Power is equivalent to work per unit of time. In 5 seconds, the hoist does W = Fs = 330 lbs × 15 ft = 4950 ft-lbs of work. The rate at which work is done (the power) is: (4950 ft-lb)/(5 sec) = 990 ft-lbs/sec Since there are 550 ft-lb/sec per horsepower, [(990 ft-lbs/sec)/(550 ft-lbs/sec-hp)] = 1.8 hp

Question:

Determine the value of x such that 10^x = 3.142.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0969.htm

Solution:

The statement 10^x = 3.142 is equivalent by de-finition to log_10 3.142 = x. Thus we must find log 3.142, using the following interpolation: We set up the proportion, (.002 / .01) = x / .0014 Cross multiply to obtain, .01x = .0000028 x = .00028 x \approx .0003 Thus log 3.142 = .4969 + .0003 = 0.4972 Therefore x = log_103.142 = 0.4972.

Question:

It is known that the sun radiates approximately 3.8 × 10^26 J of energy into space each second. Determine how much mass must be converted into energy each second, and how many years these thermonuclear reactions may continue at this rate if the sun's total mass is 2 × 10^30 kg .

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0936.htm

Solution:

The mass, m, converted to energy, E, in each second may be obtained from the relationship, E = mc^2 m = E / c2= (3.8 × 10^26 J) / {(3 × 108m/s)^2} = 4.2 × 109kg Since 1 year equals 3.1 × 107s, the mass lost in 1 year is, m = (4.2 × 10^9 kg/s) (3.1 x 10^7 s/year) . =1.3 × 10 kg/year The time t required to consume the sun's mass is, t = sun's mass / mass loss per year = (2 × 10^30 kg) / (1.3 × 10^17 kg/year) = 1.5 10^13 years

Question:

What volume (in cc) is occupied by a block of woodof dimensions25.0 m × 10.0 cm × 300 mm. All edges are 90\textdegree toone another.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E01-0008.htm

Solution:

Since all of the edges are 90\textdegree to one another, one knows that the blockis a rectangular solid. The volume of a rectangle is equal to the lengthtimes the width times the height. If one wishes to find the volume in cubiccentimeters, the lengths of all of the sides must be first expressed in centimeters. There are 100 cm in 1 m; thus, to convert meters to centimeters, thenumber of meters must be multiplied by 100 cm/1 m. 25.0m × 100 cm/1 m = 2500 cm. There are 10 mm in 1 cm; thus, to convert milli-meters to centimeters, multiply the number of milli-meters by 1 cm/10 mm. 300 mm × 1 cm/10 mm = 30 cm Solving for the volume: volume= 2500 cm × 10.0 cm × 30 cm = 7.50 × 10^5 cc.

Question:

The flywheel in a delivery truck is mounted with its axis vertical, and thus acts as a stabilizing gyroscope for the truck. Calculate the torque that would have to be applied to it when it is rotating at full speed to make it precess in a vertical plane.

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/Users/wenhuchen/Documents/Crawler/Physics/D07-0346.htm

Solution:

Figure (a) shows the situation. The flywheel is to precess about axis AA' as shown in figure (b). Originally, the angular momentum vector L^\ding{217} is as shown in figure (b), position 1. After a time ∆t, L^\ding{217} has the new value L(t + ∆t), and the angular momentum vector has gone through an angle ∆\varphi, as shown in the vector diagram. Note that we have neglected the fact that there is a component of L^\ding{217} along AA' due to the precession of the flywheel. This approximation is valid if the rate of precession, w_p , small. By the relation (dL^\ding{217}/dt) = \cyrchar\cyrt^\ding{217}(1) where \cyrchar\cyrt^\ding{217} is the net torque on the flywheel, we see that if we are to change L^\ding{217}, and thereby cause precession to occur, we must exert a torque \cyrchar\cyrt^\ding{217}. Now, the most efficient way for the torque to cause precession is if it acts in a direction perpendicular to L^\ding{217}, as shown in figure (b). As a result of (1), dL^\ding{217} is also perpendicular to L^\ding{217}, and the length of L^\ding{217} doesn't change. Hence ∆L/∆t \approx ɩ If ∆L^\ding{217} is small, and \cyrchar\cyrt^\ding{217} is always perpendicular to L^\ding{217}, \vert∆L^\ding{217}/L^\ding{217}\vert = ∆L/L = ∆\varphi or\tau \approx L(∆\varphi/∆t) = L w_p \tau \approx L w_p(2) butL = I w_s(3) where I is the moment of inertia of the flywheel about its symmetry axis and w_s is the spin angular velocity of the disc. Using (3) in (2) and w_s is the spin angular velocity of the disc. Using (3) in (2) \tau\approx I w_s w_p AlsoI = (1/2) Mr^2 where M and r are the mass and radius of the disc. Final-ly \tau \approx (1/2) Mr^2 w_s w_p

Question:

Calculate the percentage composition of aluminum sulfate, Al_2 (SO_4)_3.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0145.htm

Solution:

Because the formula of a compound is constant, the percentage composition of each element present can be calculated by using the parts by weight of each element in one molecular weight of the compound. The molecular weight of the compound can be calculated by adding to-gether the weights of the various elements contained in the compound. There are 2 atoms of aluminum present, and the molecular weight of aluminum is 27, so the contribution of the aluminum to the compound's total molecular weight is found by multiplying27 by 2. atoms of aluminum weigh 2 × 27 = 54 The weight contributions of the other elements present can be calculated in the same way. The molecular weight of sulfur is 32, and there are 3 sulfur atoms present in the compound. 3 atoms of sulfur weigh 3 × 32 = 96 The molecular weight of oxygen is 16, and there are 12 oxygen atoms present in the compound. 12 atoms of oxygen weigh 12 × 16 = 192 The weights of the three elements contained in the compound are added together to find the total molecular weight of the compound. 2 atoms of aluminum=54 3 atoms of sulfur=96 12 atoms of oxygen=192 molecular weight of A1_2(SO_4)_3 342 The percentage composition of each element present can now be calculated byconsidering what fraction by weight each element is of the total compound. The weight of two atoms of aluminum is 54 so the fraction by weight of aluminum in the compound can be found by dividing 54 by 342, the molecular weight of the compound. The percentage is then found by multiplying this fraction by 100. percentage composition of aluminum = [(weight of aluminum in compound) / (molecular weight compound)] × 100 percentage composition of aluminum = (54/342) × 100 = 15.8 %. The same method can be applied to the sulfur and oxygen. percentage composition of sulfur = (96/342) × 100= 28. 1% percentage composition of oxygen =(192/342) × 100 = 56.1% If the percent compositions of all the elements in a compound are addedtogether they will equal 100.0. percentage composition of aluminum= 15.8 percentage composition of sulfur= 28.1 percentage composition of oxygen=56.1 100.0

Question:

The period of a compound pendulum is 2 s on the earth's surface. What is its period if it is aboard a rocket accelerating upward with an acceleration of 4.3 m\bullets^-2?

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/Users/wenhuchen/Documents/Crawler/Physics/D09-0385.htm

Solution:

When on the surface the compound pendulum has a period T = 2\pi\surd(I_0/Mgh), where M is the mass of the pendulum, I is its moment of inertia about the axis of rotation, h is the distance between the center of mass and the axis of rotation and g is the accelera-tion of gravity, as shown in the figure. Under acceleration a^\ding{217} upward, the forces acting on the body when it is displaced through an angle \texttheta are those shown in the figure, the weight Mg^\ding{217} downward and the vertical force F^\ding{217} and horizontal force R^\ding{217} exerted by the pivot on the pendulum. Consider the linear accelerations, horizontal and vertical, acting on the body at its center of mass and the rotational acceleration about the center of mass. Applying Newton's Law to the horizontal and vertical motions, we get R = Ma_H(1) F - Mg = Ma_v(2) The equation of motion for the rotation is Net torque = Rh cos \texttheta - F h sin \texttheta = I_G \alpha(3) where I_G is the moment of inertia with respect to the center of mass, and \alpha is the angular acceleration. The point O thus has an acceleration, a_H horizontally, an acceleration a_v vertically, and a further linear acceleration a_v = h\alpha at right angles to OG, due to the rotation about G. But the point O does not move sideways, only upward with an accelera-tion a. Thus, there is no net horizontal acceleration a_h + h \alpha cos \texttheta = 0(4) Substituting (4) in (1), R = - Mh \alpha cos \texttheta The upward acceleration is a = a_v - h \alpha sin \texttheta ora_v = a + h \alpha sin \texttheta(5) Substituting (5) in (2) F = Mg + M (a + h \alpha sin \texttheta) = M (g + a + h \alpha sin \texttheta) and finally equation (3) becomes I_G \alpha= Rh cos \texttheta - Fh sin \texttheta = - Mh^2 \alpha cos^2 \texttheta - Mh^2 \alpha sin^2 \texttheta - M(g +a)h sin \texttheta = - Mh^2 \alpha (cos^2 \texttheta + sin^2 \texttheta)- M(g +a)h sin \texttheta = - Mh^2 \alpha - M(g +a)h sin \texttheta or(I_G + Mh^2) \alpha = I_0\alpha = - M(g +a)h sin \texttheta where we used the equation I_0 = I_G + Mh^2. But the angle \texttheta is small, and sin \texttheta can be replaced by \texttheta, \alpha = [(d^2\texttheta)/(dt^2)] = - [{M(g +a)h}/I_0-] sin \texttheta \approx - [{M(g +a)h}/I_0] \texttheta. This is the differential equation of a simple harmonic motion, therefore it follows from the theory of simple harmonic motion that the period of oscilla-tion is T' = 2\pi\surd[I_0/{- M(g +a)h}] If T is the period of the motion when the pendulum is at rest on the earth T'/T = [2\pi\surd{I_0/(M(g +a)h)}] /[2\pi\surd{I_0/(Mgh)}] = \surd[g/(g + a)] = \surd[(9.8 m\bullets^-2)/(9.8 m\bullets^-2 + 4.3 m\bullets^-2)] = 0.83 HenceT' = 0.83 T = 0.83 ×2 s = 1.67 s. Note that the result can be obtained more quickly if the idea of an accelerated frame of reference is applied. An observer in the rocket considers the point of support of the pendulum at rest relative to himself. To explain the observed equilibrium he finds it necess-ary to postulate a force Ma acting downward on the body in addition to the weight Mg^\ding{217}. Hence the pendulum actsas if the weight were M(g^\ding{217} + a^\ding{217}) instead of Mg^\ding{217}, from which the formula follows immediately. Mg^\ding{217}, from which the formula follows immediately.

Question:

20REM INTREST PROBLEM 30INPUT P, I, N 40LET A = P\textasteriskcentered (1 + I) \textasteriskcenteredN 50PRINT "P = "; P, "I = ";I, "N = "; N, "A = "; A 60END

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G25-0595.htm

Solution:

When a computer program is written, it usually contains two different types of errors. The first kind are grammatical (or syntactical) errors, the second kind are logical errors. Syntactical errors are caused by incorrectly written program statements. These errors are easily detected since a program containing such errors usually cannot be executed. In addition, the computer prints out a diagnostic message that identifies the location and nature of each error. When the program contains a logical error the computer correctly carries out the given instructions, but the wrong instructions produce incorrect results. Turning to the given problem, the FORTRAN program has the following errors: 1) FORMAT has been misspelled as FORNAT in statement 200. Since FORNAT is not part of the compiler's vocabulary, an error message is printed. The program cannot be executed because the nature of the input has not been specified (this is, of course, done by the FORMAT statement). 2) In FORTRAN, instructions must correspond exactly to a given prototype. Any deviation results in an error message. Here, the READ statement has a comma after it. In FORTRAN, READ is followed by a comma only in the unformatted input statement. Since the READ statement in the given program is formatted, there should be no comma after READ. 3) The statement A = (P\textasteriskcentered(l + I) \textasteriskcentered\textasteriskcenteredN is missing a right parenthesis. Some compilers (PL / I for example) may make the insertion, but FORTRAN is strict. An error message is sent out, and the program 'bombs'. 4) The WRITE (6,20) P,I,N,A statement contains a reference to a statement labelled 20 (the alleged FORMAT statement). But since the corresponding FORMAT statement is labelled 200, the compiler is stymied. It prints an error message. Note that all of the above errors were syntactical errors. Consider the BASIC program for solving the compound interest problem. Statement 40 is 40 LET A = P\textasteriskcentered(1 + I)\textasteriskcenteredN(1) Algebraically, the formula for the amount A from a given principal P held for N years at interest I is A = P(1 + I)^N. But statement 40 says A = P(1 + I)N Hence it should be changed to A = P\textasteriskcentered(1 + I) \uparrow N.(2) Comparing (1) and (2) one is tempted to think that the \uparrow was mistyped as \textasteriskcentered and that this is a syntax error. But the compiler will accept (1) as a valid BASIC statement. The CPU will execute it and print the results which, unfortunately, are useless to the programmer. Hence this is a logical, not a syntactical error.

Question:

The actual physical distances between linked genes bear no direct relationship to the map distances calculated on the basis of crossover percentages. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0681.htm

Solution:

In certain organisms, such asDrosophilia, the actual physical locations of genes can be observed. The chromosomes of the salivary gland cells in these insects have been found to duplicate themselves repeatedly without separating, giving rise to giant bundled chromo-somes, called polytenechromosomes. Such chromosomes show extreme magnification of any differences in density along their length, producing light and dark regions known as banding patterns. Each band on the chromosome has been shown by experiment to correspond to a single gene on the same chromosome. The physical location of genes determined by banding patterns gives rise to a physical map, giving absolute distances between genes on a chromosome. Since crossover percentage is theoretically directly proportional to the physical distance separating linked genes, we would expect a direct correspondence between physical distance and map distance. This, however , is not necessarily so. An important reason for this is the fact that the frequency of crossing over is not the same for all regions of the chromosome . Chromosome sections near thecentromereregions and elsewhere have been found to cross over with less frequency than other parts near the free end of the chromosome. In addition, mapping units determined from crossover percentages can be deceiving. Due to double crossing over (which results in a parental type ), the actual amount of crossover may be greater than that indicated by recombinant type percentages. However, crossover percentages are nevertheless invaluable because the linear order of the genes obtained is identical to that determined by physical mapping.

Question:

Describe the structures found in a cross section of the mammalian spinal cord.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0507.htm

Solution:

The mammalian spinal cord extends from the base of the brain and is tubular in shape. Along with the brain, the spinal cord makes up the central nervous system of all vertebrates. It has two very important functions: to transmit impulses to and from the brain, and to act as a reflex center. The spinal cord is protected by the vertebral column. The vertebral column is composed of segments of bone, called the vertebrae, connected to each other by cartilage. The cartilage gives it flexibility, while the bone gives it strength. It is possible to anatomically divide the spinal cord into five different regions using the vertebrae as a guide. These are (1) cervical region, (2) thoracic region, (3) lumbar region, (4) sacral region, and (5) coccygeal region. The regions of the spinal cord send out nerves which innervate different parts of the body (see Figure 1). The spinal cord is surrounded by three membranes: the dura mater, arachnoid, and pia mater. It is a hollow, tubular structure. The hollow portion, called the spinal lumen, runs through the entire length of the spinal cord and is filled with spinal fluid. A cross-section shows 2 regions, an inner mass of gray matter composed of nerve cell bodies, and an outer mass of white matter made up of bundles of axons and dendrites (see Figure 2). The coloration of the white matter region is due to the presence of white myelinated fibers. The nerve cells of the gray matter lack myelin, which is white, and hence the "natural" gray color of nerve cells is seen. Four protuberances of the gray matter are noted. The two anterior processes are called the ventral horns, and the two posterior ones are called the dorsal horns. The axons and dendrites of the white matter carry impulses between lower levels and higher levels of the spinal cord, or between various levels of the spinal cord and the brain. The spinal cord receives sensory fibers from peripheral receptors at the dorsal root. The cell bodies from which the sensory fibers arise are located in a cluster, called the dorsal root ganglion, in the dorsal root. These sensory fibers pass into the dorsal horns of the gray matter, where they synapse with interneurons in the dorsal horns and/or motor neurons in the ventral horns. Axons from the motor neurons leave the spinal cord at the ven-tral root and soon join the sensory fibers - together they constitute the spinal nerve. Spinal nerves arising from the spinal cord branch to supply various parts of the body except the head, part of the neck, the thorax, abdomen, and the upper and lower extremities.

Question:

What is the physiological basis of allergies? What measures can be taken to decrease the effect of allergies?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0367.htm

Solution:

Although immune reactions generally have a protective role in the body, they can sometimes cause damage to the body itself. Such an adverse immunological response, called an allergy or hypersensitivity reaction, results from an acquired reactivity to an antigen. The reactivity can result in bodily damage upon subsequent re-exposure to that specific antigen. The allergic response can be elicited by a variety of antigens, including pollen, dust, certain foods and bee stings. Initial exposure to the antigen leads to some anti-body synthesis (the primary response), but also to the formation of more lymphocytes which act as memory cells. Subsequent exposure to the same antigen elicits a more vigorous response (secondary response); many memory cells are stimulated to differentiate into plasma cells which produce the corresponding antibody. These particular antigens stimulate production of IgE immunoglobinswhich circulate in the blood and bind to mast cells (found in connective tissue) andbasophils(a type of white blood cell), When the antigen forms a complex with theIgEattached to the mast cell, release of histamine and other chemicals from the mast cell results. The symptoms of the allergy result from the effects of these chemicals. For example, when a previously sensitized person inhales pollen, the antigenic part of the pollen combines with theIgE-mast cell complex in the respiratory passages. The histamine released causes increased mucous secretion, in-creased blood flow and constriction of smooth muscle lining the passages. These physiological changes cause the familiar symptoms of running nose, congestion and difficulty of breathing. In some people, a very severe allergic reaction called anaphylaxis can occur which may have systemic (non-localized) symptoms. People have died from anaphylaxis resulting from a bee sting. To decrease the effect of allergies, one can ad-minister antihistamines. These only provide incomplete relief, however, since other chemicals are also released. A therapy called desensitization can help allergic people. In this process, the antigen is injected repeatedly into the person in small but increasing dosages. The theory behind desensitization is that this procedure inducesIgGsynthesis against the antigen. When the antigen is sub-sequently presented, it is more likely to becomplexedby IgG antibodies than byIgEantibodies bound to mast cells.

Question:

How does the term "growth" as used in bacteriologydiffer from the same term as applied to higher plants and animals?

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/Users/wenhuchen/Documents/Crawler/Biology/F05-0132.htm

Solution:

When a small number of bacteria are transferred into the proper medium and incubated under the appropriate physical conditions, a tremendous increase in the number of bacteria results in a short time. As applied to bacteria and microorganisms, the term "growth" refers to an increase in the entire population of cells. When we speak of the growth of plants and animals, we usually refer to the increase in size of the individual organism. The growth of bacteria involves the increase in numbers of cells over the initial quantity used to start the culture (called the inoculum) . Some species of bacteria require only a day to reach their maximum population size, while others require a longer period of incubation . Growth can usually be determined by measuring cell number, cell mass, or cell activity.

Question:

A book publisher is in the business of producing volumes in mathematics, science and the humanities. He basically uses two kinds of labor a) skilled for writing the books b) unskilled to correct the completed volumes. His average costs per book for each type of labor and each type of book is given below: Expenses Mathematics Science Humanities Skilled Unskilled 17.00 6.00 16,00 5.00 12.00 4.00 Write a program in Basic to print the costs of skilled and unskilled labor needed to produce M books in math, S books in science, and H books in humanities, if M = 1, 2, 3, 4, 5; S = 2M + 1; H = 2M - 2.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0452.htm

Solution:

We can use matrices to facilitate the solution of this problem │17.00 (M) + 16.00(2M + 1) + 12.00 (2M - 2)│(1) │6.00 (M) + 5.00(2M + 1) + 4.00(2M - 2)│ If we multiply (1) by the vector │M│ │2M + 1│we obtain: │2m - 2│ │175001620012400 ││M│ │620055004900││2M + 1 │ │2M - 2 │ =│17500M + 16200(2M+1) + 12400(2M-2)│(2) │6200M +5500(2M+1) + 4900(2M-2)│ Adding the two rows of (2) together gives the total cost of skilled and unskilled labor. Note that costs of skilled labor comprise the first row of the result in (2) , while costs of unskilled labor comprise the second row. 10DIM A (1, 2), L(1): REM L(0), L(1) DENOTE SKILLED, UN-SKILLED LABOR 15FOR I = 0 TO 1 20FOR J = 0 TO 2 25READ X 30LET A (I, J) = X 35NEXT J 40NEXT I 45PRINT "MATH", "SCIENCES", "HUMANITIES", "SKILLED LA-BOR", "UNSKILLED TOTAL COSTS" 50PRINT 55FOR M = 1 TO 5 60LET S = 2\textasteriskcenteredM-1 65LET H = 2\textasteriskcenteredM-2 70FOR I = 0 TO 1 75LET L(I) = A(I,0)\textasteriskcenteredM + A(I,1)\textasteriskcenteredS + A(I,2)\textasteriskcenteredL 80NEXT I 85PRINT M, S, H, L(0), L(l); ""; L(0)+L(1) 90NEXT M 95PRINT 100STOP 105DATA 17, 16, 12, 6, 5, 4 110END

Question:

(a) Establish the numerical relationship between the frequencyѵ_min sec^-1 of an emitted x-ray and the minimum voltage V applied to an x-ray tube. (b) What is the minimum voltage which must be applied to the x-ray tube to produce x-rays of wavelength 10^-8 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D35-1038.htm

Solution:

(a) The minimum potential required to boost the speed of the electron between the plates of an x-ray tube, to a value sufficient to excite the target atoms on the screen, is determined by the wavelength of the emitted photons (x-rays). The kinetic energy of the electron should therefore be at least as large as the energy of the outgoing photon. The electron, when moving between the plates, acquires the kinetic energy KE =eV where V is the voltage across thetuve. KE = V(volts) × 1.60 × 10^-19 joule = 1.60 × 10^-12 V erg. (This could also have been obtained directly from the knowledge that 1eV is 1.60 × 10^-12 erg.) Therefore, the minimum energy of an electron being equivalent to its minimum frequency multiplied by Planck's constant h, we find E =hѵ_m= 1.60 × 10^-12V ѵ_m= [(1.60 × 10^-12 V) / (6.625 × 10^-27)] s^-1 ѵ_m= (2.41 × 10^14 V)s^-1 . This is the required relationship betweenѵ_mand V. (b) A wavelength, \lambda, of 10^-8 cm corresponds to a frequency ѵ = c/\lambda = [(3 × 10^10) / 10^-8] see^-1. = (3 × 10^18) see^-1. In order to produce this frequency, the voltage across the tube must be at least V_min= ѵ / (2.41 × 10^14 V \bullet s) = [(3 × 10^18) / (2.41 × 10^14)]V = 12,500 volts.

Question:

Two long-winged flies were mated. The offspring consisted of 77 with long wings and 24 with short wings. Is the short- winged condition dominant or recessive?What are the genotypes of the parents?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0651.htm

Solution:

When we are not told which of the charac-teristics is dominant and which is recessive, we can deduce it from the ratio of phenotypes in the progeny. We know that 77 flies have long wings and 24 have short wings. This gives us an approximate ratio of 3 long-winged flies to every 1 short- winged fly (77)/(24) \sim (3/1) As previously noted, the three-to-one ratio signifies that dominant and recessive characteristics are most likely involved. Moreover, because there are three long-winged flies to every short-winged one, it suggests that short-winged is the recessive characteristic, and long-winged is dominant. We cannot immediately conclude that both the long--winged parents are homozygous. In fact they are not, because if they were, no short-winged offspring could have resulted in the cross. So the presence of short-winged flies (homozygous recessive) in the progeny suggests that both parents carry the recessive gene and are thus heterozygotes. Let L be the gene for long wings in flies and l be the gene for short wings in flies. In the cross between two long-winged heterozygous parents: F_11 LL:2 Ll:1 ll long wingshort wing The phenotypes of the F_1 show the three-to-one ratio of long-winged flies to short-winged flies, which concurs with the data given. Therefore the genotypes of the parents are the same, both being heterozygous (Ll).

Question:

Using the S-R (set-reset) flip-flop, design a memory cell with the following inputs: Data-in, Read/Write, Select. Output: Data-out.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0094.htm

Solution:

The memory cell, which is the basic building block of the memory unit in a computer, is capable of storing one bit of information. The memory cell is shown in Figure 1. The memory cell is used to store information (Write) or recall information (Read). To store information, connect the Select line to logic level 1, the Data-in line to the logic level to be stored, and the Read/Write line to logic level 0. Under these conditions gates G_1 and G_2 are enabled and gate G_3 is disabled. The values of the S and R inputs of the flip-flop are the Data-in and Data-in, respectively. Since the S and R inputs are complements of each other, the output of the flip-flop, Q, has the same logic level as the S input. Hence data has been written into the memory cell. After writing in the data, the Select line is connected to logic level 0. This disables all the gates so that changes in the Data-in line no longer affect the inputs to the flip-flop. Also, when the Select line is 0, both inputs to the flip-flop are 0 and the flip-flop is in its "memory" state, that is Q retains the previous value of S. To read data from the memory cell, the Select line and the Read/Write line are connected to logic level 1. Under these conditions gates G_1 and G_2 are disabled and G_3 is enabled. Hence, the value of Q appears at the Data-out line. After reading the data the Select line is connected to logic level 0.

Question:

A 60.0-lb block rests on a smooth plane inclined at an angle of 20\textdegree with the horizontal. The block is pulled up the plane with a force of 30.0 lb parallel to the plane. What is its acceleration?

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Solution:

Here three forces are acting on the block. Its weight W is 60 lb downward. The force of the plane on the block is a thrust N normal to the plane. There is a pull P parallel to the plane. The force acting on the block can be resolved into forces acting normal and parallel to the plane. The weight of the block may be resolved into components of 60.0 lb × cos 20\textdegree normal to the plane and 60.0 lb × sin 20\textdegree parallel to the plane. Since there is no motion in the direction perpendicular to the plane, forces in that direction cancel each other. Therefore, the normal component of the weight is balanced by the force N. Parallel to the plane, taking the direction of P as positive, the sum of the forces is F= 30.0 lb - 60.0 lb × sin 20o = 30.0 lb - (60.0 × 0.342)lb = 9.5 lb m= W/g = 60.0 lb/(32 ft/sec^2) = 1.87 slugs From F = ma, a= F/m = (9.5 lb) / (1.87 slugs) = 5.1 ft/sec^2. Note that if the angle were 30\textdegree, the component of the weight down the plane would be equal to the force up the plane and there would be no unbalanced force acting on the block. Hence it would not be accelerated. If the angle were greater than 30\textdegree, the block would be accelerated down the plane.

Question:

Write a function to compute the Legendre polynomials.

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Solution:

For positive values of n, the recursive definition to compute the Legendre polynomials is given by 1ifn = 0 P_n(X) =xifn = 1 ((2n - 1)xP_n-1 (x) - (n - 1)P_n-2(x)) /_nif n>1 we obtain: FunctionP(n:integer; x:real) :real; BEGIN IF n = 0 THEN p: = 1 ELSE IF n = 1 THEN p:=x ELSEp:=((2\textasteriskcenteredn-1)\textasteriskcenteredx\textasteriskcenteredp(n-1,x) - (n-1)\textasteriskcenteredp(n-2,x)) / _n END; In most of the cases, however, the recursive solution is not the best because a simple solution may be obtained by itera-tion. The Legendre polynomials are more efficiently computed in this way: FUNCTIONp(n:integer; x:real) :real; VAR prev, this, next :real; count: integer; BEGIN IF n = 0 THEN p: =1 ELSE IF n=1 THEN p: =x ELSE BEGIN prev: =1; this:=x; FOR COUNT: = 2 TO n DO BEGIN next:= ((2\textasteriskcenteredcount-1)\textasteriskcenteredx\textasteriskcentered this -( count - 1)\textasteriskcenteredprev) / count; prev:= this; this :=next END; {for} p: =next END END; {p}

Question:

A flat bottom swimming pool is 8 ft. deep. How deep does it ap-pear to be when filled with water whose refractive index is 4/3?

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Solution:

In order to see why we would expect to observe a different depth for the pool when it is filled with water, examine the figure. If no water is in the pool, light coming from a point S on the bottom of the pool will travel directly to the observer's eye. If the pool is filled with water, light emanating from point S will be refracted at P, as shown. Upon reaching the observer's eye, the light appears to be coming from Q and he perceives the depth of the pool to be the dis-tance OQ, rather than the actual depth OS. Our problem is to find the distance d. Note that, from the figure, tan \varphi_1 = (OP/d) tan \varphi_2 = (OP/8 ft). Hence (tan \varphi_1/tan \varphi_2) = (OP/d) \textbullet (8 ft./OP) = (8 ft./d) and d = [{(8 ft)tan \varphi_2}/tan \varphi_1](1) From Snell's Law, n_1 sin \varphi_1 = n_2 sin \varphi_2 where n_1 and n_2 are the indices of refraction of air and water, re-spectively. Therefore (n_1/n_2) sin \varphi_1 = sin \varphi_2(2) To calculate the tangents in (1), we must also know cos \varphi_1 and cos \varphi_2. These we may find by observing that cos \varphi = \surd(1 - sin^2 \varphi)(3) Using (2) in (3) cos \varphi_2 = \surd(1 - sin^2 \varphi_2) cos \varphi_2 = \surd[1 - (n_1/n_2)^2 sin^2 \varphi_1(4) cos \varphi_1 = \surd(1 - sin^2 \varphi_1) Hence tan \varphi_1 = (sin \varphi_1/cos \varphi_1) = {sin \varphi_1/\surd(1 - sin^2 \varphi_1)}(5) and using (2) with (4) tan \varphi_2 = (sin \varphi_2/cos \varphi_2) = [{(n_1/n_2) sin \varphi_1} / \surd{1 - (n_1/n_2)^2 sin^2 \varphi_1}](6) Substituting (5) and (6) in (1) d = (8 ft)[{(n_1/n_2) sin \varphi_1} / \surd{1 - (n_1/n_2)^2 sin^2 \varphi_1}] \textbullet [ \surd{1 - sin^2\varphi_1} / (sin \varphi_1)] d = (8 ft)(n_1/n_2) \surd[(1 - sin^2 \varphi_1) / {1 - (n_1/n_2)^2 sin^2\varphi_1}](7) Now, since we don't know the angle \varphi_1., we make an approximation. Sup-pose \varphi_1 is very small. (This means that the observer is looking almost directly down into the pool.) Then sin \varphi_1 \approx 0 and the square root in (7) becomes 1. Therefore d = (8 ft)[n_1/n_2] = (8 ft)[1/(4/3)] = 6 ft. The pool appears to be 6 ft. deep.

Question:

The density of a 25.0 % sugar solution is 1.208 g/ml. What weight of sugar would be contained in 1.00 liter of this solution ?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0156.htm

Solution:

When a solution is said to be 25.0% sugar , it means that 25.0%of the weight of the solution is made up by the sugar . Thus one can determine the weight of the sugar in this solution by multiplying the total weight of solution by .25 .Here one is not given the weight of the solution but the densityand the volume , The density is the weight 1 ml of the solution , thus the weight of 1.0 littler or 1000ml is equal to the volume 100ml times the density (1.208g/ml). weight of solution = 1000ml × 1.208 g/ml = 1208g. The weight of the sugar in the solution can now be found . weight of sugar = .25 × weight of solution = .25 × 1208 g = 302 g.

Question:

A solid cylinder of radius R rolls on a flat surface. Find the moment of inertia I_s of the cylinder about its line of contact with the surface.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0179.htm

Solution:

The definition of moment of inertia for a continuous mass distribution is dependent upon whichaxis we wish to calculate the moment axis we wish to calculate the moment of inertia about. In this case, we want to calculate I_s , the moment of inertia about an axis parallel to the symmetry axis of the cylinder, but tangent to the surface of the cylinder. To simplify the required integrations, we may equivalently calculate the moment of inertia, I_O, of the cylinder about its symmetry axis, and then employ the parallel axis theorem to find I_s. The moment of inertia I_o is (with reference to the figure) defined as: I_o = \intr^2 dm(1) where r is the perpendicular distance between the symmetry axis of the cylinder and the mass element dm. Also note that dm, the mass contained in a differential volume dv, is the mass per volume contained in the cylinder times the volume dv, or dm = [M/(\piR^2H)] r dr d\texttheta dz(2) where M is the total mass of the cylinder, and \piR^2H is the volume of the cylinder. Combining (2) and (1), we obtain I_o = \int[M/(\piR^2H)] r^3 dr d\texttheta dz(3) where the integral is over the volume of the cylinder. Performing the integral: I_o = [M/(\piR^2H)] ^H/2\int_-H/2^2\pi\int_0^R\int_0 r^3 dr d\texttheta dz I_o = [M/(\piR^2H)] [(R^4)/4] ^H/2\int-H/2^2\pi\int_0 d\texttheta dz I_o = [M/(\piR^2H)] [(R^4)/4] ^(2\pi)^H/2\int_-H/2 dz I_o = [M/(\piR^2H)] [(R^4)/4] (2\pi) (H) I_o = (1/2) M R^2(4) This is the moment of inertia about the symmetry axis. To find the moment of inertia about the axis tangent to the surface of the cylinder, use the parallel axis theorem. Mathematically, (see the diagram) I_s =I_o + M\eta^2(5) where \eta is the separation of the 2 axes. Rewriting this using (4); and noting that \eta = R, we have I_s = (1/2) MR^2 + MR^2 orI_s = (3//2) MR^2 .

Question:

The Pauli exclusion principle is the main reason why atoms do not collapse to a point.

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Solution:

The Pauli exclusion principle states that no two electrons in the same atom can be completely identical; that is, have the same values for all four quantum numbers. The quantum numbers measure the energy level of the electron, the order of increasing distance of the average electron distribution from the nucleus, the angular shape of electron distribution, (the electronic magnetism) , and the two possible orientations of electron spin. An electron in an atom is completely described by its four quantum numbers. Inherent in this description is the point of loca-tion of an electron at a specific time. If two electrons have the same four quantum numbers, they would have, the same point of location at a specific time. This means, therefore, that all the electrons would collapse to a point at a specific time. The Pauli exclusion principle states that no two electrons can have the same four quantum numbers and therefore the atom cannot collapse to a point.

Question:

How far does a wooden (spherical) ball of specific gravity 0.4 and radius 2 feet sink in water?

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Solution:

By "Archimedes' principle" we know that the ball will sink until it displaces a weight of water equal to the entire weight of the ball: (weight of ball) = (weight of displaced water) We know that: (weight of ball) = (volume) (specific gravity) (density of water) where the density of water w is measured in pounds per cubic foot. We must first calculate the volume of the part of the sphere immersed in water when the sphere sinks to a depth of h, and then solve for h from the given information (figure A). Finding this volume is equivalent to finding the volume generated by rotating the area between the curves x^2 + (y - R)^2 and y = h, about the y-axis, ( see figure (B)) We note that the general formula for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) are the coordinates of its center and r is the length of its radius. In this case the coordinates of the center are (0, R), and r = R. Recall that (see figure (C)) : dV = \pi x^2 dy = \pi[R2- (y - R)^2]^2 dy = \pi(2Ry - y^2)dy where dV is the differential cylindrical volume element, x being its radius and dy its height. Finally, we integrate dV between y = 0 and y = h. V = \pi\int^h_0 (2Ry - y^2)dy = \pi[2R (y^2/2) - (y^3/3)]^h_0 = \pi[2R (y^2/2) - (y^3/3)]^h_0 = \pi [Rh^2 - (h^3 / 3) ] = \pi [Rh^2 - (h^3 / 3) ] = \pi h^2(R - h/3) = \pi h^2(R - h/3) Hence, from the equation for the weight of the ball: (weight of ball) = (volume of submerged portion)(density of water) [(4/3) \piR^3](0.4)w = \pi h^2(R - h/2)w (4/3)(2)^3(0.4)= h^2(2 - h/3) h^3 - 6h^2 + 12.8 = 0 We find by synthetic division that h \approx 1.75 feet.

Question:

Write a complete IDENTIFICATION DIVISION for a COBOL program which calculates the weekly paychecks of the employees of a company.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0251.htm

Solution:

IDENTIFICATION DIVISION is the first part of any COBOL program. Its main purpose is to identify the program to the computer, and then it can also be used to explain to the reader what the program does. The first statement in this part of the program is the IDENTIFICATION DIVISION. In format form the statement starts from the eighth column and ends with a full stop mark (.). PROGRAM-ID. This is the second mandatory statement of IDENTIFICATIONDIVISION; it also starts from the eighth column and ends with a full stop mark. Leaving one empty space and continuing on the same line program-name follows the PROGRAM-ID statement. Program name could be any name except the reserved words such as IF, GO TO, DATA, and a hyphen must be used to separate two or more words. So, call the program PAYROLL-LIST. The third, fourth and fifth statements are not manda-tory. It is up to the programmer to decide whether the name of the author, date-written, and any extra remarks would be included or not. Following those steps, the IDENTIFICATION DIVISION, takes the complete form of fig. 1. All of the actual remarks in the REMARKS, section must come after margin B.

Question:

A constant horizontal force of 10 N is required to drag an object across a rough surface at a constant speed of 5 m/sec. What power is being expended? How much work would be done in 30 min?

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Solution:

Power is the rate of doing work, P = (∆W)/(∆t) = (F∆s)/(∆t) P = (∆W)/(∆t) = (F∆s)/(∆t) . (Note that in this problem the work reduces to the force multiplied by the distance the object is moved.) But ∆s/∆t is just the velocity. Therefore, P= Fv = (10 N) × (5 m/sec) = 50 J/sec = 50 W W= Pt = (50 W) × (1/2hr) = 25 W-hr. The work, of course, is done against the force of sliding friction.

Question:

The spherical atomic nucleus of potassium has a radius of about 4 × 10^-13 cm.; 6.02 × 10^23 of these nuclei weigh about 39 g. Calculate the approximate density of the nucleus of potassium in tons per cubic centimeter. Volume of a sphere = 4/3 \pir^3, where r is the radius.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0221.htm

Solution:

The density is defined as the mass of a substance divided by its volume. To calculate the density one must know the weight and volume of the substance. Here one is given the weight in grams but the problem asks for the density in tons, thus one must convert grams to tons. The volume of the nuclei is not given, but can be found by using the radius of a nucleus, the formula for the volume of a sphere and the number of nuclei present. 1) Converting grams to tons There are 454 g in one pound and 2000 lb in one ton, therefore, grams can be converted to tons by multiplying the number of grams by the following conversion factor 1 lb/454 g × 1 ton/2000 lb number of tons = 39 g × 1 lb/454 g × 1 ton/2000 lb = 4.3 × 10^-5 tons. 2) Calculating volume of nuclei The volume of a sphere is equal to 4/3 \pir^3 where r is the radius of the sphere. One is given that the radius of a potassium nuclei is 4 × 10^-13 cm. volume of 1 nuclei = 4/3 \pi (4 × 10^-13 cm)^3 = 4/3 (3.14)(64 × 10^-39 cm^3) = 2.6 × 10^-37 cm^3 The total volume of the nuclei is equal to the number of nuclei times the volume of one nucleus total volume = no. of nuclei × volume of one nucleus total volume of potassium nuclei = 6.02 × 10^23 × 2.6 × 10^-37cm^3 = 1.6 × 10^-13cm^3 3) The weight in tons and the volume in cubic centi-meters is now known. The density in tons per cm^3 can be calculated density = [no. of tons (= mass)] / [no. of cm^3 (=voIume)] density = (4.3 × 10^-5 tons) / (1.6 × 10^-13 cm^3) = 2.7 × 10^8 tons/cm^3.

Question:

A sign of a good computer programmer is the ability to write a program which minimizes storage usage. The IBM 360/370 computers provide us with the DSECT and CSECT statements . How do they enable us to becomebetter programmers ?

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Solution:

The DSECT (Dummy Section) is actually an indica-tor that tells the assembler that the following code defines a dummy section. This facility enables us to define and use many variables in our program without actually allocating storage space for them all at one time. This may sound somewhat strange but the procedure is rather simple. The DSECT is an assembler instruction, and it generates no ma-chine code. It tells the assembler that the storage defini-tions which follow forms a dummy section . The CSECT (Control Section) instruction indicates the end of the dummy section. For the DSECT a substitute base register is used. The register to be used is generally specified in a 'USING' statement at the beginning of the program. When the DSECT is encountered, the displacements of the labels within it are assigned, not from the beginning of the program, but from the DSECT statement. The CSECT instruction indicates the end of the dummy section. With this, the substitute base register is restored (freed), the location counter pro-ceeds as if the DSECT instruction was not there. It has the value it had previous to encountering the DSECT instruction . Instead of having storage allocation for the 'running- life' of the program , the symbols of a DSECT are allocated space as they are encountered in the body of the program. This space is then freed when usage is completed. By placing a label in the label field of the DSECT instruction, the dummy section can be named. The label field of the SCECT statement must contain the name of the program, this name is taken from the label field of the start instruc-tion. Coding of the section is generally done as follows USINGCRDMAP, 5 APTOTSUM, SLSREG CRDMAPDESECT SLSCARDDSOCL80 SLSNBRDSCL4 SLSNAMDSCL20 SLSREGDSCL2 SLSQTYDSCL6 DSCL48 PROGNAMECSECT The 'USING' statement is generally coded after the pro-gram's base register 'USING' statement. The AP statement adds a number previously placed in SLSREG to TOTSUM. The DSECT is usually coded for repetitious procedures. In many instances it is used to replace explicit coding. This is done because DSECT is considered easier to code and there are less changes of addressing errors. It is important to note that since DSECT is a dummy section only storage space can be defined in it and not con-stants. Thus, only the 'DS' statement is used for the labels.

Question:

The cross-over percentage between linked genes A and B is 40%, between B and C, 20%, between C and D, 10%; between A and C, 20%; between B and D, 10%. What is the sequence of genes on the chromosome?

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Solution:

This question allows us further practice in ordering genes. Again we rely on the visual method because it is the most convenient. From the data, we know that A and B are 40 map units apart (recall that map units are directly proportional to cross-over per-centage) . B and C are 20 units apart. In order to determine whether C is to the right or left of B, we look at the map distance between A and C. This distance is 20 units. With this information, we now know that C is to the left of B because this is the only way that A and C can be 20 units apart. If C were to the right of B, then the distance between C and A would be 60 units, which is not the case. C and D are 10 units apart. We must determine whether D is to the right or left of C. We know that B and D are also separated by 10 units. Since B and C are 20 units apart, D must be between B and C. Note: when ordering genes, we have first to establish two points (genes) on the chromosome. Using these as reference points, we make comparisons with other genes. In the case above, our reference points were A and B and we then proceeded to compare these points with C and D. Although it is helpful to pick the two most widely separated points as reference points, the choice is actually completely arbitrary. We could just as well have started with B and D as our reference points. The answer obtained would still be the same.

Question:

What function does oxygen serve in the body?

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Solution:

Oxygen is necessary for the process of cellular respiration. This process involves the catabolism of glucose to result in the production of energy, in the form of ATP, for utilization by the cell. Cellular respiration can be thought to have two parts. The first series of reactions can take place without the presence of oxygen, while the second series of re-actions is dependent on oxygen. In the second reaction sequence, oxygen acts as the final hydrogen acceptor. This occurs in the electron transport system within the mitochondria. The oxygen combines with hydrogen, and is converted to a molecule of water. If anaerobic con-ditions existed (absence of oxygen), no ATP formation will occur in the mitochondria. It should be noted however that two ATP per glucose oxidized are still produced by the first series of reactions, collectively called glycolysis. Glycolysis will occur regardless of the availability of oxygen, and is restricted to the cytoplasm; it does not occur in the mitochondria. The amount of ATP produced via glycolysis is very small com-pared to that produced by the second reaction sequence. The second reaction sequence, which occurs only in the mitochondria, is collectively referred to as oxidative phosphorylation. Since the mitochondria provide about 95% of all the ATP formed in the cell, the absence of oxygen results in less energy production. This energy is required to maintain cell structure and function, and the cell will ultimately die without it. Roughly 99% of all the molec-ular oxygen consumed by a cell is used for cellular respiration.

Question:

A 1-\muF capacitor charged to 200 V and a 2-\muF capacitor charged to 400 V are connected; the positive plate of each is connected to the negative plate of the other. Find the difference of potential and charge on each capacitor and the loss of energy that has taken place.

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Solution:

The charge Q on a charged capacitor of capacitance C is Q = CV where V is the potential difference between the plates of the capacitor. Thus the charge on the 1-\muF capacitor before connection is 1 \muf × 200 V = 200 \muC and on the 2-\muF capacitor is 2\muf × 400 V = 800 \muC. Fig. A depicts the situation before connection of \muf × 400 V = 800 \muC. Fig. A depicts the situation before connection of the two capacitors. If the two capacitors are now connected positive to negative, a charge of + 800 \muC and one of - 200 \muC are joined, as are charges of - 800 \muC and + 200 \muC. The situation just after the connection is depicted in Figure B. The composite capacitor thus has charges of \pm 600 \muC on its plates, since charge cannot be created or destroyed, but only neutralized. These charges will be shared between the individual capa-citors, with \pm Q_1 on the first capacitor, and \pm Q_2 on the second, since the charge flows from one to on the second, since the charge flows from one to the other until a common potential V_0 is achieved. Fig. C depicts this situation. Thus Q_1 = C_1 V_0 and Q_2 = C_2 V_0, where Q_1 + Q_2 - 600 \muC. Thus combining the above three equations (Q_1 + Q 2 ) = (C_1 + C_2 )V_0or ) = (C_1 + C_2 )V_0or 600 × 10^-6 C = (1 × 10^-6 F + 2 × 10^-6 F) V_0or V_0 = [(600 × 10^-6 C)/(3 × 10^-6 F)] = 200 V. \thereforeQ_1 = 200 \muC,Q_2 = 400 \muC. The initial energy of the charged capacitors minus the final energy is W = (1/2) C_1 V^2 _1 + (1/2) C_2 V^2 _2 - (1/2) C_1 V^2 _0 - (1/2) C_2 V^2 _0 = (1/2) × 10^-6 F × (200)^2 V^2 + (1/2) × 2 × 10^-6 F × (400)^2 V^2 - (1/2) × 10^-6 F × (200)^2 V^2 = 12 × 10^4 × 10^-6 J = 0.12 J. This energy is lost as heat due to the transient current in the connecting wires when the two capacitors are joined.

Question:

Input records for a sales agency include, along with identifying information for each salesman, commission sales for each of the last six months. Write a procedure division to read these cards, select the largest of six commission payments, and print out the salesman's name and his highest commission payment over the months. The commission is described as COM OCCURS 6 TIMES PIC 9(4)V99. Start your program with a flowchart.

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Solution:

This problem requires the computer to read a set of cards, find the largest commission paid to each salesman, and print the particular salesman's name and commission. The flowchart can be started by the read statement, that is a command should be put into the computer to cause it to read the set of variables. To prevent an error a conditional statement should also be used immediately so that when there are no more variables to read the process will come to halt. A complete flowchart of the problem is given in fig. 1. It should be noted that each programmer has his or her own methods in programming. Therefore flowcharts also vary from person to person, depending on the method of approach to the problem. In the flowchart given, the programmer uses a set of conditional statements to find the largest commission and once it is found, the card or the group of variables is taken out of the deck and certain information contained in it is printed. Once the flowchart is done, the rest of the work is purely mechanical, and depends on the ability to use the technical knowledge related to the particular programming language, COBOL in this case. In Fig. 2 is the procedure division of the program. Note that adding the remaining three divisions IDENTIFICATION, ENVIRONMENT, and DATA would make this program complete and ready to run. The use of the OCCURS clause sets up a table of 6 com elements. To index a certain element a statement of the form: is used. As an example, the statement COM (subs) where subs = 4 gives com the value of the fourth com element in the table.

Question:

When the higher vertebrates left the water to take on a terrestrial existence, they had to adopt mechanisms to conserve body water. One of these was the evolution of a concentrated urine. How is this achieved physiologically?

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Solution:

The higher vertebrates have evolved a final urine having a greater osmolarity than blood plasma. This is advantageous in that water can be retained within the body making the organism less likely to suffer dehydratipn. In the absence of a special renal mechanism, the urine would be highly dilute and severe water loss from the organism could become fatal. The loop of Henle is a hairpin sequent of the renal tubule having an ascending limb and a descending limb. The connecting tubule lies in close proximity and parallel to the loop of Henle so that the three parts of the renal tubule can interact with one another (see figure). The walls of these three portions differ in their permeabilities to substances such as sodium and water. Both the anatomical and physiological aspects of the renal tubule determine the concentration of the urine excreted. The ascending limb of the loop of Henle actively pumps sodium ions out of its tubular space into the interstitial fluid, with chloride following passively due to electrostatic attraction. The ascending limb is impermeable to water and thus the osmolarity of fluid inside the ascending limb decreases as the loop ascends and a concentration gradient is established between the loop and the interstitialfluid. The descending limb is permeable both to water and sodium, so that some of the sodium and chloride ions diffuse passively into its tubular space. The cycling of sodium from ascending limb to interstitial fluid to descending limb results in the establishment of a concentration gradient of sodium and chloride in the tissue fluid surrounding the loop, with the lowest concentration near the cortex and the highest con-centration deep in the medulla. (See accompanying figure) . This mechanism of flow in the mammalian kidney, called the countercurrent flow, permits the concent-ration gradient in the interstitial fluid to be main-tained; this gradient is essential in establishing the final concentration of the urine. With thisin mind, we can understand how the final urine becomes highly concentrated relative to the blood. The urine that leaves the ascending limb is not more concen-trated than the glomerularfiltrate since it has been di-luted in the ascending loop. But as the urine flows through the collecting tubule, it is essentially flowing through a concentration gradient in the interstitial fluid from the cortex to the medulla. Moreover, the collecting tubule is permeable only to water. Since the osmolarity of the inter-stitial fluid is lowest near the cortex and increases toward the medulla, the urine flowing down the collecting tubule will lose water during it scourse to the medulla. The final urine that emerges from the collecting tubule is substan-tially hypertonic to blood and remains in this same concen-tration until it is excreted to the outside.

Question:

What is the apparent loss of weight of a cube of steel 2 in. on a side submerged in water if the weight density of H_2O is 62.4 lb/ft3?

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Solution:

The situation is as depicted in the diagram. Assume that a diver's hand exerts an upward force H on the steel, which is just large enough to keep the block in equilibrium. In this case, from Newton's Second Law, the net force on the block is zero and B + H - mg = 0(1) where B is the buoyant force of the water on the steel. By Archimedes' Principle, B is equal to the weight of water displaced by the block. Hence, B = \rho_w gV(2) B = \rho_w gV(2) where V is the block's volume, and \rho_w is the density of water. Therefore, solving (1) for H, and using (2), H = mg - B = mg - \rho_wgV But m may be written as m = \rho_sV(3) where \rho_s is steel's density. Finally, then H = \rho_sgV - \rho_wgV But H is the apparent weight of the steel, since this is the force we exert on the block to keep it in equi-librium. The weight of the block outside the water is given by (3) as mg = \rho_sgV The difference between "apparent" and "actual" weights is then ∆ = H - mg = (\rho_sgV - \rho_wgV) - \rho_sgV \vert∆\vert = \rho_wgV which is B. Hence, \vert∆\vert = (62.4 lb/ft^3) (2 in.)^3 = (62.4 lb/ft^3) (1/6 ft)^3 \vert∆\vert = (62.4/216) lb = .29 lb.

Question:

Discuss feedback control involved inglycolysisand the TCA cycle.

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Solution:

Glycolysisand the TCA cycle are both regulated by means of allostericcontrol of the enzymes involved.Aliosteryinvolves regulation of theactivity of an enzyme by the binding of a molecule at a site other than theactive site. The molecule that is bound changes the conformation of theenzyme, thereby changing its activity. If the molecule regulates an enzymaticreaction involved in its own synthesis, this regulation is called feedbackcontrol. Inglycolysis, one control point occurs at the conversion of fructose- 6-phosphate to fructose 1,6-diphosphate, catalyzed by the enzyme phosphofructokinase. This enzyme can beallostericallyinhibited by either ATP or citrate.This explains why the rate ofglycolysisdecreases as soon asaerobic respiration begins in yeast; once the electron transport chain beginsto operate, much more ATP is formed per glucose molecule. This causesan accumulation of ATP in more than sufficient amounts for cellularmetabolism. This large amount of ATP inhibits phosphofructokinaseand therefore inhibitsglycolysis. Sinceglycolysisis inhibited, so is ATP production, for the end products ofglycolysiswould normallyenter the Krebs cycle in order to produce more ATP. This makes sense, since the cell no longer needs to utilize glucose at such a high rate inorder to generate sufficient energy. The cell needs only to allow glycolysisto proceed at a rate such that the amount ofglycolyticproducts enteringthe Krebs cycle will yield an appropriate supply of ATP. Feedback inhibitionprovides the mechanism for this control. The inhibition by citrate, aKrebs cycle intermediate, works much the same way. In the Krebs cycle, a major control point occurs at the conversion of isocitrateto \alpha-ketoglutarate, catalyzed by the enzymeisocitrate dehydrogenase. This enzyme isallostericallyinhibited by either ATP or NADH.If too much ATP is being synthesized, or too much NADH producedto be oxidized by the electron transport chain, the reaction is blockedand the whole cycle ceases. Some citrate is accumulated which actstoallostericallyinhibitphospho-fructokinase, thus shutting down glycolysisas well. Withglycolysisshut down,pyruvatewill not build up or beconverted to lactate or ethanol (which would normally occur if only the TCAcycle werestopped). Allostericcontrol allows for a high degree of efficiency and cooperationinglycolysisand the TCA cycle. In addition, thereareother controlpoints in both these pathways and in other pathways as well, all servingto keep the cell's metabolic processes in balance.

Question:

How do the fork ferns differ from the true ferns? What unusual feature is present inPsilotum?

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Solution:

The fork ferns (Psilophyta) and true ferns (Pterophyta) are classified separately primarily because of the difference in their leaf structures . The leaves of the fork ferns are small, simple, scale--like, and are considered mere epidermal outgrowths ratherthentrue leaves. The leaves of the true ferns are relatively large, elaborate, vein-containing, and usually compound, being finely divided intopinnae. The fork and true ferns also differ in that the well- developed roots present in the true ferns are lacking in the fork ferns. The latter have instead, numerous unicellular rhizoids which grow off the rhizome. The rhizome of the fork ferns is usually found in association with a fungus. Sporangia of the fork ferns are borne in axils of some of the scale-like leaves; sporangia of most true ferns are, however, carried on the undersurfaces of the leaves. Psilotum, one of the two existing genera of fork ferns, is of particular interest to botanists because both its gametophyte and sporophyte have vascular tissues. Whereas the gametophyte of the bryophytes , seed plants, and almost all lower vascular plants lack a vascular system, the gametophyte ofPsilotumhas a stele complete with xylem and phloem surrounded by an endodermis.

Question:

How are oxidation and reduction reactions related? How does an oxidase enzyme differ from a dehydrogenase enzyme?

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Solution:

An oxidation reaction involves the removal of electrons from an atom or molecule. The reverse process, called a reduction reaction, involves the addition of electrons to an atom or molecule. Every oxidation reaction must be accompanied by a corresponding reduction reaction. The electrons given off by the substance oxidized (also called the reductant or reducing agent), must be accepted by the substance reduced (also called the oxidant or oxidizing agent). A simple example of an oxidation reaction is the oxidation of iron: Fe\textdegree \textemdash\textemdash\textemdash> Fe^2+ + 2e- This involves the actual removal of electrons from the iron atom. The reverse process is a reduction reaction. Fe^2+ + 2e- \textemdash\textemdash> Fe\textdegree A not-so-obvious oxidation reaction is the oxidation of methane (CH_4): CH_4 + 2O_2 \textemdash\textemdash> CO_2 + 2H_2O This involves the formation of bonds in which electrons are drawn toward a more electronegative (electron attracting) atom or atoms which is essentially the loss of electrons required in oxidation. Both carbon and hydrogen are considered to be oxidized, since in the new the electrons are removed by oxygen farther from the C and H atoms then they were in the C-H bonds of methane. Oxygen is strongly electronegative and is a common agent in oxidation reactions. Oxidation and reduction processes are important as a means of liberating and storing energy. This can be seen in the oxidation of succinate to fumarate by succinic dehydrogenase. The removal of two hydrogen atoms (with their corresponding electrons) from succinate forms fumarate. Succinate is therefore oxidized. The two hydrogen atoms are transferred to a molecule of flavin adenine dinucleotide (FAD) to form FADH_2. FAD is therefore reduced. This coupling of an oxidation to a reduction reaction traps the energy given off during the oxidation of succinate and uses it to synthesize FADH_2, an energy- requiring reduction process. FADH_2 can subsequently under-go an oxidation reaction and release this temporarily stored energy to another molecule. Oxidation processes thus liberate energy, while reduction processes act to store chemical energy. Most organisms obtain energy by enzymatic reactions which involve the flow of electrons from NADH (reduced nicotinamide adenine dinucleotide) to oxygen, the final electron acceptor. The electrons are transferred by a system of enzymes located in the mitochondria called the electron transport system. An oxidase is an enzyme which removed electrons, while a dehydrogenase is an enzyme which removes hydrogen ions (H+) and their associated electrons. Cytochrome oxidase, the final enzyme in the electron trans-port chain, transfers electrons directly to O_2 to form H_2O. Succinate dehydrogenase, in the citric acid cycle, transfers electrons and hydrogen ions from succinate to FAD, yeilding fumarate and FADH_2 as a result.

Question:

With a certain current in circuit 1 of the figure, a flux of 5 × 10^-4 weber links with circuit 2. When circuit 1 is opened the flux falls to zero in 0.001 sec. What average e.m.f. is induced in circuit 2 ?

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Solution:

The induced e.m.f. \epsilon in circuit 2 opposes the reduction of the magnetic flux \varphi linking it by setting up its own flux in the direction of \varphi. (see figure) . The average rate of decrease of flux in circuit 2 is [(∆\varphi)/(∆t)] = [(5 × 10^-4 w)/(.001 sec)] = 0.5 weber/sec . The average induced e.m.f. is therefore, by faraday's Law, \epsilon̅ = - [(∆\varphi)/(∆t) = - 0.5 volt since 1 volt = 1 weber/sec .

Question:

An oil drop carries a net charge of three times the electronic charge and has a radius of 10^\rule{1em}{1pt}4 cm. What is its terminal velocity when it falls between two horizontal plates kept at a potential difference of 1000 V and 2 cm apart, the positive plate being upper-most? The densities of the oil and of air are 800 kg\bulletm^\rule{1em}{1pt}3 and 1.29 kg\bulletm^\rule{1em}{1pt}3 and the visocity of air is 1.80 x 10^\rule{1em}{1pt}s N\bullets\bulletm^\rule{1em}{1pt}2.

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Solution:

Between two parallel plates, separated by a distance d and maintained at a difference of potential V, the electric intensity E is E = V/d. The electro-static force acting on the drop is upward, and of magnitude F_E = qE = qV/d. This follows from the defintion of E (q = 3e^\rule{1em}{1pt}).Three other forces are acting on the drop; its weight downward and two upward forces, the viscous retarding force f, due to the surrounding air, and the buoyant upthrust B (see figure). When the terminal velocity is achieved, the forces balance, by definition of terminal velocity. The magnitude of the viscous retarding force is given by Stoke's law: f = 6\pi\etarv\bullet r is the radius of the drop, \rho its density, \eta the viscosity of air and v its terminal velocity. The buoyant force is equal to the weight of air dis-placed by the drop (Archimede' s principle). The volume of the (spherical) drop is 4/3 \pir^3 and the density of air is \sigma. The weight of the dis-placed air (equal to B) is then (4/3\pir^3)\sigmag = B. The weight of the drop is W = mg = (4/3\pir^3)\rhog, where \rho is oil's density. Therefore W = B = f + FE (4 / 3)\pir^3 \rhog = (4 / 3)\pi r^3 \sigmag + 6\pi\etarv + qV / d,(1) Hence, v = [(4 / 3)\pir^3g(\rho \rule{1em}{1pt} \sigma) \rule{1em}{1pt} (qV / d)] /6\pi\etar = [(4 / 3)\pi× 10^\rule{1em}{1pt}18m^3 ×9.8m \bullet s^\rule{1em}{1pt}2 (800 \rule{1em}{1pt} 1\bullet 29)kg \bullet m^\rule{1em}{1pt}3 {-(3 ×1 \bullet 6× 10^\rule{1em}{1pt}19 C×103v) / 0.02 m)}] / [6\pi×1.80×10\rule{1em}{1pt}5N\bullets\bullet m^\rule{1em}{1pt}2×10^\rule{1em}{1pt}6 m] = [{(3.27 \rule{1em}{1pt} 2.40)×10^\rule{1em}{1pt}14 } / (3.40 × 10^\rule{1em}{1pt}10 )] m \bullet s^\rule{1em}{1pt}1 = 2.56 × 10\rule{1em}{1pt}5m \bullet s^\rule{1em}{1pt}1. In the classical Millikan oil drop experiment, the same set up is used to determine the electronic charge. A microscope is used to find the terminal velocity v of the drop. Equation (1) is then used to solve for q.

Question:

The accompanying figure shows the wavelength ranges for the different colors in the visible spectrum. If potassium- containing materials emit light of frequency 7.41 × 10^14 sec-^1, what color flame would you expect to see when you heat a potassium compound such as potassium chloride in a Bunsen burner flame?

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Solution:

The color that you can expect will depend on the wavelength of the light emitted, as indicated by the diagram. Given the frequency of the light emitted, the wavelength can be determined. The frequency (ѵ) , wavelength (\lambda), and speed of light (c) are related by the equation ѵ\lambda = c. c = 3.00 × 10^10 cm/sec. Solving for \lambda: (7.41 ×10^14) \lambda = 3.00 × 10^10. \lambda = (3.00 × 10^10) / (7.41 × 10^14) = 4.05 × 10^-5 cm. To convert this into \AA; remember that there are 10^8 \AA per centimeter. Consequently, 4.05 × 10-^5 cm = (4.05 × 10-^5) (10^8) = 4050 \AA. According to the diagram, this \lambda corresponds to the violet region.

Question:

Give a pseudo code to do a tree search, using an ALGOL- like language.

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Solution:

We make the following assumptions: Each node n of the tree has 3 fields 1st field is: Data [n], i.e., the number or data in the sort list. 2nd field is: Left [n] , i.e., pointer to the root of the left sub-tree. (the pointer = 0 if there is no left sub-tree) 3rd field is: Right [n], i.e., pointer to the root of the right sub-tree (the pointer = 0 if there is no right sub-tree) The placement of the different data items in the tree is as follows: If the first data item is x, then, if the second data item in the list is greater than x it is placed in the left subtree, and if the second item is less than x it is placed in the right sub-tree. i.e., Data [n] < Data [x] for all nodes x which are elements of the left sub- tree (n), and, Data [n] > Data [x] for all nodes x which are elements of the right sub-tree (n) We are now ready to form a strategy to search a tree, as follows: 1) Begin the search of the sort tree at the root for a node whose data field contains the argument, i.e., the value or number which we want to search 2) If there is such a node, set m to point to it and set a variable 'treesearch' to the value 'True' 3) Otherwise, if there is no such node then set the variable 'treesearch' to 'false'. The ALGOL-like pseudo-code is given below:-- procedure Treesearch (root, arg, m) / \textasteriskcentered arg is the data we are searching for, and m is a pointer to the node we are on \textasteriskcentered / if root = 0 then Treesearch : = false / \textasteriskcentered we are checking if there is still any root at all \textasteriskcentered / else begin d: = Data [root] if arg = d then begin m: = root Treesearch: = True end; else if arg > d then Treesearch: = Treesearch (left (root), arg, m); else Treesearch: = (right (root), arg, m) end;

Question:

It was once thought that a neutron is made up of an electron and proton held together by Coulomb attraction. Assume the neutron radius is 10^-15 m. (a) According to the uncertain-ty principle, find\Deltapfor such an electron, (b) The lowest average momentum such an electron could have would be (1/2)\Deltap. What would be the corresponding energy? (c) What is the electrostatic potential energy of an electron 10^-13 cm from a proton? (d) From the calculations in (b) and (c) does it appear likely that a neutron could be made up of an electron and a proton?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0988.htm

Solution:

(a) The uncertainty principle states that\Deltap\Deltax\cong (h/4\pi) , where h is Planck's constant,\Deltapis the un-certainty of the momentum of a particle, and\Deltaxis the un-certainty in the position of the particle. If we assume that the electron can be found anywhere within the neutron, then\Deltaxwill be of the order of the neutron radius 10^-15 m. p can then be found using the uncertainty prin-ciple . \Deltap= (h/4\pi\Deltax) = [(6.63 × 10^-34 Joule-sec)/(4\pi 10^-15 m)] = 5.28 × 10^-20 kg-m/sec (b) We must find a relation between energy E and momentum p. Since E = mc^2 = [m_0/\surd{1 - (v^2/c^2)}]^c(2) where m_0 is the rest mass of the electron, and v its speed, we have, upon squaring both sides of the equation, E^2[1 - (v^2/c^2)] = (m_0 c^2)^2 E^2 = [(E^2 v^2)/c^2] + (m_0 c^2)^2 = (E^2/c^4)c^2v^2 + (m_0 c^2)^2 = [(m^2 c^2)/c^4]c^2 v^2 + (m_0 c^2)^2 E^2 = p^2 c^2 + (m_0 c^2)^2.(1) Here, we used the definition of momentum as p =mv. We have m_0 = 9.11 × 10^-31 kg, c = 3 × 10^8 m/sec and p = (\Deltap/2) = [(5.28 × 10^-20)/2] (km/sec) = 2.64 × 10^-20 (km/sec). Upon substitution into (1), the result becomes E = \surd[(2.64 × 10^-20 (k-m/sec)^2 {3 × 10^8 (m/sec)}^2 + (9.11 × 10^-31 kg)^2 {3 × 10^8 (m/sec)}^4] = 7.46 × 10^-12 Since1eV= 1,6 × 10^-19 J = 7.46 × 10^-12 J × [(1eV)/(1.60 × 10^-19 J)] = 4.67 × 10^7eV. (c) By the definition of V V = [1/(4\pi\epsilon_0)] [{(e)(e)}/r] = [[(9.0 × 10^9 {(N - m)/coul^2} {(1.60 × 10^-19 coul)^2}] / {10^-13 cm (1m/10^2 cm)}] = 2.30 × 10^-13 J = 2.30 × 10^-13 J × [1eV/(1.60 × 10^-19 J)] = 1.44 × 10^6eV. (d) No. The energy of the electron is much greater than the energy available to bind the electron to the proton to form the neutron.

Question:

Associate a medicinal property for each of the following families of compounds: (a)Phenothiazines, (b) Barbiturates, and (c) Amphetamines.

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Solution:

(a)Phenothiazinesare one major type of tranquilizers used to treat psychoses. In general, tran-quilizers are compounds which act as depressants of the central nervous system with highly selective action on brain function. (b) There exist drugs that induce relaxation and sleep, which are sometimes used to treat patients with organic and emotional disorders. The barbiturate family is the most common and widely used for this sedative and hypnotic purpose. (c) Amphetamines belong to the class of anti-depressant drugs. These drugs, as their name implies, are used to treat depression disorders, one of the conditions associated with mental illness. They stimulate the central nervous system, and are used to bring about an increase in alertness, elevation of blood pressure and increase in heart action.

Question:

An LED used to display a decimal number has the following configuration. Design a decoder which will accept a four bit binary number, whose value will range from 0_10 to 9_10, and decode it to a seven bit number to be accepted by the LED. the LED will then display the decimal form of the four bit number.

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Solution:

Begin designing the decoder by listing a truth table. The inputs to the decoder will be A, B, C and D with A the most significant bit. The outputs are a, b, c, d, e, f and g. decimal number Input ABCD Output abcdefg 0 0000 1111110 1 0001 0110000 2 0010 1101101 3 0011 1111001 4 0100 0110011 5 1010 1011011 6 0110 1011111 7 0111 1110000 8 1000 1111111 9 1001 1111011 10 1010 XXXXXXX 11 1011 XXXXXXX 12 1100 XXXXXXX 13 1101 XXXXXXX 14 1110 XXXXXXX 15 1111 XXXXXXX Notice that all of the inputs greater than 9_10 have don't-care outputs. This is possible because the inputs will not be greater than 9_10. The don't care states will help in minimizing the expressions. Evaluate the sum of products form for each output, a(A, B, C, D) = \sum (0, 2, 3, 5, 6, 7, 8, 9) + d b(A, B, C, D) = \sum (0, 1, 2, 3, 4, 7, 8, 9) + d c(A, B, C, D) = \sum (0, 1, 3, 4, 5, 6, 7, 8, 9) + d d(A, B, C, D) = \sum (0,2, 3, 5, 6, 8, 9) + d e(A, B, C, D) = \sum (0, 2, 6, 8) + d f(A, B, C, D) = \sum (0, 4, 5, 6, 8, 9) + d g(A, B, C, D) = \sum (2, 3, 4, 5, 6, 8, 9) + d where d represents the don't care conditions d(A, B, C, D) = \sum10, 11, 12, 13, 14, 15 Construct a 4 variable K-map to minimize each Boolean expression. Note: X corresponds to don't care conditions The logic diagram of figure 1 is constructed from the minimized expressions a through g.

Question:

As you know 9 bits can address 512 words. Show how on a PDP-8 machine with 4096_10 words in memory, any word can be addressed using 9 bits (extend the memory addressing capa-bilities) .

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Solution:

When there were 512_10 words in memory, 9 bits were required to address any word location in the memory. Hence the Instruction Counter register (IC) and Memory Address Register (MAR) were to store 9 bits. The word length we want is 12 bits long therefore, the Memory Buffer Register (MBR) is 12 bits long. The scheme was as follows: When a word had an instruction in it, the first three bits specified operation code and the remaining 9 bits were sufficient. Now as there are 4096 words IC and MAR have to be of 12 bits. The MBR is of 12 bits as word length is 12 bits. Direct addressing of any word in the 4096 words will use a total 12 bits of word and no room (bits) is left for the OP-CODE. The scheme to extend the memory addressing capabilitiesis that is that the 4096_10 words memory is divided into pages. Each page has 128_10 (200_8) words. This is a logical division and not a physical division of memory as shown in Fig. 1. As each page has 128_10 words, 7 bits can address any word on the same page, but as there are 32 pages we need 5 bits to address a page. 5 bits for page + 7 bits for words on page = 12 bits and we end up still with the same problem of having only a total of 12 bits where the first 3 bits are for the OP-CODE and only 9 bits for address. To solve this, two bits, bit 3 and bit 4 are used as I bit and P bit respectively - as illustrated in Fig. 2 When the P bit is 1, the word is on the current page, where the instruction was, and the next address is found by concatenating the 7 address bits and the first five bits of the instruction counter. For example, if we are on page 2 with the IC pointing to 0410_8, the instruction at this location is 1352: 1352 = 001011101010. The first three bits 001 specify the ADD instruction to the accumulator ; the address of the operand is found by observing bits. Bit 4 = 1 i.e. P bit is 1, hence the operand is on the current page and its address is: first 5 bits IC concatenate 7 bits of address 0410_8 = 000100001000 Address = 000101101010 = 0552_8 and 552_8 is the word which is on page 2. If P bit = 0, the page referred is always Page 0 and the address is formed by 5 zeros followed by a 7 bit address in MBR. As an example, assume we are in page 2 with IC point-ing to 0410_8. Now suppose the instruction at this location was 1152_8 i.e. 1152_8 = 001001101010. Since P bit = 0, it references page 0, the OP-CODE 001 part specifies the ADD instruction and the address of the operand is 000001101010 = 0152_8 where this a word on page 0. By doing the above we have extended our addressing capabilities to address 256_10 words, 128 word on page 0 + 128 words on any current page. Yet the memory has 4096 words. A mechanism to address from one page to another is established using bit 3 as an indirect bit. When I = 0 it is Direct address and the address is ob-tained, depending on the P bit and the 7-address bits. When I = 1 it is Indirect address, i.e. the address specified by the address bit is not the address of the op-erand but the contents of the given address (location) are the address of the operand. For example, suppose data 25_10 is stored in location 100_10 and 100_10 is stored at 15_10. Then indirect addressing will be ADD I15 I being present, data at location 15 i.e. 100 is not to be added to accumulator; rather, the data of 100 has to be used as an address. At this address data = 25 is stored; this is to be added to accumulator. Now in the PDP-8 when I bit = 1 there are two condi-tions of the P bit. If I = 1, P = 0 then the page is zero page. On page zero, go to the location specified by 7 - address bits. The contents of this location, a 12 bit word, is the address of the operand. If I = 1, P = 1 then the page is the current page as given by instruction counter IC. Suppose we are on page 2 where the IC is pointing to 0410_8 brought in the MBR instruction1752_8, 1752_8 = 001111101010 OP-CODE = 001 is the ADD instruction. I bit = 1, P bit = 1 As in indirect addressing, the location specified by address bits has the address of the operand we need to find the lo-cation specified. As P = 1 the address of the location = first 5 bits of IC concatenated with 7 address bits = 000101101010 = 0552_8. Now location 0552_8 has the full 12 bit address of operand. These 12 bits can be the location of any of the 4096 words in memory. Thus by using the concept of I bit and P bit the addressing capability of memory is increased.

Question:

Modify the program of the previous problem to handle overflowif the capacity of the tank is finite.

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Solution:

Each time VOL is computed, check to see if it exceeds the given capacityCAP. If it does, indicate this and stop the simulation: DATA T/0.0/ READ N, TFIN, Q, VOL, CAP PRINT T, VOL REALN = N DT + (TFIN - T)/REALN DO 50 I = 1,N T = T + DT VOL = VOL + Q {_\ast} DT IF(VOL.GT. CAP) GO TO 60 PRINT T, VOL 50CONTINUE STOP 60PRINT 'OVERFLOW' STOP END

Question:

Two gases,HBrand CH_4, have molecular weights 81 and 16, respectively. TheHBreffuses through a certain small opening at the rate of 4 ml/sec. At what rate will the CH_4 effuse through the same opening?

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Solution:

The comparative rates or speeds of effusion of gases are inversely proportional to the square roots of their molecular weights. This is written rate_1 / rate_2 =\surdMW_2 /\surdMW_1 For this case =rate_HBr/ rate_(CH)4 =\surdMW_(CH)4 / \surdMW_HBr One is given therate_HBr, MW_(CH)4 andMW_HBrand asked to find rate _(CH)4 Solving for rate_(CH)4 : rate_HBr/ rate(CH)4=-\surdMW_(CH)4/ -\surdMW_HBrrate_HBr= 4 ml/sec. rate_(CH)4 = ? (4 ml/sec) / (rate_(CH)4) = \surd16 / \surd81MW_(CH)4 = 16 MW_HBr= 81 rate_(CH)4 = [(4 ml/sec) × \surd81] / \surd16 = [(4 ml/sec) × 9] / 4 = 9 ml/sec.

Question:

What velocity must a rocket be given so that it escapes the gravitational field of the earth?

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Solution:

In order to escape the earth's gravitational field, the rocket must travel an infinite distance from the earth. The change in potential energy, ∆V, of the rocket as it goes from the earth's surface to infinity is ∆V =GM_Em_R[1/R_E - 1/r] =GM_Em_R[1/R_E - 1/\infty] = (GM_Em_R)/(R_E). The gravitational acceleration at the surface of the earth is written as g = (GM_E)/(R^2_E) = 9.8 m/sec^2. Rewriting ∆V so that g appears in the expression, we have ∆V = (GM_E)/(R^2_E) ×R_Em_R=gR_Em_R. This is the potential energy that the rocket must acquire if it is to escape the earth's pull. This energy comes from the conversion of kinetic energy to potential energy. There-fore the initial kinetic energy given to the rocket must at least equal, this change in potential energy. Equating the two energy expressions, with R_E = 6.36 × 10^6 meters, ∆V = gR_Em_R= (1/2) m_Rv^2 v^2 = 2g R_E = 2 × 9.8 × 6.36 × 10^6 v= 1.12 × 10^4 m/sec. This minimum velocity needed to escape the earth's gravita-tional field is known as the escape velocity.

Question:

Explain how the following procedures are useful in debugging a program: a) tracing a function b) Stop control.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0380.htm

Solution:

When a function returns improper or unexpected results, the programmer can investigate what is happening in each instruction of his program by tracing its execution, using the trace function. Consider the following program: \nablaDOOMED [1]S \leftarrow X1, X2, ..... , XN [2]\DeltaS \Delta \Delta [3]S[\Delta] \Delta [4]INVS \leftarrow \divS [5]PS1S2 \leftarrow ((S(I) - S(I - 1)) \div (INVS(I) - INVS(I - 1)) [6]\rightarrow [8] × ʅ(PS1S2 < 0) [7]'THE NEGATIVE PRODUCTS ARE:'; PS1S2 [8]\rightarrow [10] × ʅ(PS1S2 < 100) [9]'THE POSITIVE PRODUCTS LESS THAN 100:'; PS1S2 [10]'THE POSITIVE PRODUCTS GREATER THAN 100:'; PS1S2 [11]\nabla If we run this program through, only negative products are printed. To see what is wrong we can put the function in trace mode, i.e. T\DeltaDOOMED \leftarrow 1 2 3 4 5 6 7 8 9 10 Where 1,2...,10 are the statement numbers. But the program is quite long. The mistake probably lies in statement [5]. Hence, instead of trying to trace the action of the system through the whole function we can concentrate on [5]. Key in TADOOMED \leftarrow [5]. The output is DOOMED DOOMED[5]-XIX2 DOOMED[5]-X3X2 DOOMED[5]-X4X3 \bullet\bullet \bullet\bullet \bullet\bullet DOOMED[5]-XNXN-1 Since all the products are negative, there is something wrong with instruction [5]. Converting into algebraic form: PS1S2 = [X_1 -X_2] / [(1/X_1) - (1/X_2)] = [X_1 - X_2] / [(X2- X_1) / (X_1X_2)] = [(X1- X_2) / (X_2 - X_1)] (X_1X_2) = [(X1- X_2) / -(X_1 - X_2)] (X_1X_2) = -(X1X_2). Statement [5] always results in a negative product. To en-sure correct calculations we can change [5] to [5]PS1S2 \leftarrow ((S(I) - S(I - 1)) \div (INVS(I - 1) - INVS(I)). b) The stop control mode halts program execution at a specific instruction so the programmer can examine it. For example, suppose we wish to inspect [6] of DOOMED. To activate the stop control mode, key in S\DeltaDOOMED \leftarrow 6 DOOMED The system responds with [6] \rightarrow [8] × ʅ(PS1S2 < 0). To end the stop control mode, key in S\DeltaDOOMED \leftarrow 10.

Question:

In the reaction, CaCO_3(s) \rightarrow CaCO(s) + CO_2 (g) at 950\textdegreeC and CO_2 pressure of 1 atm, the ∆H is found to be 176 kJ/mole. Assuming that the volume of the solid phase changes little by comparison with the volume of gas generated, calculate the ∆E for this reaction.

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Solution:

To solve this problem, you must know that ∆E, the change in energy, can be related quantitatively to the change in enthalpy, ∆H, the pressure (P) and the change in volume (∆V), by the formula ∆E = ∆H - P∆V. This equation means that if heat is added to a system at constant pressure, part of it goes into increasing the internal energy of the system (∆E) and the rest is used to do work on its surroundings, P∆V. Therefore, the total of both ∆E and P∆V is the heat content of the system, ∆H. To solve the problem, therefore, you need only substitute in the values for P, ∆V and ∆H. ∆V = V_products - V_reactants There is little change in the volume of the solid by comparison with the gas. This means, therefore, that ∆V is approximately equal to V_gas \bullet V_gas can be found from the equation of state; PV = nRt or V = nRt/P, where n = number of moles, P = pressure, R = universal gas constant and t = temperature in Kelvin (Celsius temp + 273 ). From the reaction, 1 mole of CO_2 gas is produced, thus n = 1. R is defined as being equal to 8.31 J mol^-1 deg^-1. You are given P and t. Therefore V = [1(8.13) (1223)] / [1] = 10.2 kJ Recalling, ∆E = ∆H - P∆V, you have ∆E = 176 - 1(10.2) = 166 kJ.

Question:

In the circuit shown in Fig. 1, find V_1 and the voltage v across both sources if i_s is given as 12A.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1094.htm

Solution:

In Fig. 2 observe that i = i_s + 0.6v_1 = i_1 + i_2. Since i_1 and i_2 both flow through a branch with the same equivalent resistances, they must be equal. Thus we have i = i_s + 0.6v_1 = 2i_1 = 2i_2. We can write v_1 = i_1R where R = 2-\Omega. Since i = 2i_1 we have i1= (1/ 2)i = (1 / 2) (i_s + 0.6 v_1) v_1 = i_1R = (1 / 2) (i_s + 0.6 v_1)R Substituting 2-\Omega of R and 12A for i_s, solve for v_1 v_1 = (1 / 2)(12 + 0.6v_1) 2 v_1 = (12 + 0.6v_1) 0.4v_1 = 12 givingv_1 =12 / 0.4 = 30V. v is found by calculating i and the equivalent resistance R_eq across the terminals of the two sources From Fig. 3 we find v = R_eq , i = 7.5(30) = 225V.

Question:

Calculate ∆H\textdegree_r for the combustion of methane, CH_4. The balanced reaction is CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) ∆H\textdegree_f in Kcal/mole = -17.89 , 0 , - 94.05 , 2(-68.32)

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Solution:

∆H\textdegree_r is the Standard Enthalpy Change of Reaction. Standard conditions are defined as 25\textdegreeC and 1 atm. Enthalpy is the heat content of a system. If the overall change in enthalpy is negative, then heat is given off to the surroundings and the reaction is called exothermic. When the change is positive, heat is absorbed and the reaction is endothermic. Endothermic compounds are often unstable and can sometimes explode. An endo-thermic compound, however, is a more efficient fuel because, upon combustion, it yields more heat energy. ∆H\textdegree_r is calculated using the enthalpies of formation, ∆H\textdegree_f. The sum of enthalpies of formation of products minus the sum of the enthalpies of formation of reactants, where each product and reactant is multiplied by its molar amount in the reaction as indicated by the coefficients, gives the value of ∆H\textdegree_r. In other words, ∆H\textdegree_r = \sum∆H\textdegree_f _products - \sum∆H\textdegree_f _reactants ∆H\textdegree_f for elements is always zero. In this reaction, therefore, ∆H\textdegree_f for O_2 is zero. \sum∆H\textdegree_f _products = - 94.05 - 136,64 \sum∆H\textdegree_f _reactants = - 17.89 Kcal/mole ∆H\textdegree_r = - 94.05 - 136.64 - (- 17.89) = - 212.8 Kcal/mole CH_4, burned This reaction is exothermic, because the ∆H\textdegree_r = a negative value.

Question:

A missile is fired radically from the surface of the earth (of radius 3.4 × 10^6 m) at a satellite orbiting the earth. The satellite appears stationary at the point where the missile is launched. Its distance from the center of the earth is 25.4 × 10^6 m. Will the missile actually hit the satellite?

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Solution:

To an observer outside the planet (see fig. A), the earth and the missile (of mass m) on the sur-face are rotating about the axis of the planet with angular velocity \omega_0. When the missile is fired radical-ly from the surface, its distance from the center of the earth increases and thus its moment of inertia (I = mr^2) about the rotation axis increases also. There are no forces with a moment about the rotation axis acting on the missile (the gravitational force of attraction acting on it is exerted along the rotational axis and has no moment). The net torque \cyrchar\CYRG acting on the missile is then zero. According to the rigid body analogue of Newton's second law, if L is the magnitude of the angular momentum of the missile, then \cyrchar\CYRG = (dL)/(dt) But\cyrchar\CYRG = 0 and0 = (dL)/(dt) or L = constant at all times. But L = mvr where v is the tangential velocity of the missile. Since v = \omegar where \omega is the angular velocity of the missile, then L = m\omegar^2 = I\omega. Thus since L and m are constant at all times, as the missile moves farther away from the earth and closer to the satellite (i.e., r increases) then \omega must decrease. We are given that the satellite appears stationary at the point where the missile is fired. Thus the radius vector passing through the launching pad and the satellite continues to rotate with angular velocity \omega_0. The missile has an angular velocity \omega which drops more and more from the value \omega_0 as the missile rises. At the height of the satellite, the moment of inertia of the rocket about the axis of rotation is I_1 = mr^2_s = m × (25.4 × 10^6 m)^2 ,whereas, at the launching pad, its moment of inertia is only I_2 = mr^2_e = m × (3.4× 10^6 m)^2 .Thus, finally, if \omega is the satellite's angular velocity at height r_s, and \omega_0 its angular velocity at launching (equal to that of the earth) then I_1\omega= L = I_2\omega_0 \omega/\omega_0= (I_2)/( I_1) = (mr^2_e)/(mr^2_s) = [(3.4× 10^6 m)^2]/[(25.4 × 10^6 m)^2 ] = 0.018. The missile thus moves further and further from the vertical as it rises and will miss the satellite (unless the missile is fitted with a homing device). To an observer on the planet, the departure of the missile from the vertical is, of course, also observed and is explained in terms of the Carioles force associated with a rotating frame of reference.

Question:

Outline a breeding procedure whereby a true breeding strain of red cattle could be established from a roan bull and a white cow.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0657.htm

Solution:

This problem deals with establishing a pure line having a trait which is codominant. A pure line is a strain of organisms that when inbred, always produce offspring having the same phenotype. They are homozygous and can never be heteroyzgous because when heterozygotes are inbred, they produce offspring of more than one phenotype, (which is not true breeding). Color in cattle is determined by two alleles which are codominant. Neither allele is dominant or recessive to the other. In the heterozygous state, each allele acts to produce its own gene product independent of the other, and the total product is a sum of the contributions from each allele. Phenotypically, this usually results in a trait which is intermediate between the two dominant alleles. In this case, red and white color are codominant characteristics, and roan color is the intermediate phenotype of a heterozygote. Let R respresent the gene for red color in cattle and W the gene for white color. By definition then, red cattle have the genotype RR, and white cattle have the genotype WW, and roan cattle are RW. (Note that we no longer use capital and lower case forms of the same letter to represent the alleles, since neither is dominant or recessive). Now since red cattle by definition must be homozygous, any red cattle produced will breed true when inbred. Therefore, in order to establish a pure line of red cattle from a roan bull and a white cow, all we need do is obtain male and female red cattle among the off-spring. Then these red cattle should only be bred among themselves. In the cross between the roan bull and the white cow: No red cattle were obtained in the F_1. The only way to obtain red cattle would be to cross two of the roan cattle from the F_1. In the mating then: In the F_2, 1/4 of the offspring will be red, and thus homozygous for red color. They will always breed true.

Question:

A plane parallel plate capacitor consisting of two metal circular plates 5 cm in radius separated 1 mm in air, is charged to 300 stat-volts, whereupon it is connected in parallel to another similarly charged capacitor (positive terminals connected together and negative terminals connected) (see Figure A). How much energy would be re-- leased if the combinations were discharged by a short circuit?

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Solution:

For a plane parallel plate capacitor: C = KA / 4\pid This result can be obtained as follows. According to the definition of capacitance C, C = Q / V where Q is the total charge on one plate, and V is the potential difference between the plates. According to Gauss's law, if the Gaussian Surface is constructed as shown in Figure B, then (K / 4\pi) \varphiE= (K / 4\pi)EA = Q(1) in the CGS system. This relation holds because the electric field E is a constant in the parallel plate capacitor. Q is the charge enclosed by the Gaussian Surface. It is also the total charge on either plate, due to the construc-tion of the surface. Also, for a parallel plate capacitor V = Ed(2) Therefore combining (1) and (2), we get the result for the capacitance of the parallel plate capacitor. \therefore C_1 =\pir^2 / 4\pid = 5^2 / (4 × 1 mm × 1cm / 10 mm) = 25 / .4 = 62.5 stat\rule{1em}{1pt}farads. Recalling that the energy stored in a capacitor is W = (1 / 2)QV = (1 / 2) CV^2 = (1/2)(Q^2 / C) and choosing the second form because C and V are known W_1 = (1 / 2)C_1V_1^2 = (1 / 2)(62.5)(300)^2 = 31.25(90,000) =2,820,000 ergs But the total energy stored in the two capacitors is W_1 + W_2 = 2W_1 \thereforeW = 2(2,820,000) = 5,640,000 ergs =.564 joulesAns. This is the energy available and able to be released if the combination were discharged by a short circuit. Solution in Mks Units: Data: plate radius = 5 × 10^\rule{1em}{1pt}2 m d =10^\rule{1em}{1pt}3 m V = 9 × 10^4 volts C = (1 / 4\pik)(A / d) = (\pi25 × 10^\rule{1em}{1pt}4) / (4\pi9 × 10^9 × 10^\rule{1em}{1pt}3) = 6.94 x 10^11 farads But 1 farad = 9 × 10^11 stat\rule{1em}{1pt}farads \thereforeC = 6.94 × 10^\rule{1em}{1pt}11 × 9 × 10^11 = 62.5 stat-faradsCheck. W = (1 / 2) C_1V_1^2 = (1 / 2)6.94 x 10^\rule{1em}{1pt}11 × 81 × 108 = 28.1 × 10^\rule{1em}{1pt}2 joules But 1 joule = 10^7 ergs. \thereforeW = 28.1 × 10^\rule{1em}{1pt}2 × 10^7 = 2.81 x 10^6 ergs Total W = 2 × 28.1 × 10^\rule{1em}{1pt}2 = .562 joules = 5.62 × 10^6 ergsCheck.

Question:

(a) Find the current in a circuit consisting of a coil and capacitor in series, if the applied voltage is v = 110 volts, frequency f = 60 cycles/sec; the inductance of the coil is L = 0.80 henry; the resistance of the coil is R = 50.0 ohms; and the capacitance of the capacitor is C = 8.0 \muf. (b) Find the power used in the circuit.

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0772.htm

Solution:

(a) The coil can be represented by an inductance and resistance in series (see figure A). The effective impedance of the inductor and capacitor depend on frequency. Since the inductor and capacitor cause the current to either lead or lag the voltage by a phase shift of 90\textdegree, the impedance Z is the vector addition of their combined reactance and the pure resistance (see figure B). X_L = 2\pifL = 2\pi(60) (0.80) ohms = 300 ohms X_C = (\rule{1em}{1pt}1/2\pifC) = {\rule{1em}{1pt}1/(2\pi × 60 × 8.0 × 10^\rule{1em}{1pt}6)} ohms = \rule{1em}{1pt} 330 ohms X_L + X_C = 300 - 330 = \rule{1em}{1pt} 30 ohms Z = \surd{R^2 + (X_L + X_C)^2} = \surd{(50)^2 + (300 - 330)^2} ohms = \surd{50^2 + (\rule{1em}{1pt} 30)^2} ohms = 58 ohms I= (V/Z) = (110 volts/58 ohms) = 1.9 amp (b) The power is equal to P = I^2R or P = VI cos \textphi where is the angle between the impedance and the pure resistance in the circuit, (see figure b) cos \textphi = (R/Z) = (50 ohms)/(58 ohms) = 0.86 P= VI cos \textphi = 110 volts × 1.9 amp × 0.86 = 180 watts. Using P = I^2R as a check, P = I^2R = (1.9 amp)^2 × 50 ohms = 180 watts.

Question:

Determine the power required to maintain a temperature difference of 20\textdegree C between the faces of a glass window area 2 m^2 and thickness 3 mm. Why does a much lower power suffice to keep a room with such a window at a temperature 20\textdegreeC above the outside? The thermal conductivity of glass is 25 × 10^-4 cal\bullets^-1\bulletcm^-1\bulletC deg^-1.

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Solution:

The equation appropriate to thermal conductivity is dQ/d\cyrchar\cyrt = - k A dt/dx(1) where dQ/d\cyrchar\cyrtis the rate of heat transfer across a cross section of area A of a slab of material of thermal conductivity k. dt/dx is the temperature gradient in the material. Now, if we assume that the heat transfer is occurring as a steady state process, the temperature of each point of the slab will be time independent. If this is the case, dQ/d\cyrchar\cyrtis the same at all cross- sections of the slab. However, by (1), this means that dt/dx is constant at all cross-sections. Hence dt/dx decreases linearly along the slab. Using the figure, dt/dx = (t_2 - t_1)/(x_2 - x_1) But x_2 - x_1 = L, the slab thickness, whence dt/dx = (t_2 - t_1)/L anddQ/d\cyrchar\cyrt= -k A [(t_2 - t_1)/L] Since t_2 < t_1 , we may rewrite this as dQ/d\cyrchar\cyrt= k A [(t_1 - t_2)/L] In this equation, t_1 and t_2 are the temperatures of the slab at positions x_1 and x_2. (See figure.)In this particular case, t_1 = t_2 + 20\textdegreeC and dQ/d\cyrchar\cyrt= 25 × 10^-4 cal\bullets^-1\bulletcm^-1\bulletC deg^-1 × 2 × 10^4 cm^2 × 20 C deg/0.3 cm = 3.33 × 10^3 cal \bullet s^-1. Here we have used the fact that 2 m^2 = 2 × 10^4 cm^2 and3 mm = .3 cm These conversions follow from the definitions of a cm. and mm. Since 1 cal = 4.19 Joules dQ/d\cyrchar\cyrt= 3.3 × 10^3 cal\bullets^-1 = (3.3 × 10^3 cal\bullets^-1)(4.19 J\bulletcal^-1) dQ/d\cyrchar\cyrt= 13.95 × 10^3 J\bullets^-1. Thus 3.33 × 10^3 cal = 13.95 × 10^3 J are required per second to replace the lost heat. The power required is thus 13.95 kilowatts (kW). A much smaller power than this is required in practice because the inner surface of the window and the air in contact with it drops in temperature because of the heat loss through the glass. Heat is conducted from the rest of the room through air, the thermal conductivity of which is very low. The inner surface of the window is thus not maintained at a temperature 20\textdegreeC above the outside. A similar effect will occur on the outside of the window. The temperature difference across the window may well drop to only a few degrees, in which case only a fraction of the above power needs to be supplied, giving a much more reasonable figure for the heat that needs to be supplied per second.

Question:

Aspaceship moving away from the Earth at a velocity v_1 = 0.75 c with respect to the Earth, launches a rocket (in the direction away from the Earth) that attains a velocity V_2 = 0.75 c with respect to the spaceship. What is the velocity of the rocket with respect to the Earth?

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Solution:

Newtonian relativity states that the laws of mechanics are the same in all inertial reference frames but that the laws of electrodynamics are not. Since such a theory was not acceptable, the theory of relativity was developed and resolved this problem. One of the statements of this new theory was that the velocity of light is independent of any relative motion between the light source and the observer. Experiments have shown this to be true. In support of this statementEinstein showed that ordinary mechanical velocities do not add algebraically. Instead, for the addition of two velocities v_1 and v_2 , the sum is V = [ (v_1 + v_2) / {1 + (v_1v_2 / c^2)}] For small velocities (v < < c), this reduces to V = v_1 + v2, as in Newtonian mechanics, since the term v_1v_2 / c^2 is very small compared to one and can be ignored. Since the velocities in this problem are large, the velocity of the rocket with respect to the Earth is V = [ (v_1 + v_2) / {1 + (v_1v_2 / c^2)}] = (0.75c + 0.75c) / [1 + {(0.75c)(0.75c) / c^2 }] = 1.5c / (1 + 0.5625) = 0.96c Therefore, in spite of the fact that the simple sum of the two velocities exceeds c, the actual velocity relative to the Earth is slightly less than c .

Question:

Why would one call a flock of birds a \textquotedblleftsociety\textquotedblright?

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Solution:

A society is a group of animals that belongs to the same species and is organized in a cooperative manner. The animals are usually bound together by re-ciprocal communication, which leads to the cooperative behavior. A flock of birds can be considered an elementary form of society called a motion group. Motion groups in-clude schools of fish and herds of mammals. Birds within a flock communicate with one another in order to stay together as they move from place to place. The coopera-tive behavior displayed by a flock of birds enables them to more efficiently detect and avoid predators. It is more likely that a flock of birds will detect a predator than a solitary bird. Many birds have special signals, such as alarm cries and wing movements, that seem to alert the entire group to the presence of a predator. Membership in a group also affords protection once the predator attacks the group. For example, starlings usually fly in a wide, scattered pattern, but assemble quickly in tight formation when attacked by a falcon, their natural predator. The falcon usually attacks by swooping at high speed into the group. By assembling into a tight group, the starlings make it difficult for the falcon to hit one bird without collid-ing into others and possibly injuring itself. Other cooperative behavior patterns which may be performed by a flock of birds are the discovery of food and the coordination of reproductive activities.

Question:

Discuss the citric acid cycle as the common pathway of oxidation of carbohydrates, fatty acids and amino acids.

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Solution:

The oxidation of the above substances all yield products which either enter or are intermediates of the citric acid (also known as Krebs and TCA, tricarboxylic acid) cycle. Glucose, a carbohydrate, undergoes a series of reactions in glycolysis to be oxidized to pyruvic acid. Pyruvic acid reacts with coenzyme A to yield acetyl coenzyme A. It is acetyl coenzyme A which enters the citric acid cycle (see Figure). Acetyl coenzyme A is also formed from the oxidation of fatty acids. A repeating series of reactions cleaves the long carbon chain of a fatty acid into mole-cules of acetyl coenzyme A. Finally, amino acids are metabolized by one of several reactions to products capable of participating in the reactions of the TCA cycle. The first step in the oxidation of all amino acids is a deamination reaction by which an amino group (\ElzbarNH_2) is removed from the amino acid. For three amino acids, this is the only reaction required for conversion into a compound entering the TCA cycle directly. Alanine undergoes deamination to form pyruvate which is converted into acetyl coenzyme A. The deamination of glutamic acid yields \alpha-ketoglutaric acid,while that of aspartic acid yields oxaloacetic acid. Both these compounds are intermediates of the TCA cycle. The TCA cycle is thus the final common pathway by which carbohydrates, fatty acids and amino acids are metabolized.

Question:

A small metallic ball is charged positively and negatively in a sinusoidal manner at a frequency of 60 cps. The maximum charge on the ball is 10^-6coul. What is the displacement current due to this alter-nating charge? What would the displacement current be if the charging frequency were 10^6 cps?

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Solution:

The charge on the ball can be written as q = q_0 sin\omegat where q_0 is the maximum charge on the ball (10^-6coul), \omega = 2\pif (f = 60 cps) and the sine function expresses the sinusoidal variation of the charge. According to Gauss's law, the total electric flux through a Gaussian surface about the sphere is given by \varphi_E= \int E^\ding{217} \bulletds^\ding{217} = q/\epsilon_0 = q_0 sin\omegat/\epsilon_0 The displacement current i_0 is defined as i_0 \equiv \epsilon_0 [(d\varphiE)/dt] = \omegaq_0 sin\omegat=dq/dt(1) It is called a displacement current because even though there is no actual movement of charge from one position to another, the mathematical form of i_0 has the form of an actual charge flow. (See equation (1)). The maximum value of the displacement current occurs when \vertcos\omegat\vert = 1. Then i_(D)max = \omegaq_0 = 2\pifq_0 At 60 cpsi_(D)max = 2\pi × 60 sec^-1 × 10^-6coul = 3.77 × 10^-4 amp If the frequency were 10^6 cps, then i_(D)max = 2\pi × 10^6 sec^-1 × 10^-6coul = 6.28 amp

Question:

Explain how the planarian changes its body shape.

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Solution:

In the planarian, three kinds of muscle fibers can be differentiated: an outer, circular layer of muscle just beneath the epidermis; an inner, longi-tudinal layer; anddorsoventralmuscles that occur in strands. Contraction of the longitudinal muscles causes constriction of the body, whereas contraction of the circular muscles causes an elongation of the body. Other alterations of body shape are produced by contraction of the dorsoventral muscle strands. Thus, by coordinated contraction and relaxation of these muscles, the body shape of a planarian can be varied.

Question:

Calculate the normality of a solution containing 2.45 g of sulfuric acid in 2.00 liters of solution. (MW of H_2SO_4 = 98.1.)

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Solution:

Normality is defined by the following equation. Normality = [(grams of solute)/(equivalent weight × liters of solution)] In this problem, one is given the grams of solute (H_2SO_4,) present and the number of liters of solution it is dissolved in. One equivalent of a substance is the weight in which the acid contains one gram atom of re-placeable hydrogen. This means that when the acid is dissolved in a solution, and it ionizes, that in one equivalent weight, one hydrogen atom is released. The equivalent weight for acids is defined as: equivalent weight = [(MW)/(no. of replaceable H)] When H_2SO_4 is dissolved in a solution, there are two replaceable H as shown in the following equation: H_2SO_4 \leftrightarrows 2H^+ + SO-_4 This means that in calculating the equivalent weight for H_2SO_4 the molecular weight is divided by 2. equivalent weight of H_2SO_4 = [(98.1 g)/(2 equivalents)] = 49 g/equiv The normality can now be calculated. Normality = [(grams of solute)/(equiv. wt × liters ofsoln)] Normality = [(2.45 g)/(49 g/equiv × 2.0 liters)] = 0.025 equiv/liter.

Question:

What is the illumination 5 feet from a 75 candlepower lamp?

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Solution:

Illumination is defined as flux divided by area [E = (F/A)] . Making the simplifying assumption that our lamp is a point source, the total flux that it 4\piI where I is intensity of the source .If we construct a sphere enveloping the point source with the point source at its center , and having a radius of 5 ft., we canfind the illumination at this distance. Hence , E = F/A =4\piI /4\pir2= I / r2 And Illumination= 75 candlepower / (5 × 5) ft^2 = 3 foot-candles .

Question:

A solution of 10.0 g of HF in 500 g H_2O freezes at - 1.98\textdegreeC. Calculate the degree of ionization of HF. (M.W. HF = 20.0 The Freezing point depression of H_2O is 1.86\textdegree.)

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Solution:

When an acid is dissolved in solution, it dissociates into ions. For example, when one mole of HF is dissolved in H_2O, one mole of H^+ and one mole of F^- are present after complete dissociation. The freezing point of a solution is dependent on the number of particles in the solution, which means that the freezing point of a solution will be lowered more by a compound which ionizes than by the same amount of a compound which does not ionize. The degree of ionization of a compound is a measure of what percent of the compound is ionized when it is placed in a particular solution. In this problem, one is told that when 10 g of HF is added to 500 g of H_2O the original freezing point of the water (0\textdegree) is lowered to - 1.98\textdegreeC. The freezing point depression is related to the concentration of particles in the solution by the statement freezing point depression =molality× freezing point constant The freezing point constant of water is 1.86\textdegree. This means that one mole of a substance (except those substances which ionize) dissolved in 1000 g of water lowers the freezing point 1.86\textdegree. Since HF ionizes, we cannot use itsmolalityin this equation. You can use the effectivemolality, which is the sum of themolalitiesof H^+, F^- and HF. Themolalityof H^+ and F^- will be equal to the degree of ionization of the HF. The concentrations of H^+ and F^- will be equal because, when HF ionizes, one H^+ and one F^- will be formed. This reaction is given by the equation HF \leftrightarrows H^+ + F^- To solve for the degree of ionization of HF, one must: (a) find the molalityof HF as if it were not ionized, (b) define a variable for the molalitiesof H^+ and F^- ; here, x will be used, (c) find themolalityof HF after ionization is taken into account, (d) find the effectivemolalityof the species, (e) use the effectivemolalityin the freezing point depression equation to solve for x, themolalityof H^+ and F^- . (a) Themolalityof HF before ionization. Themolalityis defined as the number of moles of solute in 1000 g of solvent. The number of moles of HF can be found by dividing the number of grams present by the molecular weight. number of moles = [(number of grams)/(molecular weight)] moles of HF = [(10.0 g)/(20.0 g/mole)] = 0.5 moles Themolalitycan now be found by dividing the number of moles of HF by the number of kg of water present. molality= [(no. of moles of HF)/(no. of kg of H_2O)] molality= [(0.5moles)/(0.500 kg)] = 1.0 moles/kg (b) x =molalityof H^+ =molalityof F^- (c) After ionization of xmolalof H^+ and F^-, themolalityof HF will be 1.0 - x. (e) The effectivemolalityof the species will be equal to the sum of themolalitiesof all of the species present. The effectivemolality=molalityof H^+ +molalityof F^- +molalityof HF effectivemolality= x + x + (1.0 - x) = 1.0 + x. To solve for x, the concentration of H^+ and F^-, the effectivemolality of the species will be used in the freezing point depression equation. freezing pointdepreesion= eff.molality× freezing pt constant The freezing point depression in this case is 1.98\textdegree and the freezing point constant of water is 1.86\textdegree. Solving the equation 1.98\textdegree = (1.0 + x) × 1.86\textdegree 1.0 + x = (1.98\textdegree/1.86\textdegree) x = .06 The percent of ionization is themolalityof H^+ and F^- divided by the molalityof the unionized HF multiplied by 100. degree of ionization = [(molalityof ion)/(molalityof unionized species)]× 100 degree of ionization =(0.06/1.00) ×100 = 6% Hence, the HF in this solution is 6 % ionized.

Question:

For the general reaction A + B \rightarrow C + D, what is the effect on the number of collisions between A and B of (a) tripling the concentration of each, (b) quadrupling the concentration of each?

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Solution:

For any given chemical reaction, the number of collisions is proportional to the rate. The Law of Mass Action states that the speed of a chemical reaction is proportional to the product of the concentrations of the reacting molecules. The number of collisions in the above reaction depends upon the concentration of A and B. If the concentration of A is tripled the number of collisions is tripled, then if the concentration of B is tripled the number of collisions will triple again. Thus the number of collisions is increased 3 × 3 times or 9 times. When the concentrations of A and B are quadrupled a similar method is used. The number of collisions is increased 4 × 4 or 16 times.

Question:

How many Btu are required to raise the temperature of 2 pounds of water from 32\textdegreeF to 212\textdegreeF?

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Solution:

The temperature rise is 180 degrees Fahrenheit. The number of Btu necessary is 2 × 180\textdegree = 360 Btu, since one Btu is required to raise the temperature of one pound of water one degree Fahrenheit.

Question:

A group of students were invited to taste phenylthiocarbamide (PTC). The ability to taste PTC is inherited by a single pair of genes and tasting (T) is dominant to non-tasting (t). Among 798 students, 60.4 percent were tasters, a) Calculate the allelic frequency of T and t. b) How many of the students were TT? Tt? tt?

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Solution:

Putting this problem in terms of the Hardy- Weinberg principle, let p represent the frequency of the T allele and q the frequency of the t allele. We know that p^2 + 2 pq + q^2 = 1 describes the distribution of phenotypes in the population, where p^2 equals the frequency of TT, 2 pq equals the frequency of Tt, and q^2 equals the frequency of tt. Note that the sum of ail proportions, or frequencies, equals 1. a) We are told that 60.4% of our given population are tasters. Since tasting is dominant to nontasting, tasters can have one of two genotypes, TT and Tt. This means that the sum of the frequencies of the TT and Tt genotypes is .604. Since all frequencies total 1, we can infer that the frequency of nontasters (tt) in the sample of students is 1- .604 or .396. Thusq^2 = .396, and q = \surd.396 or .629. Since p+q = 1, p = 1-.629 or .371. The frequency of T is therefore .371, and that of t, .629. b) frequency TT = p^2 = (.371)^2= .137 frequency Tt = 2p = 2(.371)(.629)= .467 frequency tt = q^2 = (.629)^2= .396 1.0 1.0 To convert the actual number of students carrying each genotype, we multiply the corresponding genotypic frequencies (or proportions) by the total sample size (i.e., 798): TT = (.137) (798) = 109 Tt = (.467) (798) = 373 tt = (.396) (798)=316 798

Question:

The average concentration of particulate lead in a city with highly polluted air is 5\mug/m^3. If an adult respires 8500 liters of air daily and if 50 per cent of the particles below 1\mumin size are retained in the lungs, calculate how much lead is absorbed in the lungs each day. Of the particu-late lead in urban air, 75 per cent is less than 1\mumin size.

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Solution:

To solve this problem (1) determine the total percentage of particulate lead remaining in the lungs, (2) determine the amount of lead absorbed into the lungs. The problem states 75 % of particulate lead is less than 1\mumin size, therefore, 25 % of the lead is greater than 1\mumin size and does not remain inside the lungs. Of the 75 % particulate lead less than 1\mumin size, only 50 % of this is retained in the lungs. The total percentage of lead retained is 37.5 %. The total volume of air-intake is 8500 liters, so that total volume of air intake = (8500) (10^-3 m^3/liter) = 8.5 m^3 The total amount of lead-intake is amount of lead = (8.5 m^3)(5 \mug/m3) = 42.5\muglead. The total amount of lead absorbed is amount of lead absorbed = (42.5\mug)(0.375) = 15.9\mug.

Question:

How was it proven that DNA replication is semi-conservative?

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Solution:

Three possible models v/ere considered to account for DNA replication, the process by which the two complementary strands of the double helical DNA replicate to form new complementary strands. In the first model, called conservative replication, the two strands of parental DNA replicate to yield the unchanged parental DNA and a newly synthesized DNA. The second model, semi-conservative replication, claims that the replication of one DNA molecule yields two hybrids, each composed of one parental strand and one newly synthesized strand. Finally, the third model, dispersive replication, has the parental strand breaking at intervals. These parental segments then combine with new segments to form the daughter strands(see Figure 1). Meselson and Stahl's classic experiments proved that DNA replicates by a semi-conservative mechanism. They grew E. coli cells for several generations on a medium which contained "heavy" nitrogen, ^15N. Thus, all the nitrogen bases of the E. coli DNA were labeled with ^15N instead of the normal or "light" nitrogen, ^15N. The ^14N - DNA of these cells could be separated from the ^14N - DNA by equilibrium centrifugation in a cesium-chloride gradient . because the ^15N - DNA has a significantly greater density than the ^14N - DNA. Meselson and Stahl took the "heavy" (^15N containing) cells and placed them in a 11 light" (lf\textasteriskcenteredN containing) medium. The cells were allowed to grow for several generations and were analyzed at various generation times. From the samples collected, the DNA was examined by determining its buoyant density with cesium-chloride density gradients. It was found that after one generation, the isolated DNA showed only a single band falling midway between the places where the heavy and light bands lie in a cesium-chloride density gradient, This is to be expected if the first generation DNA is a hybrid of one parental strand and one newly synthesized strand. After two generations, the isolated DNA exhibited two bands, one equal to the hybrid density of the first generation and one having a density equal to that of normal "light" DNA. After many generations passed, a \textquotedblleftlight" density band of DNA predominated. These observations are consistent with those expected from the hypothesis of semi-conservative replication but not with the other two models (see Figure 2) .

Question:

What is meant by a gene pool, balanced polymorphism, and genetic load?

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Solution:

Population geneticists visualize a population of organisms as a collection of all the genes possessed by the individuals of the population. Such a sum total of genes is termed a gene pool. Each allele in the gene pool has a certain probability of being expressed in the population. The frequency of occurrence of a given genotype in the population is given by the product of the frequencies of its alleles in the gene pool. A gene pool is said to be in equilibrium if the frequency of each individual allele remains constant from generation to generation; that is to say, when there is no change in the genetic composition of the population over a period of time. The term polymorphism refers to the existence in the same interbreeding population of two or more distinct phenotypic forms of a genetically determined trait. The human blood groups 0, A, B, and AB are a classic example of polymorphism. Balanced polymorphism is a state in which the different forms of the polymorphic genotype are maintained together in equilibrium in the population over a period of time, Such a balance can be achieved and maintained through a variety of means, involving, what are often complex genetic-environmental interactions. One condition which produces balanced polymorphism is heterozygote superiority or heterosis. Here the heterozy-gote (Aa) has a survival advantage over both the dominant (AA) and recessive (aa) homozygotes. Thus both alleles are maintained in the population, and neither can eliminate the other. A classic example of this is sickle-cell anemia in a malarial environment. Contrary to what one might expect, the sickle-cell trait is maintained at a relatively high frequency in the population despite its obvious harmful effects. Individuals carrying the recessive sickle-cell gene in the heterozygous state have resistance to malarial infection. When both malaria and sickle-cell anemia are strong selection factors, the heterozygote has superiority, since normal homozygotes are subject to malaria and those homozygous for the sickle-cell gene have sickle-cell anemia. Thus the recessive gene is maintained along with the normal gene, and their traits are kept in balance. Often, a given species maintains different forms in the population because one or the other is favored at varying times in the year. The polymorphism gives the species an overall survival ability throughout the year. In geographic polymorphism, the species population may maintain a variety of forms, each of which is advantageous under specific environment conditions; one form may do better in one area, another in a different area. On the whole, the species is thus able to maintain itself in a variety of areas and environmental conditions, allowing it to become more widespread. Balanced polymorphism is important in evolution because it maintains a certain amount of variability in the population to permit further adaptive changes. However, although a certain amount of variability is advantageous, too great a genetic variety in the population can be harmful. Geneticists reason that different forms have differing effects upon the overall fitness of a population. Given that there is one optimal phenotype, the greater the number of other suboptimal types present, the less the average fitness of the population will be. The presence of these suboptimal forms imposes a condition of genetic load in terms of a reduction in average population fitness. It is possible that the population could be driven to extinction as a result of too heavy a genetic load.

Question:

Find the logarithm of 30,700.

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Solution:

First express 30,700 in scientific notation. 30,700 = 3.07 × 10^4. 4 is the characteristic. To find the mantissa, see a table of common logarithms of numbers The mantissa is 4871. Thus log 30,700 = 4 + .4871 = 4.4871.

Question:

Plot the following function on a K-map and simplify in SOP and POS forms. F(A,B,C,D) = \sum(0,2,5,8,10,13,14,15) + X(1,11,12). By using the simplified expression, determine the output when a redundant input occurs.

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Solution:

The K-map is shown in figure 1. The "X" mark at locations one, eleven, and twelve denotes the redundancies. First simplify in the SOP form. Minterms at 13,14, and 15 and the redundancy at 12 combine to form AB. Minterms at 0,2,8, and 10 combine to formBD. Minterms at 5 and 13 combine to form BCD. The SOP form is: F = AB +BD+ BCD To simplify F into the POS form, simplifyFin SOP form and then use DeMorgan's Law. From the K-map it is seen that the 0's at 6 and 7 combine to formABC. The 0 at 4 and the redundancy at 12 combine to form BCD. The 0's at 3 and 9 and the redundancies at 1 and 11 combine to formBD. The SOP ofFis F=ABC + BCD+BD Apply DeMorgan's Law toFgives the POS form of F. F = (A+B+C) (B+C+D) (B+D) The table of fig. 2 shows the output when redundancies occur.

Question:

Calculate the electrostatic force on q1due to q_2(F_12) and the force on q_2 due to q_1( F_21 ) for the case illustrated in the figure.

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Solution:

Using Coulomb's law and doing the calculation in the CGS system, of units we have F_12 ={(+3 atat C)(+5 stat C)} / (5cm)^2 = + 0.6 dyne (to the left) F_21 = {(+5 stat C) (+3 stat C)} / (5 cm)^2 = + 0.6 dyne (to the right) Both forces have the same magnitude (Newton's third law for electrostatis forces) and are repulsive.

Question:

Produce a circuit for coding a binary coded decimal digit input into the Hamming code.

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Solution:

A binary coded decimal digit has four input lines D_0,D_1,D_2,D_3. The number K of parity bits required with 4 information input lines = 3 (see the previous problem). Let these 3 parity check bits be C_1, C_2, C_3 Hence, the coded information will be of the type:C_1 C_2 D_3 C3D_2 D_1 D_0 C_1 C_2 D_3 C3D_2 D_1 D_0 (see problem 3 of this chapter). And, let us name these 7 lines as: d_1 d_2 d_3 d_4 d_5 d_6 d_7 Now, by the theory of Hamming code, one must have an even parity between C_1, D_0, D_2, D_3; C_2, D_0, D_1, D_3; andC_3, D_0, D_1, D_2. In other words, one must ensure that the exclusive-or sum of each of the four variables in the above 3 sets equals to 0. i.e.,C_1 \oplus D_0 \oplus D_2 \oplus D_3 = 0, C_2 \oplus D_0 \oplus D_1 \oplus D_3 = 0, C_3 \oplus D_0 \oplus D_1 \oplus D_2 = 0, The circuit for generating C_1, C_2, and C_3 can nowbe conceived by noting that the parity check bit should be a 1 if the exclusive-or of the other noting that the parity check bit should be a 1 if the exclusive-or of the other three variables = 1 (i.e., there is an odd number of variables which are equal to 1), and the parity check bit should be = 0 if the exclusive-or of the other three variables = 0 . Hence, the circuit can be drawn as shown in Fig. 1. Note in the above figure, that all gates A to F are 2-input exclusive-or gates. Gate A gives the exclusive-or of D_2 and D_3.Output of gate A is 1 if only of D_2 or D_3 is a 1 . This output is again exclusive-ored with D_0 .If the output of A was a 1, then, if D_0 is 1, the output of B will be 0, i.e., C_1 = 0, as required. In each case it can be verified that the check bits C_1, C_2 ,C_3 take on values 0 or 1 so as to maintain an even-parity in their respective groups.

Question:

Explain the operation of NOR gates, and generate OR, NOT, and AND operations using NOR gates. Draw symbol diagrams, and explain the functions of each system with the help of truth tables.

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Solution:

The NOR gate is derived by combining the functions of an OR operator followed by a NOT operator. The key to remembering the function of a NOR gate is the first row of the truth table of fig. 1; ab a + b f(a, b) 00 01 10 11 0 1 1 1 1 0 0 0 Fig. 1f(a, b) =a+b The output of a NOR gate is 1. if and only if both inputs are simultaneously 0. The block diagram of a NOR gate is shown in fig. 2. NOR gates may also be used to generate AND, OR, and NOT operations. As seen in fig. 3, an OR gate can be constructed using two NOR gates, When the inputs a and b into gate 1 are both 0, the output of gate 1 (which is the input of gate 2) is 1. Therefore, the output of gate 2 is 0, following the truth table given for NOR gates. When one or both inputs are 1 the output of the OR gate is 0. The truth table of fig. 3(a) illustrates this principle throughly; ab F(a, b) 00 01 10 11 0 1 1 1 Fig, 3(a)f(a, b) = a + b Construction of a NOT gate using NOR gates: Operation of the circuit in fig. 4 is illustrated by its truth table given in fig. 5; a f(a) 0 1 1 0 Fig. 5 Fig. 6 shows the construction of an AND gate using NOR gates. 1st and 2nd NOR gates act as NOT gates and supplya, andbinto the third NOR gate. The output of this gate is then;by the use of DeMorgan's Law.

Question:

What is the rotational inertia about an axis through the What is the rotational inertia about an axis through the center of a 25-kg solid sphere whose diameter is 0.30 m?

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Solution:

For a rigid body rotating with angular speed \omega about a fixed axis, the kinetic energy is K = (1/2) mv^2 = (1/2) m(\omegar)^2. Each particle of this body can be considered as contributing to the total kinetic energy. The angular velocity, \omega, of all the particles is the same, but their distance r from the axis of rotation varies. Therefore, the total kinetic energy can be written as K = (1/2) (m_1r^2_1 + m_2r^2_2 + ... ) \omega^2 = (1/2)\sum(m_ir^2_i)\omega^2 where the summation is taken over all the particles in the rigid body. The rotational inertia, I, is defined as I = \sum m_ir^2_i. As can be seen from the above equations, the ro-tational energy of a body, for a given angular speed \omega, depends on the mass of the body and the way that mass is distributed around the axis of rotation. Since most rigid bodies are not composed of discrete point masses but are continuous distributions of matter, the summation for I in the above equation becomes an integration. Let the body be divided into infinitesimal elements of mass dm at a distance r from the axis of rotation. Then the rotational inertia is I = \int r^2 dm where the integral is taken over the whole body. For a solid sphere of radius R, dm = \rho dV where \rho is the density of the sphere and dV is an infinitesimal volume. For dV, take a circle of radius l and of thickness r d\texttheta, where l is the distance from the axis of rotation. We have dV= (2\pil) (dr) (rd\texttheta) = 2\pi (r sin \texttheta) r drd\texttheta = 2\pi r^2 sin \texttheta dr d\texttheta \rho= m/v = m/[(4/3)\piR^3] Then I= \int l^2 dm = \int (r sin \texttheta)^2 \rhodv = _\texttheta=0\int^\pi _r=0\int^R (r^2 sin^2 \texttheta) [m / (4/3)\piR^3](2\pi r^2 sin \texttheta dr d\texttheta) = _\texttheta=0\int^\pi _r=0\int^R {(3m)/(2 R^3)} r^4 sin^3 \texttheta dr d\texttheta = _\texttheta=0\int^\pi ({(3m)/(2 R^3)} (r^5/5) sin^3 \texttheta]^R_r=0 ) d\texttheta = 3/10 m R^2^\pi\int_\texttheta=0 sin^3 \texttheta d\texttheta = 3/10 m R^2^\pi\int_\texttheta=0 sin \texttheta (1 - cos^2 \texttheta) d\texttheta Let x = - cos \texttheta and dx = sin \texttheta d\texttheta. Then I= 3/10 m R^2^\pi=-cos\pi\int_x=-cos0\textdegree- (1 - x^2) dx = 3/10 m R^2 {x - (x^3/3)}]^1_-1 = 3/10 m R^2 [1 - (1/3) + 1 - (1/3)] = [(3/10) mR^2][4/3] = 2/5 mR^2 For the given sphere, the mass is 25 kg and the radius is 0.30 m/2 = 0.15 m. Its rotational inertia is then I= (2/5)mR^2 = (2/5) (25 kg) (0.15 m)^2 = 0.22 kg-m^2.

Question:

The Berkeley synchrocyclotron was designed to accelerate protons to a kinetic energy of 5.4 x 10^-11 J, which corresponds to a particle speed of about 2 × 10^8 m/s. At this speed the relativistic mass increase is important and must be taken into account in the design of the accelerator. Calculate the percentage increase in the proton mass en- countered in this instrument.

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Solution:

The relation between relative mass, m, and rest mass, m_0, is given by m = m_0 / \surd(1 - v2/ c^2)' = m_0 / \surd[1 - {(2 × 10^8) / (3 × 10^8)}^2] = m0/ \surd{1- (4/9)} = m_0 / {\surd(5/9)} = 3m0/ \surd5 = 1.34 m_0 . The percentage increase in mass is the fractional increase in mass times 100, or [(m - m_0) / m_0] × 100 = [(1.34m_0 - m_0 ) / m_0] × 100 = 34 per cent. The proton mass at a speed of 2 × 10^8 m/s is 34 per cent greater than its rest mass.

Question:

An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to be I_1. When he joins a second, identical cell in series with the first, the current becomes I_2 . When he connects the cells in parallel, the current through the circuit is I_3. Show that the relation he finds between the currents is 3I_2 I_3 = 2I_1 (I_2 + I_3).(See figures (a), (b) and (c)).

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Solution:

Let the emf of any of the cells be \epsilon and its internal resistance be r. Let the external circuit have a resistance R. When a single cell is used (see figure (a)) R and r will be in series. Hence, the equivalent circuit resistance R_eq is R_eq = R + r Ohm's Law yields I_1 = [(\epsilon_net )/(R_net )] = [(\epsilon)/(R_eq )] = [(\epsilon)/(R + r)] If two identical cells are connected in series, their emf's act in the same sense. Hence, again using Ohm's Law (see figure (b)) I_2 = [(\epsilon_net )/(R_net )] = [(2\epsilon)/(R + r)] When the cells are connected in parallel, since they are identical, by the symmetry of the arrangement identical currents I_0 must flow through each cell. Further, since no charge accumulates at point A in this circuit, (by conservation of charge) I_3 = I_0 + I_0 = 2I_0 Considering the passage of current through either cell, we have V_AB = \epsilon - I_0 r = \epsilon - (I_3 /2)r. This follows because, each charge passing through \epsilon is raised in potential an amount \epsilon. However, by Ohm's Law, each charge also loses potential l_0 r by crossing the battery's internal resistance. When we consider the passage of current through the external circuit, then V_AB = I_3 R, by Ohm's Law. Hence I_3 R = \epsilon - (I_3 r/2) Rewriting the three equations obtained, we find that R + r = (\epsilon/I_1 ), R + 2r = (2\epsilon/I_2 ), and R + (r/2) = (\epsilon/I_3 ) We must eliminate all resistance variables to obtain the given formula. Solving the first equation for r, and substituting this value in the second equation, we obtain r = (\epsilon/I_1 ) - R R + 2 [(\epsilon/I_1 ) - R] = (2\epsilon/I_2 ) or- R = (2\epsilon/I_2 ) - (2\epsilon/I_1 ) R = 2\epsilon [(1/I_1 ) - (1/I_2 )](1) Inserting the calculated value of r in the third equation R + (1/2) [(\epsilon/I_1 ) - R] = (\epsilon/I_3 )or(1/2) R = (\epsilon/I_3 ) - (1/2) (\epsilon/I_1 ) R = \epsilon [(2/I_3 ) - (1/I_1 )](2) Dividing equations (1) and (2) gives 1 = [{2[(1/I_1 ) - (1/I_2)]} / {(2/I_3) - (1/I_1)}] (2/I_3 ) - (1/I_1 ) = (2/I_1 ) - (2 /I_2 ) or(2/I_3) = (3/I_1) - (2/I_2) Multiplying both sides by I_1 I_2 I_3 2I_1 I_2 = 3I_2 I_3 - 2I_1 I_3 or3I_2 I_3 = 2I_1 (I_3 + I_2 ) \therefore2I_1 I_2 - I_2 I_3 = 2I_2 I_3 - 2I_1 I_3 .\therefore3I_2 I_3 = 2I_1 (I_2 + I_3 )

Question:

Theketoneacid (CH_2CO_2H)_2CO undergoes a first-order decomposition in aqueous solution to yield acetone and carbon dioxide: (CH_2CO_2H)_2CO \rightarrow (CH_3)_2CO + 2CO_2 (a) Write the expression for the reaction rate, (b) The rate constant k has been determined experimentally as 5.48 × 10^-2/sec at 60\textdegreeC.Calculate t_1/2 at 60\textdegreeC. (c) The rate constant at 0\textdegreeC has been determined as 2.46 × 10^-5/sec. Calculate t_1/2 at 0\textdegreeC. (d) Are the calculated half-lives in accord with the stated influence of temperature on reaction rate?

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Solution:

(a) For a chemical decomposition, the rate of the reaction is equal to the product of the rate constant (k) and the concentration of the compound decomposing. Thus, Rate = k [(CH_2CO_2H)_2CO] (b)Because the rate is only proportional to [(CH_2CO_2H)_2CO], the reaction is first-order. For a first- order reaction,. the half-life (t_1/2) is related to k by the following equation: t_1/2 = 0.693 / k Solving for t_1/2 : t_1/2 = (0.693) / (5.48 × 10^-2 /sec) = 12.65 sec. (c) One can solve for t_1/2 at 0\textdegreeC using the same equation. t_1/2 =(0.693) / (2.46 × 10^-5/sec)= 2.82 × 10^4 sec. (d) In general, the speed of a chemical change is approximately doubled for each ten degrees rise in tem-perature. The temperature rises 60\textdegree, from 0\textdegreeC to 60\textdegreeC, therefore, the rate should double six times or the ratio of the t_1/2, at 0\textdegreeC to the t_1/2 at 60\textdegreeC is 2^6. 2^6 = 64. (t_1/2 0\textdegree) /(t_1/2 60\textdegree) = (2.82 × 10^4 sec) / (12.65 sec) = 2.23 × 10^3 This is much greater than the expected ratio of 64.

Question:

Explain why the hydrolysis of an ester to yield alcohol and acid is best carried out in basic rather than acidic solution. Use equations to illustrate your answer.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0774.htm

Solution:

The best way to solve this problem is to consider what products will be formed with acid and base. The general formula of an ester is . The carboxylic acid and alcohol are generated. But esters are generated from carboyxlic acid and alcohols by the use of H+. In other words, there exists nothing to prevent the reaction from going back in the other direction. That is, you have the following equilibrium: Let us consider the reaction with base. With the basic reaction, you generate an alcohol and a carboxylate anion, not an acid. This is the key. If the base is KOH, is produced. If NaOH you have Due to the fact that the carboxylate anion, which forms a salt, is generated, the reaction is prevented from going in the opposite direction. As such, your yields upon hydrolysis are greater with the base than with acid.

Question:

A neutron traveling at a speed of 2 × 10^3 m/s collides with a nucleus and rebounds with a speed of 3 × 10^2 m/s. (a) Determine the work done by the force of interaction between the nucleus and neutron and (b) estimate the strength of the force if the distance over which the collision occurred is 2 × 10^5 m. The mass of a neutron is 1.67 × 10^-27 kg.

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Solution:

The mass of the neutron, m = 1.67 × 10^-27 kg, its initial speed v_0 = 2 × 10^3 m/s, its final speed, v = 3 × 10^2 m/s, and the distance over which the particles interacted, d = 2 × 10^-15 m, are the known observables. The work done by the force of the interaction of the particle, W, and the force F are the unknown observables. (a) Assume that the nucleus experiences no change in velocity due to the collision with the neutron. Then the total change in energy of the system is due only to the change in the kinetic energy of the neutron. This change in energy is equal to the work W. W = ∆KS = (1/2) m(v^2 - v^2_0) W = (1/2) (1.67 × 10^-27 kg) (3 × 10^2 m/s)^2 - (1/2) (1.67 × 10^-27 kg)(2 × 10^3 m/s)^2 = 7.5 × 10-2 3J - 3.34 × 10^-2 1 J = - 3.27 × 10-2 1J The minus sign indicates that the work done by the force decreased the kinetic energy of the neutron. (b) Although it is not stated that the force is constant during the collision, assume that it is and also assume that it is parallel to the displacement vector. Using these assumptions, the force can be determined from F = W/d = (-3.75 × 10-2 1Nm)/(2 × 10^-15m) = - 1.64 × 10^-6 N

Question:

A cart of mass 5 kg moves horizontally across a frictionless table with a speed of 1 m/sec. When a brick of mass 5 kg is dropped on the cart, what is the change in velocity of the cart?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0300.htm

Solution:

Assume that the brick has no horizontal velocity when it is dropped on the cart. Its initial horizontal momentum is therefore zero. Since no external horizontal forces act on the system of cart and brick, horizontal momentum must be con-served. We can say, for the horizontal direction, m_cv_ci+m_bv_bi=m_cv_cf+m_bv_bf Since the final velocities of the brick and cart are the same, m_cv_ci= (m_c +m_b)v_f Substituting values, v_f= (m_cv_ci)/(mc+m_b) = [(5 kg) (1 m/sec)]/[(5 kg + 5 kg)] = .5 m/sec The change in velocity of the cart is v_f-v_ci_ = (0.5 - 1.0)m/sec = - 0.5 m/sec.

Question:

At t_1 = 0 an automobile is moving eastward with a velocity of 30 mi/hr. At t_2 = 1 min the automobile is moving north-ward at the same velocity. What average acceleration has the automobile experienced?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0015.htm

Solution:

Since velocity is a vector quantity, vector addition must be used to solve this problem. Geometrical-ly, when two vectors are added, the tail of the second vector is placed at the head of the first and the re-sultant vector is drawn from the tail of the first to the head of the second. To find the difference between the two velocities, we write v_2\ding{217} - v_1\ding{217} = \Deltav\ding{217} Changing the expression above into one including only addition: v_2\ding{217} = \Deltav\ding{217} + v_1\ding{217} This is shown in the accompanying vector diagram. The magnitude of \Deltav is (refer to the figure and use the Pythagorean theorem) \Deltav = \surd[(30 mi/hr)^2 + (30 mi/hr)^2] = \surd[1800 (mi/hr)^2] = 42.4 mi/hr The magnitude of the average acceleration is a= (\Deltav/\Deltat) = [(42.4 mi/hr)/60 sec] = 0.71 (mi/hr)/sec The direction of \Deltav, and hence the direction of a is, from the figure, in the direction northwest.

Question:

Explain and exemplify internal and external parasitism.

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0766.htm

Solution:

Parasites can be classified as external or internal parasites. The formerlive on the outer surface of their host, usually feeding on the hair, feathers, scales or skin of the host or sucking its blood. Internal parasites maylive in the various tubes and ducts of the host's body, such as the digestivetract, respiratory passages or urinary ducts. They may also bore intoand live embedded in tissues such as muscle or liver. However, pathogenslike viruses and some bacteria are believed to actually live insidethe individual cells of their host. External parasitism may have evolved from what was once a commensalrelationship. In some cases, internal parasitism may have evolvedfrom external parasitism. External parasites might have wandered intoone of the body openings of their host, such as the mouth nasal openings, or anus. Most of them cannot survive in the different environmentthey encounter there. But it is possible that in the course of thousandsor millions of years some wander-ers might have had the geneticconstitutions enabling them to survive in the new habitat they foundinside the host's body. From such a beginning, the specializations thatordinarily characterize the more advanced internal parasites could haveevolved. Other internal parasites may have evolved from free living formsthat were swallowed or inhaled. Internal parasitism is usually marked by much more extreme specializationsthan external parasitism because the habitats available insidethe body of a living organism are completely unlike those outside it. The unusual internal conditions have resulted in adaptations different from thoseseen in free-living forms. For example, internal parasites have frequentlylost individual organs or whole organ systems that would be essentialto a free-living species. Tapeworms, for instance, have no digestivesystem. They live in their host's intestine, where they are bathed byproducts of the host's digestion, which they can absorb directly across theirbody wall without having to digest anything themselves. Loss of certain structures is not the only sort of special adaptation commonlyseen in internal parasites. They often have body walls highly resistantto the de-structive enzymes and antibodies of the host. Moreover, theyfrequently possess specialized structures such as the specialized headsof tapeworms with hooks and suckers that enable them to anchor themselvesto the intestinal wall. One of the most striking of all adaptations of internal parasites concernstheir life cycles and repro-ductive capability. Individual hosts do notlive forever. If the parasitic species is to be perpetuated, a mechanism isneeded for leaving one host and penetrating another. Thus, at some pointin their life cycle, all internal parasites must try to move from one hostindi-vidual to another. But this is seldom simple. Rarely can a parasite movedirectly from one host individual to another. It is not unusual for a life cycleto include two or three intermediate hosts and a free living larval stage. But such a complex development makes the chances that any one larvalparasite will encounter the right hosts in the right sequence exceedinglypoor. The vast majority die without completing their life cycle. As a result, internal parasites characteristically produce huge numbers of eggs. Although they may be deficient in some organs through specializationthey usually have extremely well-developed reproductive structures. In fact, some internal parasites seem to be little more than a sac of highly efficient reproductive organs.

Question:

Ribonuclease is 1.65% leucine and 2.48% isoleucine by weight. Calculate its minimum molecular weight.

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Solution:

If Ribonuclease is 1.65% by weight leucine, .0165 times the minimum molecular weight of ribonuclease is equal to the molecular weight of leucine. (MW of leucine = 130) MW of Leu = MW of Ribonuclease × .0165) MW of Ribonuclease = (130) / (.0165) = 7.89 × 10^3 If Ribonuclease is 2.48% by weight isoleucine (MW = 131), one can solve for the molecular weight of ribonuclease using a similar equation as the one used for leucine. MW of Ribonuclease= (MW of Ile) / (.0248) = (131/.0248) = 5.29 × 10^3 Therefore, to cover both conditions, the minimum molecular weight must be 7.89 × 10^3.

Question:

Discuss the possible sequence of events in the evolution of photosynthesis.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0217.htm

Solution:

There are two types ofphotophosphorylationcyclic andnoncyclic. It has been proposed that cyclicphotophosphorylationis the more primitive of the two processes and perhaps was the earliest form of energy production. It might have developed at a time when the earth's atmosphere contained little or no oxygen gas. Cyclicphotophosphorylation probably served as a means of synthesizing ATP for primitive organisms living in an anaerobic environment. The next evolutionary step was per-haps the development ofnoncyclicphotophosphorylation, such as that found in certain bacteria today. In these bacteria, NADP^+ molecules are the electron acceptors but the electron donors that restore the excited chlorophyll molecules to ground state are molecules such asthiosulfateor succinate , not water. The third step, achieved by green plants, was probably the ability to use water molecules innoncyclic photophosphorylation . By evolving such an ability, green plants were able to live nearly everywhere and were not restricted to places where thiosulfate orsuccinateare found. Then, as plants spread and proliferated, they released into the atmosphere oxygen gas, a by - product of their noncyclic photophosphorylation. The accumulation of oxygen made possible the further biochemical evolution of animals and other organisms that use aerobic respir-ation (oxygen) for survival and active growth.

Question:

What potential difference V in an electron gun is required to accelerate an electron that was initially at rest to a final speed of 1 × 10^7 m/s?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0625.htm

Solution:

The electron starts from rest at the negative plate and ac-celerates straight across to the positive plate and then escapes through a small hole to form the electron beam. (See figure). By the work- energy theorem the change in kinetic energy of the electron is equal to the work done on the electron by the net force acting on it. Looking at the diagram, the net force on the electron is that due to the elec-tric field E^\ding{217}. By definition, E^\ding{217} =F^\ding{217}/q where F^\ding{217} is the force on charge q. Since q = \rule{1em}{1pt}e, the electronic charge, F^\ding{217} = qE^\ding{217} = \rule{1em}{1pt}eE^\ding{217} Hence, the work done on the electron is W= ^b\int_a F^\ding{217}\bullet dr^\ding{217} = ^b\int_a \rule{1em}{1pt}eE^\ding{217} \bullet dr^\ding{217} where the integral is evaluated over a path between points a and b. Using the work-energy theorem, (1/2) mv^2 _b \rule{1em}{1pt} (1/2)mv2a= \rule{1em}{1pt}e^b\int_a E^\ding{217}\bullet dr^\ding{217}. Here, v_b and vaare the electron speeds at points a and b, and m is its mass. Since the electron starts from rest, va= 0 and 1/2 mv^2b= \rule{1em}{1pt}e^b\int_aE^\ding{217}\bullet dr ^\ding{217}(1) By definition, \rule{1em}{1pt} ^b\int_a E^\ding{217}\bullet dr ^\ding{217} = v_b\rule{1em}{1pt}va(2) where v_b ,v_aare the potentials at points b and a. Then, using (2) in (1) 1/2 mv2_b = e(v_b\rule{1em}{1pt}_ v_a) Hence v_b\rule{1em}{1pt}_ va= mv2b/ 2e Using the given speed and the known values of electron mass and charge, v_b\rule{1em}{1pt}_ va= [(9.11 x 10^\rule{1em}{1pt}31kg) (10^14 m^2/ s^2)] / [2(1.6× 10\rule{1em}{1pt}19coul)] v_b\rule{1em}{1pt}_ va= (9.11 x 10^\rule{1em}{1pt}17 joules) / (3.2 × 10^\rule{1em}{1pt}19 Coul) But1 (Joule / coul) = 1 volt v_b\rule{1em}{1pt}_ va= 2.8 × 102volts

Question:

Liquid naphthalene normally freezes at 80.2\textdegreeC. When 1 mole of solute is dissolved in 1000 g of naphthalene, the freezing point of the solution is 73.2\textdegreeC. When 6.0 g of sulfur is dissolved in 250 g of naphthalene, the freezing point is 79.5\textdegreeC. What is the molecular weight of sulfur?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0241.htm

Solution:

In order to determine the molecular weight of sulfur, we must determine how many moles of sulfur corre-sponds to 6.0 g. When 1 mole of solute is added to 1000 g of naphthalene, the normal freezing point is lowered by 7\textdegreeC, from 80.2\textdegreeC to 73.2\textdegreeC. This corresponds to a concentration of 1molal(1 mole solute / 1000 g solvent) = (1molal). The ratio of sulfur to naphthalene in the sulfur-naphthalene solution is (6.0 g sulfur) / (250 g) naphthalene = 0.024. Thus, adding 6.0 g of sulfur to 250 g of naphthalene is equivalent to adding 24 g of sulfur to 1000 g of naphthalene (because 24 g sulfur / 1000 g naphthalene is also 0.024). The observed freezing point depression in the sulfur- naphthalene solution is 0.7\textdegreeC (from 80.2\textdegreeC to 79.5\textdegreeC). This is the freezing point depression one would obtain by adding 24 g of sulfur to 1000 g of naphthalene. But this is one- tenth the lowering one would obtain by adding 1 mole of sulfur (the lowering would then be 7.0\textdegreeC), hence 24 g of sulfur must correspond to one-tenth of a mole. Thus, the apparent molecular weight of sulfur is (24 g / 0.1 mole) = (240 g / mole).

Question:

Iron (III) oxide is reacted with carbon monoxide in a blast furnace to produce iron. The balanced reaction is: Fe2O_3 + 3CO + 2Fe + 3CO_2 What volume of CO at STP is required to completely use up 31.94 kg of iron oxide? (MW of Fe_2O_3 = 159.7, MW of CO = 28.)

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Solution:

From thestoichiometryof the above reaction, one sees that 1 mole of Fe_2 O_3 requires 3 moles of CO to react. Thus, three times as many moles of CO will react with a given number of moles of Fe_2 O_3 . Once the number of moles of CO is known, one can use the fact that 1 mole of any gas at STP has a volume of 22.4 l to determine the volume of CO required. The number of moles of Fe_2 O_3 available for the re-action is the weight of Fe_2 O_3 in grams divided by its molecular weights : moles Fe_2 O_3 = [(weight Fe_2 O_3) / (MW Fe_2 O_3)] = [(31940 g) / (159.7 g/mole)] = 200 moles. The number of moles of CO needed to completely react is 3(200 moles) = 600 moles. The volume at STP of 600 moles of CO is the number of moles of CO times the volume of 1 mole: volume CO at STP = (moles CO) (22.4 l/mole) = (600moles) (22.4 l/mole) = 13.440 l.

Question:

Two vessels, filled with liquids at temperatures T_1i and T_2i are joined by a metal rod of length L, cross-section A and thermal conductivity k. The masses and specific heats of the liquids are m_1, m_2 and c_1, c_2 respectively. The vessels and rod are thermally insulated from the surrounding medium. What is the time t required for the temperature difference to be halved?

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/Users/wenhuchen/Documents/Crawler/Physics/D16-0544.htm

Solution:

Let the vessels have instantaneous temperatures T_1 and T_2 (as shown in the figure) with T_1 > T_2. After they are joined by the metal rod, heat flows from vessel 1 to vessel 2 at a rate heat/sec = dQ/dt = kA[(T_1 - T_2)/L].(1) As heat is transferred, the temperatures of the vessels tend to equalize. Therefore the rate of heat transfer de-creases in time. The temperature variations dT_1 and dT2caused by the transfer of heat can be obtained by noting that the same heat dQ which leaves vessel 1 must enter vessel2. Therefore dQ = c_1m_1(-dT_1) = c_2m_2dT_2 where (-dT_1) indicates a decrease in T_1. Substituting for dQ in (1), we get [-c_1m_1dT_1]/dt = [{kA(T_1 - T_2)}/L]and(2) [c_2m_2dT_2]/dt= [{kA(T_1 - T_2)}/L](3) Let us rearrange equations (2) and (3) as follows dT_1/dt = - [kA(T_1 - T_2)] / [c_1m_1L_1],(4) dT_2/dt =[kA(T_1 - T_2)] /[c_2m_2L_2].(5) Then, if we subtract (5) from (4), we obtain the rate of change of the temperature difference T_1 - T_2 with time: (dT_1/dt) - (dT_2/dt)= [d(T_1 - T_2)]/dt = -(kA/L)[(1/c_1m_1) + (1/c_2m_2)](T_1 - T_2). The solution of the above differential equation gives the time dependence of (T_1 - T_2): (T_1 - T_2) = (T_1 - T_2)_initial e-(kA/L)[(1/c1m1) + (1/c2m2)]t where (T_1 - T_2)_initial is the temperature difference at t = 0. The time t required for (T_1 - T_2) to equal one half of its initial value (T_1 - T_2)_initial is e-(kA/L)[(1/c1m1) + (1/c2m2)]t= [(T_1 - T_2)(t)]/[(T_1 - T_2)_initial] = (1/2). Taking the natural logarithm of both sides of the above equation, we get -(kA/L)[(1/c_1m_1) + (1/c_2m_2)]t = 1n (1/2) = - 1n 2. ort = L/[kA(1/c_1m_1) + (1/c_2m_2)] ln 2 = (Lc_1c_2m_1m_2)/[kA(m_1c_1 + m_2c_2)] 1n 2.

Question:

A specific enzyme is found in the endoplasmic reticulum of a cell. The code for this enzyme is contained in a molecule of DNA in the nucleus. Start with this DNA and trace all steps involved in the production of the enzyme and its movement to the endoplasmic reticulum.

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Solution:

This question deals with a series of processes beginning with transcription (production of mRNA), followed by translation (production of a polypeptide chain) and ending with the passage of the polypeptide into the endoplasmic reticulum. Inside the nucleus, transcription of the DNA is mediated by RNA polymerase, an enzyme composed of five polypeptide subunits comprising the core, plus a loosly bound protein known as the sigma factor. The DNA undergoes a localized unfolding in the vicinity of the gene to be transcribed. The weak hydrogen bonds are broken and the two strands separate. Only one of the strands within a particular genome serves as a template for transcription by RNA polymerase. The sigma factor is responsible for re-cognizing the promoter region and attaches the enzyme core, to the DNA where transcription is initiated. Without the sigma factor RNA polymerase would bind at any place along the template and transcription would proceed on both strands. Once the core has been properly attached, the sigma factor can be released to become associated with another core polymerase molecule. Several core molecules may be trans-cribing a certain genome simultaneously. RNA polymerase selects precursor ribonucleotides complementary to the DNA template. That is, adenine on the template is paired with uracil, cytosine is paired with guanine, and thymine with adenine. The precurser ribonucleotides consist of a base attached to a ribose tri-phosphate molecule. Other enzymes catalyze phosphate bond formation between adjacent ribonucleotides, and the new strand of mRNA is held together by these bonds in the back-bone. The polymerization of the mRNA is driven by the energy of hydrolysis of the triphosphate of the precursor ribonuc-leotide. The direction of transcription is specific. RNA polymerase reads from the 5' end of the DNA, with subsequent synthesis of mRNA in the 3' to 5' direction. The newly synthesized single-stranded mRNA peels away from the template allowing a new RNA polymerase molecule to attach or the DNA strand to reunite. Now the synthesized mRNA has to move into the cytoplasm where protein synthesis occurs. The mRNA is able to diffuse into the cytoplasm because the nuclear membrane surrounding the nucleus has numerous rather large openings (nuclear pores) which allow the mRNA to pass through. Once the mRNA gets into the cytoplasm, a ribosome attaches to the 5' end of the mRNA and translation begins. Several ribosomes may bind sequentially one mRNA strand, forming a polyribosome or polysome. The sequence of genetic information present in the mRNA directs the order of in-corporation of amino acids into the protein chain. But before the amino acids can line up in the correct sequence, several things must happen. Amino acids have to be oriented in the correct sequence on the ribosome. This is achieved by transfer RNA, or tRNA, whose function is to transport amino acids to the ribosome and translate the genetic code of the mRNA. To be transported, an amino acid must first be activated. This occurs when the amino acid, which is bound to a loading enzyme, joins with a molecule of ATP, forming the activated amino acid.The activated amino acid is then transferred to the correct tRNA, which carries it to the ribosome, (see fig. 1) The high energy bond originally derived from the activation by ATP is retained in the amino acid \sim tRNA complex. The activating enzyme is now free to bind another amino acid. The amino acid must be transferred to a tRNA molecule which is specific for the amino acid carried by the activating enzyme. Therefore, an activating enzyme must be able to re-cognize both a specific amino acid and a tRNA specific for that amino acid. There are at least twenty different ac-tivating enzymes and each one is specific for one amino acid and tRNA. There are also at least 20 different tRNA's, each one is specific for one amino acid or the DNA strands to reunite. The correct sequencing of the amino acids is specified by the codon, a sequence of three bases on the mRNA. Each group of 3 bases codes for an amino acid. The tRNA has an anticodon, also a nitrogenous base triplet, which is comple-mentary code to the codon specific for the amino acid carried by that tRNA. It is through the complementation of the codon and anticodon that the right order of amino acids is obtained. As each amino acid gets into the right position in the sequence, an enzyme, peptidyl transferase, forms the peptide bond between two adjacent amino acid residues, (see fig.2). The reaction is driven by the splitting of the high energy bond between the amino acid and the tRNA. As more and more amino acids are linked together, a polypeptide chain is formed. When translation is complete, this chain forms the enzyme that the DNA coded for. Ribosomes are composed of two subunits, the 30S and 50S subunits. These two subunits will dissociate upon reaching the stop codon at the 3' end of the mRNA and the newly synthesized polypeptide chain is released. The sub-units can reassociate at the 5' end of the mRNA, forming a functional 70S complex, and translation of the mRNA may proceed again. The formation of the polypeptide begins with the amino terminal end. In bactelrial polypeptides, the amino terminal residue is always N-formy1-methionine, in which a hydrogen atom on the amino group of methionine has been replaced by a formyl group. The function of this modified residue is not entirely clear, but it may be involved in the protection of the growing polypeptide from digestion by various peptidases. To get the enzyme into the endoplasmic reticulum, transport across the membrane of the endoplasmic reticulum has to occur. The actual mechanism of this transport is not known but there is a fairly sound hypothesis to account for it. This hypothesis states that when the enzyme is first formed on the ribosome, it has not assumed its three- dimensional shape, but remains an unfolded chain. In this conformation, the unfolded chain is transported across the membrane into the interior of the endoplasmic reticulum. We have to recall the fact that polyribosomes are associated with the endoplasmic reticulum. So the polypeptide chain made on theribosome is next to the membrane of the endo-plasmic reticulum. Once it gets across the membrane and reaches the interior, the polypeptide chain becomes folded into its normal three-dimensional configuration. Because it isnow folded, it can no longer squeeze out through the membrane the way it came in. A summary of the transfer of genetic information ispresented below:

Question:

Given that ∆H\textdegree_CO2(g) = \rule{1em}{1pt} 94.0, ∆H\textdegree_CO(g) = \rule{1em}{1pt} 26,4, ∆H\textdegree_H2O(l) = 68.4 and ∆H\textdegree_H2O(g) = -57.8 Kcal/mole, determine the heats of reaction of (1) CO(g) + (1/2)O_2 (g) \rightarrow CO_2 (g), (2) H_2(g) + (1/2) O_2 (g) \rightarrow H_2O(l) and (3) H_2O(l) \rightarrow H_2O(g).

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Solution:

To solve this problem, you need to know what heat of reaction means and how to determine it quantitati-vely. The heat of reaction may be defined as the heat released or absorbed as a reaction proceeds to completion. It is measured quantitatively from the heats of formation of products minus heats of formation of reactants. Namely, heat of reaction, ∆H\textdegree = ∆H\textdegree _products - ∆H\textdegreereactants. (Note: By convention, the ∆H\textdegree of any element is defined as being equal to zero.) You proceed as follows: (1) ∆H\textdegree = ∆H\textdegree_CO2(g) -∆H\textdegree_CO(g) - (1/2) ∆H\textdegree_O2(g) = - 94 - (- 26.4) - (1/2) (0) = - 67.6 Kcal/mole. (2) ∆H\textdegree = ∆H\textdegree_H2O(l) - ∆H\textdegree_H2(g) - (1/2) ∆H\textdegree_O2(g) (2) ∆H\textdegree = ∆H\textdegree_H2O(l) - ∆H\textdegree_H2(g) - (1/2) ∆H\textdegree_O2(g) = - 68.4 - 0 - (1/2) (0) = - 68.4 Kcal/mole. (3) ∆H\textdegree = ∆H\textdegree_H2O(g) -∆H\textdegree_H2O(l) = - 57.8 - (- 68.4) = + 10.6 Kcal/mole.

Question:

The density of a liquid can be measured by means of a "pyknometer" (see the figure). The pyknometer is a glass vessel with a ground glass stopper having a narrow hole along its axis. If you are given the mass, m_p, of the empty pyknometer, the mass m_p+l, of the pyknometer when filled with the liquid and the mass m_p+w of the pykno-meter filled with distilled water (all masses being de-termined at the same temperature), calculate the density of the liquid.

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Solution:

Let V be the volume of liquid contained by the pyknometer. Then V = m_\omega/\rho_\omega = m_l/\rho_l(1) where \rho_\omega and \rho_l are the water and the liquid densities respectively, and m_\omega and m_l are the masses of the water and the liquid in the pyknometer. Since m_\omega = m_p+\omega - m_p m_\omega = m_p+\omega - m_p m_l = m_p+l - m_p equation (1) becomes V = (m_p+\omega - m_p)/(\rho_\omega) = (m_p+l - m_p)/(\rho_l), giving \rho_l in terms of \rho_\omega: \rho_l = [(m_p+l - m_p)/(m_p+\omega - m_p)] \rho_\omega.

Question:

Develop a flow-chart for evaluating the Fourier coefficients of the periodic waveform f(t) in the given figure. The values measured for f(t) are f(0) = 0, f(0.05) = 0.25, f(0.1) = 0.3, f(0.15) = 0.5, f(0.2) = f(0.25) = f(0.3) = 1.0, f(0.35) = 0.55, f(0.4) = 0.3, f(0.45) = 0, f(0.5) = -0.25, f(0.55) = f(0.6) = -0.3, f(0.65) = -0.2, f(0.7) = -0.1, f(0.75) = f(0.8) = f(0.85) = f(0.9) = f(0.95) = 0 .

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0535.htm

Solution:

The waveform in the figure can be expressed as a Fourier series: f(t) = a0 + ^\infty\sum_n=1 (a_n cos n\omegat + b_n sin n\omegat)(1) where, a0 = (1/T) ^T\int0 f(t) dt(2) a_n = (2/T) ^T\int0 f(t) cos n\omegat dt(3) b_n = (2/T) ^T\int0 f(t) sin n\omegat dt.(4) Equations (2)-(4) are the Fourier coefficients. We approximate the integrals by dividing f(t) into small pulses of width ∆T and summing: a0 = (1/T) ^K\sum_m=1 f(m∆T) ∆T(5) a_n = (2/T) ^K\sum_m=1 f(m∆T) cos (n(2/T)m∆T) ∆T(6) b_n = (2/T) ^n\sum_m=1 f(m∆T) sin (n(2/T)m∆T) ∆T(7) Equations (5) - (7) must be evaluated for each (harmonic) value of n. The period T of the waveform is broken into (K-1) parts such that (K-1)∆T = T. Then using the values found for f(m∆T), m = 1 to K and setting a limit for n, say 20, a computer program can be developed for evaluating the Fourier coefficients from the flow-chart. Note, finally, that a subroutine can process the coefficients found by the program into magnitude and phase form for the plotting of spectral lines.

Question:

At the melting point of a solid (or the freezing point of a liquid), the free energies of the solid state and the liquid state are equal, ∆G = 0. Likewise, at the boiling point of a liquid, where there is an equilibrium between the liquid and vapor phases, the free energy is equal in the two states. Calculate the change in entropy for the following process at 0\textdegreeC if the heat of fusion of H_2 O = 80 cal/g. H_2 O (s) \rightarrow H_2 O (l).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0532.htm

Solution:

The free energy, ∆G, can be expressed as ∆G = ∆H - T∆S, where ∆H, in this case, is the heat of fusion, T is the absolute temperature and ∆S is the change in the entropy. One is given that ∆G = 0 for this process and thus, ∆H - T∆S = 0. One is told that the process takes place at 0\textdegreeC. To use the free energy equation, the tem-perature must be in \textdegreeK. To convert from \textdegreeC to \textdegreeK, add 273 to the temperature in \textdegreeC, T = 0\textdegree + 273 = 273\textdegreeK ∆H is given as 80 cal/g. When solving for the entropy, convert ∆H to cal/mole. There are 18 g of H_2 O in one mole, thus the ∆H can be converted by multiplying it by 18 g/mole ∆H = 80 cal/g × 18 g/mole = 1440 cal/mole It is now possible to solve for ∆S. 0 = ∆H - T∆S∆H = 1440 cal/mole 0 = 1440 cal/mole - 273\textdegreeK × ∆S - 1440 cal/mole = - 273\textdegreeK × ∆S [(- 1440 cal/mole)/(- 273\textdegreeK)] = ∆S = 5.27 cal/mole - \textdegreeK

Question:

Manufacturers use soda-lime glass for containers. This type of glass can be prepared by melting sodium carbonate (Na_2CO_3), limestone (CaCO_3), and sand (SiO_2) together. The equation for this reaction is Na_2CO_3 + CaCO_3 + 6SiO_2 \ding{217} Na_2O\bulletCaO\bullet6SiO_2 + 2CO2 soda-lime glass If a manufacturer wanted to make 5000 wine bottles, each weighing 400 g, how many grams of sand would be required?

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Solution:

From the reaction equation, it can be seen that 6 moles of SiO_2 are consumed for every mole of soda-lime glass formed. Thus, one can determine the amount of sand needed to form the 5000 bottles once the number of moles of glass that makes up these bottles is found. To solve: (1) determine the weight of the bottles and the number of moles of soda-lime glass that this weight is equivalent to. (2) Calculate the number of moles of sand needed and from this its weight. Solving: (1) weight of bottles = (400 g / bottle) (5000 bottles) = 2.0 × 106 g. The molecular weight of soda-lime glass is 478 g / mole. number of moles = [(2.0 × 10^6 g) / (478 g / mole)] = 4.18 × 10^3 moles of glass. (2) For each mole of glass formed, 6molesof sand must react. Thus, no. of moles of sand = 6 × 4.18 × 10^3 = 2.51 × 10^4 moles. The molecular weight of sand is 60.1 g / mole. Hence, the amount of sand needed = (2.51 × 10^4 moles) (60.1 g / mole) = 1.51 × 10^6 g.

Question:

"Since plants cannot move about as freely as animals, they have no need for a skeletal system." Do you agree or disagree with this statement? Justify your answer.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0223.htm

Solution:

A skeletal system not only provides for locomotion but also for support. While it is true that most algae are small, aquatic, and have little need for a skeletal system, the land plants do need some skeletal structures to hold their leaves in position to receive sunlight and to help the stems stand erect on land. The land plants meet these requirements in one of two ways: either the cell wall can be very thick, as in the woody stems of trees and shrubs, and serve directly for the sup-port of the plant body, or the cell wall can be rather thin yet provide support indirectly by way of turgor pressure. Trees and shrubs have in their stems tough, woody cells called tracheids and vessels, whose structures are adapted for support in addition to conduction of water and minerals. These cells secrete a very thick wall made up largely of cellulose. Cellulose is a polysaccharide of repeating molecules of glucose attached end to end. The cellulose molecules are bound together in a complex matrix to form long, thin microfibrils. The microfibrils intertwine to form fine threads or fibrils, which may in turn coil around one another like strands in a cable. Cellulose molecules arranged in such an organized pattern form fibers as strong as steel cables of equivalent thickness. The cell walls of tracheids and vessels are also impregnated with lignin which adds to the strength of cellulose. Lignin is a highly branched, rigid organic molecule of repeating subunits. It is often deposited between the cellulose microfibrils in the walls of tracheids and vessels when the protoplast has ceased to function, and serves as a strengthen-ing element because of its regular, rigid strucure and its resistance to decomposition. In addition, land plants have phloem cells which contain thick fibers in their walls to help support the trunk. Non - woody plant cells are hot equipped with thick walls but manage to support their shape and form with the aid of turgor pressure. Turgor pressure is the pressure exerted by the contents of the cell (such as organic mate-rials and inorganic salts dissolved in water) against a cell wall which is slightly elastic and stretchable. Turgor can be maintained by a plant cell because it is naturally in a hypotonic medium. In such a medium, water tends to enter the cell, passing osmotically from a region of higher water concentration to a region of lower water concentration. The pressure resulting from a difference in the concentration of water molecules between two re-gions is known as osmotic pressure. The influx of water causes the cell to swell and the cell contents to exert a hydrostatic pressure, or turgor pressure, against the cell wall. Equilibrium is reached when the two opposing forces of hydrostatic pressure (built up in response to cell wall resistance) and osmotic pressure are equalized. Under such opposing pressures, the plant cell remains stiff or turgid The importance of turgor pressure in supporting the form of the plant body can be demonstrated by placing a crisp lettuce leaf in salt water. The lettuce leaf, after being in the salt water for some time, becomes limp and flaccid. In this hypertonic solution, water follows the osmotic gradient from the leaf into the salt solution. Without the accumulation of water inside the cell, no turgor pressure can be exerted against the walls of the cell. As a result, the cell loses its stiffness and be-comes flaccid. For although the cells of the leaf still have their cellulose walls, these walls are alone unable to provide sufficient support. (See Figure.)

Question:

Pilocarpineis a drug that stimulates the nerve endings of parasympathetic nerves. What effects would you expect this drug to have on (a) the digestive tract, (b) the iris of the eye, and (c) the heart rate?

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Solution:

The parasympathetic system consists exclusively of motor fibers originating from the brain and emerging via the third, seventh, ninth and tenth cranial nerves, and of fibers originating from the sacral region of the spinal cord and emerging by way of the spinal nerves in that region. The parasympa-thetic system therefore has nerves from two separate regions - the brain and the sacral region of the spinal cord. (a) The ninth cranial nerve, theglossopharyngeal, innervates the salivary gland. It stimulates the salivary gland to release digestive enzymes . A more important cranial nerve, thevagus, sends branches to the stomach, duodenum and the pancreas. This nerveenchancesmotility and peristalsis of the digestive tract. (Motility and peristalsis of the digestive tract should not be confused with each other. Motility of the tract causes food to mix and churn but no movement of it. Peristalsis moves the food down the digestive tract). It also stimulates the secretion of digestive enzymes by digestive glands. For example, stimulation of the pancreas by the vagusnerve causes secretion of the pancreatic juice. Thus the glossopharyngeal andvagustogether act to promote digestion. When pilocarpine is present, the parasympathetic fibers are stimulated. The result is that digestion is facilitated or enhanced. (b) The third cranial nerve supplies the muscles of the iris which controls the size of the pupil of the eye. Excitation of this nerve causes constriction of the pupil, by contraction of the circular smooth muscle. Sincepilocarpineexcites the parasympathetic system, its presence would cause the pupil to reduce in size. (c) The parasympathetic branch supplying the heart is part of the vagus nerve. Its action is to weaken and slow down the heart beat. Pilocarpine intro-duced into the blood stream would activate the parasympa-thetic nerve endings, causing the heart beat to slow and weaken .

Question:

Explain and discuss why the branch and link register (BALR) instruction is usually one of the most important instruc-tions of the IBM/360-370 computers?

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Solution:

The branch-and-link register instruction has the register-to-register (RR) format, and as such, it has a single register for each of its operands. When this instruc-tion is executed, it places the address of the next instruc-tion in storage in the register specified as the first op-erand, and then branches of the address are specified in the second register. To illustrate the function of this instruc-tion, let us assume that the instruction is stored in bytes 5000 and 5001 (the instruction is 2 bytes long). As the first operand we have Register 3 and as the second - Regis-ter 8, which contains the binary equivalent of 2048. When this instruction is executed, the address 5002 is placed in Register 3 and the computer then goes to location 2048, the address specified by Register 8. Execution now proceeds from this location. However, if a zero is specified as the second operand of the BALR instruction, no branching takes place. The address of the next instruction in memory is placed in op-erand 1 as usual, but since the operand 2 is zero, no branch-ing takes place. Instead, processing of the next sequential instruction occurs. For example, let's use the addresses of Fig. 1 for Fig. 2. On execution, address 5002 is placed in Register 3, and pro-cessing of the next sequential instruction begins. As we can see this instruction also begins at location 5002. It is this feature of the BALR instruction that makes it one of the IBM/360-370 computers' most important instructions. By this procedure a base address is loaded into the base regis-ter of the program. In the example (2), Register 3 is the base register. If a program is loaded into storage and a proper address is not loaded into the base register, the program will not be executed as intended. As a result, a BALR instruction with zero as its second operand is normally the first executable instruction of a program.

Question:

A myelinated nerve can conduct impulses of velocities of up to 100 meters per second, whereas unmyelinated nerve conducts at velocities of 20 to 50 meters per second. Explain this difference.

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Solution:

The big discrepancy between the rates of conduction of a myelinated and an unmyelinated nerve lies in the morphology of the nerves. Myelinated nerves are bundles of axons, each enveloped in a sheath of fatty substance, known as myelin. Axons of unmyelinated nerves have no myelin sheath, and are for this reason sometimes referred to as bare axons. A myelin sheath around a neuron is a highly insulating covering which prevents the flow of ions between the cell interior and the exterior of the membrane. Hence no action potential will be generated at the myelinated regions of an axon. More importantly, the insulating sheath prevents the loss of charges when Na^+ ions flow from one region to another. At successive nodes of Ranvier there are gaps in the sheath where there is no insulation. At these points, free ionic communication between the inside and outside of the membrane is possible. Thus, when Na^+ ions arrive at the gaps, the membrane is depolarized and an action poten-tial is set off. Therefore, impulses are generated only at the nodes, and nerve impulses leap from one node to the next. This kind of conduction, by "jumping" from node to node, is called saltatory transmission. The "jumping" is responsible for the more rapid impulse conduction by myelinated nerves, as compared with unmyelinated nerves. The latter lack the extensive insulation that makes saltatory conduction possible. In myelinated fibers the key permeability changes take place at the nodes where the insulationg myelin sheath is interrupted. Between the nodes, the flow of Na^+ ions carries the impulse which "jumps" from node to node along the axon.

Question:

Consider the experimental apparatus, illustrated in figure A which is used to measure the deflection of an electron. The source of electrons is a heated filament and the detector consists of a screen painted with a material which phosphoresces when struck by energetic particles. The magnetic field is confined to the spherical region. During a particular run of this experiment, an electron enters the magnetic field at a distance of 30 cm away from the detecting screen and travels 25 cm after leaving the field. If the strength of the magnetic field is 0.18 g/coulomb-sec and the beam deflection on the screen is 4.0 mm, what is the velocity of the electrons in the beam?

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Solution:

We are looking for a relationship between the magnetic field strength H, the deflection of the electron and the electron velocity v. Such a relationship is pro-vided by the equation of motion of an electron in a magnetic field, v = Her/m, where e is the electronic charge (e = 1.602 × 10^-19 coulomb), r is the radius of curvature, H is the strength of the magnetic field, and m is the electron mass (m = 9.11 × 10^28 g). In order to apply this equation to solve for v, we must first determine r. In the following discussion to determine r, we will make reference to figure B (which is not drawn to scale for reasons of clarity). In the problem, we are given that the distance from the point at which the electron beam leaves the field to the deflecting screen is 25 cm. The point at which the electron leaves the field is D. H and D lie on the same vertical line, hence the distance from D to the screen is the same as the distance from H to the screen, or HJ = 25 cm. Since the electron de-flection is small, we can assume that the path of the electron is not far from the path the electron would have in the absence of a field. Hence, we can assume that HI is much smaller than CI and HJ, and that FG is much smaller than FB and BG. If we make this assumption that HI is negligible, then IJ = HJ - HI \cong HJ = 25 cm. The distance from the point at which the electron enters the field (point C) to the screen is 30 cm, or CJ = 30 cm. Then Cl = CJ - IJ = 30 cm - 25 cm = 5 cm. Denoting the center of the circle containing the magnetic field by G, we have CG = GI = (1/2)CI = 5 cm/2 = 2.5 cm. Using our assumption on the smallness of HI, we obtain GH = GI - HI \cong GI = 2.5 cm. In summary, so far we have obtained the following distances: CG = GI \cong GH = 2.5 cm,IJ \cong HJ = 25 cm. The strategy from this point on will be as follows. We will use the similarity of triangle GJE and triangle GHD to determine HD. Then, using the similarity of tri-angle CHD and triangle CFB, we will obtain the ratio BF/CF, and from this, the radius of curvature, r. There is a theorem from elementary plane geometry which states that when a triangle is cut by a line parallel to one of its sides, the two resulting tri-angles are similar. Since HD and JE are both drawn vertically, and are therefore parallel, triangle GJE is similar to triangle GHD, Hence [(GH)/(GJ)] = [(HD)/(JE)]or HD = JE × [(GH)/(GJ)] = 4 mm × [(2.5 cm)/(27.5 cm)] = 0.4 cm × [(2.5 cm)/(27.5 cm)] = 0.036 cm. Next we note that, using the theorems of plane geometry, it can be shown that triangle CHD is similar to CGB. Then [(HD)/(CG)] = [(CD)/(BG)] Because r = BG, we solve this ratio for BG. We have al-ready determined HD and CG and can find CD by using the Pyhthagorean Theorem, thus, we can find BG. Solving for CD: CD^2 = CH^2 + HD^2 = (5.0 cm)^2 + (0.036 cm)^2 = 25.001 cm^2; CD = 5.00 cm. Solving for BG; BG = [(CD \textbullet CG)/(HD)] = [(5.0 cm) (2.5 cm)] / (0.036 cm) = 347.2cm. We can now substitute r into the equation of motion of an electron in a magnetic field. We thus obtain the electron velocity, v = (Her/m) = [(0.18 g/coulomb-sec × 1.602 × 10^-19 coulomb × 347.2 cm)] /(9.11 × 10^-28 g)] = [(1.10 × 101 0cm/sec)].

Question:

What is the approximate boiling point at standard pressure of a solution prepared by dissolving 234 g ofNaClin 500 g of H_2O?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0246.htm

Solution:

The boiling point of a solution is acolligativeproperty. This means that it depends on the ratio of number of solute particles to number of solvent particles ormolality.Molalityis defined as the number of moles of solute divided by the number of kilograms of solvent. When water is the solvent the boiling point constant, K_b,0 is 0.52\textdegree. This means that when one mole of solute is in1 kg of H_2O the boiling point of water is raised 0.52\textdegree. boiling pt elevation =kb×molalityof solute One can find the boiling point elevation once themolalityof the NaCl is found. To determine themolalityone must first determine the number of moles present. This is done by dividing the number of grams present by the molecular weight. (MW ofNaCl= 58.5.) no. of moles = (234g) - (58.5) = 4.0 moles There are 4.0 moles ofNaClin 500 g, or 0.5 kg, of H_2O. Themolalitycan now be found. Molality = (moles of solute) / (kg of solvent) molality = (4.0moles) / ( 0.5 kg) = 8.0m NaCl is an electrolyte which means that when it is placed in H_2O it dissociates into its ions. NaClt Na+ +Cl" This means that for everyNaClpresent, there are two ions formed. There are thus twice as many moles of particles as molecules ofNaCl present. The effectivemolalityis, therefore, twice the truemolality. effectivemolality= 2 × 8.0 m = 16.0 m One can now solve for the boiling point elevation, boiling point elevation = (0.52\textdegree) × (16.0 m) = 8.32\textdegree The boiling point of the water is raised 8.32\textdegree. It was originally 100\textdegreeC, therefore the new boiling point is 100\textdegreec +8.32\textdegree = 108.32\textdegreeC.

Question:

What is a polypeptide chain and how is it related to the proteins in a cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0623.htm

Solution:

A polypeptide chain consists of linked units called amino acids. Each amino acid consists of a carboxyl group (C terminal), an amino group (N-terminal), and a side chain which varies with each amino acid. Structurally an amino acid is represented as follows: Amino acids can be linked together by peptide bonds, which form between the N terminal of an amino acid on a C terminal. An example of a peptide bond is illustrated below: When many amino acids are linked together by peptide bonds, we have a chain of amino acids known as a polypeptide chain. Proteins are actually-polypeptide chains. A protein molecule may be composed of one such chain, usually folded and cross-linked within itself to form a more compact and stable structure than a very long expanded chain. Some protein molecules are made of combinations of two or more polypeptide chains. The human protein hemoglobin, for instance, is composed of four polypeptides, two identical alpha chains and two identical beta chains. Each alpha chain has 141 amino acids, and each beta chain has 146. An important distinction must be made between poly-peptides and proteins at the genetic level. A given gene can code for only one polypeptide. Many proteins, however, are composed of a number of polypeptides. When a protein is composed of more than one polypeptide chain, each chain may be coded for and made separately. These chains aggregate only after their individual syntheses to form the final protein. In addition, some proteins exist in their functional form only after certain changes have taken place in the original polypeptide, such as the cleavage or loss of certain portions of the chain. Thus a given gene does not necessarily code directly for an active, functioning protein.

Question:

What roles does glucose play in cellmetabolism ?

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Solution:

Glucose, a six carbon monosaccharide, is the primary source of energyfor all cells in the body. The complete oxidation of one molecule of glucoseyields 36 ATP molecules in a cell. The energy stored in the form ofATP can then be used in a variety of ways including the contraction of a muscleor the secretion of an enzyme. The supply of glucose to all cells mustbe maintained at certain minimal levels so that every cell gets an ade-quateamount of glucose. Brain cells, unable to store glucose, are the firstto suffer when the blood glucose concentration falls below a certain criticallevel. On the other hand, muscle cells are less affected by changes inthe glucose level of the blood because of their local storage of glucose asglycogen. In certain cells, glucose can be converted to glycogen and can be storedonly in this form, because glucose will diffuse into and out of a cell readily. Glycogen is a highly branched polysaccharide of high molecular weight, composed of glucose units linked by \alpha-glycosidicbonds. Once the conversionis complete, glycogen remains inside the cell as a storage substance. When glucose is needed, glycogen is reconverted to glucose. In addition to being stored as glycogen or oxidized to generate energy, glucose can be converted into fat for storage. For example, after a heavymeal causes the supply of glucose to exceed the immediate need, glucosemolecules can be transformed into fat which is stored in the liver and/oradipose (fat) tissue. This fat serves as an energy supply for later use.

Question:

An athlete in his run-up for a pole vault can achieve a speed of 30 ft/s. What is the maximum possible record for the pole vault likely to be?

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Solution:

At the end of the run with a velocity of 30 ft/s, the athlete possesses kinetic energy of amount E_k = (1/2) × m × 30^2 ft^2/s^2, where m is his mass. By causing rotation about the end of his pole he transforms this kinetic energy into potential energy. The mass of the pole is negligible in comparison with that of the man and need not be considered. Further, the pole must not be made of a material which can boost the athlete's energy by its elastic springiness. The most favorable case occurs when the athlete plans his jump in such a way that, as he clears the bar, he has negligible kinetic energy left. Thus if h is the height by which the athlete's center of gravity alters in the jump, we have, by conservation of energy, (1/2)m × 30^2 ft^2/s^2 =mgh.\therefore h = (30^2 ft^2/s^2)/(2g) = 14 ft (3/4) in. But if the athlete is very tall, his center of gravity may be as much as 3 ft 9 in. from the ground during his run-up. Hence the final height of his center of gravity above the ground is maximally 17 ft 9 3/4 in. His center of gravity is located inside his body and the bar must be lower than his center of gravity by roughly half the thickness of his body. If we assume a reasonably thin athlete, a minimum of 4(1/2) in. must be subtracted from the height previously mentioned to allow for clearance. The maximum possible record for the pole vault would appear to be 17 ft 5(1/4) in. (as measured by the height of the bar). The present world record is 17 ft 4 in. (It should be noted that fiber-glass poles do not meet the conditions about elasticity stated above.)

Question:

Why do ecologists not attempt to divide the ocean intobiome typeslike those on land? How do they insteaddistinguish thedifferent parts of the ocean?

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Solution:

Because of the vastness and relative uniformity of the ocean, and thecomparatively rapid mixing of the organisms within it, the various regionsof the ocean cannot be divided into distinct biome types as are terrestrialregions. Even though the ocean is charac-terized by many differenttypes of habitats, these habi-tats are all interconnected, and some organismsmay move freely from one to the other. It seems more appropriate, therefore, to describe an ocean byzonationrather than by biometypes. The oceans and seas cover 71 per cent of the earth's surface, but, exceptfor the shallow water of the margins, they are far less complex in structureand productivity than the land. The continental margin gradually slopesseaward to depths of 150 to 200 meters and then slopes steeply to depthsof 3000 meters or more. This slope is known as the continental shelf. The continental shelf then gives way to the abyssal ocean floor. The accompanying figure depicts thezonationin an ocean. Waters overthe continental shelf comprise theneriticzone, and these beyond the edgeof the shelf comprise the main part of the ocean basin, called the oceaniczone. The floor of the ocean basin, called the abyssal plain, rangesfrom 3000 to 5000 meters in depth and may be marked by such featuresas sea mounts, ridges and trenches. The oceanic zone is further distinguishableinto four zones. The upper part of the open ocean into whichenough light can penetrate is called theeuphoticzone. Below this zoneis thebathyalzone, which ranges to a depth of perhaps 2,000 meters. Below thebathyalzone, the waters over the abyssalplain .(ocean floor) form the abyssal zone. Trenches and valleys of the ocean floor form thetidal zone, which is generally below 6,000 meters. Theeuphoticzone, usually about 100 meters in thickness, has enoughpenetrated sunlight effective for photosynthesis. Below this is a transitionzone where some light penetrates. From the transition zone downto the ocean floor, total darkness prevail and this region constitutes theaphoticzone. The exact boundary between the lighted zone and the deeplightless zone at any given locality depends upon the intensity of the sunlight, which in turn depends upon the latitude, and upon the turbidity of thewater. Usually no light penetrates below 600 meters. The physical environment of the ocean is most diverse in the shallowwater along its margin. Here in theneritic(coastal) zone exists the greatestvariation in temperature, salinity, lightintesityand turbulence. Also, here, sunlight can penetrate to the bottom of the water body. For this reason, here occurs the most luxuriant forms of both plant and animal life. The edge of the ocean, which rises and falls with the tide, is the intertidal zoneor littoral zone. It extends out into the water to the depths at which thewater is no longer stirred by tides and waves. The region after the intertidalzone is called thesubtidalzone and farther on is the lowerneritic region. In general, the density of living organisms in the open ocean is muchless than that found in theneriticregion, but because of its large areaand volume, the total biomass in the open ocean is greater.

Question:

The following reaction 2H_2S(g) \rightleftarrows 2H_2(g) + S_2(g) was allowed to proceed to equilibrium. The contents of the two-liter reaction vessel were then subjected to analysis and found to contain 1.0 mole H_2S, 0.20 mole H_2, and 0.80 mole S_2 . What is the equilib-rium constant K for this reaction?

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Solution:

This problem involves substitution into the equilibrium constant expression for this reaction, K = {[H_2 ]^2 [S_2]} / [H_2S]^2 . The equilibrium concentration of the reactant and products are [H_2S] = 1.0 mole/2 liters =0.50M, [H_2] = 0.20 mole/2 liters = 0.10M, and [S_2] = 0.80 mole/2 liters = 0.40M , Hence, the value of the equilibrium constant is K = {[H_2 ]^2 [S_2 ]} / [H_2S ]^2 = {(0.10)^2 (0.40)} / (0.50)^2 = 0.016 for this reaction.

Question:

A sample of gas occupies 14.3 l at 19\textdegreeC and 790 mm Hg. Determine the number of moles of gas present, what volume will this same amount of gas occupy at 190\textdegreeC and 79.0 mm Hg?

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Solution:

This problem involves an application of the ideal gas equation, PV = nRT, where p = pressure, V = volume, n = number of moles, R = gas constant, T = absolute temperature, we will solve this equation for n and then use this value of the number of moles, and the ideal gas equation, to obtain V at the new temperature and pressure. For our gas, we initially have V = 14.3 l, T = 19\textdegreeC = 292.15\textdegreeK, P = 790 mm Hg × 1 atm/760 mmatm/760 mm Hg = 1.039 atm and R = 0.0821 l-atm/\textdegreeK-mole. Solving the ideal gas equation for n, we obtain n = (PV/RT) = [(1.039 atm × 14.3 l)/(0.0821 l-atm/\textdegreeK-mole × 292.15\textdegreeK)] = 0.6194 mole. Under the new conditions, we have T = 190\textdegreeC = 463.15\textdegreeK and P = 79.0 mm Hg = 79.0 mmHg × 1 atm/760 mm Hg = 0.1039 atm. Solving the Ideal gas equation for V, the new volume is V = (nRT/P) = [(0.6194 mole × 0.0821 l-atm/\textdegreeK-mole × 463.15\textdegreeK) / (0.1039 atm)] = 226.68 l.

Question:

An ionic bond is established between positive ion A and negative ion B. How would one expect the strength of the bond to be affected by each of the following changes: (a) Doubling the charge on A, (b) Simultaneously doubling the charge on A and B, (c) Doubling the radius of B, and (d) Simultaneously doubling the radius of A and B?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0618.htm

Solution:

When two ions come together to form an ionic bond, energy is released. The more energy released, the stronger the ionic bond. The amount of energy released is determined by the equation E = 1.44 (q_1q_2 / r) where E = energy in electron volts, q_1 = charge on the positive ion, q_2 = charge on the negative ion, and r = the distance between the nuclear centers. If one changes one parameter, another must adjust to maintain this e-quality. Solving: (a) If the charge on A, the positive ion, is doubled, the energy released is doubled. That is, 2E = 1.44 [(2q_1q_2)/ r] is equivalent to E = [(1.44 q_1q_2) / r]. If E is doubled, the strength of the chemical bond is doubled. (b) Using a similar line of reasoning, it is found that if one doubles both the positive and negative charges, four times E is released. The strength of the bond is, thus, 4 times as great as the original bond. (c) If one increases the radius of the ions then, the distance between their nuclei must also increase. To maintain the equality, E must decrease because r is inverselypropotionalto E. Therefore, the bond strength decreases. When the ionic radius of one ion is doubled, E is halved. (d) Both radii are doubled. E is diminished by a factor of 4 and the strength of the bond is decreased by a factor of 4.

Question:

An Inclined plane 5 meters long has its upper end 1 meter above the ground. A load of 100 newtons is pushed up the plane against a force of friction of 5 newtons. What is the effort, the work input, the work output, the AMA, the IMA, and the efficiency?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0289.htm

Solution:

We first construct a diagram (see diagram). A vector is drawn vertically downward representing the force of gravity. Orienting our axes so that the y-axis coincides with the inclined plane, we resolve the gravitational force into its x and y components. This gives us a right triangle in which angle A' is equal to angle A, which the inclined plane forms with the ground, since they both are complements of the same angle. Therefore, since both large and small right triangles have equal angles, they are similar, so that we can write a pro-portion : 1 m/5 m = x/(100 nt) x = 20 nt Therefore, the force that the load exerts parallel to the inclined plane is 20 nt. This, plus the frictional force, is the effort. Hence, Effort = Gravitational force along plane + frictional force E = 20 nt + 5 nt = 25 nt Work is force × distance when the force is in the direction of the distance. Therefore, the work input (W_i) is: W_i = 25 nt × 5 m = 125 joules AMA (actual mechanical advantage) is: AMA = R/E where R is resistance (load), and E is effort. Then, AMA = (100 nt)/(25 nt) = 4 The IMA (imaginary mechanical advantage) is the ratio of the length of the plane to its height: IMA = (5 m)/(1 m) = 5 Efficiency is output work over input work. Efficiency= (W_O)/(W_i) = (100 joules)/(125 joules) = (AMA)/(IMA) = 4/5 = 0.80 = 80%

Question:

Perform the following conversions from a) 1 0 0 1 1 0 1 0 b) 1 0 1 \textbullet 1 0 1 c) 0 \textbullet 0 0 1 1 base 2 to base 8, 10, and 16 .

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Solution:

a) To convert base 2 to base 8, we use the triad method where three digits are taken together and converted to equivalent decimal value; this results in base 8 conversion Similarly for hexadecimal conversion 4 bits are taken together. Zeros are assumed for most significant grouping. For binary to decimal conversion 10011010 1×2^7 + 0×2^6 + 0×2^5 + 1×2^4 + 1×2^3 + 0×2^2 + 1×2^1 + 0×2^0 128 + 0 + 0 + 16 + 8 + 0 + 2 + 0 = 154_10 b)i) 101 \textbullet 101_2 = 5 \textbullet5_8 ii) (0101/5)\textbullet (1010_2/A) = 5 \textbullet A_16 iii) 101 \textbullet 101_2 (1×2^2 + 0×2^1 + 1×2^0) \textbullet (1×2^-1 + 0×2^-2 + 1×2^-3) = 4 + 0 + 1 + 0.5 + 0 + 0.125 = 5 \textbullet 625_10 = 0 \textbullet 1875_10

Question:

A boy leaning over a railway bridge 49 ft high sees a train approaching with uniform speed and attempts to drop a stone down the funnel. He releases the stone when the engine is 80 ft away from the bridge and sees the stone hit the ground 3 ft in front of the engine. What is the speed of the train?

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0087.htm

Solution:

Applying the equation applicable to uniform acceleration, x - x_0 = v_0t + (1/2) at^2, to the dropping of the stone 49 ft from rest under the action of gravity, we can find the time t the stone is in motion. The initial velocity of the stone v_0 is zero. The distance the stone travels, x - x_0 = 49 ft. Therefore, 49 ft = 0 + (1/2) (32 ft/sec^2) (t^2) \thereforet = \surd[(2 × 49 ft)/(32 ft/s^2)] =7/4 s. In the time of 7/4 s it takes the stone to drop, the engine has moved with uniform speed u a distance of (80 - 3) ft. \thereforeu = d/t = (77 ft)/(7/4 sec) = 44 ft/sec = 30 mph.

Question:

Why is the drop in blood pressure important in retaining fluid within the capillaries?

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0387.htm

Solution:

As the blood pressure through the capillaries, there is a pressure resulting from the beating of the heart, which tends to force the plasma through the capillary membrane and into the interstitial fluid. This pressure is called the capillary. or hydrostatic pressure. There are also two other factors determining whether fluid will move out of the blood or not: 1) the interstitial fluid pressure and 2) the plasma colloid osmotic pressure. The interstitial fluid pressure is the hydrostatic pressure exerted by the interstitial fluid. It has been measured to be about - 7 mm Hg , and so tends to move fluid outward through the capillary membrane. The plasma colloid osmotic pressure arises from the proteins found in the plasma. The plasma proteinsdo not diffuse readily through the capillary membrane. Therefore, there is a concentration gradient with four times as much protein in the plasma as in the interstitial fluid. An osmotic pressure develops since water tends to pass through the capillary membrane from the interstitial fluid (which has a higher concentration of water and a lower concentration of protein molecules) to the blood plasma. The plasma colloid (\textquotedblleftcolloid\textquotedblright because the protein solution resembles a colloidal solution) osmotic pressure is approximately 23.5 mm Hg. The pressure, unlike the interstitial fluid pressure, opposes the capillary pressure by tending to move fluid inward through the capillary. An analysis of the force at the arteriole end follows: Forces moving plasma outward: (mm Hg/25) Capillary pressure25 Interstitial fluid pressure7 Total outward pressure32 Forces moving plasma inward: Plasma colloid osmotic pressure23.5 Total inward pressure23.5 Net outward force32 - 23.5 = 8.5 At the arteriole end, then, there is a net outward filtration pressure of about 8.5 mm Hg. If this force were constant throughout the length of the capillary, there would be a large net loss of fluid outward. There is, however, a drop in capillary pressure (to 9 mm Hg) at thevenuleend due to the filtration of fluid along the capillary. Therefore netreabsorptionof the lost fluid occurs at thevenuleend as follows: Forces moving plasma outward: mm Hg Capillary pressure9 (Negative) interstitial fluid pressure7 Total outward pressure16 Forces moving plasma inward: Plasma colloid osmotic pressure23.5 Total inward pressure23.5 Net inward force23.5 - 16 = 7.5 At thevenuleend, there is areabsorptionpressure of 7.5 mm Hg This reabsorptioncauses about nine-tenths (7.5/8.5) of the fluid that had filtered out at the arteriole end to be recovered. The slight excess of filtration will be recovered by the lymphatic system (see next question).

Question:

A boy drops a rubber ball from a height of 4 feet and lets it bounce. If the ball loses 10% of its energy, due to friction of com-pression on each bounce, how many bounces will it take before the ball will only rise 2 feet above the ground?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0221.htm

Solution:

When the boy first lets go of the ball, its energy is purely potential and is given byE_p=mghwhere h is the ball's original height above the ground. (In this problem h = 4 feet). When the ball hits the ground its energy is purely kinetic. This is also the case when the ball just leaves the ground and begins its upward flight. As it rises its energy gradually changes from kinetic to potential. We note from the above equation that the potential energy is directly pro-portional to the height the ball rises. Thus, each time the ball bounces and loses 10% of its energy, it rises to 9/10 of its previous height. After the first bounce the ball rises (9/10) 4 feet, then after 2 bounces it rises (9/10) (9/10) 4 feet = (9/10)^2 4 feet. We can see that after n bounces maximum height of ball = (9/10)^n 4 feet Thus, we set the expression for maximum height equal to 2 feet and solve for n: (9/10)^n 4 feet = 2 feet (9/10)^n = 1/2 n log (9/10) = log (1/2) n[log 9 - log 10] = - log 2 n[0.9542 - 1.0000] = - 0.3010 - 0.0458n = - 0.3010 n = 6.55 We must round this off to 7 since physically we cannot have a fraction of a bounce.

Question:

What is the evidence that the chemical compounds of a cell arein a "dynamic state"?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0071.htm

Solution:

The term "dynamic state" is meant to indicate that the chemical compoundsof a cell are cons-tantly being broken down and synthesized. At one time, it was thought that since an adult organism fails to change in outwardappearance over a short period of time, its molecular constitution isalso unchanging. That is, the cellular compounds are stable and remain inthe cell for long periods of time without being degraded and replaced, theonly exception being those compounds used as "fuel" for energy, whichby necessity must undergo catabolism. However, experimental evidence has shown that this belief in the staticstate of the cell's compounds is false. The cell's compounds are constantlychanging - being broken down and replaced by newly synthesizedcompounds. Radio- activelylabelledamino acids, fats, and carbohydratescan be injected into laboratory animals in order to demonstratethe dynamic nature of the cell's environment. If the cell were static, thelabelledmolecules would only be broken down for fuel. However, it has been observed that thelabelledamino acids are incorporatedinto proteins and that thelabelledfatty acids into fats(triglycerides). The marked proteins and fats are then degraded by the bodyto be replaced by the incorporation of new amino acids and fatty acids. All these metabolic processes occur without any visible change in bodysize. The only exception to the dynamic state of the cell is its DNA molecules, which contain genes, the body's basic units of genetic infor-mation. DNA is extremely stable in order that the here-ditary characteristics, for which it codes, have little or no chance of changing. Not only is the cell dynamic on the molecular level, but on the cellularlevel as well. Indeed, 2.5 million red blood cells alone are producedand destroyed per second in the human body.

Question:

A copper bar is 8.0 ft long at 68\textdegreeF and has a linear expansivity of 9.4 × 10^\rule{1em}{1pt}6/F\textdegree, What is the increase in length of the bar when it is heated to 110\textdegreeF?

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0445.htm

Solution:

Change in this object's dimensions is pro-portional to the change in temperature and the original length. Therefore the change in length of the bar is \DeltaL = L_0\alpha\Deltat= (8.0 ft)(9.4 × 10^\rule{1em}{1pt}6/F\textdegree)(110\textdegreeF \rule{1em}{1pt} 68\textdegreeF) = 0.0032 ft.

Question:

Given that the masses of a proton, neutron, and electron are 1.00728, 1.00867, and .000549amu, respectively, how much missing mass is there in ^19_9F (atomic weight = 18.9984)?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E19-0711.htm

Solution:

The total number of particles in ^19_9F and their total weight can be calculated. The amount of missing mass in ^19_9F will be the difference of this calculated weight and the given atomic weight of ^19_9F. The subscript number 9, in ^19_9F, indicates the atomic number of fluorine (F) . Because the atomic number equals the number of protons, there are 9 protons in F. The superscript, 19, indicates the total number of particles in the nucleus. Since the nucleus is composed of protons and neutrons, and there are 9 protons, there are 10 neutrons present. In a neutral atom, the number of electrons equals the number of protons. Thus, there are 9 electrons. The total number of particles is, thus, 28. The mass and quantity of each particle is now known. Calculating the total weight contribution of each type of particle: Protons:9 × 1.00728= 9.06552 Neutrons:10 × 1.00867= 10.0867 Electrons:9 × 0.000549=0.004941 Total mass= 19.1572. It is given that the mass of the fluorine atom is 18.9984. Therefore, the missing mass is 19.1572 - 18.9984 = 0.1588amu.

Question:

In aquatic plants, which haveprotoplasmsthatare hypertonic in comparison to the water in which they live, the amount of water in the plant cells remains at a constant level . Explain how this is regulated.

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/Users/wenhuchen/Documents/Crawler/Biology/F10-0261.htm

Solution:

Plants living in fresh water have a protoplasm more concentrated in solutes than their surroundings i.e. their surroundings are hypotonic. Accordingly by osmosis, water tends to enter the cells of these plants in order to equalize the concentration of solutes inside and outside the cells. Although plant cells do not have contractile vacuoles (theosmoregulatory organelle in amoeba) to pump out excess water, they do have a firm cellulose wall to prevent undue swelling. As water enters a plant cell, an internal pressure known asturgorpressure is built up within the cell which counterbalances the osmotic pressure and prevents the entrance of additional water molecules. Hence theturgorpressure is pressure that is generated against the cell wall to regulate the influx of water. Thus water plant cells are protected against excessive swelling and bursting.

Question:

A chemist wants to produce chlorine from molten NaCl. How many grams could be produced if he uses a current of one amp for 5.00 minutes?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0577.htm

Solution:

When electrical energy is used to produce a chemical change, it is called electrolysis. Here, electricity is being used to produce chlorine from NaCl. Michael Faraday discovered the laws that illustrate this process quantitatively. You must make use of them to answer this question. You know that chlorine exists as Cl_2 gas. In NaCl, however, it is in the form of the Cl^- ion. Therefore, to obtain Cl_2 from NaCl, you must cause the following reaction to occur: 2Cl^- \rightarrowCl_2 + 2e^- A loss of 2 electrons is involved. Faraday's law states that the amount of gas produced is proportional to the amount of electricity. One mole of electrons is defined as the faraday (F) and equals 96,500 coulombs of electricity. From the equa-tion,2Cl^- \rightarrowCl_2 + 2e^-, you see that 1 mole of Cl_2 is produced for every 2 faradays of electricity. We can determine the amount of Cl_2, by relating it to the number of Faradays present. First determine the number of coulombs. You used one ampere, which is just equal to 1 coulomb/sec, for 5 minutes. Therefore, number of coulombs = (1.00 coulomb/sec) (5.00 minutes)(60 sec/min) = 300 coulombs. Since there are 96,500 coulombs per faraday, then the number of faradays = (300 coul) / (96500coul/F) = 0.0311F. Recalling that 1 mole of Cl_2 is obtained for every two faradays, you have (0.00311 / 2) = .00155 moles of chlorine formed, which translates into 0.111g of Cl_2. Atomic weight Cl_2 = 70.9 g/mole. wt. of Cl_2=(mole)(g/mole) weight of Cl_2 = .00155 moles × 70.9g/mole = 0.111g .

Question:

The teeth of different species of vertebrates are special-ized in a variety of ways and may be quite unlike those of man in number , structure, arrangement, and function. Give examples .

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0481.htm

Solution:

The teeth of snakes, unlike those of man, are very thin and sharp, and are usually curved backward. These teeth do not serve the function of mechanical breakdown of food; they serve in capturing prey. Snakes swallow their prey whole. The teeth of carnivorous mammals, such as cats and dogs, are more pointed than those of man; the canines are longer, and the premolars lack flat, grinding surfaces, being more adapted for cutting and shearing. On the other hand, herbivorous animals such as cows and horses may lack canines, but have very large, flat premolars and molars. Notice that sharp pointed teeth, poorly adapted for chewing, seem to characterize meat eaters, whereas broad flat teeth, well-adapted for chewing, seem to characterize vegetarians. In vegetarians, the flat teeth serve to break up the indigestible cell walls of the ingested plant, allowing the cellular contents to be exposed to the action of di-gestive enzymes . Animal cells do not have the indigestib-le armor of a cell wall and can be acted upon directly by digestive enzymes; hence, there is no need for flat, grinding teeth.

Question:

Write a program which reads an input text and prints out the number of occurences of each adjacent letter pair which appears within the text.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0418.htm

Solution:

PROGRAM NUMBEROFOCCURENCESOFADJACENTPAIRS (INPUT, OUTPUT ) ; (\textasteriskcentered This program, reads an input text and \textasteriskcentered) (\textasteriskcentered prints out the number of occurences of \textasteriskcentered) (\textasteriskcentered each adjacent letter pair which appears within \textasteriskcentered) (\textasteriskcentered the text. This program assumes that the \textasteriskcentered) (\textasteriskcentered letters A to Z are 26 consecutive values \textasteriskcentered) (\textasteriskcentered of the type char.\textasteriskcentered) TYPE letter = 'A'. . 'Z'; VAR COUNT: Array [Letter, Letter] of o. . Maxint; FIRST, SECOND ; LETTER ; Thisch, Lastch : char; Last was letter : Boolean; BEGIN \textasciigraveFor First: ='A' To 'Z' DO For second: ='A' To 'Z' DO COUNT [First, second]:= o; While not EOF (input) DO BEGIN Last was letter : = False; (\textasteriskcentered At the start of a line \textasteriskcentered) While not EOLN (Input) DO BEGIN Read (Thisch); write (Thisch) If (Thisch>='A') and (Thisch<='Z') Then if last was letter Then count[Lastch, Thisch]:=count[Lastch, Thisch] + 1 Else lastwasletter: = True Else lastwasletter:=False; lastch:=Thisch End; Read1n; write1n End; write1n; write1n; write1n ('occurences of letter pairs......'); For First: ='A' to 'Z' DO For second : = 'A' to ' Z ' DO IF COUNT [First, second]<>o Then write1n (First, second, , count[First, second]:4) End.

Question:

A car battery supplies a currentI of 50 amp to the starter motor. How much charge passes through the starter in (1/2) min?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0652.htm

Solution:

Current (I) is defined as the net amount of charge, Q, passing a pointper unit time. Therefore, Q = It = (50 amp) (30 sec) = 1500 coul .

Question:

A chemist mixes nitric oxide and bromine to form nitrosy1 bromide at 298\textdegreeK, according to the equation 2NO_(g)+ Br_2(g) \rightleftarrows 2NOBr_(g) . Assuming K = 100, what is the quantity of nitrosy1 bromide formed, if the reactants are at an initial pressure of 1atm? R = 0.0821 liter-atm./mole\textdegreeK.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0310.htm

Solution:

You are given the equilibrium constant for this reaction and asked to calculate the quantity of nitrosy1 bromide produced. The first step is to write out the equilibrium expression and equate it with the given value. For the general reaction , xA+yB\rightarrowzC, K is defined[C]^2 / { [B]^y [A]^x } , where the brackets represent concentrations. For this reaction , K = [NOBr]^2/ { [NO]^2[Br_2 ] } = 100 . To find out how muchNOBris produced, you would have to know how many moles of NO and Br_2 were reacted. Once this is known, you can find the number of grams produced. You know that the equi-librium expression is based on concentration of reactants and prod-ucts. Concentration is expressed in moles per liter. This means that if the volume of theNOBrand its concentration is known, you can find moles, since concentration X volume (liters) = moles. Let us represent the concentration as (N / V) = (moles) / (Volume) . Thus, the equilibrium expression becomes K = [(N_NOBr/ V)^2] / [(N_NO / V)^2 (N_(Br)2 / V)]. Let x = moles ofNOBrformed. Then, x moles of NO and x/2molesof Br_2 are consumed, since the coefficients of the reaction show a 2:2:1 ratio amongNOBr:NO :Br_2 . The equilibrium expression becomes 100 = (N^2_NOBrV) / (N^2 _NO N_(Br)2 ) = {x^2 V} / { (2\Elzbar x)^2 (1\Elzbar .5x) } . If x moles ofNOBrform, and you started with 2 moles of NO, then, at equilibrium, you have left 2\Elzbarx moles of NO . You started with only 1 mole of Br_2 and (1/2)x moles of itformNOBr; thus you have 1\Elzbar.5x moles left. Therefore, you need to determine only the volume to find the quantity NOBrformed. V can be found from the equation of state, PV = NRT , where P = pressure, V = volume, N = moles, R = universal gas constant, and T = temperature inkelvin(celsiusplus 273\textdegree). You are told that the reactants are under a pressure of 1 atm. at 298\textdegree K . N = 3, since the coefficients inform you that a relative sum of 3 moles of reagents exist. You know R. Thus , V = NRT / P = { (3) (.0821) (298) } / {1} = 73.4 liters . Now that V is known, the equilibrium expression becomes { x^2 (73.4)} / {(2\Elzbar x)^2 (1\Elzbar 0.5x)} = 100 . Solving for x, you obtain x = .923 moles = moles ofNOBrformed. Molecular weight = 110. Grams produced = .923 × 110 = 101.53g.

Question:

If a nuclear reactor is designed to deliver 1 MW. of heat energy continuously, how many fission events must occur each second to sustain this power level? How much uranium-235 would be consumed each year?

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Solution:

It is known that the amount of energy released per fission event is about 200 MeV. In joules, this energy becomes E = (2.0 × 10^8 ev) × (1.6 × 10^-19 J/ev) = 3.2 × 10^-11 J. The total number, n, of nuclei undergoing fission each second to produce 1 MW = 10^6 W of power is n = power / Energy per fission = (1.0 × 106J/sec) / (3.2 × 10^-11 J/nuclei) = 3.1 × 10^16 nuclei/sec. Since 1 year = 365 days/year × 24 hrs/day × 60 min/hr × 60 sec/min 7 \approx 3.1 × 10^7 sec . , the total number N of uranium-235 atoms consumed in one year is N = (3.1 × 10^7 sec/year ) × (3.1 × 10^16 nuclei/sec) = 9.6 × 10^23 nuclei/year. The mass m of one uranium-235 atom is approximately m = (2.35 × 10^2 nucleons) × (average mass of one nucleon) = (2.35 × 10^2 nucleons) × (1.67 × 10^-27 kg/nucleon) = 3.92 × 10^-25 kg. The total mass M of, uranium fuel consumed in one year is M = mn = (3.92 × 10^-25 kg/atom) × (9.6 × 10^23 atoms/year) = 3.8 × 10^-1 kg/year.

Question:

How is the respiratory quotient obtained and how can it be usedto determine when a person is in an advanced state of starvation?

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Solution:

The respiratory quotient (RQ) is a quantitative measure of the type offoods being used by the body. If we measure both the oxygen consumptionand carbon dioxide release by the body, we can determine whethercarbohydrates, fats, or proteins are being used for respiratory needs. This measure, called the respiratory quotient, is the ratio of carbon dioxideemitted to the amount of oxygen taken in: RQ = {CO2[produced]} / {O2[consumed]} During aerobic respiration, one molecule of glucoseutilisessix moleculesof oxygen and releases six molecules of carbon dioxide. This 1:1 ratio gives carbohydrates a respiratory quotient of 1.0. Fats, however, havea lower respiratory quotient. This indicates that more oxygen is necessaryto break down a fat than a carbohydrate molecule. Oxygen is neededto convert the fat (or fatty acid) into intermediate products before it canenter the respiratory process. Free fatty acids are first formed from fatsenzymatically, and are then oxidized to acetyl-CoA, which enters the TCA cycle.The overall equation forpalmiticacid (a 16-carbon fatty acid) oxidationcan be written: palmitate+ 23 O2 + 129 Pi + 129 ADP \textemdash> 16 CO2 + 145 H2O + 129 ATP The respiratory quotient is therefore (16/23) = 0.70 This is typical for most fats. The RQ of proteins lies between those of carbohydrates and fats- usuallyabout 0.80. The oxidation of two molecules ofalaninerequires six moleculesof oxygen and liberates five molecules of carbon dioxide accordingto the following equation: 2 C3H7O2N + 6 O2 \textemdash\textemdash> (NH2)2 CO + 5 CO2 + 5 H2O. This gives a respiratory quotient of 5/6, or 0.80 for proteins. Normally, if an individual is consuming all three types of food, his RQ will be about .85. After a meal heavy in carbohydrates, however, a person'sRQ will approach 1.0. If he stops eating, his RQ will decrease to .70 as his body uses its fatty deposits as an energy source. Further starvation(about 2 weeks) will tend to increase his RQ to 0.80 as the fatty depositsbecome exhausted and the proteins of the body are used. This is dangerousand can result in consumption and atrophy of muscles and organs. Therefore, a person in an advanced state of starvation will have anRQ of about 0.80.

Question:

Write a BASIC program to compute the inverse and the transpose of a 3×3 matrix A. Apply the program to the case where A =\mid 1-12 \mid \mid 201 \mid \mid 031 \mid

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G20-0502.htm

Solution:

The inverse of a matrix A is another matrix B such that, when multiplied by A, gives an identity matrix. The statement in BASIC which results in forming of an inverse matrix B of matrix A is as follows: MATB = INV (A) A transpose C of the matrix A is a matrix with rows identical to the respective columns of A, and columns identical to the respective rows of A =\mid 123 \mid \mid 7110 \mid \mid-645 \mid Then the transpose of A is: \mid 17-6 \mid \mid 2114 \mid \mid 305 \mid The corresponding BASIC statement is: MAT C = TRN(S) After generating the inverse and the transpose matrices of A, the following program also checks if B is really an inverse by multi-plying A and B to form matrix I. If I comes out to be an identity matrix, then B is indeed an inverse of A. The program looks as follows: 100DIM A (3, 3), B (3, 3), C (3, 3), I (3, 3) 110MAT READA 120MAT B = INV (A) 121PRINT 122PRINT 130MAT PRINT B: 140MAT C = TRN (A) 141PRINT 150MAT PRINT C: 160MAT I = A\textasteriskcenteredB 161PRINT 170MAT PRINT I; 800DATA 1, -1, 2, 2, 0, 1, 0, 3, 1 999END

Question:

In BASIC the RND function is used to generate a random number between 0 and 1. What statements would you use to: (a) generate integral random number between 0 and X - 1? (b) generate integral random number between 1 and X? (c) simulate the toss of a cubical die? (d) simulate a binary random choice taking values 1 or 2?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0242.htm

Solution:

(a) 1\O LET T = INT(X\textasteriskcenteredRND) (b) 2\O LET V = INT(1 + X\textasteriskcenteredRND) (c) 3\O LET W = INT(1 + 6\textasteriskcenteredRND) (d) 4\O LET N = INT(1 + 2\textasteriskcenteredRND)

Question:

The solubility product constant of magnesium hydroxide is 8.9 × 10-^12 , calculate its solubility in (a) water and (b) .05M NaOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E11-0389.htm

Solution:

Whenever an ionic solid is placed in water, an equilibrium is established between its ions and the excess solid phase. The solu-bility constant,K_sp, measures this equilibrium. For the general re action AxBy\rightleftarrowsxA+ +yB- , the K_sp is defined as being equal to [A+]^x [B-]^y . The concentration of solid is always constant, no matter how much is in contact with the ions. This means the solid phase will not appear in the equilibrium constant expression. For this problem, part (a), you have Mg(OH)_2(s) \rightleftarrows Mg^2+ + 2OH- . The K_sp= [Mg^2+] [OH- ]^2 = 8.9 × 10-^12 . You are asked to find these concentrations. From the chemical equa-tion, you find 2 moles of OH- will be generated per mole of Mg^2+ . Thus, at equilibrium, if the concentration of Mg^2+ = x (mol/liter), then the concentration of OH- = 2x(mol/litei). (Note: The dissociation of water contributes some OH-, but this amount is very small, and can be ignored.) Thus, you have [Mg^2+] [OH- ]^2 = x(2x)^2 = 8.9 × 10-^12 =K_sp. Solving for x, you obtain x = 1.3 × 10-^4 mol/liter. Thus, 1.3 × 10-^4 mole of Mg(OH)_2 dissolves per liter of water, producing a solution of 1.3 × 10-^4 M Mg^2+ and 2.6 × 10-^4 M OH- . To find the solubility in .05MNaOH, part (b), perform the same process, except you must realize that theNaOHsupplies OH- in addition to the OH- that dissolves from the salt. This means, as such, [Mg^2+ ] [OH- ]^2 =K_sp= (x)(2x + .050)^2 = 8.9 × 10-^12 . If you solve for x, you obtain 3.6 × 10-^9 mol/liter.

Question:

Obelia, a marine hydrozoan, exists in a polypoid and medusoid form. Explain how these stages are formed in the life cycle of the organism.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0280.htm

Solution:

The coelenterate Obelia is a marine colonial consisting of a group of polypoid individuals living together in close association. Within the colony are two types of polyps serving different functions. There are nutritive (or feeding) polyps and reproductive polyps. All the polyps of the colony are derived from an ancestral hydralike polyp by asexual budding. These buds fail to separate and the new polyps remain attached by hollow stemlike connections to the colony. Each reproductive polyp has a central core, the blastostyle. When mature, the blastostyle buds off an entirely new body form, an umbrella- shaped medusa. In the process of budding, the medusa breaks free of the blastostyle and swims out through the opening of the blastostyle called the gonotheca. The medusa of Obelia has a fringe of tentacles on its edge, and hanging down from the center, similar to the handle of an umbrella, is the manubrium containing the mouth. The gastrovascular cavity extends from the mouth in the manubrium into four radial canals, on which are located the reproductive organs, or gonads. Medusae also reproduce sexually. There are two types of medusae, each bearing either testes, the male gonads which pro-duce sperm, or ovaries, the female gonads which produce eggs. Sperm and eggs are released into the surrounding sea water. Fertilization takes place in the sea water, and a zygote is formed. Following early embryonic cleavage, a free-swimming larva, called a planula, is formed which is ciliated and lacking a gastrovascular cavity or mouth. After a free-swimming existence lasting for several hours to several days, the planula attaches to an object and develops into a polypoid colony, thus completing the life cycle. The alternation of a polypoid and medusoid stage is an example of "alternation of generations", or meta-genesis. Asexually and sexually reproducing stages al-ternate with each other. It must be noted that it is different from the alternation of generations that occurs in plants in that both polyp and medusa are diploid, as in all multicellular animals, the only haploid stage in the life cycle are the gametes.

Question:

What power is needed to move a 3000-lb car up an 8.0\textdegree incline with a constant speed of 50 mi/hr against a frictional force of 80 lb?

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Solution:

In order to just be able to move the car up the incline at constant velocity, there must be no net force on the car. Looking at the figure, we see that the net force on the car, acting down the plane, is F_net = mg sin 8\textdegree + f. If there is to be no resultant force on the car, we must act on it with a force equal in magnitude to F_net, but acting up the plane. Therefore, F_app = mg sin 8\textdegree + f = (3000 lb) (.139) + 80 lb = 497 lb Now, the power expended in moving an object equals the time rate of change of the work done on the object. Hence, P = dW/dt ButW = \intF_app^\ding{217} \bullet ds^\ding{217} P = d/dt [\intF_app^\ding{217} \bullet ds^\ding{217}] In our case, F_app^\ding{217} is constant, and P = [d/dt] F_app^\ding{217} \bullet\intds^\ding{217} = [d/dt] F_app^\ding{217} \bullet s^\ding{217} Since F_app^\ding{217} is constant in time also, we obtain P = F_app^\ding{217} \bullet ds^\ding{217}/dt = F_app^\ding{217} \bullet v^\ding{217} Since ds^\ding{217}/dt is defined as the velocity of the object we're moving. In this problem, F_app^\ding{217} and v^\ding{217} are parallel and in the same direction and P = F_app^\ding{217} \bullet v^\ding{217} = F_app v = (497 lb) (50 mi/hr) But 1 mi/hr = 1.48 ft/sec andP = (497 lb) (50) (1.48 ft/sec) = 3.7 × 10^4 ft \textbullet lb/sec

Question:

A 3200-lb car traveling with a speed of 60 mi/hr rounds a curve whose radius is 484 ft. Find the necessary centripetal force.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0161.htm

Solution:

A Centripetal force is a force which results when a particle executes circular motion with constant speed. It is called centripetal because it points to the center of the circle. Note that although the speed of the particle is constant, its velocity is not, because the latter is continually changing in direction. As a result, the centripetal force is responsible for changing the velocity of the particle. Using Newton's Second Law, we may write F = ma(1) where F is the net force acting on the mass, m. Because this is uniform circular motion, a = v^2/R , where v is the speed of the particle in a circular orbit of radius R. Inserting this result in (1), F = mv^2/R(2) Equation (2) gives the centripetal force needed to accelerate m. In order to use this formula, we must trans-form the weight, mg, given in the question as 3200 lb., into a mass by dividing by 32 ft/sec.^2 . Then, using (2) F = [(3200/32)sl] [(88 f/s)^2/(484 ft)] = 1600 lb.

Question:

Write a BASIC program to search a data statement for a given name which is typed in when the program is executed.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0235.htm

Solution:

This problem introduces the notion of string variables and of comparisons between strings. String variables are denoted by any valid variable name with $ at the end, for example: A$, B$, C1$,..., etc. One can test strings for equality by means of the ' = ' and '< >' operators. When the search is finished, if the user wishes to search for another name, the RESTORE statement is used. 1\O REM PROGRAM SEARCHES THE DATA STATEMENT 2\O REM FOR A CERTAIN NAME 3\O PRINT "TYPE NAME, THEN HIT RETURN" 4\O INPUT A$ 5\O FOR J = 1 TO 5 6\O READ B$ 7\O IF B$ = A$ THEN 11\O 8\O NEXT J 9\O PRINT "CAN'T FIND NAME" 1\O\O GO TO 12\O 11\O PRINT B$; 'WAS FOUND" 12\O PRINT "TO CONTINUE TYPE YES; ELSE - NO." 13\O INPUT C 14\O IF C$ < > "YES" THEN 18\O 15\O DATA "JOHN", "BOB", "JILL", "MARY", "FRED" 16\O RESTORE 17\O GO TO 3\O 18\O END

Question:

Suppose the coefficient of friction between a horizontal surface and a moving body is \mu. With what speed must a body of mass m be projected parallel to the surface to travel a distance D before stopping?

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Solution:

In this problem, we actually want to describe the motion of m, for we want it to travel a distance D before stopping. Our task is to determine the initial velocity which the mass must have in order for this to be possible. We are seeking to describe the properties of the motion of m, such as its acceleration. Therefore, we apply Newton's Second Law, F = ma, to both the vertical and horizontal directions for the mass m (see the free body diagram). In the horizontal direction, Newton's Law becomes F_x = ma_x(1) where F_x represents the sum of all forces acting horizontally on m, and a_x represents the resulting acceleration in the x direction due to these forces. From the figure, we see that f, the force of friction, is the only horizontal force acting on m. Therefore, substitut-ing into equation (1) we find - f = ma_x(2) where the minus sign appears because f acts in the negative x direction. (We are taking a_x to be positive in the positive x direction). Writing Newton's Second Law for the vertical direction, we find F_y = ma_y(3) where F_y is the sum of all forces acting on m in the y direction, and a_y is the resulting acceleration of m in the y direction. The only forces which act on m in the vertical direction are N, the normal force of the surface which pushes on m, and mg, the weight of the mass, which points downward. Substituting into equation (3), we obtain N - mg = ma_y(4) where the minus sign indicates that the two forces point in opposite directions. Furthermore, note that a_y must be zero since the mass never rises off the surface on which it slides. Substituting this into equation (4), we have N = mg(5) Now, the frictional force law is given by f = \muN(6) where \mu is the coefficient of sliding friction between m and the surface, and N is the magnitude of the normal force. Substituting equation (5) into equation (6), we obtain f = \mumg(7) Inserting this into equation (2), -\mumg = ma_x(8) Solving for a_x a_x = -\mug(9) Note that because \mu and g are constants, a_x is also constant. Hence we may use the kinematical equations for constant acceleration to describe the position of the mass. The equation needed is V^2_f - V^2_o = 2a_x(x_f - x_o)(10) where x_o and v_o are the initial position and velocity of m, respectively, and x_f, v_f are the final position and velocity of m, respectively. For this problem: x_o = 0x_f = D v_o = ?v_f = 0(11) v_f is 0 because the mass is at rest after travelling to its final position, which is D. Substituting these values into equation (10), 0 - v^2_o = 2a_x(D - 0)(12) or-v^2_o = 2a_xD(13) Then, substituting equation (9) into equation (13), we obtain v^2_o = 2\mugD orv_o = \surd(2\mugD)(14) for the initial velocity of m needed so that it may travel a distance D before stopping.

Question:

In the Citric Acid Cycle (also known as the Krebs Cycle) , citrate is converted to isocitrate. From the algebraic sign of ∆G\textdegree for the isomerization, what is the favored or spontaneous direction of the reaction? Calculate ∆G\textdegree for the following reaction and explain what bearing this reaction will have on the isomerization of citrate and the operation of the Krebs Cycle: Isocitrate^3- + (1/2) O_2 (g) + H^f \rightleftarrows \alpha-ketoglutarate^2- + H_2O (l) + CO_2 (g) ∆G_f in (Kcal / mole) are for citrate^3- = - 279.24; isocitrate^3- = -277.65; H^f = 0, O_2 = 0; \alpha-ketoglutarate^2- = -190.62; H_2O(l) = - 56.69; CO_2(g) = - 94.26.

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Solution:

The Krebs Cycle is a series of oxidations aided by enzymes that leads to citric acid and eventually to CO_2, H_2O and ATP. To find ∆G of the isomerization, set up the equation citrate^3- \rightleftarrowsisocitrate^3- and use the Gibbs free energy of formation values given. ∆G\textdegree_reaction= \sumG\textdegree_f,_products - \sum∆G\textdegree_f,_reactants = (-277.65 Kcal/mole) - (- 279.24 Kcal/mole) = + 1.59 Kcal/mole. Thus, isomerization of citrate to isocitrate is not a spontaneous process, since ∆G\textdegree for the reaction is positive. To find ∆G of the following reaction: Isocitrate^3- + (1/2)O_2 (g) + H^f \rightleftarrows \alpha-ketoglutarate^2- + H_2O (l) + CO_2 (g) Use ∆G\textdegree_reaction = \sum∆G\textdegree_f,_products - \sum∆G\textdegree_f,_reactants = [∆G\textdegree_f,CO_2 + ∆G\textdegree_f,H_2O + ∆G\textdegree_f,_ketoglutarate] - [∆G\textdegree_f, H^f + ∆G\textdegree_f, (1/2O_2) + ∆G\textdegree_f,_Isocitrate] = (- 94.26 - 56.69 - 190.62) - (0 + 0 - 277.65) = -63.92 (Kcal /mole), ∆G\textdegree for the oxidation and decarboxylation of isocitrate is -63.92 (Kcal / mole); this large free energy decrease will favor the constant removal of isocitrate and permit the cycle to move forward despite the non-spontaneity of the isomerization.

Question:

State what is wrong with the following flowchart entries, and suggest better ways of writing them:

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Solution:

a) When writing flowcharts, it is bad practice to use the equal sign to mean "store the integer 7 in location B." You should use equal signs only in decision boxes when comparing two quantities. Instead, a better way of writing this box requires an arrow; thus we have b) As written, this decision box is meaningless. When the question "Is Q odd?" is asked, an appropriate answer would be "yes" or "no", or perhaps "true" or false". Equal signs should be used when two things are compared, not when you question the status of one item. c) This box has a trivial error, but one that should be clarified. The author of this statement probably meant "store the integer 3 in location N." To correct this, write the following: d) The colon is used in flowcharts to compare two items. The pos-sibilities are M > N, M \geq N, M = N, M < N, or M \leq N, and these cases should be taken into consideration. One possible interpretation is the following:

Question:

Three researchers studied 1 mole of an ideal gas at 273\textdegreeK in order to determine the value of the gas constant, R. The first researcher found that at a pressure of 1 atm the gas occupies 22.4 l. The second researcher found that the gas occupies 22.4 l at a pressure of 760 torr. Finally, the third researcher reported the product of pressure and volume as 542 cal. What value for R did each researcher determine?

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Solution:

This problem is an application of the ideal gas equation, PV = n RT, where P = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute tem-perature. Specifically, we are trying to determine R from the relation R = PV/nT. All three researchers worked with one mole of gas (n = 1) at T = 273\textdegreeK. Their results are as follows: First researcher: P = 1 atm, V = 22.4 l. R = (PV/nT) = [(1 atm × 22.4 l)/(1 mole × 273\textdegreeK)] = 0.0821 [(l-atm)/(\textdegreeK-mole)] Second researcher: P = 760 torr = 760 torr × [(1 atm)/(760 torr)] = 1 atm, V = 22.4 l. R = (PV/nT) = [(1 atmt × 22.4 l)/(1 molex 273\textdegreeK)] = 0.0821 [(l-atm)/(\textdegreeK-mole)] Third researcher: PV = 542 cal = nRT = 1 mole (R)(273\textdegreeK) R = (PV/nT) = [(542 cal)/{(1 mole) (273\textdegreeK)}] = 1.99 cal/mole \textdegreeK.

Question:

How does a facultative anaerobe differ from anobligatory anaerobe?

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Solution:

Organisms that can liveanaerobicallyare divided into two groups. The obligatory, or strict, anaerobe cannot use oxygen and dies in the presenceof oxygen. These include denitrifying bacteria of the soil, which areresponsible for reducing nitrate to nitrogen, and methane-forming bacteria, which produce marsh gas. Some obligatory anaerobes are pathogenicto man; these includeClostridiumbotulinum, responsible for botulism, a fatal form of food poisoning;Clostridiumperfringens, which causesgas gangrene in wound infections; and Clostridiumtetani, which causesthe disease tetanus. The facultative anaerobes can live either in the presence or absenceof oxygen. Under anaerobic conditions, they obtain energy from afermentation process; under aerobic conditions, they continue to degradetheir energy sourceanaerobically(viaglycolysis) and then oxidizethe products of fermentation using oxygen as the final electron acceptor. Yeast will grow rapidly under aerobic conditions but will still continueto live when oxygen is removed. It reproduces more slowly but maintainsit-self by fermentation. Winemakers take advantage of this behaviorby first aerating crushed grapes to allow the yeasts to grow rapidly. They then let the mixture stand in closed vats while the yeasts convertthe grape sugaranaerobicallyto ethanol.

Question:

Describe the various types and functions pertaining to the epithelial tissues of animals.

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Solution:

Epithelial tissues form the covering or lining of the internal and external body surfaces. Epithelial tissues include, for example, the outer part of the skin, the linings of the digestive tract, the lungs, the urogenital tract, the covering of the body cavity and so forth. The cells that make up the epithelial tissues are packed closely together, thus providing a continuous protective barrier between the underlying cells and the outside world. This protection keeps the body safe from mechanical injury, from harmful chemicals and bacteria, and from drying. Epithelial cells may have other functions, such as absorption, secretion, and sen-sation. Types of epithelial tissue A. squamous epitheliumB. cuboidal epithelium C. columnar epitheliumD. ciliated columnar epithelium E. sensory epitheliumF. glandular epithelium The surface of the epithelial cell that is exposed to air or fluid often becomes specialized. This surface commonly bears cilia, hairs or fingerlike processes. It may also be covered with waxy or mucous secretions. The opposite surface rests upon other cell layers. Epithelial cells are generally grouped into four categories according to their shape and function: squamous, cuboidal, columnar, and glandular epithelial cells. Squamous cells are much broader than they are thick and have the appearance of thin, flat plates. They are found on the surface of the skin and the lining of the mouth, esophagus and vagina. Cuboidal epithelium (cube-shaped cells) is found in kidney tubules. The cells of columnar epithelium are long and thin, resembling pillars or columns. The stomach and intestines are lined with columnar eoithelium. Some columnar cells have cilia on their surfaces. The function of the cilia is to move sub-stances past the cell. The respiratory tact is character-ized by ciliated columnar epithelium. Finally, a gland may consist of one or many epithelical cells. Exocrine glands have ducts, while endocrine glands do not. Examples of secretions from glandular epithelium include wax, sweat and milk. Epithelial tissue may further be classified as simple or stratified. Simple epithelium is one cell layer thick. Stratified epithelium consists of several layers of cells. Both simple and stratified epirhelium may be squamous, cuboidal, or columnar.

Question:

The principle of negative feedback, where an increase in the output of a system acts on the system to halt its further production, is manifested in the menstrual cycle. Where?

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Solution:

The first event in a menstrual cycle following the menstrual phase (flow phase) is an increase in the secretion of FSH by the anterior pituitary as a result of stimulation by the hypothalamus. This hormone stimulates the development of the follicles in the ovaries, which contain the eggs. After some time, one of the follicle gains predominance. At this point, only this one follicle continues to grow. Under the combined influence of both FSH and LH, the latter of which is maintained, at a low level in the blood by the follicular cells, the cells surrounding the egg within the follicle begin to secrete estrogen. As more and more estrogen is produced, the increased level of this hormone in the blood exerts an inhibitory effect on the FSH-stimulating center in the hypothalamus. This results in a decrease in the level of FSH produced by the anterior pituitary. This is the negative feedback mechanism between estrogen and FSH. High concentration of a secreted substance (estrogen) turns off the original stimulus for its production (FSH), so that an excessive amount of the secreted substance (estrogen) would not be made. This occurs roughly 8 to 10 days following the menstrual phase, or 13 to 15 days after the start of the cycle. While the high level of estrogen inhibits FSH production, it activates the LH-stimulating center in the hypothalamus, ultimately inducing a surge of LH into the circulatory system, which initiates ovulation. After ovulation the ovarian follicle with the egg discharged develops into the progesterone-secreting corpus luteum. The corpus luteum promotes the plasma progesterone level to such a state that progesterone begins to exert ah inhibitory effect on the LH-stimulating center in the hypothalamus. This is the negative feed-back mechanism between progesterone and LH. The high progesterone concentration suppresses further release of LH so that another ovulation is prevented; hence no other corupus luteum is formed to contribute more progesterone to the circulatory system.

Question:

How many moles are contained in 196 g of H_2SO_4?

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Solution:

To calculate the number of moles of a given compound from a certain number of grams of that compound one should use the following relation: number of moles =(numberof grams) / (molecular weight) When determining the molecular weight of a compound one must first know the atomic weights of the elements within the compound. One then determines the molecular weight by adding together the weights of the individual elements making up the compound. The molecular weights of the elements contained in H_2SO_4 are MW of H = 1, MW of S = 32, and MW of O = 16. If there is more than one atom of a particular element, the molecular weight of the element must be multiplied by the number of atoms present in the compound. Calculating the molecular weight of H_2SO_4 H 2 1.0 =2 S 1 32 =32 O 4 16 =64 Total =98 The molecular weight of H_2SO_4 if is thus 98 g. The number of moles of H_2SO_4 in 196 g can now be calculated. number of moles = (196 g) / (98 g/mole) = 2.0 moles.

Question:

Write the reaction equation for the formation of the corresponding ether from methanol. How many ethers are possible when a mixture of CH_3OH and CH_3CH_2OH reacts with the acid H_2SO_4?

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Solution:

The formation of ethers from alcohols requires acid. In acid solution, the hydroxyl group of an alcohol is protonated to form the oxonium ion : A free electron pair on the oxygen of another molecule of alcohol reacts with the oxonium ion to form an ether. This is accompanied by water loss : If you had a mixture of CH_3OH and CH_3CH_2OH, three ethers would result. Upon the addition of acid, methanol molecules can react with other methanol molecules to form methyl ether. It can also react with ethanol (CH_3CH_2OH) to form methyl ethyl ether. The ethanol molecules can react with other ethanol molecules to form ethyl ether. (Ethanol can react with meth-anol, but this yields methyl ethyl ether.) Thus, there are 3 possible ethers: methyl ether, ethyl ether and methyl ethyl ether.

Question:

For the block and tackle shown in the figure (a) find the displacement ratio. (b) What force, F, must be exerted on the free end of the rope to lift a 200 lb load?

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Solution:

(a) When F\ding{217} pulls down the rope by an amount L, pulley 2 moves up by (1/2)L (as shown in the figure) since the shortening of the rope is shared by the two segments of rope that hold the pulley. Therefore, the ratio of the displacement of load to the displacement of the rope is [(1/2)L]/L = (1/2). (b) From the figure, we see that the load is held up by a force 2T\ding{217} where T\ding{217} is the tension in the rope. Hence, in order to lift the load, the minimum tension should satisfy W= 2T orT= (1/2)W where W is the weight of the load. F\ding{217} is equal to T\ding{217} as long as the rope does not break since the stress in the rope is caused by the action of F\ding{217}. We have F= T = W/2 = (200 lb)/2 = 100 lb.

Question:

Explain why the following statements or expressions are illegal in Pascal.

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Solution:

a) Program name is an identifier. It must start with a letter and continue with letters or digits only. No hyphen is allowed. b) 2.5 is an invalid label. A label must be an integer. c) Constants are defined, not assigned. The correct form is CONST K = 10.5; d) 'New' is a reserved word in Pascal. e) Theindextypeof an array must be a finitesubrange. 'Integer' is not a subrange. f)Anindextypemay not be asubrangeof real numbers. There are infinitelymany real numbers between 3.5 and 8.5. g)Each record definition must end with 'END' statement. h)illegalassignment. Should be y: =7.5; i )Function odd(x) returns abooleanvalue, while p is declared as an integer. Its type must be changed tobooleanfor the statement to be correct.

Question:

Use the quadratic formula to solve for x in the equation x^2 - 5x + 6 = 0.

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Solution:

The quadratic formula, x = [{\rule{1em}{1pt}b \pm\surd(b^2 - 4ac)} / {2a}], is used to solve equation in the form ax^2 + bx + c = 0. Here a = 1, b = \rule{1em}{1pt}5, and c = 6. Hence x = {\rule{1em}{1pt} (\rule{1em}{1pt}5) \pm\surd[(\rule{1em}{1pt}5)^2 - 4\bullet1\bullet6)]} / {2 \bullet 1} = [{5 \pm\surd(25 - 24)} / {2}] = [(5 \pm\surd1) / 2] = [(5 \pm1) / 2] = [(5 + 1) / 2]or[(5 \rule{1em}{1pt} 1) / 2] = 6/2or4/2 = 3or2 Thus the roots of the equation x^2 - 5x + 6 = 0 are x = 3 and x = 2.

Question:

H. Hertz produced radio waves whose wavelength was about 3 m. What was the frequency of the oscillating electric charges responsible for this electromagnetic radiation?

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Solution:

Hertz produced radio waves by use of an induction coil. This alternating current generator was connected in series with a metal loop containing a gap in it (see the figure). A second loop (the receiver) with a similar gap was placed a few feet away from the first loop (the transmitter). The induction coil created an arc discharge across the gap of the transmitting circuit. A similar discharge appeared across the gap of the receiving circuit. Apparently, the disturbance at the first circuit was transmitted to the second circuit. This disturbance was shown to propagate with the speed of light c. The effect of the AC generator is to cause the electric charge within the wire in the transmitting circuit to oscillate about an equilibrium position. An oscillating charge undergoes acceleration and emits electromagnetic radiation with a frequency equal to the oscillating frequency of the charge. The electro-magnetic wave disturbance is transmitted through space. When it reaches the. receiving circuit, it causes the charge in the wire to oscillate. An AC current is then created. An arc discharge is then developed across the gap due to this current. If the wavelength of the wave is given, the frequency of the wave can be found by use of the following relation. c = f\lambda Since the value of the wavelength, \lambda = 3m, and the speed of light, c = 3 × 10^8 m/s, are known, the frequency is then f = (c/\lambda) = {(3 × 10^8m/s)/3m} = 1 ×10^8 Hz This frequency falls within the present-day FM radio band. The frequency of the source (i.e. the oscillating charge) is the same as that of the wave, so the current (i.e. the movement of charge) is oscillating at a frequency of 1 × 10^8 Hz also.

Question:

Change the Subroutine of the previous problem into a FUNCTION SUBPROGRAM and explain the procedure.

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Solution:

In FORTRAN, there is another kind of subprogram, called the FUNCTION SUBPROGRAM. The Function Subprogram may replace the subroutine if the result obtained by the SUBROUTINE is a single value. In order to get a function subprogram from the subroutine of the previous problem, 3 changes have to be made. FUNCTIONIGCD(I,J) that is, the word SUBROUTINE is replaced by FUNCTION. Parameter K is not being used anymore because the name of the function will be assigned the value of the greatest common divisor. The divisor value obtained by the subprogram is an integer; therefore the name of the FUNCTION subprogram must begin with the corresponding letter (i.e., I,J,...,N). time it appears in the statement, i.e., statement 4, (K = J), has to be changed into IGCD = J. parameters . It was observed in the Subroutine program GCD that the values of I and J were changed, therefore some modifications must be made before this subroutine can become a function subprogram. The simplest modification would be to assign the values of I and J to two independent variables, say M and N, and then use them in place of I and J. Therefore, the modified FUNCTION SUBPROGRAM will take the following form: FUNCTION IGCD (I,J) M = I N = J IF (M - N) 1,4,2 1 L = N N = M M = L 2 L = M - (M/N){_\ast}N IF( L)3,4,3 3 M = N N = L GO TO 2 4 IGCD = N RETURN END

Question:

Calculate, E\textdegree, ∆G\textdegree, and K for the following reaction at 25\textdegreeC. (1/2) Cu (s) + (1/2) Cl_2 (g) = (1/2) Cu^2+ + Cl^-.

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Solution:

The magnitude of the standard electrode potential, (E\textdegree) is a measure of the tendency of the half-reaction to occur in the direction of reduction. The standard EMF (Electromotive Force) of a cell is equal to the standard electrode potential of the right - hand electrode minus the standard electrode potential of the left - hand electrode (by convention) , E\textdegree = E\textdegree _right - E\textdegree _left The reaction taking place at the left electrode is written as an oxidation reaction and the reaction taking place at the right electrode is written as a reduction re-action. The cell reaction is the sum of these two reactions. Thus, Oxidation:(1/2) Cu (s) = (1/2) Cu^2+ + e Reduction:(1/2) CI_2 (g) + e = Cl^- Since the oxidation reaction is to occur at the left electrode and the reduction at the right electrode, the cell is written as Cu \vert Cu^2+ \vert \vert Cl^- \vert Cl_2 (g) \mid Pt E\textdegree = E\textdegreeCl- \mid Cl2 \mid pt- E\textdegree _Cu^2_+\midCu E\textdegree can be found on any Standard Electrode Potential Table. E\textdegreeCl- \mid Cl2 \mid pt= 1.360 E\textdegree _Cu^2+ \midCu= 0.337; Thus, E\textdegree = 1.360 - 0.337 E\textdegree = 1.023 V Now that E\textdegree is known, ∆G\textdegree, standard Gibbs free energy, i.e., the energy available to do useful work, can be calcul-ated from ∆G\textdegree = - n F E\textdegree, where n = number of electrons transferred and F = the value of a Faraday of electricity, (23, 060 cal/V-mole). From the redox half-reactions written, one sees that 1 electron is being transferred. Consequent-ly, n = 1. Substituting these values one obtains ∆G\textdegree = - (1) [23, 060 (cal/V-mole)] (1.023 V) = - 23, 590 cal. The equilibrium constant, K, can be calculated using the Nernst equation for unit activity. In other words, E\textdegree = .0591/N log K. Substituting the previously calculated values for E\textdegree and n, 1.023 V = .0591 log K Solving,log K = 17.31orK = 2 × 101 7.

Question:

At constant volume, the heat capacity of gas differs from the heat capacity at constant pressure. Why?

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Solution:

Heat capacity is defined as the quantity of heat that will result in a temperature rise of 1\textdegreeC per mole of material. When heat is added at constant volume, all the heat goes into increasing the kinetic energy of the mo-lecules. However, with constant pressure, the volume can expand. This means the added heat must do two things. It must be able to increase the kinetic energy of the molecules AND do work against the pressure to expand the volume. The constant pressure case would require more heat. In fact, the difference between C_p andC_v, where C_p = heat capacity at constant pressure, andC_v= heat capacity at constant volume, is the universal gas constant R. In summary, then, the difference stems from the fact that C_p must increase kinetic energy and do work to expand the volume, whileC_vonly has to increase the kinetic energy.

Question:

What difference can be observed in carbohydrate metabolism between liver and skeletal muscle cells?

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Solution:

Both liver cells and muscle cells have the ability to store excess glucose as glycogen. Glucose is not transformed directly into glycogen. The first step in this conversion involves the change of glucose into glucose 6-phosphate. This ATP-dependent reaction is catalyzed by hexokinase. After several reversible intermediate steps, glucose 6-phosphate molecules are con-densed to form the polysaccharide glycogen. The major difference in carbohydrate metabolism between liver and muscle cells is the presence of the enzyme, glucose 6-phosphatase in liver cells. This enzyme converts glucose 6-phosphate to free glucose. The unphosphorylated glucose can then enter the circula-tory system. The presence of glucose 6-phosphatase allows the liver cells to break down glycogen into glucose molecules and release the latter into the bloodstream. Thus, one of the functions of the liver is to regulate glucose level in the blood by releasing or storing the monosaccharide as the situation demands. Muscle cells are unable to release free glucose and thus cannot regulate the blood glucose level because it lacks glucose 6-phosphatase. However, muscle cells can still store glucose as glycogen due to the presence of hexokinase.

Question:

The owner of a car and a helpful passer-by attempt to pull the former's car from the field Into which it has skidded. They attach two ropes to the front of the chassis symmetrically, each rope being 1 ft from the center point, C, and exert pulls of 200 lb and 150 lb in parallel directions, both at an angle of 30\textdegree to the horizontal (see the figure). To what point of the chassis must a tractor be attached and what horizontal force must it exert to produce an equivalent effect?

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Solution:

The resultant force of the two pulls exerted by the men must be R^\ding{217} of magnitude (200 + 150) lb = 350 lb, in the same direction as either of the forces, i.e., at 30\textdegree to the horizontal. Only the horizontal component of this force is doing useful work in pulling the car from the field. This component has magnitude R cos 30 = 350 lb × \surd3/2 = 303.1 lb. This is the force that the tractor must exert. The point of attachment of the tractor must be the point O at which the line of action of the resultant R^\ding{217} cuts the front of the chassis. The forces that the two men exert on the chassis produce a net torque \cyrchar\cyrt about the center point C. The point of action O ofR^\ding{217} must lie at a distance from C, along the front of the chassis, such that R^\ding{217} produces a net torque equal to \cyrchar\cyrt: \cyrchar\cyrt = (150 lb)(1 ft) - (200 lb)(1 ft) = (350 lb)x x = (-50 ft-lb)/(350 lb) = - (1/7)ft where x is the distance of O from C. All counterclock-wise torques are taken as positive. Since x is negative, we see that R produces a clockwise torque about C. This tells us that O must be to the left of C (above C in the figure). Thus the point of attachment of the tractor is 6/7 ft from A, that is, 1/7 ft from the center point of the front of the chassis.

Question:

Sunlight is the ultimate energy source on earth. Energy from sunlight is not returned to its source but is transformed to other forms of energy which are closely tied together in an energy cycle. Describe the energy cycle.

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Solution:

The energy cycle starts with sunlight being utilized by green plants on earth. The Kinetic energy of sunlight is transformed into potential energy stored in chemical bonds in green plants. The chemical bonds are synthesized by the process of photosynthesis. The potential energy is released in cell respiration and is used in various ways. Thus, fundamental to the energy cycle is the ability of energy to be transformed. Inorganic (nonliving) matter in the ecosystem is closely tied to organic (living) matter in the energy cycle. For example, photosynthesis requires carbon dioxide from the air and water and minerals from the soil to occur besides sunlight. These are the nonliving components of photosynthesis. Chlorophyll in green plants captures the sunlight, and organic substances (i.e., glucose) are generated from inorganic ingredients via a series of enzymatic reactions in the plant cell. Chlorophyll, enzymes, and other cellular components form the living part of photosynthesis. In the energy cycle (Figure), some of the food synthesized by green plants are broken down by the plants for energy, and consequently carbon dioxide and water are released. These again become available for green plants in capturing more energy of the sunlight. Some of the synthesized compounds are used in building the bodies of the plants and are hence stored as potential energy until the plants die. The bacteria and fungi of decay break down the bodies of the dead plants, using the liberated energy for their own metabolism. In these processes, carbon dioxide and water are released, and the minerals go back into the soil. These substances are thus recycled. Animals which feed on plants utilize a part of the energy from the food in cell respiration, with a release of carbon dioxide and water which are again recycled. Some of the minerals in the plant food are excreted by the animals and are thus available to be reused. Animals which feed on other animals utilize some energy from the food in building their own bodies. They break down some of their stored food to yield energy for daily activities such as locomotion. Food degradation is accompanied by release of carbon dioxide and water, which are returned to the ecosystem. When animals die, their bodies decay and all of the materials that were used in the construction are restored to a state which can be reused by the action of the bacteria and fungi of decay. It must be remembered that at no point in the cycle is energy destroyed. The energy from sunlight is not destroyed but is transformed into heat, chemical or mechanical energy.

Question:

The development of a parasitic mode of existence is a remarkable example of an adaptation that has evolved to permit one organism to exist at the expense of an-other. Among the flatworms the flukes are parasitic. Describe the body structure of a fluke, its modifica-tions for a parasitic existence, and its life cycle.

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Solution:

The adult fluke is roughly one inch long, and its body is quite flat. The worm is covered by a cuticle that is secreted by the underlying cells. The cuticle is one of its adaptation for parasitism, for it protects the worm from the enzymatic action of the host. With the development of the cuticle, the cilia and sense organs so characteristic of other flatworms have disappeared. The mouth is located at the anterior end and is surrounded by an oral sucker. Another sucker, the ventral sucker, is located some distance posterior to the mouth. The ventral sucker is used to attach the fluke to the body of its victim, the host. The anterior sucker, with the aid of the muscular pharynx, is used to withdraw nutrients from the host. The mouth opens into a muscular pharynx, and just back of this, the alimentary tract branches into two long intestines that extend posteriorly almost the entire length of the body. By far the most complicated aspect of these worms is their reproductive cycle. The fluke is hermaphroditic, that is, it contains both male and female sexual organs. There is no copulatory structure in the flukes and re-production occurs through self fertilization. Following fertilization, the eggs are expelled from the fluke, which at this point is within the host's body; they then exit from the host's body in the feces. Hatching of the eggs in the oriental liver fluke occurs only when the eggs are eaten by certain species of freshwater snails, and takes place in the digestive system of the snail. Here, the egg hatches into a ciliated, free-swimming miracidium (see Figure). Within the digestive system of the snail, which is called the first intermediate host, the miracidium develops into a second larval stage, called a sporocyst. Inside the hollow sporocyst, germinal cells give rise to a number of embryonic masses. Each mass develops into another larval stage called a redia, or daughter sporocyst. Germinal cells within the redia again develop into a number of larvae called cercariae. The cercaria possesses a digestive tract, suckers and a tail. The cercaria is free swimming and leaves the snail. If it comes in contact with a second intermediate host, an invertebrate or a vertebrate, it penetrates the host and encysts. The encysted stage is called a metacercaria. If the host of the metacercaria is eaten by the final vertebrate host, the metacercaria is released, migrates, and develops into the adult form within a characteristic location in the host, usually the bile passages of the liver. Man, the final host, usually gets the fluke by eating a fish which contains the en-cysted cercaria.

Question:

Body temperature on the average is 98.6\textdegreeF. What is thison (a)theCelsius scale and (b) the Kelvin scale?

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Solution:

(a) One converts \textdegreeF to \textdegreeC by using the following equation. \textdegreeC = 5/9 (\textdegreeF\Elzbar32) \textdegreeC = 5/9 (98.6\Elzbar 32) = 5/9 (66.6) = 37.00\textdegreeC. (b) \textdegreeC can be converted to \textdegreeK by adding 273.15 to the Celsius temperature. \textdegreeK = \textdegreeC + 273.15 \textdegreeK = 37.0 + 273.15 = 310.15\textdegreeK.

Question:

In the figure \epsilon_1 = 12 volts, r_1 = 0.2 ohm; \epsilon_2 = 6 volts, r_2 = 0.1 ohm; R_3 = 1.4 ohms; R_4 = 2.3 ohms; compute (a) the current in the circuit, in magnitude and direction, and (b) the potential difference V_ac .

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Solution:

(a) From conservation of energy we know that the sum of the changes in potential (voltage changes) around any closed loop must equal zero. Therefore the current (which is conventionally taken as the flow of positive charge) is in a clockwise direction because \epsilon_1 > \epsilon_2 ; but let us choose it as going counterclockwise to show that it doesn't make any difference. Note that each battery has an internal re-sistance. Starting at point A and traversing counterclock-wise yields the following equation: - \epsilon_1 - ir_1 - iR_4 - iR_3 - ir_2 + \epsilon_2 = 0 i (r_1 + r_2 + R_3 + R_4 ) = \epsilon_2 - \epsilon 1 i = [(\epsilon2- \epsilon_1 )/(r_1 + r_2 + R_3 + R_4 ) = (-6/4) = -1.5 amp. The negative value for the current merely means that we chose the wrong direction (i.e., the current flows clockwise) (b)We may use either a clockwise or counterclockwise path from point A to point C to find V_AC . Since we are only concerned with differences in potential, let us assume a zero potential of A and then traverse the loop clockwise from A to C. Taking into account the clockwise flow of cur-rent this yields, -\epsilon_2 - ir_2 - iR_3 = -6 - (1.5) (.1) - (1.5) (1.4) = -8.25 volts This means that the potential is 8.25 volts lower at point C than at point A. If we go from point A to point C in a counterclockwise direction we must remember to use the negative value for the current. This path yields: -\epsilon_1 - ir_1 - iR_4 = -12 - (-1.5) (.2) - (-1.5) (2.3) = -12 + .3 + 3.45 = -8.25 volts. Again, we see that the potential at point C is 8.25 volts lower than that at point A.

Question:

What are the common editing procedures in APL? Give examples.

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Solution:

Errors have to be corrected before a program be-comes operational. Editing procedures help in making the corrections. These procedures are: i ) Open a function, display all lines of the func-tion, then close the function. ii) Open a function, replace a line that contains an error with a new line, then close the function. iii) Open a function, list all its instructions and remain in definition mode. iv) While in programming mode, direct program control to a specific line number. v) While in programming mode, direct program control to a specific line and then eliminate that line. vi) Insert an instruction between two other instruc-tions. To illustrate these procedures, consider the following program: \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]'ENTER BASE' [4]B \leftarrow [] [5]A \leftarrow .5 × H × B [6]'THE AREA IS:';A i) To display the total program (assuming it has been keyed in) key in \nablaTRIANGLE [ [] ]\nabla. This command causes a listing of the program to be made at the users terminal. Note that [6] has an error. ii) 'THEAREA...' should be 'THE AREA...'. To make this correction, type in \nablaTRIANGLE [6] 'THE AREA is:'; A\nabla. iii) To remain in definition mode, type \nablaTRIANGLE [ [] ]. (the \nabla has not yet been added). iv) Suppose we want to add more information so as to calculate the perimeter of the triangle. Then [3] and [4] can be deleted. Key in [3] and depress ATTN and RETURN keys. This action deletes [3]. Similarly, key in [4] and depress ATTN and RETURN keys. The program now looks like this: \nablaTRIANGLE [ [] ] \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]A \leftarrow .5 × H × B [4]'THE AREA IS:'; A [5] v,vi) Of course, the program does not make any sense right now. We must key in the three sides, replace B in [3] by the side corresponding to B and add a line for cal-culating the perimeter. To add the three sides: \nablaTRIANGLE[2.1] [2.1]'ENTER LEFT SIDE' [2.2]L \leftarrow [] [2.3]'ENTER BASE' [2.4]B \leftarrow [] [2.5]'ENTER RIGHT SIDE' [2.6]R \leftarrow [] [2.7]P \leftarrow L + B + R [2.8]\nabla Note that there is no need to change B. However, since P should be printed, add another line after [6]: \nablaTRIANGLE[6.1] [6.1]'THE PERIMETER IS:';P [6.2]\nabla The command NUM will renumber the statements. The fin-al result is: \nablaTRIANGLE [1]'ENTER HEIGHT' [2]H \leftarrow [] [3]'ENTER LEFT SIDE' [4]L \leftarrow [] [5]'ENTER BASE' [6]B \leftarrow [] [7]'ENTER RIGHT SIDE' [8]R \leftarrow [] [9]P \leftarrow L + B + R [10]A \leftarrow .5 × H × B [11]'THE AREA IS:'; A [12]'THE PERIMETER IS:'; P \nabla

Question:

Describe the output of the following program. L = 0 DO 1 K = 1, 25

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Solution:

This problem is a good example of the usage of the CONTROL STATEMENT "DO". DO statements are always of the form DO ri= j, k, or DO ri= j, k, l, where r-statement number,i- variable name,j,k, l,-variable names or positive integers. A DO statement is interpreted as follows: do all FORTRAN statements down to and including the one numbered r, first using the value ofi= j, theni= j + l,i= j + 2l, and so on, untilibecomes larger than k. If l is missing, it is assumed that l = 1. Therefore, the program takes L = 0 and K = 1 as initial values and then, according to DO statement, executes the formula L = L + K{_\ast}K 25 times with the value of K being increased by 1 each time. The program then writes the value of L which is finally obtained. Arithmetical execution of the program is as follows: L = 0 + 1 \bullet 1 = 1 L = 1 + 2 \bullet 2 = 5 L = 5 + 3 \bullet 3 = 14 L = 14 + 4 \bullet 4 = 30 L = 30 + 5 \bullet 5 = 55 L = 55 + 6 \bullet 6 = 91 L = 91 + 7 \bullet 7 = 140 L = 140 + 8 \bullet 8 = 204 L = 204 + 9 \bullet 9 = 285 L = 285 + 10 \bullet 10 = 385 L = 385 + 11 \bullet 11 = 506 L = 506 + 12 \bullet 12 = 650 L = 650 + 13 \bullet 13 = 819 L = 819 + 14 \bullet 14 = 1015 L = 1015 + 15 \bullet 15 = 1240 L = 1240 + 16 \bullet 16 = 1496 L = 1496 + 17 \bullet 17 = 1785 L = 1785 + 18 \bullet 18 = 2109 L = 2109 + 19 \bullet 19 = 2470 L = 2470 + 20 \bullet 20 = 2870 L = 2870 + 21 \bullet 21 = 3311 L = 3311 + 22 \bullet 22 = 3795 L = 3795 + 23 \bullet 23 = 4324 L = 4324 + 24 \bullet 24 = 4900 L = 4900 + 25 \bullet 25 = 5525 5525 is the desired result, which will be printed.

Question:

A hydrogen atom consists of a proton and an electron separated by about 5 × 10^-11 m. If the electron moves around the proton in a circular orbit with a frequency of 10^13 sec^-1 , what is the magnetic field at the position of the proton due to the moving electron?

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Solution:

The motion of the electron is equivalent to an electric current. That is, the charge e moves by a point on the orbit in time T, where T is the period of revolution. Thus the equivalent current is i = [(charge pass a point)/(time)] = e/t = ef . The magnetic field at the center of the circular loop of radius R is obtainedfrom the Biot - Savart Law; dB^\ding{217} = (\mu_0i / 4\pi) [(dl^\ding{217} × R^\ding{217})/R^3](1) where the permeability constant\mu_0 = 4\pi × 10[-7\omega/(a-m)] . As shown in the diagram, all the Infinitesimal contributions to B^\ding{217} from the infinitesimal circuit elements are in the direction per-pendicular to the plane of the orbit. In this case we may neglect the vector nature of (1) and obtain dB = (\mu_0i / 4\pi) [(dlR sin 90 o )/(R^3)] = (\mu_01 / 4\pi) (dl/R^2) since R^\ding{217} and dl^\ding{217} are perpendicular to each other. Therefore, the total magnetic field is the sum of the infinitesimal contributions dB or B = \intdB = (\mu_0i / 4\pi) (1/R^2 ) \intdl = (\mu_01 / 4\pi) [(2\piR)/R^2] = (\mu_0 / 2) (i/R) = 2\pi × 10[-7\omega/(a-m)] × [(1.6 × 10^-19 coul × 10^13 sec^-1 ) / (5 × 10^-11 m)] = 2.0 × 10^-2 (\omega/m^2) .

Question:

What are the five principles of evolution?

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Solution:

There are now five principles of evolution on which all biologists agree. First, evolution occurs more rapidly at some times than at others. At the present geological time it is occurring rapidly, with many new forms appearingand many old ones becoming extinct. Secondly, evolution does notproceed at the same rate in all types of organisms. For example, the lampshellshave remained unchanged for at least five hundred million years, while several species of man have appeared and become extinct in thepast few hundred thousand years. Thirdly, most new species do not evolvefrom the most advanced and specialized forms already living, but fromrelatively simple, unspecialized forms. Thus, mammals did not evolve fromthe large, specialized dinosaurs but from a group of rather small and unspecializedreptiles. Likewise, the vascular plants did not derive from thebryophytes but from lower and more primitive plants, like green algae. Fourthly, evolution is not always from the simple to the complex. There are manyexamples of "regressive" evolution, in which a complex form has givenrise to simpler ones. For instance, many wingless birds have evolvedfrom winged ones, and legless snakes from reptiles with appendages. This is because mutations occur at random, and if there is someadvantage for a species in having a simpler structure, any mutations thathappen to occur for such conditions will tend to accumulate by natural selection. Those mutations that result in a complicated delirious structure willbe selected against. Fifthly, evolution occurs in populations (not in individuals) by the processes of mutation, migration, natural selection, and geneticdrift.

Question:

Find thedeBrogliewavelength corresponding to an electron with energy 1, 10^4 , or 10^5eV. Neglect any corrections connected with the theory of relativity.

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Solution:

DeBroglieknew that light has a dual, wave-particle nature given by the relationship \lambda = h/p which relates the wavelength \lambda of a light wave with the momentum p of its photons through Planck's constant h.DeBrogliereasoned that matter also has a wave-particle nature, and that the wavelength of the matter wave was given by the same equation used for light with p the momentum of the particle of matter. We then have \lambda = h/mv The energy given for the electron is its kinetic energy K. This is equal to K = (1/2) mv^2 orv = \surd(2K/m) We can therefore say \lambda = [h/{m\surd(2K/m)}] = \surd[h^2/(2Km)] where m is the mass of the electron. Substituting values, we find \lambda = \surd[{6.6 × 10^-34 joule/sec^2} / {(2) × (KeV) × (1.6 × 10^-19 joule/eV) × (9.1 × 10^-31 kg)}] = \surd(150/K) 10^-10 m = \surd(150/K) A where K is in election volts and \lambda in Angstroms. ForK1= 1eV \lambda_1 = \surd(150/1)A = 12.2 A ForK_2 = 10^4eV \lambda_2 = \surd(150/10^4)A = 0.122 A ForK_3 = 10^5eV \lambda_3= \surd(150/10^5)A = 0.039 A

Question:

A brass plug has a diameter of 10.000 cm at 150\textdegreeC. At what temperature will the diameter be 9.950 cm?

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Solution:

It is observed experimentally that when a sample is exposed to a temperature change \DeltaT, the sample experiences a change in length \DeltaL, proportional to \DeltaT and L, the original length of the sample. This (approximate) result may be written as: \DeltaL =\alphaL\DeltaT(1) where \alpha is the coefficient of linear thermal expansion. Solving (1) for \DeltaT, we find that \DeltaT = (\DeltaL/ \alphaL \alphaL )(2 )(2 ) is the change in temperature required to implement a change in length \DeltaL. Substituting the given data into (2), we obtain: \DeltaT = [{(9.950 - 10.00) cm.}/{(19 x 10^\rule{1em}{1pt}6 /\textdegreeC)(10.000 cm.)}] T= \rule{1em}{1pt} 260\textdegree C But we want the final value of T, not \DeltaT. Since \DeltaT =T_f\rule{1em}{1pt} T_O T_f = \DeltaT + T_O T_f = 150\textdegree C \rule{1em}{1pt} 260\textdegree C = \rule{1em}{1pt} 110\textdegree C. This is the value to which the temperature must be lowered in order to shrink the diameter of the plug to 9.950 cm.

Question:

Phosphate can be removed from sewage effluent by adding magnesium oxide and ammonium ion according to the re- action 5H_2O + H_2PO^-_4 + MgO + NH^+_4 \rightarrow Mg(NH_4)PO_4 \bullet 6H_2O. The magnesium ammonium phosphate hexahydrate (Mg(NH_4)PO_4 \bullet6H_2O) precipitates out. In theory, how much of this could be produced if a city processes 5 million gal. (41.7 million lb.) daily of wastewater containing 30 ppm H_2PO^-_4? (1 pound = 453.59 g.)

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Solution:

The best way to approach this problem is to calculate how much H_2PO^-_4 is present in grams. Once the weight in grams is known one can calculate from the stoichiometry of the equation, the number of moles of Mg(NH_4)PO_4 \bullet 6H_2O that will precipitate. The wastewater has 30 ppm (parts per million) of H_2PO^-_4. In other words, 30 lbs. of H_2PO^-_4 are present in one million lbs. of wastewater. The city produces 41.7 million lbs. of wastewater. Thus, 30(41.7) = 1,251lbs. ofH_2PO^-4should be present. Because 1 pound = 453.59 grams; 1,251 lbs = 567,454 grams. Compute how many moles of H_2PO^-_4 are present. A mole is defined as grams (weight)/molecular weight (M.W.). The M.W. of H_2PO^-_4 = 96.97 g/mole. The number of moles of H_2PO^-4is =(567,454 g)/(96.97 g/mole) = 5852 moles. According to the reaction equation, one mole of H_2PO^-4generates one mole of precipitate. The M.W. of the precipitate = 245 g/mole. Recalling the definition of a mole, the amount of Mg(NH_4)PO4\bulletH_2O = (5852) (245) = 1.433 × 10^6 g = 1.433 × 10^3 kg.

Question:

Calculate the theoretical quantity of chlorine obtainable by the electrolysis of 2.0 kg of a 20% sodium chloride solution. What other products would be obtained and what would be the weight of each?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0560.htm

Solution:

The equation for this reaction is as follows: 2NaCl + 2H_2O (electric / current)\rightarrow 2NaOH + H_2 + Cl_2 . One mole of Cl_2 will be formed from every 2 moles of NaCl reacted. Therefore, to find the quantity of Cl_2 formed one must first know the amount of NaCl reacted. This can be determined by solving for the weight of NaCl in the solution and dividing this number by the molecular weight to find the number of moles present. Since the solution is 20% NaCl, it means that 20% of its weight is made up by NaCl. weight of NaCl = 0.20 × 2.0 kg = 0.4 kg . Therefore, there are 0.4 kg or 400 g of NaCl present. One can now solve for the number of moles present by dividing 400g by the molecular weight (MW of NaCl = 58.5) no. of moles = (400g) / (58.5 g /mole) = 6.84 moles . From the equation one sees that (1 /2) of this amount is equal to the number of moles of Cl_2 formed. no. of moles of Cl_2 = (1 / 2) × 6.84 moles = 3.42 moles. The weight of this number of moles of Cl_2 will be equal to the mole-cular weight of Cl_2 times the number of moles. (MW of Cl_2 = 71.0) weight of Cl_2= 3.42 moles × (71.0g / mole) = 242.8g. From the reaction one can also see that if 2 moles of NaCl react, 2 moles of NaOH are formed. Here, 6.84 moles of NaCl is reacted, therefore, 6.84 moles of NaOH are formed. The weight of this quantity is equal to the number of moles × the molecular weight. (MW of NaOH = 40). weight of NaOH = (40g / mole) × 6.84 moles = 273.6g. It is also seen from the reaction, that if 2 moles of NaCl are reacted, 1 mole of H_2 is formed. Thus, if 6.84 moles of NaCl are reacted, 3.42 moles of H_2 are formed. The weight of H_2 can then be found. (MW of H_2 = 2) weight of H_2 = (2g / mole) × 3.42 moles = 6.84g .

Question:

Two conductors of the same length and material but of different cross-sectional area are connected (a) in series, and (b) in parallel. When a potential differ-ence is applied across the combinations, in which conductor will the heating be greater in each case?

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Solution:

The resistance of each conductor has the form R =\rhol/A. That is, R is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area of the conductor. \rho is the constant of proportionality (the resistivity). Since the resistivity and length are the same in each case then R_1 = \rho l/A_1 and R_2 = \rho l/A_2 or R_1/R_2 = A_2/A_1 . (a)When the conductors are in series, the same current passes through each, by definition of a series connection. Hence, the ratio of the heating produced (or the power developed) in the wires is (H_1 /H_2 ) = [(i^2 R_1 )/(i^2 R_2 )] = (R_1 /R_2 ) = (A_1 /A_2 ) . The heating is thus greater in the conductor with the smaller cross-sectional area. (b)When the conductors are in parallel, different currents pass through them but the potential difference across each is the same, by definition of a parallel connection. Hence, (H_1' /H_2' ) = [(V^2 /R_1 )/(V^2 /R_2 )] = (R_1 /R_2 ) = (A_1 /A_2 ) . We made use of the alternate form of power development, P = IV = I^2 R = V^2 /R . In this case, the heating is greater in the conductor with the larger cross-sectional area.

Question:

A free particle, which has a mass of 20 grams is initially at rest. If a force of 100 dynes is applied for a period of 10 sec, what kinetic energy is acquired by the particle?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0229.htm

Solution:

In order to calculate the kinetic energy we must compute the final velocity acquired by the particle: v = at +v_Owherev_Ois the initial velocity. Since we are told that initially the particle is at rest, the expression for velocity becomes v = at. Now, from Newton's Laws we know that a = F/m. Substituting this for a yields v = [F/m]t = [(100 dynes) / (20 g)] × (10 sec) = 50 cm/sec Then, KE = (1/2)mv^2 = 1/2 × (20 g) × (50 cm/sec)^2 = 25,000 ergs How much work was done by the applied force? The dis-tance moved is s = (1/2)at^2 = (1/2)[F/m]t^2 = 1/2 × [(100 dynes) / (20 g)] × (10 sec)^2 = 250 cm so that the work done is W = Fs since the force and displace-ment are in the same direction. W = (100 dynes) × (250 cm) = 25,000 ergs Thus, the work done is transformed entirely into the kinetic energy of the particle.

Question:

What structural advantages enable mitochondia to be efficient metabolic organelles?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0081.htm

Solution:

Mitochondria are called the "powerhouses" of the cell because they are the major sites of ATP production in the cell. They contain the enzymes involved in both the Krebs (citric acid) cycle and the electron transport system, which work together to furnish 34 of the 36 ATPs produced by the complete oxidation of glucose. Another 2 ATPs are produced by glycolysis, which occurs in the cytoplasm, making the total 36 ATPs /oxidized glucose. Pyruvate from glycolysis traverses both the outer and inner mitochondral membranes easily because the membranes are totally permeable to pyruvate. Once inside the mitochondrion's matrix (see figure) pyruvate is converted by the enzyme pyruvate dehydrogenase to acetyl coenzyme A. The coenzyme can then enter the Krebs Cycle which occurs in the matrix where all the enzymes that catalyze the cycle's reactions are present (except succinic dehydrogenase which is located on the inner membrane). Conveniently bound to the surface of the cristae, the greatly folded inner membrane of the mitochondria, are the enzymes of the electron transport system. Since the cristae jut into the matrix, the distance needed to travel by FADH_2 and NADH, generated by the TCA cycle, to the enzymes of the electron transport system is reduced, and, since the surface area of the inner membrane is so greatly enhanced by its cristae, the probabilities of NADH and FADH_2 encount-ering the electron transport system's enzymes which are con-cerned with ATP production become sharply increased. Also cont-ributing to the efficiency of this organelle's ATP producing ability is the nature of the cristae's enzymes: they are actually multi-enzyme complexes which group together a number of sequentially acting enzymes responsible for electron transport and oxidative phosphorylation. These assemblies enhance the efficiency of respiration, since the product of one reaction is located near the enzyme of the subsequent reaction.

Question:

To what volume must a liter of oxygen be expanded if the mean free path of the molecules is to become 2m? The molecules of oxygen have a diameter of 3 A. Assume that the gas starts at STP.

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Solution:

The mean free path L is the average distance between successive collisions of a gas molecule and is given by L = 1/(\pi\surd2nd^2) = (0.707)/(\pind^2) where n is the number of molecules per unit volume and d is the molecule's diameter. If L is 200 cm, then n = (0.707)/(L\pid^2) = [0.707]/[200 cm × \pi(3 × 10^-8)^2 cm^2]. The gas equation ispV= n_0RT, where n_0 is the number of moles present in the volume V. But n_0/V is the number of moles of gas per unit volume. Multiplying it by Avogadro's number, N_0, (the number of molecules in one mole of a substance) yields the number of molecules of oxygen in a unit volume. Therefore, (n_0/V)N_0 = nandn_0/V = n/N_0. Substitution into the gas equation gives P = (n/N_0)RT = [0.707 × 8.3 × 10^7 dynes \textbullet cm \textbullet mole^-1 \textbullet K deg^-1 × 273 K deg]/[200 cm × (3 × 10^-8)^2 cm^2 × 6.02 × 10^23 mole^-1)] = 0.047 dyne \bullet cm^-2. But since the temperature remains unchanged, the expansion takes place according to Boyle's law. P_1V_1 = P_2V_2 The gas starts at STP (standard temperature and pressure). This corresponds to a temperature of 0\textdegreeC and a pressure of 1 atm. Thus 1 liter changes to a volume V_2 while the pressure changes from 1atmto 0.047 dyne \bullet cm^-2 . To keep the units consistent, use is made of the fact that 1atm= 1.013 × 10^6 dynes \bullet cm^-2 and 1 liter = 10^3 cm^3 . Then 1.013 × 10^6 dynes \textbullet cm^-2 × 10^3 cm^3 = 0.047 dyne \textbullet cm^-2 × V2 \thereforeV_2 = [(1.013 × 10^9)cm^3]/[0.047] = 2.155 × 10^10 cm^3.

Question:

Describe the mode of action of enzymes. What factors affect enzyme activity?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0067.htm

Solution:

An enzyme (E) combines with its substrate (S) to form an intermediate enzyme-substrate complex (ES), which then decomposes into reaction products (P) and the free enzyme, as seen in the equation below. E + S \textemdash> ES\textemdash> E + P enzyme-substrate complex An enzyme causes the substrate upon which it is acting to be much more reactive than when it is free. One postulate accounting for this is that the enzyme holds the substrate in a position which strains and weakens the substrate's molecular bonds. This weakening of the bonds within the substrate makes them easier to cleave and results in a general lowering of the energy of acti-vation of the reaction. This postulate is extremely simplistic - the actual forces at work are much more numerous and complex. When the substrate binds to the enzyme, it combines with only a relatively small part of the enzyme molecule the active site. Information about the active site, such as its location and the nature and sequence of amino acids in it, provides an indication of the mechanism of binding and catalysis. The binding of the substrate to the enzyme's active, site depends on many forces: hydrogen bonding, the interaction of hydrophobic (water-repelling) groups, and the electrostatic interaction between charged groups on the amino acids. Many active sites also contain metal ions which aid in binding the substrate or expediting the catalytic reaction by withdrawing or stabilizing electrons. For example, the enzyme carboxypeptidase, which hydrolyzes polypeptide bonds of proteins in food, contains a zinc atom in its active site. The electrophilic (electron-attracting) zinc atom coordinates electrons from the carbonyl of the peptide bond, weakening the bond for attack by a specific amino acid of the enzyme at the active site. Such a mechanism, however, is beyond the scope of elementary biology and one would require a good course in biochemistry to understand fully. Some enzymes, the regulatory or allosteric enzymes, have two binding sites: an active site and a regulatory site Regulatory enzymes are a key controlling factor in me-tabolic pathways. If the end product of a pathway is in excess, it inhibits the action of the regulatory enzyme by binding to its regulatory site. The end product shuts off the catalytic activity of the active site by altering the arrangement of the enzyme's polypeptide chains, thus deforming and inactivating the enzyme (see diagram below) . This feedback mechanism is known as end-product inhibition and is important in preventing the accumulation of unwanted substcincss. There are several factors which affect enzyme activity, one of which is temperature. High temperatures of 50\textdegreeC or above can inactivate or denature most enzymes. Upon denaturation, the structure of the enzyme is per-manently altered resulting in an irreversible loss of activity. When most organisms are exposed to high temp-eratures, death occurs due to enzyme inactivation and the resulting loss of metabolic activity. Enzymes are usually not denatured by freezing, but their activity is decreased or disappears. This loss of activity is tempo-rary and the activity reappears upon exposure to normal temperatures. Most enzymes have an optimal temperature range. At temperatures below 50\textdegreeC, enzymatic reactions double for each temperature rise of 10\textdegree. Enzymes are also affected by and can be denatured by changes in pH. Although most enzymes have an optimum pH a- round neutrality, (pH 7) some require an acidic medium and others require an alkaline medium. For example, although both pepsin and trypsin works optimally at pH 8.5. The de-pendence of enzyme activity on pH is explained by the pre-sence of ionizable groups on the protein molecules of the enzyme. pH controls, in part, the number of positive and negative charges on the enzyme molecule, which consequently affect activity. A final factor affecting enzyme activity is the presence of enzyme poisons. Cytochrome oxidase, an enzyme involved in respiration, is inactivated by minute amounts of cyanide. Death from cyanide poisoning thus results from the inhibition of cytochrome enzymes. Other enzyme poisons include iodoacetic acid, flouride and lewisite.

Question:

How can the stack be used with subroutines?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0138.htm

Solution:

The stack can be used to store return addresses and any other contents of registers that need to be saved before going to a subroutine. Consider a computer with ac-cumulator ACC and general purpose registers X and Y. The program in fig. 1 shows how the stack is used to call a sub-routine which will store the sum of the contents of the ACC and register X into register Y. The main program includes steps 14 through 18. and subroutine 1 starts at step 57 and stops at step 60. At location 15 the ACC contents is pushed into the stack. Note that the value of the ACC remains unchanged even though it was pushed on the stack. During the Jump to Subroutine 1 the return address is stored in the stack. At location 58 the sum of [ACC] + [X] is stored in the ACC. At location 60 the return address is taken off of the stack and put back in the PC, thus the computer has "returned" to the main pro-gram. After returning, the ACC is popped, hence the ACC con-tains its previous value. As you can see stacks are a valuable tool for storing registers while they are temporarily being used for other purposes.

Question:

Write a FORTRAN program that generates the prime numbers in the 1-2000 interval, using an ancient method known as the SIEVE of ERATOSTHENES.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G22-0547.htm

Solution:

A prime is a number which has only 2 divisors: 1 and itself. There is an ancient method for finding the primes from 2 (the smallest prime) to some number N, called the SIEVE of ERATOSTHENES. This method works as follows: First, the integers from 2 to N are written down. Then, the multiples of the smal-lest available integer, 2, greater than this integer, are crossed out. This step, obviously, eliminates all the even numbers except 2. The procedure is then repeated for the next smallest available integer-3, and then next - 5, because 4 is not available, since it was eliminated during the first step, and so on... Thus, the initial list is scanned repeatedly, and the appropriate integers are crossed off, until the next least remaining integer considered is greater than \surdN. At this point, all the remaining integers are indeed all the primes between 2 and N. The program is created based upon the above discussion. First, an array of integers from 2 to 2000 is generated. Next, this array is scanned repeatedly, deleting the appropriate integers by replacing them with zeros. At the point when the considered integer is greater than \surd2000 , all the nonzero elements of the array are placed left justified within the array and then printed. The program looks as follows: DIMENSION N(2000) K = 2000 RK = K KROOT = SQRT (RK) KK = K - 1 CGENERATE ARRAY OF INTEGERS FROM 2 TO 2000 DO 10 I = 1,KK 10N(I) = I+1 J = 1 CSECTION OF REPEATED ELIMINATION PROCESS 15INC = N(J) JJ = J + INC DO 20 L = JJ, K, INC 20N(L) = 0 25J = J + 1 IF (N(J).LE.O) GO TO 25 IF (N(J).LE.KROOT) GO TO 15 CPLACE THE PRIMES LEFT JUSTIFIED CIN THE ARRAY N. J = 0 DO 45 I = 1, KK IF (N(I).LE.O) GO TO 45 J = J + 1 N(J) = N(I) 45CONTINUE COUTPUT THE PRIMES WRITE (6,50) (N(I), I = 1,J) 50FORMAT (1H, 10I6) STOP END

Question:

How much heat is required to change 25 kg of ice at -10 \textdegreeC to steam at 100\textdegreeC?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0471.htm

Solution:

We have 4 separate situations which we must consider in this problem. First, the ice is heated to its melting point, which involves a change in temperature. Next, the ice changes state to form water during which there is no change of temperature. Then, as heat is added the water reaches its boiling point and any further addition of heat serves only to finally change the state of the water to steam and will not raise the temperature of the boiling water. Note that the specific heat of ice is different from that of water. Heat to raise temperature of ice to its melting point =m_ic_i∆t_i= 25 kg (0.51 Cal/kgC\textdegree) [0 - (-10\textdegreeC)] = 128 Cal. Heat to melt ice = m_1L_v = 25 kg (80 Cal/kg) = 2000 Cal. Heat to warm water to its boiling point =m_wc_w∆t_w= 25 kg(1.0 Cal/kg C\textdegree) (100\textdegreeC - 0\textdegreeC) = 2500 Cal Heat to vaporize water =m_wL_v= 25 kg(540 Cal/kg) = 13,500 Total heat required128 Cal 2,000 2,500 14,000 19,000 Cal Note that in this summation 128 is negligible and may be disregarded, since there is a doubtful figure in the thou-sands place in 14,000.

Question:

A charge of 1 C flows from a 100-V outlet through a 100-W light bulb each second. How much work is done each second by this electrical source?

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0561.htm

Solution:

The voltage, V = 100 V, and the charge, q =1 C, are given. By definition of potential V = W/q where W is the work done in moving the charge q through a potential difference V. Then W = qV = (1C)(100 V) = 1 × 10^2 J Alternatively, since the power rating of the light bulb is given as 100 watt, it dissipates 100 joules of energy per second. Therefore, for each second, the source must supply 100 joules of energy (or do 100 joules of work per second) to provide the energy dissipated by the light bulb resistance.

Question:

What is the power factor of a circuit if the following meter readings prevail? Ammeter reads .15 amperes, volt-meter reads 115 volts, and wattmeter reads 15.9 watts.

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/Users/wenhuchen/Documents/Crawler/Physics/D24-0796.htm

Solution:

The ammeter and voltmeter read root mean square (RMS) values of the actual current and voltage. If V = V0cos\omegat andI = I0cos(\omegat+ \rho) are the actual current and voltages through the circuit, then the instantaneous power (i.e. the instantaneous rate at which energy flows in the circuit) is P = VI = V0I0cos\omegatcos(\omegat+ \rho) = V0I0 (cos^2\omegatcos\rho -cos\omegatsin\omegatsin \rho) Nowcos\omegatsin\omegat= (1/2) sin 2\omegat, and the time average of sin 2\omegat is zero (since its graph is symmetric about the horizontal axis). The average of cos^2\omegatis 1/2 , for its graph is symmetric about the value of 1/2. This is to say that its value is greater than 1/2 as often as it is less than 1/2 , and by the same amount. Therefore the average power is: P= V0I0 (cos^2\omegat\leftrightarrowcos\rho\rule{1em}{1pt}cos\omegatsin\omegatsin \rho) = V0I0 [(1/2)cos\rho - 0] cos \rho is called the power factor of the circuit. It is a measure of the phase relation between the current and voltage in the circuit. According to the definition of the RMS value of a function, I_RMS = \surd[(1/2\pi) ^2\pi\int_0 I^2dt] = \surd[(1/2\pi) ^2\pi\int_0 I^20 cos^2 (\omegat+ \rho)dt] = (1/\surd2)I0 Similarly, V_RMS = \surd[(1/2\pi) ^2\pi\int_0 V^2dt] = \surd[(1/2\pi) ^2\pi\int_0 V^20 cos^2\omegatdt] = (1/\surd2) V0 HenceP= 1/2 (\surd2 I_RMS) (\surd2 V_RMS)cos\rho = I_RMS V_RMScos\rho The wattmeter reads the average power of a circuit. Then given I_RMS = .15 amp, V_RMS = 115 vP= 1.59 watt, the power factor is cos\rho = [P/(I_RMS V_RMS] = {15.9w/[(.15 a) (115 v)]} = 0.92

Question:

How many nuclei are there in 1kg of aluminum?

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0911.htm

Solution:

The mass m of a single aluminum atom ^27_31A1, is 26.98153amu. 1a.m.u. corresponds to the mass of a proton or 1.66 × 10^-27 kg, therefore m = [{26.98amu} {1.66 x 10^-27(kg/amu)}] = 4.48 x 10^-26kg This mass value is also approximately equal tothe mass of an aluminum nucleus since the electron mass is so much smaller than the mass of the atom. Therefore, the number of aluminum nuclei is N = [{1.0 kg} / {4.48 x 10^-26 (kg/nuclei)}] = 2.23 x 10^-25 nuclei.

Question:

A uranium nucleus under certain conditions will spontaneously emit an alpha particle, which consists of two pro-tons and two neutrons. If the nucleus was initially at rest and the speed of the emitted alpha particle is 2 × 10^7 m/s, what is the "recoil" speed of the nucleus? The nuclear mass is 3.9 × 10\Elzbar25kg and the alpha-particle mass is 6.7 × 10\Elzbar27 kg.

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Solution:

The "nuclear\textquotedblright force responsible for the particle emission is not explained, but this information is not needed if it is recognized that the total momentum of the system is zero. The mass of the recoiling nucleus m_1 = 3.9 ×10\Elzbar25kg; the mass of the alpha particle m_2 = 6.7 × 10\Elzbar27kg ; and the velocity of the alpha particle v_2 = 2 × 10^7 m/s are the known observables. As shown in the figure, the momentum of the initial state is zero, since the uranium atom is initially at rest. Therefore, the vectorial sum of the momenta of the frag-ments in the final state is also zero. (We are viewing the \alpha-decay of the uranium atom from the center of mass frame of the motion.) The speed of fragment x is expected to be smaller than that of the \alpha-particle since x is more massive than \alpha. Both speeds are therefore non-relativistic and we are allowed to use the non-relativistic formulae. From the conservation of momentum, we have P^\ding{217}_f = P^\ding{217}_i = 0 P^\ding{217}_1 + P^\ding{217}_2 = 0 orm_2v^\ding{217}_2 + m_1 v^\ding{217}_1 = 0 Thereforev^\ding{217}_1 =\Elzbar (m_2 / m_1 ) v^\ding{217}_2 . We see that v^\ding{217}_1 and v^\ding{217}2have opposite directions. The magni-tude of v^\ding{217}_1 is V_1 = (m_2 v_2 / m_1 ) = [{(6.7 × 10\Elzbar27kg)(2 × 10^7 m/s.)} / (3.9 ×10\Elzbar25kg )] = 3.4 × 10^5 m/s After emission, the alpha particle and nucleus move apart as free particles according to Newton's first law because the force causing the separation quickly becomes negligible.

Question:

Compute the electric field and the electric potential at point P midway betweentwo charge, Q_1 = Q_2 = +5 statC, separated by 1 m.

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Solution:

The magnitude of a test charge is +1 unit of charge. The forces on a test charge placed midway between two identical charges Q_1 and Q_2 are F_1^\ding{217} = (Q_1 / r^2_1 ) ȓ_12 F_2^\ding{217} = (Q_2 / r^22)ȓ21= (\rule{1em}{1pt}Q_2 / r^22)ȓ_12 where the unit vector ȓ_12points from to Q_1 to Q_2, r_1, r_2 are the distances between the test charge and and Q_1 and Q_2, F^\ding{217}_1 and F^\ding{217}_2 are equal in magnitude but opposite in direction. The net force is therefore zero at P, F = F_1 + F_2 = 0. The force on a unit charge gives the electric field strength at that point, then the electric field E^\ding{217} at P is also zero. Although the electric field at P is zero, this does not imply that the electric potential is also zero. The total potential \varphi_E _total is the sum (the algebraic sum since potential is a scalar) of the potentials due to Q_1 and Q_2: \varphi_E,1 = (Q_1 / r_1) = 5 / 50 = 0.1 statV \varphi_E,2 = Q_2 / r_2 = 5 / 50 = 0.1 statV Therefore, \varphiE, total= \varphi_E,1 + \varphiE,2= 0.2 statV Notice that if either Q_1 or Q_2 is changed from +5 stateC to \rule{1em}{1pt}5 stateC, the electric potential will vanish but the electric field will not. Therefore, the fact that either the field or the potential is zero in any particular case does not necessarily mean that the other quantity will also be zero; each quantity must be calculated separately.

Question:

What is the structure of a Pascal program?

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Solution:

A Pascal program consist of the following sections outputfiles. Programexample (input, output): (All underlined words are reserved words and are required in a given situation .) 2.Label Declarations (optional) Specifies labels used in the program. Label 5 , 20; 3.Constant definitions (optional) Defines constants used in the program. CONSTmax = 80, tax = 0.0825; 4.TYPE Definitions (optional) Defines data structures used in the program TYPE fixed= integer ; string= PACKED ARRAY [1 . .MAX ]of CHAR 5.Variable Declarations Declares variables used in the program and their type. VAR X :integer ; TEST :BOOLEAN ; NAME :STRING ; 6.Subroutine Declarations (optional) Declares function and procedures internal to the mainprogram . Each subroutine may contain including othersubroutines . PROCEDURESexample 1 (varparam1: real); FUNCTIONEXAMPLES 2 (param1: integer): integer; 7.Main body of the program always has the following form: BEGIN \bullet \bullet \bullet \bullet \bullet Some Other Statements \bullet \bullet \bullet END .

Question:

Calculate ∆H for the reaction at 25\textdegreeC. CO(g) + (1/2)O_2 (g) \rightarrow CO_2 (g)∆E = - 67.4Kcal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0508.htm

Solution:

The change in enthalpy (∆H) is defined by the first law of thermodynamics as the sum of the change of the internal energy (∆E) and the work done by a particular system. Work is defined as the product of the pressure (P) and the change in volume (∆V) of a system, where only pressure-volume work is done. Much of the experimentation done by chemists is pressure-volume work because the work is usually done in an open container in the laboratory. The pressure, in this case, is constant and the volume of the materials is allowed to change. ∆H = ∆E + P∆V In this problem, one is given ∆E but not P or ∆V, so that one must calculate P∆V in some other way. The Ideal gas Law is defined as: P∆V = ∆nRT, where P is the pressure, ∆V is the change in volume, ∆n is the change in the number of moles reacting, R is the gas constant (1.99 cal/mole-\textdegreeK), and T is the absolute tem-perature. Here, one knows R and can calculate T and ∆n. Therefore, ∆nRTcan be substituted for P∆V in the equation to find ∆H. ∆n is found by subtracting the number of moles reacting from the number of moles formed as products. Here, 1 mole of CO and 0.5 mole of O_2 form 1 mole of CO_2. ∆n = 1 mole CO_2 - (1 mole CO + 0.5 mole O_2) = - 0.5 mole. The absolute temperature is found by adding 273 to the temperature in \textdegreeC. T = 25 + 273 = 298\textdegreeK. Therefore, ∆nRT= (- 0.5 mole) × 1.99 cal/mole-\textdegreeK × 298\textdegreek = - 0.3 Kcal. One is now ready to find ∆H, via substitution. ∆H = ∆E + ∆nRT ∆E = - 67.4 Kcal ∆nRT= - 0.3 Kcal ∆H = - 67.4 Kcal + (- 0.3 Kcal) = - 67.7 Kcal.

Question:

A steel bar, 20 ft long and of rectangular cross-section 2.0 by 1.0 in., supports a load of 2.0 tons. How much is the bar stretched?

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Solution:

The Young's modulus of the metal bar is the ratio of longitudinal stress , F/A, to tensile strain ∆L/L (see the figure) Y = (F/A)/(∆L/L) = (FL)/(A∆L) Here, A is the bar's cross-sectional area. Therefore, the elongation ∆L of the bar is ∆L = (∆FL)/YA ∆F = 2.0 tons = 2.0 ton × (2000 lb)/(1 ton) = 4000 lbs. A= (2.0 in. × 1.0 in.) = 2.0 in.^2. Young's modulus for steel is 29 × 10^6 lb/in.^2. ∆L = [(4.0 × 10^3 lb)(20 ft)]/[(29 × 10^6 lb/in.^2)(2.0 in.^2)] = 0.0014 ft = 0.017 in.

Question:

Balance the following by filling in missing species and proper coefficient: (a) NaOH + _____\rightarrow NaHSO_4 + HOH; (b)PCI_3 +__ HOH \rightarrow _____ + 3HCl, (c) CH_4 + ____ \rightarrowCC1_4 + 4HCl.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0167.htm

Solution:

To balance chemical equations you must remember thatALLatoms (and charges) must be accounted for. The use of coefficients in front of compounds is a means to this end. Thus, (a)NaOH + ____ \rightarrow NaHSO_4 + HOH On the right side of the equation, you have 1 Na, 3 H' s, 5 O' s, and 1 S. This same number of elements must appear on the left side. However, on the left side, there exists only 1 Na, 1 0, and 1 H. You are missing 2 H' s, 1 S, and 4 O' s. The missing species is H_2 SO_4, sulfuric acid. You could have anticipated this since a strong base (NaOH) reacting with a strong acid yields a salt (NaHSO_4) and water. The point is, however, that H_2 SO_4 balances the equation by supplying all the missing atoms. (b)PC1_3 + 3HOH \rightarrow ____ + 3HCl. Here, the left side has 1 P, 3Cl' s, 6 H' s, and 3 O' s. The right has 3 H' s and 3 C1' s. You are missing 1 P, 3 O' s and 3hydrogens. Therefore, P (OH)_3 is formed. (c)CH_4 + ____ \rightarrow CCl_4 + 4HCl Here, there are 1 C, 8Cl's, and 4 H's on the right and 1 C and 4 H' s on the left. The missing compound, therefore, contains 8 C1' s and thus it is 4 CI2. One knows that it is 4 CI_2 rather than CI_8 or 8CI because elemental chlorine gas is a diatomic or 2 atom molecule.

Question:

In over 90% of cases, the concentration of ethanol (C_2H_5OH, density = 0.80 g/ml) in blood necessary to produce intoxication is 0.0030 g/ml. A concentration of 0.0070 g/ml is fatal. What volume of 80 proof (40% ethanol by volume) Scotch whiskey must an intoxicated person consume before the concentration of ethanol in his blood reaches a fatal level? Assume that all the alcohol goes directly to the blood and that the blood volume of a person is 7.0 liters.

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Solution:

The difference between a fatal concentration of ethanol and an intoxicating concentration of ethanol is 0.0070 g/ml - 0.0030 g/ml = 0.0040 g/ml. We must determine the total volume of ethanol in the blood which corresponds to a concentration of 0.0040 g/ml and then calculate the volume of Scotch whiskey which will provide this concentration. The total mass of ethanol in blood needed to raise the concentration from an intoxicating to a fatal level is equal to concentration of ethanol × volume of blood = 0.0040 g/ml × 7 liters = 0.0040 g/ml × 7000 ml = 28 g of ethanol. Dividing this mass by the density, we obtain the corresponding volume of ethanol, or 28 g/0.80 g/ml = 35 ml of ethanol. The amount of Scotch whiskey that must be consumed must provide 35 ml of ethanol. But the Scotch whiskey is only 40% ethanol, or 0.40 ml ethanol/ml Scotch. Let v denote the volume of Scotch in ml. Then [ratio of ethanol to Scotch ] × volume of Scotch = volume of ethanol 0.40 ml ethanol/ml Scotch × v ml Scotch = 35 ml ethanol or, v = [(35 ml ethanol)/(0.40 ml ethanol/ml Scotch)] 88 ml Scotch. Thus, under our assumptions, an intoxicated person must drink 88 ml of 80 proof Scotch whiskey (about 3 ounces) before ethanol reaches a fatal level in his blood.

Question:

A force F = 10 newtons in the +y-direction is applied to a wrench which extends in the +x-direction and grasps a bolt. What is the resulting torque about the bolt if the point of application of the force is 30 cm = 0.3 m away from the bolt?

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Solution:

Torque is calculated from the relation: \cyrchar\cyrt^\ding{217} = r^\ding{217} × F^\ding{217} where \cyrchar\cyrt stands for torque, F stands for the force, and r denotes the distance from the origin, about which the torque is calculated, of the point of application of the force. In this problem we use the bolt as our origin about which we calculate the torque (see the Figure above). Then, \cyrchar\cyrt^\ding{217} = 0.3 m \^{\i} x 10N\^{\j} = 3 N\bulletm ( \^{\i} × \^{\j}) = 3 N \bullet m k˄ where \^{\i}, \^{\j}, and k˄ are the unit vectors in the +x, +y, and +z directions respectively.

Question:

What arrangement of the integers 1, 2, 3, 4, and 5 in a five- element array will yield the longest sort time (also known as the worst-case sort time) for the insertion sort? Gen-eralize the answer you obtain for N integers.

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Solution:

As developed earlier, in the insertion sort method, the list is scanned until an element is found to be out of order. The unordered element is stored temporarily, and the scan reverses direction until it reaches the place where the unordered element should be. The rest of the list's elements are moved up one place, and the sort continues. If the list of integers is presented initially in re-verse order, the insertion sort will have to make the great-est number of comparisons. Let us walk through the steps that must be taken: pass after 1st pass4, 5, 3, 2, 1 after 2nd pass4, 3, 5, 2, 1 after 3rd pass3, 4, 5, 2, 1 after 4th pass3, 4, 2, 5, 1 after 5th pass3, 2, 4, 5, 1 after 6th pass2, 3, 4, 5, 1 after 7th pass2, 3, 4, 1, 5 after 8th pass2, 3, 1, 4, 5 after 9th pass2, 1, 3, 4, 5 after 10th pass1, 2, 3, 4, 5 Ten comparisons must be made to get the list in order. We can generalize from this result to obtain a worst-case equation for the insertion sort. The maximum number of comparisons for the insertion sort is N (N - 1) / 2 This can be seen by writing the series in the order given and in the reverse order: (N - 1) + (N - 2) + (N - 3) + . . . + 3 + 2 + 1 1 + 2 + 3 + . . . + (N - 3) + (N - 2) + (N - 1) Adding term by term, we find that twice the sum of the series is N + N + N + ... + N + N + N Since N-l terms are in the series above, we can say that twice the sum of the series is N (N - 1), so, dividing by 2, the sum of the series is N (N -1) / 2.

Question:

What fraction of the total space in a body-centered cubic unit cell is unoccupied? Assume that the central atom touches each of the eight corner atoms of the cube.

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Solution:

To determine the percent of the unit cube that is unoccupied by the atoms making up the lattice, one must subtract the volume taken up by the atoms from the volume of the cube. This volume is then divided by the volume of the cube and multiplied by 100 to find the percent of unoccupied space. 1) Determining the volume of the atoms: One must first determine the number of atoms contributing to the unit cube. In this lattice there are 8 corner atoms and 1 atom in the center. Corner atoms contribute 1/8 of their volume to the cube. The atom in the center contributes its entire volume. no. of atoms in unit cube = (1/8 atom / corner × 8 corners) + 1 atom (in center) = 2 atoms. Thus, the volume in the unit cube taken up by the atoms is equal to the volume of two atoms. The radius of these atoms is taken to be 1. The volume of a sphere is 4/3 \pir^3. volume of 2 atoms = 2 × 4/3 \pi(1)^3 = 8.38 2) Volume of the cube: The corner atoms are assumed to be touching the central atom. The diagonal of the cube can be visualized as shown in Figure 2. BecauseABis shown to be 4r or 4 the side of the cubeACcan be found using the Pythagorean Theorem. OnceACis known the volume of the cube can be found. From geometry, it is known that CB=AC× \surd2 AB^2 =AC^2 +CB^2AB= 4 4^2 =AC^2 + 2(AC)^2CB=AC× \surd2 4^2 = 3AC^2 4/\surd3 =AC The volume of the cube is equal to the length of the side cubed. volume of cube = [4/(\surd3)]^3 3) The space in the cube is equal to the volume of the spheres subtracted from the volume of the cube. vol. of space = vol. of cube - vol. of spheres vol. of space = 12.32 - 8.38 = 3.94 percentage of cube taken up by space = [3.94/12.32] × 100 = 32%.

Question:

Using the tables of standard electrode potentials, arrange the fol-lowing substances in decreasing order of ability as reducing agents: Al, Co, Ni, Ag, H_2, Na .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0582.htm

Solution:

The tables of standard electrode potentials list substances according to their ability as oxidizing agents. The greater the stan-dard electrode potential, E\textdegree, of a substance, the more effective it is as an oxidizing agent and the less effective it is as a reducing agent. From the table of standard electrode potentials, Al^3+ + 3e^-\leftrightarrowsAl(s) E\textdegree = -1.66v Co^2+ + 2e^-\leftrightarrowsCo(s) E\textdegree = -0.28v Ni^2+ + 2e^-\leftrightarrowsNi(s) E\textdegree = -0.25v Ag^+ + e^-\leftrightarrowsAg(s) E\textdegree = +0.80v 2H^+ + 2e^-\leftrightarrowsH_2(g) E\textdegree = 0v Na^+ + e^-\leftrightarrowsNa(s) E\textdegree = -2.71v Thus, in increasing ability as oxidizing agents, Na^+ < Al^3+ < Co^2+ < Ni^2+ < H^+ < Ag^+ But if Na^+ has a greater tendency to oxidize (gain electrons) than Al^3+ then, from looking at the reverse reactions, the "conjugate oxidant" Na must have a greater tendency to reduce (lose electrons) than the "conjugate oxidant" AI . Thus, Na, is a better reducing agent than Al, and so on. The substances, in order of decreasing ability as reducing agents, are therefore Na > Al > Co > Ni > H_2 > Ag .

Question:

What are the functions of the stem? How are stems and roots differentiated?

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Solution:

The stem is the connecting link between the roots, where water and minerals enter the plant, and the leaves, where organic foodstuffs are synthesized. The vascular tissues of the stem are continuous with those of the root and the leaves and provide a pathway for the transport of materials between these parts. The stem and its branches support the leaves so that each leaf is exposed to as much sunlight as is possible. The stem of a flowering plant also supports flowers in the proper orientation to enhance reproduction and later, seed dispersal. Along the stem are growing points where theprimordiaof leaves and flowers originate. Some stems have cells which carry out photosynthesis, others have cells specialized for the storage of starch and other nutrients. Roots and stems are structurally quite different. Stems have an epidermis covered with a protective layer ofcutinwhile the epidermis of roots has nocutinbut gives rise tohairlikeprojections called the root hairs. Stems, but not roots, have nodes, which are junctions where leaves arise. Stems may have lenticels for "breathing" while roots do not. The tip of a root is always covered by a root cap whereas the tip of a stem is naked unless it terminates in a bud. Thedicotstem typically contains separate rings of xylem and phloem, with the xylem central to the phloem, whereas in the roots, phloem tubes lie between the arms of the star-shaped xylem tissues.

Question:

It has been determined experimentally that two elements, A and B react chemically to produce a compound or compounds. Experimental data obtained on combining proportions of the elements are: Grams of A Grams of B Grams of compound Experiment 1 06.08 04.00 10.08 Experiment 2 18.24 12.00 30.24 Experiment 3 3.04 2.00 5.04 (a)Which two laws of chemical change are illustrated by the above data? (b) If 80 g of element: B combines with 355 g of a third element C, what weight of A will combine with 71 g of element C? (c) If element B is oxygen, what is the equivalent weight of element C?

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Solution:

(a) If one adds the weight of A to the weight of B and obtains the weight of the compound formed, the Law of Conservation of Mass is illustrated. This law states that there is no detectable gain or loss of mass in a chemical change. Using the data from the experiments described, you find the following: For experiment 16.08 g A + 4.00 g B = 10.08 g of compound 6.08 + 4.00 = 10.08 For experiment 218.24 g A + 12.00 g B =30.24 g of compound 18.24 + 12.00 = 30.24 For experiment 303.04 g A +2.00 g B =5.04 g of compound 3.24 +2.00 = 5.04 From these calculations one can see that the Law of Conservation of Mass is shown. Another important law of chemistry is the Law of Definite Proportions. This law is stated: When elements combine to form a given compound, they do so in a fixed and invariable ratio by weight. One can check to see if this law is adhered to by calculating the ratio of the weight of A to the weight of B in the three experiments. If all of these ratios are equal, the Law of Definite Proportions is shown. For experiment 16.08 g A /4.00 g B = 1.5 For experiment 218.24 g A / 12.00 g B =1.5 For experiment 303.04 g A /2.00 g B =1.5 The Law of Definite Proportions is illustrated here. (1) From the Law of Definite Proportions, one can find the number of grams of B that will combine with 71 g of C. After this weight is found, one can find the number of grams of A that will react with 71 g of C by finding the amount of A that reacts with that amount of B. It is assumed that the amount of A that reacts with 71 g of C is equal to the amount of A that will react with the amount of B that reacts with 71 g of C. The amount of A that will react with this amount of B can be found by remembering from the previous section of this problem, that A reacts with B in a ratio of 1.5. (1) Finding the amount B that would react with 71 g of C. One is told that 80 g of B reacts with 355 g of C. By the Law of Definite Proportions, a ratio can be set up to calculate the number of grams of B that will react with 71 g of C. Let x = the number of grams of B that will react with 71 g of C. (80g B) / (355 g C) = (x g B) / (71 g C) (80g B) / (355 g C) = (x g B) / (71 g C) x= (71 × 80) / (355) = 16. x= (71 × 80) / (355) = 16. 16 grams of B will react with 71 g of C. (2) It is assumed that the same amount of A that will react with 16 g of B will react with 71 g of C. Therefore, using the fact that the ratio of the amount of A that reacts to the amount of B is equal to 1.5 (this fact was obtained in part (1)), one can calculate the amount of A that will react with 71 g of C. Let x = the number of grams of A that will react with 16 g of B . (xg A) / (16g B) = 1.5 x = 16 × 1.5 = 24 g. 24 g of A will react with 16 g of B or 71 g of C. (c) In finding the equivalent weight of C when B is taken to be oxygen, the Law of Definite Proportions is used again. The equivalent weight of oxygen is 8. Knowing that 16 g of B react with 71 g of C, one can set up the following ratio x = weight of C if the weight of B is taken to be 8. (71 g C) / (16 g B) = (xg C) / (8g B) x = (8 × 71) / (16) = 35.5 g. The equivalent weight of C when B is taken to be oxygen is 35.5 g.

Question:

If an electron is projected into an upward electric field with a horizontal velocity v_0, find the equation of its trajectory.

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0627.htm

Solution:

The direction of the field is upward (see the figure) and the force on e is eE^\ding{217}, where e is the electronic charge, and E^\ding{217} is the electric field intensity. Since e < 0 the force on the electron points downward. The initial velocity is along the positive x-axis. The only force is in the - y- direction, therefore the acceleration along the x-axis is zero. The y acceleration is a^\ding{217}_y = F^\ding{217}y/m = eE^\ding{217}/m,(1) wheree < 0. After a time t, the position of the electron will be x = x_0 + v(0)xt + 1/2 a_xt2 y = y_0 + v(0)yt + 1/2 a_yt2 (x_0, y_0) and (v_(0)x, v_(0)y) are the components of the initial position and velocity of the particle. If we take x_0 = 0, y_0 = 0 as the starting point of the particle, and note that a_x = 0, v_(0)y = 0, we may write x = v_(0)xt y = 1/2 a_yt^2(2) substituting (1) in (2) x = v(0)xt y = [eE^\ding{217} / 2m] t^2(3) Writing E^\ding{217} in terms of a unit vector (\^{\j}) in the direction of the y - axis E^\ding{217} = E \^{\j} Substituting this in the equation for y appearing in (3) for, we obtain y = (eEt^2 / 2m)\^{\j}(4) Solving the x equation of (3) for t, and sub-stituting this in (4), t = x / v(0)x y = (eEx2/ 2mv_(0)x^2)\^{\j}(5) Note that e < 0, and since all the other quantities in (5) are positive, y < 0 which is the equation of a parabola. The motion is the same as that of a body projected horizontally in the earth's gravitational field. The deflection of electrons by an electric field is used to control the direction of an electron stream in many electronic devices, such as the cathode-ray oscilloscope.

Question:

In the casualty department of a hospital, it is necessary to raise or lower the examination table without disturbing the patient. This is accomplished by mounting the table on a screw jack which has a pitch of (1/2) in. The raising of the table is accomplished by applying a force of 12.5 lb tangentially at the end of a lever 12 in. long and rotating the lever in a circle. Find the efficiency of this machine if patient and table together have a weight of 480 lb.

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Solution:

When the lever is rotated through one complete circle, the table is raised by one pitch of the screw. The work done on the machine by the operator is the force applied times the distance traveled in the direction of the force. Thus W_1= F × 2\piR, where R is the radius of the circle swept out by the lever. W1= 12(1/2) lb × 2\pi × 1 ft = 25\pi ft \textbullet lb. The table and patient acquire additional potential energy, since their height above the ground is increased. The additional energy is their combined weight times the extra height. Thus W_2 = 480 lb × (1/24) ft = 20 ft \textbullet lb. The efficiency of the machine is the energy gained by the table divided by the energy supplied. Thus the efficiency is E = W_2/W_1 = (20 ft \bullet lb)/(25\pi ft \bullet lb) = 0.255or E = 25.5%.

Question:

Outline a method, giving the precautions to be aware of, by which an enzyme could be used to measure the amount of a given substrate.

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0069.htm

Solution:

An enzyme catalysed reaction can be represented by the equation: E + S \rightleftharpoons ES \rightleftharpoons E + P By increasing the concentration of substrate and keeping the concentration of enzyme constant, we can increase the rate of catalysis: the above reaction will shift toward the right. We can react a fixed amount of enzyme with different substrate concentrations to determine the corresponding reaction velocity. The reaction velocity must be an initial velocity, that is, the rate must be measured very soon after the start of the reaction. If the rate is measured over a long period, the measured rate may not be the true value since the reaction would tend to slow down because the amount of substrate becomes used up as the reaction proceeds. The plot of reaction velocity V, as a function of the substrate concentration [S], for a constant amount of enzyme is shown below: We now know that a specific substrate concentration corresponds to a specific rate. We can then react an unknown amount of substrate with the same amount of enzyme used to determine our original plot. By measuring the initial rate of the reaction, we can then read the corres-ponding concentration of substrate from our plot. For example, if we measured a rate V_1 we know that the 'unknown' concentration of substrate is S_1 . An important precaution must be noted: at very large substrate con-centrations, the plot is no longer linear. The rate increases very slowly at large [S] until it reaches a maximum velocity. The slow-down of rate increase occurs when the amount of enzyme is limiting. At this point, (point E_2 on the plot) most of the catalytic sites on the enzyme molecules are filled, so increasing the amount of substrate present has little effect on increasing the rate. Therefore, at very high [S], the rate becomes independent of the amount of substrate present. It is therefore critical that the unknown concentration of substrate be low enough to validly correspond to a specific rate. One can measure the initial rate of reaction bystopping the reaction after a short period of time, usually by adding certain inhibitors or poisons. To the reaction mixture may then be added a substance which forms a colored complex with one of the reaction products. The denser the color, the more product formed within the measured time interval, and therefore, the higher the reaction rate. The density is related to how much light the colored sample absorbs when a beam of light is shone on the sample. (An instrument called a spectrophotometer is used to measure absorbance of light by a liquid sample.) The measured absorbance is compared to the absorbance of known amounts of product. Hence, when the absorbance measured is known, we will know how much product was formed by the reaction and since we know the time the reaction was stopped, we know how much product is produced per unit time, that is, the rate. The substrate concentration can then be read off the graph from its corresponding velocity.

Question:

In what respects is Euglena like a plant and in what respects is it like an animal?

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Solution:

Euglena is a single-celled organism. Like plants, Euglena contains chlorophyll in chloroplasts and is capable of photosynthesis. Like animals, Euglena lacks a cell wall, is capable of active motion and ingests organic compounds as food. Botanists classified Euglena and its relatives as algae (phylum: Euglenophyta). Zoologists classified the euglenoids in the phylum Protozoa (class Phytomastigophora). Euglenoids are also often classified as a separate phylum in the kingdom Protista. Most euglenoids are capable of photosynthesis, but require certain organic nutrients as well as in-organic salts. Some are completely autotrophic, and some do not possess chlorophyll and are completely hetero- trophic. Of the heterotrophs, some are saprophytic, feeding on dead organic materials, and others are holozoic, capturing and ingesting other organisms. The euglenoids all have two flagella, by means of which they are capable of active motion. The euglenoid cell is delimited by a cell membrane within which there is a series of flexible, interlocking proteinaceous strips termed the pellicle. The flexible pellicle permits Euglena to change its shape, providing an alternate means of locomotion for mud-dwelling forms. As is true for protozoans, but not for plants, an area of the plasma membrane is specialized for the ingestion of food. This area is near the base of the flagella and is called the gullet. Unlike higher plants, but similar to many algae, the euglenoids have a stigma or an eyespot that functions as a light detector. The non--photosynthetic euglenoid species lack the stigma. Unique to the euglenoids is the storage product paramylum. This compound is a carbohydrate polymer unlike plant starch or animal glycogen. The paramylum is produced in an organelle called a pyrenoid. The cell division cycle of Euglena is also different from that of plants or animals. During division the nucleolus does not dis-appear, and the nuclear membrane remains intact. The euglenoids have a mixture of plant and animal characteristics, in addition to having their own unique features. They have no close relatives among the algae and few among the protozoans.

Question:

Write a pseudocoded program to find an approximation of pi using the Monte Carlo method. This method is an example of a system which uses random digits to solve a problem that does not, by itself, have ran-domness.

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Solution:

First, we must assume that there is a subprogram called RANDOM available in the user's compiler. Different compilers have different means of obtaining random digits, so we omit a discussion of the actual process of randomization. Monte Carlo methods typically use random digits to solve problems which have no inherently random properties. For the approximation of pi, we need the dart board analogy. We have a square board with an inscribed circle, as in Fig. 1. Suppose we have monkeys that have been trained to toss darts at the board. Now, assuming all the darts land somewhere on the board, and assuming that a dart can land anywhere with equal likelihood, we ask: what is the probability that a dart will land inside the circle? If you remember the equations for the areas of a circle and a square (A = \pi(L/2)^2 and A = L^2, respectively), you can discover that the probability equals the ratio of the circle's area to the square's area, or \pi/4, which is about 0.7854. For our example, we do not know \pi. However, if we could get the monkeys to throw a large number of darts at the board, we could compute the proportion of hits falling inside the circle. From this proportion, we can find an approximation for \pi. The following pseudocode generates randomly the X- and Y- coordinates of each "dart" thrown at the board. Since these coordinates lie between 0 and 1, darts appear only in the positive quadrant of the square (the top right corner). However, the full figure can be built from four identical pieces, so the result remains the same. We just multiply \pi/4 by 4 to get the desired result. Also note that if we increase the number of darts thrown, the degree of deviation from the expected result decreases. This axiom is called in statistics the "law of large numbers". We are testing for 4-digit accuracy in this problem. Our output consists of the number of darts, the approximate value of \pi, and the deviation from the expected result. INTEGER REAL C END PROGRAM I, M, N, KOUNT APPROX, X, Y N \leftarrow 0 KOUNT \leftarrow 0 OUTPUT MESSAGE 'INPUT M: NUMBER OF DARTS TO BE TOSSED' INPUTM DO WHILE M > 0 DO FOR I \leftarrow 1 TO M X \leftarrow RANDOM Y \leftarrow RANDOM CHECK TO SEE IF THE DART HAS LANDED INSIDE CIRCLE IF(X\textasteriskcenteredX) + (Y\textasteriskcenteredY) < 0 THEN KOUNT \leftarrow KOUNT + 1 END IF - THEN END DO FOR I N \leftarrow N + M APPROX \leftarrow FLOAT (KOUNT) / FLOAT (N) \textasteriskcentered4.0 OUTPUT N,APPROX, APPROX - 3.1416 OUTPUT MESSAGE 'IF YOU WISH TO TOSS MORE DARTS, INPUT M' INPUT M END DO WHILE

Question:

How are fungi important to man?

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Solution:

Fungi are divided into four groups:Oomycetes(Oomycota)- the egg fungi,Zygomycetes(Zygomycota)\rule{1em}{1pt}Zygo-spore-forming fungi, Ascomycetes (Ascomycota)-the sac fungi, andBasidiomycetes (Basidiomycota)-the club fungi. Fungi are both beneficial and detrimental to man. Bene-ficial fungi are great importance commercially.Ascomycetes, or sac fungi, are used routinely in food production. Yeasts, members of this group, are utilized in liquor and bread man-ufacture. All alcohol production relies on the ability of yeasts to degrade glucose to ethanol and carbon dioxide, when they are grown in the absence of oxygen. Yeasts used in alco-hol production continue to grow until the ethanol concentration reaches about 13 percent. Wine, champagne, and beer are not concentrated any further. However, liquors such as whiskey or vodka are then distilled, so that the ethanol concentration reaches 40 to 50 percent. The different types of yeasts used in wine production are in part responsible for the distinctive flavors of different wines. Bread baking relies on CO_2 pro-duced by the yeasts which causes the dough to rise. Yeasts used in baking and in the brewing of beer are cultivated yeasts, and are carefully kept as pure strains to prevent contamination. Sac fungi of genusPenicilliumare used in cheese production. They are responsible for the unique flavor of cheeses such as Roquefort and Camembert. The medically important antibiotic penicillin is also produced by members of this genus. CertainAscomycetesare edible. These include the delicious morels and truffles. The club fungi, orBasidiomycetes, are of agricultural importance. Mushrooms are members of this group. About 200 species of mushrooms are edible while a small number are poi-sonous. The cultivated mushroom, Agaricus campestris, differs from its wild relatives, and is grown commercially. Fungi are often of agricultural significance in that they can seriously damage crops. Members of theOomycetes, also known as water molds, cause plant seedling diseases, downy mildew of grapes, and potato blight (this was the cause of the Irish potato famine). Mildew is a water mold that grows parasitically on damp, shaded areas. Rhizopusstolonifer, a member of theZygomycetes, is known as black bread mold. Once very common, it is now controlled by refrigeration and by additives that inhibit mold growth. TheAascomyceteClaviceps purpurea causes the disease ergot, which occurs in rye and other cereal plants and results in ergot poi-soning of humans and livestock. This type of poisoning may be fatal. The disease caused in humans is called St.Vitus's dance Visual hallucinations are a common symptom of this disease. Lysergic acid is a constituent of ergot and is an intermediate in the synthesis of LSD. The "dance macabre" of the Middle Ages is now believed to have been caused by ergot poisoning. Basidiomycetes are also responsible for agricultural dam-age. Certain club fungi are known as smuts and rusts. Smuts damage crops such as corn, and rusts damage cereal crops such as wheat. Bracket fungi, another type of club fungi, cause enormous economic losses by damaging wood of both living trees and stored lumber. Fungi are also important to man because of the diseases they cause in man and livestock.Candidaalbicanscauses a throat and mouth disease, "thrush", and also infects the mucous membranes of the lungs and genital organs. Many skin diseases are caused by fungi, including ringworm and athlete's foot.

Question:

What is the rotational inertia of a 50-lb cylindrical flywheel whose diameter is 16 in.?

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Solution:

To find the rotational inertia of the flywheel, consider a mass element consisting of a thin cylindrical shell of radius r, thickness dr, and length L, as shown in the figure. Then dV= (2\pir) (dr) (L) density,\rho= m/V = m/(\piR^2L) anddm= \rhodV = m/(\piR^2L)(2\pir dr L) = (2m/R^2) r dr The moment of inertia is given by I = \intr^2 dm = \int(r^2)[(2m/R^2) r dr] = 2m/R^2 ^R\int_r=0 r^3 dr = (2m/R^2)(r^4/4)]^R_r=0 = (1/2)mR^2 For the given cylindrical flywheel, m= w/g = (50 lb)/(32 ft/sec^2) = 1.6 slugs R= 8.0 in = 2/3 ft I = (1/2)mR^2 = (1/2)(1.6 slugs) (2/3 ft)^2 = 0.35 slug-ft^2

Question:

Obtain the 15's and 16's complement of the following hexadecimalnumbers: a) FFOb) 1234c) ABCDd) 0000

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Solution:

Hexadecimal numbers arise frequently in computer applications because16 is the fourth power of 2. This means that each hexadecimal digitcan be represented uniquely by four bits. a) The 15's complement of FFO is found by subtracting it from FFF. FFF - FFO 00F Since radix complement = radix complement minus one + 1, the 16's complementhere is: 00F + 1 =010 . b) The 15's complement of 1234 is: FFFF - 1234 EDCB The16 'scomplement is then EDCB + 1, or EDCC. To make this more comprehensible, convert into the decimal system: 10000_16, = 16^4 + 0 × 16^3 + 0 × 16^2 + 0 × 16^1 + 0 × 16^0 = 65,536_10 1234_16 = 1 × 16^3 + 2 × 16^2 + 3 × 16^1 + 4 × 16^0 = 4660_10 Then the 16's complement (decimal equivalent) is: 65,536 - 4,660 60,876 Now, EDCC_16 = 14 × 16^3 + 13 × 16^2 + 12 × 16^1 + 12 × 16^0 = 60,876_10as above. Thus EDCC_16 is the 16's complement of 1234_16. c) The radix minus one complement of ABCD is: FFFF - ABCD 5432 The radix complement of ABCD is 5433, d) The radix complement of 0000 is 10000 while the radix complement minus one is FFFF.

Question:

One pair of genes for coat color in cats is sex-linked. The gene B produces yellow coat, b produces black coat, and the heterozygous Bb produces tortoise-shell coat. What kind of offspring will result from the mating of a black male and a tortoise-shell female?

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Solution:

Sex determination and sex linkage in cats is similar to that found in man, and indeed, in most animals and plants that have been investigated. So for cats, we can assume that if a gene is sex-linked, it is carried on the X chromosome. We can also assume that the Y chromosome carries few genes, and none that will mask the expression of a sex-linked gene or an X chromosome. Female cats, like female humans, are XX, and male cats are XY. Let X^B represent the chromosome carrying the gene for yellow coat, and X^b represent the chromosome carrying the gene for black coat. The male parent in this problem is black, so his genotype must be X^bY. The female is tortoise-shell, which means that she is carrying both the gene for yellow color and the gene for black color. Her genotype is X^BX^b. In the cross: Phenotypically, the offspring consist of: 1/4 tortoise-shell females (X^BX^b), 1/4 black females (X^bX^b), 1/4 yellow males (X^BY),and 1/4 black males (X^bY). Note that there can never be a tortoise-shell male, because a male can carry only one of the two possible alleles at a time.

Question:

Black-headed gulls remove broken eggshells from their nests immediately after the young have hatched. Explain the biological significance of this behavior.

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Solution:

Behavior that contributes to the survival and reproduction of the animal is adaptive. Since those animals which demonstrate adaptive behavior survive longer and successfully raise more offspring, it is more likely that the behavior patterns of these animals will be continued throughout further generations. The eggshell - removing habit of Black-headed gulls is such adaptive behavior. The gulls do not remove only broken shells, but also any conspicuous object placed in the nest during the breeding season. The more conspicuous the object, the more likely it is to be removed. The gulls seem to be discarding conspicuous objects as a defense mechanism against visual predators. When inves-tigators placed conspicuous objects along with eggs in nests, these nests were robbed of eggs (by other gulls) more often than nests having only eggs. Thus eggshell- removing behavior is significant in that it reduces the chances of a nest being robbed, thus enhancing the survi-val of offspring. The gulls that evoke this behavioral pattern will successfully raise more offspring than those who do not, and thus this behavioral pattern will tend to be passed on.

Question:

A refrigerator which has a coefficient of performance A refrigerator which has a coefficient of performance one-third that of a Carnot refrigerator is operated between two reservoirs at temperatures of 200\textdegreeK and 350\textdegreeK. It absorbs 500 J from the low-temperature reservoir. How much heat is rejected at the high- temperature reservoir?

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Solution:

The coefficient of performance of a Carnot refrigerator is defined as the ratio of the heat extracted from the cold source and the work needed to run the cycle. Hence, E_c= Q_1/(Q_2 - Q_1)orequivalentlyT_1/(T_2 - T_1) where Q_1 is the heat absorbed at temperature T_1 and Q_2 is the heat rejected at the higher temperature T_2. Here all temperatures T are to be expressed in Kelvin degrees. The actual refrigerator has thus a coefficient of performance Q_1/(Q_2 - Q_1) = (1/3) [T_1/(T_2 - T_1)] or(Q_2 - 500 J)/(500 J) = [{3(350 - 25 0)\textdegreeK}/{250\textdegreeK}] \thereforeQ_2 = 1100 J.

Question:

A long solenoid of length լ and cross-sectional area A is closely wound with N_1 turns of wire. A small coil of N_2 turns surrounds it at its center, as in the figure. Find the mutual inductance of the coils.

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Solution:

If a current I in a coil is changing at a rate [(dI)/(dt)], then a voltage across the coil will be induced in the direction to oppose the change, V_coil = - L [(dI)/(dt)] where L is the self inductance of the coil. Similarly, if the time varying magnetic flux of one coil links another coil, then it will induce a voltage in the second coil equal to V'_coil= -M [(dI)/(dt)](1 ) where I is the current in the first coil. The constant M is the mutual inductance of two coils. Its value is determined only by the geometry and positioning of the coils. In our problem, the magnetic field in the solenoid (away from the edges) is given by B = [(\mu_0 N_1 I_1 )/(լ)] where I_1 is the current in the solenoid coil. The flux \varphi through the N_2 coil is BA, where A is the area of the second coil. Then the voltage induced in the second coil by I 1 is, by Faraday's Law, V1 2= - N_2 [(d\varphi)/(dt)] = - [(\mu_0 N_1 N_2 )/(լ)] [dI/dt] Therefore we see by comparison with (1) that the mutual inductance of this system is M = [(\mu_0 N_1 N_2 )/(լ)] . If լ = 0.50 m, A = 10 cm^2 = 10^-3 m^2 , N_1 = 1000 turns, N_2 = 10 turns, M = [{(4\pi × 10^-7 henry/m) (10^-3 m^2 ) (1000 turns) (10 turns)} / (.5 m)] \approx 25 -10^-6 henry \approx 25 \muh .

Question:

50 ml of rhubarb juice is titrated against 0.25 NNaOH. 20 ml ofNaOHsolution is required for neutralization. Assuming the acidity of the juice is due to oxalic acid (H_2C_2O_4) determine (a) the weight of oxalic acid per liter of juice, (b) normality of the juice.

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Solution:

(a) For the neutralization to occur, there must be the same number of OH^- ions as there are H^+ ions present. H^+ + OH^- \rightleftarrows H_2O The normality of a base is defined as the number of equivalents of base in one liter of solution. An equi-valent is the weight of an acid or base that produces one mole of H^+ or OH^- ions, respectively. WhenNaOH ionizes, there is 1 OH^- ion formed by eachNaOHthat ionizes. NaOH\leftrightarrows OH^- + Na^+ Therefore, in a 0.25 NNaOHsolution there are 0.25 equivalents of OH^-, which, in this case is also the number of moles. 20 ml is 0.02 liters. The number of moles ofNaOHpresent will equal the number of liters of solution times themolarityof the solution. As pre-viously indicated, normality equalsmolarity. Thus, no. of moles = no. of liters × normality = 0.02 liters × (.25 moles/liter) = 0.005 moles. Thus, there must be 0.005 moles of H^+ ions from the oxalic acid to neutralize theNaOH. When oxalic acid ionizes, there are 2 H^+ ions formed for each molecule ionized. H_2C_2O_4 = 2H^+ + C_2O_4 Therefore, each mole of oxalic acid ionizes 2 moles ofNaOH. Thus, for neutralization to occur, only one half as much oxalic acid is needed as NaOH. There are 0.005 moles ofNaOH, present, therefore, 0.0025 moles of oxalic acid is needed to neutralize it. One now knows that there are 0.0025 moles of oxalic acid in the 50 ml of rhubarb juice. One can find the number of moles in 1 liter by multiplying 0.0025 moles/50 ml by the conversion factor 1000 ml/1 liter. no. of moles/liter of oxalic acid = [(0.0025moles)/(50 ml)] × (1000 ml/1 liter) = 0.05 moles/liter. Since there are 0.05 moles of H_2C_2O_4, in 1 liter of juice, one can find the weight of H_2C_2O_4 in this quantity of juice by multiplying 0.05 moles by the MW of H_2C_2O_4. (MW = 90). weight of H_2C_2O_4, in 1 liter = 0.05 moles × 90 g/moles = 4.5 g (b) For titrations, the following relation is found: N_acidV_acid=N_baseV_base whereN_acidis the normality of the acid,V_acidis the volume of the acid, N_baseis the normality of the base, andV_baseis the volume of the base. Here, one is givenN_base,V_base, andV_acid. One is asked to findN_acid. N_b= .25 NN_aV_a=N_bV_b V_b= 20 mlN_a × 50 ml = .25 N × 20 ml N_a = ? V_a= 50 mlN_a = [(0.25 N × 20 ml)/(50 ml)] = 0.10 N.

Question:

Discuss the various ways in which energy is expended within the human body.

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Solution:

Energy is primarily expended during the anabolic reactions of the cell; that is, energy is required for the synthesis of macromolecules. Some organisms ingest simple chemical substances and combine them to form the complex molecules needed for the building of new protoplasm. Others, the higher organisms, ingest complex organic materials, degrade them, and use the simple products to synthesize the complex organic compounds that they need. In both cases, growth results. Energy is also expended during motion. Motion includes the contraction of muscle tissue, which allows an animal to either move itself (locomotion) or part of its body (the bending of a limb). Muscle contraction also occurs in the diaphragm during breathing, in the heart and in the walls of the digestive organs. Energy is also required during movement of materials across the cell membrane. During active transport, molecules move from a region of low concentration to one of high concentration,a process which requires energy. Active transport is thought to involve a carrier molecule driven by the hydrolysis of ATP, the energy-releasing source. Another energy-requiring process is heat production. During metabolic activity, the heat given off is important to "warm-blooded" animals (homeotherms) in keeping body temperature constant despite changes in evironmental temperatures. Even "cold-blooded" animals(poikilotherms) whose body temperature is determined by the environmental temperature, must release heat in order to survive extremely cold conditions. Homeotherms, which include birds and mammals, generally eat more food per body weight than do poikilotherms, which include fish, amphibians, reptiles, and invertebrates. This fact reflects the need to furnish more energy to maintain body temperature. Both anabolism and muscle contraction require the energy obtained from the splitting of the energy-rich terminal phosphate group of ATP, which has the structure: ATP is produced during oxidative phosphorylation - the process by which the flow of electrons in the electron transport system of respiration transfers the released energy to be stored in ATP. However, certain compounds, such as the hormone thyroxine, can uncouple phosphory-lation from the electron flow so that the energy is not trapped as ATP, but is released as heat. This mechanism is important when more heat is needed by an organism than is provided by its normal metabolic activity.

Question:

Explain what occurs during the human "birth process" and when these phenomena take place.

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Solution:

The actual process of birth is preceded by the gestation period. The gestation period or duration of pregnancy varies in different species. It normally lasts about 280 days in humans (the time of the last menstrual period to the birth of the baby). Little is known about the factors that initiate the birth process, or parturition, after gestation is complete, but it is believed that hormon-al factors play an important role. It has been found ex-perimentally in mice that the level of progesterone in the blood declines sharply, while that of estrogen rises short-ly before parturition. The start of parturition is marked by a long series of involuntary uterine contractions called "labor pains" Labor is divided into three periods. The first lasts about twelve hours. The fetus is moved down toward the cervix which becomes dilated to enable the fetus to pass through. The amnion ruptures, releasing the amniotic fluid through the vagina. The second period is the birth, where the fetus passes through the cervix and vagina and is "delivered." This stage takes about twenty minutes to an hour, and consists of combined involuntary uterine contractions and voluntary contractions of the abdomen by the mother. The last stage of labor begins after birth and lasts for ten to fifteen minutes. The placenta and the fetal membranes are loosened from the uterine lining by another series of contractions and expelled as the after--birth. Sometimes, a woman will require more help than usual during childbirth. Drugs such as oxytocin and prosta-glandins will help increase the uterine contractions. Special forceps can be used by an obstetrician to pull the infant through the birth canal. When the canal through which the baby passes is too small, a caesarean delivery is performed. This is an operation in which the abdominal wall and uterus are cut open from the front and the baby is removed.

Question:

Calculate the volume of O_2 necessary to burn 50 liters of CO completely. The balanced reaction is: 2CO + O_2 \rightarrow 2CO_2 Also, calculate the volume of CO_2 formed.

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Solution:

This is a volume-volume problem and the technique shown applies only to gases. It is assumed that all the gases must be at the same temperature and pressure. No molecular weights or molecular volumes are needed. One solves this problem by setting up a ratio between the actual volumes present and the mole volumes of the re-acting compounds as shown in the equation below. [(volume CO present) / ( mole volume CO)] = [(volume O_2 needed)/(mole volume O_2 )] The mole ratio for this reaction is 2 : 1, therefore, the mole volume ratio is 2 : 1. Substitute, to obtain [(50 liters) / (2 liters)] = [(volume O_2 ) / (1 liter)] volume O_2 = 25 liters. To find the volume of CO_2 produced, set up a similar proportion between CO and CO_2 or O_2 and CO_2. The mole ratio for CO and CO_2 is 1 : 1 and for O_2 and CO_2 is 1 : 2. There-fore use the equation [(volume reactant) / (mole volume reactant)] = [(volume CO_2 ) / (mole volume CO_2 )] The values for CO as the reactant are: (50/2) = [(volume CO_2 ) / (2)] volume CO_2 = 50 liters. The values of O_2 as the reactant are: (25/1) = [(volume CO_2 ) / (2)] volume CO_2 = 50 liters.

Question:

When mercury is vaporized at its boiling point at standard pressure, the entropy change is 20,7 cal/mole - \textdegreeK, Determine the boiling point of Hg if the heat of vaporization is 65 cal/g.

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Solution:

When a process occurs at constant temperature, the change in entropy, ∆S, is equal to the heat absorbed divided by the absolute temperature at which the change occurs, i.e. , ∆S = (∆H/T), where ∆H is the heat of vaporization (in this, case) and T is the absolute temperature. T will be equal to the boiling point of mercury in this problem. Using this equation, one can solve for T after either converting ∆H to cal/moles or converting ∆S to cal/g. Here, one will convert ∆H from cal/g to cal/mole by multiplying ∆H by the molecular weight of Hg. (MW of Hg = 200.6.) ∆H in cal/moles = 65 cal/g × 200.6 g/mole = 13039 cal/mole One can now solve for T. T = [(∆H)/(∆S)]∆H = 13039 cal/mole ∆S = 20.7 cal/mole - \textdegreeK T = [(13039 cal/mole)/(20.7 cal/mole - \textdegreeK)] = 630 \textdegreeK T in \textdegreeC = 630 - 273 = 357\textdegreeC.

Question:

Calculate the weight of iron in 350 pounds of Fe_2O_3. First, calculate the percent of iron in the compound.

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Solution:

The percent composition of iron in the compound Fe_2O_3 can be calculated by first finding the molecular weight of the compound. After the molecular weight of the compound is found, the percent of this weight that is made up by the iron is found by dividing the weight of the iron by the molecular weight of the compound and then multiplying this fraction by 100. The molecular weight of iron is 55.8, so that two atoms of iron weigh 2 × 55.8 = 111.6. The molecular weight of oxygen is 16.0, so that 3 atoms of oxygen weigh 3 × 16.0 = 48.0 The molecular weight of Fe_2O_3 is found by adding the weights of the iron and oxygen together. weight of iron=111.6 weight of oxygen=48.0 molecular weight of Fe_2O3=159.6 The percent composition of iron = {(weight of iron in compound ) / (molecular weight of compound)} × 100 percent composition of iron = (111.6 / 159.6 ) × 100 = 69.9 Since the percent composition by weight of iron in the compound is known, the weight of iron in 350 pounds of this compound can now be found. Because the percent compo-sition of iron in this compound is 69.9, the percent of iron in 350 pounds of Fe_2O_3is also 69.9 . Weight of iron in 350 lb of Fe_2O3= 350 × 0.699 = 245 lb.

Question:

Many organs develop in the embryo without having to function at the same time, but the heart must function while undergoing development. Describe the steps in the formation of the heart.

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Solution:

In its early stage of development, the heart consists only of a simple tube, formed from the fusion of two thin-walled blood vessels located below the head. In this early condition, the human heart resembles a frog heart. A frog heart consists of four "chambers" arranged in series: the sinus venosus, which receives blood from the veins, an atrium, a ventricle, and the arterial cone, which leads to the aortic arches. Since the tube grows faster than the points to which its front and rear ends are attached, it bulges to one side. The ventricle then twists in an S-shaped curve down and in front of the atrium, coming to lie ventral to it as it exists in the adult. Gradually, the sinus venosus becomes incorporated into the atrium as the atrium grows around it, and most of the arterial cone is merged with the wall of the ventricle. The heart becomes partitioned into a right side and a left side in all air-breathing vertebrates. This takes place rather early in development, long before the lungs have begun to function at the time of birth. No such division occurs in the hearts of vertebrates which breathe in water. Their hearts pump just one kind of blood, namely, "venous blood" (blood low in oxygen). Aeration takes place in the gills, and only "arterial blood" (blood rich in oxygen) is distributed to the body. The advent of lungs necessitated a radical change in the arrangement of the circulation. The heart now receives two sorts of blood: venous blood from the body, and arterial blood from the lungs. It is impor-tant that the two streams be kept separate so that the venous blood is sent to the lungs for aeration and only the arterial blood is distributed to the body. The heart begins separating into four chambers at an early stage. By the end of the second month of de-velopment, the two ventricles are completely separated. The septum separating the atria, the foramen ovale, however, is not completely fused until the fetus is born. If this septum remains open at birth, the baby is termed a "blue baby\textquotedblright due to the flow of deoxygenated blood through the body. Surgical correction is performed for this condition.

Question:

Determine the de Broglie wavelength of an electron (mass = 9.11 × 10^-28 g) having a kinetic energy of 100eV.

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Solution:

Using the value for the kinetic energy of the electron, we will calculate its momentum and substitute this into the de Broglie equation to get the wavelength of the corresponding matter wave. The kinetic energy, k, of a particle of mass, m, moving with velocity v is k = (1/2) mv^2 and its momentum, p, is p =mv. Kinetic energy and momentum are related as follows: k = (1/2) mv^2 = (1/2m) m^2 v^2 = (1/2m) (mv)^2 = (p^2/2m) = k or p = (2mk)^1/2. The kinetic energy of our electron is k = 100eV = 100eV× 1.602 × 10^-12 erg/eV= 1.602 × 10^-10 erg. Hence, its momentum is: p = (2mk)^1/2 = (2 × 9.11 × 10^-28 g × 1.602 × 10^-10 erg)^1/2 = (.2919 × 10^-36)^1/2 erg^1/2 g^1/2 = (.2919 × 10^-36)^1/2 (g - cm^2/sec^2)^1/2 g^1/2 = 5.403 × 10^-19 g-cm/sec. The de Broglie equation is: \lambda = h/mv= h/p, where h is Planck's constant (h = 6.626 × 10^-27 erg/sec) = 6.626 × 10^-27 g-cm^2/sec. Hence, the de Broglie wave-length of a 100eVelectron is: \lambda = (h/p) = [(6.626 × 10^-27 g-cm^2/sec)/(5.403 × 10^-19 g-cm/sec)] = 1.226 × 10^-8 cm = 1.226 × 10^-8 cm × 10^8 \AA/cm = 1.226 \AA.

Question:

Suppose a 15-g bullet is fired into a 10-kg wooden block that is mounted on wheels and the time required for the block to travel a distance of 45cm is measured. This can easily be accomplished with a pair of photocells and an electric clock. If the measured time is 1 sec, what is the muzzle velocity of the bullet?

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Solution:

In this example, the bullet comes to rest in the block and imparts its momentum to the block. Since the block travels 45 cm in one second, the recoil velocity of the block is 45 cm/sec, and from momentum conservation we have m_1v_1 = m_2v2 (Here, we do not have a negative sign because both velo-cities are in the same direction. Also, we take m_2 to be 10 kg, that is, we neglect the mass of the bullet embedded in the block.) Then, v_1 = (m_2v_2)/(m_1) = [(10^4g ) × (45 cm/sec )]/15g = 3 × 10^4 cm/sec = 300 m/sec \cong 985 ft/sec

Question:

The hydrogen-filled dirigible "Hindenberg" had a gas volume of 1.4 × 10^8 liters. Assuming that it was filled to a pressure of 1 atm at 0\textdegreeC, how much energy was released when it burned to gaseous products at a reaction temperature of 25\textdegreeC? The standard heat of formation of water vapor at 25\textdegreeC is - 57.8 Kcal/mole.

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Solution:

The solution to this problem involves calcu-lation of the heat of reaction for the combustion of hydrogen and the number of moles of hydrogen in the dirigible. The combustion reaction is written H_2 (g) + (1/2) O_2 (g) \rightarrowH_2O (g) . The heat of reaction at 25\textdegreeC (298\textdegreeK) is written ∆H\textdegree_298 = sum of heats of formation of products - sum of heats of formation of reactants = ∆H\textdegreef, H2O- [∆H\textdegreef, H2+ ∆H\textdegreef, O2] Since the heats of formation of O_2 and H_2 are zero by definition, ∆H\textdegree_f, O2 = ∆H\textdegreef, H2= 0 Kcal/mole, the heat of reaction is equal to the heat of formation of water vapor: ∆H\textdegree_298 = ∆H\textdegreef, H2O=- 57.8 Kcal/mole. Since ∆H\textdegree_298is negative, the reaction is exothermic and - 57.8 Kcal are released per mole of water produced. But since one mole of water is formed per mole of hydrogen burned, - 57.8 Kcal are released per mole of hydrogen burned. To calculate the number of moles of hydrogen present in the dirigible, we make use of the fact that at 0\textdegreeC and 1 atm pressure, one mole of gas occupies 22.4 l (molar volume of gas). Then number of mole of H_2 = (volume of dirigible) / (number of moles of H_2) = (1.4 × 10^8l) / (22.4 l/mole) = [(1.4 × 10^8) / (22.4)] moles The heat released is then the absolute value of the heat of reaction (per mole) multiplied by the number of moles of gas. heat released = \vert∆H\textdegree_298\vert × number of moles of H_2 = 57.8 Kcal/mole × [(1.4 × 10^8_ ) / (22.4)] moles = 360 × 10^6 Kcal. (We use the absolute value of ∆H\textdegree298because the word "released\textquotedblright already takes into account the fact that the re-action is exothermic and therefore the sign of ∆H\textdegree_298 is negative.) Note that we have assumed that all of the hydrogen is burned.

Question:

Why is it that animals and plants of England and Japan are very similar despite the fact that they lie on nearly the opposite sides of the world?

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Solution:

The observation that animals and plants of England and Japan closely resemble each other is explained by the fact that England and Japan belong to the same biogeographic realm. A realm is a geographic division separated from other realms by major physical barriers, and characterized by the existence of certain unique organisms. Inside a realm there may be variations of climate and topography, but organisms are nevertheless able to pass more or less freely from one region of the realm to another. It must be understood that some species have more ready physical access to a region suitable for habitation than others. What may be an effective barrier for one species may be a possible but difficult route for another or an easily crossed part for a third. Thus, keeping in mind the fact that different species have varying access to widely separated habitats, like England and Japan, and realizing that historically, geogrophic and environmental conditions were different than they are today, we can begin to understand the underlying concept to this question. England and Japan both belong to the Palearctic realm (see Figure). By definition then, the animals of these two countries are free to migrate from one country to the other, either by way of the sea or the air. Though plants themselves cannot migrate, their seeds can be carried by wind, water or migratory animals to distant locations. In other words, there are no absolutely impassable physical barriers between England and Japan, and genes can flow from one country to the other. Therefore England and Japan share many common groups of animals and plants.

Question:

Memory Reference Instructions: 013415 I OP CODE ADDRESS Register-Reference Instruction (Indirect): Input Output Instruction: The computer operates in four different cycles: Fetch Cycle (read instruction) Indirect cycle (read address of operand) Execute cycle (read operand) Interrupt where each cycle will execute no more than four steps. Design a block diagram of the control unit for this com-puter.

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Solution:

The control unit is shown in Figure 1. The F and R registers are used to denote which of the four cycles the computer is operating in. The 2 × 4 decoder associated with these registers generates four control signals C_0 - C_3. Each cycle has its own control signal specified by the table of Figure 2. Registers FR Decoder Output Computer Cycle 00 C_0 Fetch Cycle 01 C_1 Indirect Cycle 10 C_2 Execute Cycle 11 C_3 Interrupt Cycle Each cycle is divided into four steps. Each step is designated by the output of the two-bit binary step counter, SC. The counter starts at 00 when the computer enters a cycle and the computer stays in this cycle until the counter reaches 11. At this point, the computer enters another cycle and the counter returns to 00. Each step is decoded into one of four timing signals, t_0 - t_3. The S bit is used to start and stop the computer. The user starts the computer by putting a logic level 1 in the S register. This enables the SC decoder and the timing signals. The user stops the computer by putting the S register to 0. This disables the decoder and all the timing signals t_0 - t_3-. The OPR register contains the three bit op-code specified in the instruction format. Each of the eight possible op-codes is decoded into eight separate control signals q_0 - q_7. Each signal specifies one of eight instruc-tions from the instruction set of the computer. The I bit is bit 1 of the instruction format. It is used to designate whether the instruction is: (1) In-direct addressing (I = 1), when q_7 = 0 (op-code \not = 111); (2) Register reference (I = 0), or I/O (I = I), when q_7 = 1 (op-code = 111). All of these signals (C_0 - C_3; t_0 - t_3; q_0 - q_7; I) go to the control logic section which then carries out the actual Instruction through the control functions.

Question:

The arithmetic logic unit (ALU) in Figure 1 can add the contents of A and B, complement the contents of A or B, and logically AND the contents of A and B. Show how the ALU can: a)Increment b)Decrement c)Subtract d)Logically OR e)Shift left f)Clear

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Solution:

(a) The contents of register A can be incre-mented by clearing register B and letting C_i = 1. When the addition operation is performed the value of sum be-comes A + B + C_i. = A + C_i. = A + 1. Thus, A is incremented. (b)The contents of A can be decremented by adding nega-tive ones. Negative one in two's complement is all 1's; thus, to decrement A: let the contents of B be all 1's, C_i = 0, and add. (c)Subtraction is simply adding a negative number. Thus, to form A - B, get the two's complement of B and add to A. The two's complement of a number is found by complementing the number and adding one. Thus, to subtract B from A: complementBthrough the ALU using theBfunction input. The complement function will place the result (B) in storage area SUM. Put the contents of SUM back into B. Then, to complete the complementing process, add 1 to the new B by zeroing A and making C_i = 1. Again, this result will be placed in SUM. This new number in SUM now repre-sents the fully complemented B (or -B). With this, we can store the complement in Register B. Let us now place the number which B (or more precisely, the contents of B) is to be subtracted from in Register A. Hence, by adding the contents of the two registers, we obtain the difference of the two numbers, represented in two's comple-ment form. (d)Logically ORing A and B can be carried out using DeMorgan's Law. A + B =A ̅ \textbullet B ̅ Thus, to OR A and B: complement A and put the result back into A, complement B and put the result back into B, AND A and B and put back in A, then complement A. (e)Shifting-left is the same as multiplying by two in base two, just like shifting left is the same as multi-plying by ten in base ten. Instead of multiplying by two, however, just add A to A. Thus, to shift the con-tents of register A left: put the contents of A into B (registers A and B now hold the same numbers) , let C_i = 0 and add. (f)Clearing a register (making it all zero's) is done by the following formula: 0 = AA ̅ Thus, to clear register A:complement A through the ALU and put back into B, AND registers A and B, put sum (which is zero) into A.

Question:

Discuss the role of hemoglobin in the transport of oxygen and of carbon dioxide.

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Solution:

Hemoglobin is the pigment in red blood cells that is responsible for transporting nearly all the oxygen and some of the carbon dioxide in the body. In arterial blood, oxygen is present in two forms. It is either physically dissolved in the blood plasma or chemically bound to hemoglobin. Because oxygen is relatively in-soluble in water, and blood plasma is comprised primarily of water, only 3 ml. of oxygen can be dissolved in 1 liter of plasma at normal oxygen pressure (100 mm Hg on average). Hence, approximately 2% of the oxygen in the blood is dissolved in the plasma; the rest is transported via hemo-globin. In the lungs, oxygen enters the capillaries, and diffuses into the red blood cells where it binds to hemoglobin. Four oxygen molecules are attached to four iron atoms in a single hemoglobin molecule (see Figure). The chemical reaction between oxygen and hemoglobin is usually written as O_2 + Hb \rightleftarrows HbO_2 (oxyhemoglobin) The reaction goes to the right in the lungs and to the left in the body tissues. Hemoglobin combined with oxygen (HbO_2) is called oxyhemoglobin; when not combined (Hb), it is called reduced hemoglobin or deoxyhemoglobin. Because there is a finite number (four) of binding sites for oxygen on the hemoglobin molecule, there is a maximum amount of oxygen which can combine with hemoglobin. When hemoglobin exists as both Hb and HbO_2, it is said to be partially saturated. When it exists as only HbO_2, it is said to be fully saturated. The combination of oxygen with hemoglobin and its release from oxyhemoglobin are controlled by the concen-tration of oxygen present and, to a lesser extent, by the amount of carbon dioxide present. For example, the percentage of saturated hemoglobin undergoes little change between the lungs and the arteries, because the oxygen concentration in the lungs is very similar to that of the arteries. However, in the tissues, the concentration of oxygen is low, so by the law of mass action, the reaction will go to the left. As a result, oxygen will break off from the oxyhemoglobin to diffuse into the tissues. Carbon dioxide reacts with water to form carbonic acid, H_2CO_3, which then dissociates: CO_2 + H_2O \rightleftarrows H_2CO_3 \rightleftarrows H^+ + HCO_3^- (bicarbonate) An increase in the concentration of CO2drives the reaction to the right, increasing the H^+ concentration and hence the acidity of the blood. The oxygen carrying capacity of hemoglobin decreases as the blood becomes more acidic. Thus, more O_2 is released to the tissues under conditions of increasing acidity. This shows that the combination of hemoglobin and oxygen is controlled, in part, by the amount of CO_2 present. This results in a very efficient transport system. In the lung capillaries, oxygen is taken up by the hemoglobin due to the effects of high oxygen tension and low CO_2 tension. The situa-tion is reversed in the tissues. There, CO_2 pressure is high and oxygen pressure is low, so oxygen dissociates from oxyhemoglobin and diffuses into the tissues. It is important to realize that the principal factor that determines the mode of hemoglobin dissocia-tion and rate of diffusion is the partial pressure (P_gas) of each particular gas. As is true for oxygen, the quantity of carbon dioxide that can physically dissolve in blood is quite small. Carbon dioxide can undergo the following re-action: CO_2 + H_2O \rightleftarrows H_2CO_3 (carbonic acid) This reaction would go quite slowly if it were not catalyzed by the enzyme carbonic anhydrase. The quan-tities of both dissolved carbon dioxide and carbonic acid are directly proportional to the partial pressure of CO_2. In this case, as in the case with oxygen, we are only concerned with the pressure of the gas in question. This is because in a mixture of gases, each one acts indepen-dently of the others, and exerts the same pressure it would if it were present alone. The actual quantity of carbonic acid in the blood is small, because, as we saw previously, it dissociates into H^+ and HCO_3^- ions. These ions are quite soluble in the blood. Thus, the addition of carbon dioxide to blood plasma results ultimately in the produc-tion of hydrogen and bicarbonate ions. Carbon dioxide is transported primarily as bicarbonate ions to the lungs, where it is excreted as carbon dioxide. Carbon dioxide can also react with proteins, particularly hemoglobin, to form carbamino compounds. CO_2 + Hb \rightleftarrows HbCO_2 (carbamino hemoglobin) Carbon dioxide diffuses from the tissues into the blood. In the blood, some (8%) of the CO_2 stays dissolved, and some (25%) reacts with hemoglobin to form HbCO_2. The largest fraction (67%) is converted to H^+ and HCO_3^-. This occurs primarily within the red cells because they contain large quantities of carbonic anhydrase, whereas the plasma does not. This dissociation into H^+ and HCO_3^- explains why tissues and capillaries, where CO_2 concentration is high, have a hydrogen ion concentration higher than that of arterial blood. This also explains the increase in H^+ concentration as metabolic rate increases. The CO_2 it-self passes from the tissues to the blood, and then to the lungs by diffusing from a region of high CO_2 tension to one of low CO_2 tension.

Question:

A stone of mass 100 grams is whirled in a horizontal circle at the end of a cord 100 cm long. If the tension in the cord is 2.5 newtons, what is the speed of the stone?

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Solution:

First, we shall calculate the acceleration of the object, and from that we may calculate its veloci-ty. Firstly, F = mg 2.5 newtons = 100 gm × a Newtons have the units (kg - m)/(s^2), and 100 gm = 0.100 kg, so that 2.5[(kg - m)/(s^2)] = 0.100 kg × a a = 25 m/s^2 Alsoa = v^2/r for uniform circular motion, where a is the linear acceleration and v is the linear velocity of the object. Therefore, 25(m/s^2) = v^2/(1m) v = 5 (m/s) .

Question:

A young scientist is watching a group of new-born ducklings. On the first day, he observes them to crouch in fear at the sight of falling leaves. He also sees them "practice" flying. Over the next few days he observes that they do not react to the falling leaves and can now fly expertly. He concludes that the ducklings have learned to ignore falling leaves and have also learned to fly. Why is he partly incorrect?

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Solution:

The scientist is incorrect because he uses the term "learning" to describe both processes. Learning is the adaptive change of behavior as a result of experience The animal modifies its behavior to respond to stimuli in ways that promote its survival and reproduction; that is, in ways that adapt it to its environment. One of the simplest forms of learning is called habituation. Habituation is the gradual decline in response to repeatedly presented stimuli not associated with reward or punishment. Young ducklings initially show fear and crouch when presented with any overhead flying objects, such as falling leaves. But the ducklings cease to respond to the leaves since experience has shown them that they are insignificant. (no associated reward or punishment) The duckling, however, is not "learning" to fly. One of the factors that complicates the study of learning is maturation. Maturation is the automatic disclosing of a behavioral process due to physiological development. The flight of birds is a prime example of maturation. As the scientist observed the flying attempts of the ducklings, he termed it "learning", thinking that the ducks flew better (change in behavior) because of this apparent "practice" (experience). But flying can be achieved solely by maturation, with no practice involved. This has been shown by experiments with pigeons. Pigeons were raised in narrow tubes that prevented them from moving their wings and practicing flying. However, when the pigeons were released at an age when they would have already been flying, they were able to fly as well as control pigeons raised under normal conditions. It is not practice that enables the flight of a young bird to improve; it is progressive maturation. The development of the bird's nervous and muscular systems enables it to improve its flying. The bird has not "learned" to fly since experience is unnecessary. Hence the young scientist is incorrect in this respect.

Question:

Large deficits of water can be only partly compensated for by renal conservation; drinking is the ultimate compensatory mechanism. What stimulates the subjective feeling of thirst which drives one to drink water?

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Solution:

The feeling of thirst is stimulated both by a low extracellular fluid volume and a high plasmaosmolarity. The production of ADH by the hypothalamus is also stimulated by these factors. The centers which mediate thirst are located in the hypothalamus very close to those areas which produce ADH. Should this area of the hypothalamus be damaged, water in-take would stop because the sensation of thirst would be impaired. Conversely, electricalexcitation of this area stimulates drinking. There has been much speculation that, because of the similarities between the stimuli for ADH secretion and thirst, the receptors which initiate the ADH controlling reflexes are identi-cal to those for thirst. These are the osmoreceptorsre-ferred to in the previous solution. In addition to hypothalamic pathways, there are also other pathways controlling thirst. For example, dryness of the mouth and throat causes profound thirst, which is relieved by moistening. There is also a learned control of thirst; the quantity of fluid drunk with each meal is, in large part, a learned response determined by past experience.

Question:

What is the de Broglie wavelength of a radium alpha particle (a helium nucleus,He^2+) having an energy of 4.8MeV(million electron volts)?

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Solution:

The de Broglie wavelength, \lambda, of a particle of mass, m, moving at speed, v, is given by \lambda = (h/mv), where h is Planck's constant (h = 6.6 × 10^-27 erg-sec). The mass of an alpha particle is 66 × 10^-24 g. The energy of an alpha particle is given by E = (1/2) mv^2 = 4.8 MeV . Since 1MeVis equal to 1.6 × 10^-6 erg, we have: (1/2) mv^2 = 4.8 × 1.6 × 10^-6 erg. We will employ this value of the energy in order to obtain the value ofmvwithout having to find the velocity of the particle. (mv)^2 = 2 [(1/2) mv^2] m = 2 (4.8 × 1.6 × 10^-6 erg) (6.6 × 10^-24 g) = 1.0 × 10^-28 erg-g = 1.0 × 10^-28 g^2 cm^2/sec^2, where we have used 1 erg = 1 g - cm^2/sec^2. Taking the square root of both sides gives: (mv)^2 = 1.0 × 10^-28 g^2 cm^2/sec^2 mv= 1.0 × 10^-14 g cm/sec. Substituting this value into the expression for the de Broglie wavelength gives: \lambda = (h/mv) = [(6.6 × 10^-27 erg-sec)/(1.0 × 10^-14 g-cm/sec)] = [(6.6 × 10^-27 g-cm^2/sec)/(1.0 x 10^-14 g-cm/sec)] = 6.6 × 10^-13 cm.

Question:

Find the wavelength (\lambda) of the first transition, from the first excited state to the ground state in the Lyman and Paschen series. These series show the lines of the emission spectra of H_2 gas. The Lyman series defines its lowest energy level n_1 as being equal to 1, the Paschen series defines n_1 as 3.

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Solution:

When an atom is in its ground state, its electrons are in the lowest energy state possible. When energy is added to the system, its electrons are promoted to a higher energy level. These electrons are said to be in the excited state. After a short time, the electrons fall back to lower energy levels and eventually to their ground states. The same amount of energy that was needed to excite the electron is released when the electron falls back to its ground state. The Paschen and Lyman series differ in the value of their lowest energy states. The Paschen series is the spectra of electrons falling from excited states to the third energy level (n_1 = 3) , whereas the Lyman series is the spectra that is exhibited when the electrons fall from excited states to the first energy level (n_1 = 1). This is illustrated in the accompanying figure. Wavelength and the energy levels of the transition are related in the Rydberg - Ritz equation, (1/\lambda) = R [(1/n_1 ^2) - (1/n_2 ^2)], where \lambda is the wavelength, R the Rydberg constant, and n_1 and n_2 are the energy levels of the transition. The electron falls from n_2 to n_1, thus, n_2 is greater than n_1. Solving for \lambda: Lyman series: In the Lyman series n_1 = 1. The first transition is from n_2 = 2 to n_1 = 1. One solved for \lambda by using the Rydberg - Ritz equation. (1/\lambda) = (109, 678 cm^-1) [1/(1)^2 - 1/(2)^2], (1/\lambda) = 82258 cm^-1 \lambda = [(1)/(82258 cm^-1] = 1.21 × 10^-5 cm. Wavelengths are usually expressed in angstroms. 1 \AA = 10^-8 cm. (1.21 × 10^-5 cm) [(1 \AA)/(10^-8 cm)] = 1210 \AA Paschen series: The first transition in the Paschen series is from n_2 = 4 to n_1 = 3 because the base energy level in this series is 3. Here again, one uses the Rydberg - Ritz equation to solve for \lambda. (1/\lambda) = 109678 cm^-1 [1/(3)^2 - 1/(4)^2] = 5332 cm^-1 \lambda = [(1) / (5332 cm^-1)] = 1.87 × 10^-4 cm. Converting to angstroms: 1.87 × 10^-4 cm ×[(1 \AA) / (10^-8 cm)] = 18700 \AA.

Question:

In a hydrogen bomb, deuterium nuclei fuse to produce a helium nucleus. In the process, 5MeV of energy is re-leased per deuterium. If, 1/7000 of natural H is D, how many liters of gasoline (C_8H_18) are needed to be burned to produce the same amount of energy as derived from 1 liter of water. Assume the following: the heat of combustion of gasoline is 5100 KJ/mole the density of gasoline is .703 g/cm^3; leV = 1.60 × 10^-19 J.

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Solution:

The answer to this question involves several calculations. To determine the amount of gasoline which will yield an equal amount of energy upon combustion as one liter of water, first compute how much energy can be obtained from each. For gasoline, this entails calcula-tion of the number of moles per liter of the substance. Multiplication by the heat of combustion yields the total energy. For water, first determine, how many deuterium ions are present per liter and then, using conversion factors, multiply this by the energy re-leased per deuterium ion fused. To obtain the energy per liter of water, first find the number of deuterium ions. To obtain this, you need to first find the number of atoms of H per liter of water, since the density of water is defined as one and density = mass/volume, 1 liter H_2 O =1000 g H_2 O. The molecular weight of one mole of water is 18. Therefore, the number of moles of water in one liter is (1000) / (18) = 55.5 [(moles water) / (liter)].There are two moles of H atoms, per mole of water, since its molecular formula is H_2 O. Therefore, the number of moles of H atoms = [(55.5 moles water) / (liter)] × [(2 mole H atoms) / (moles H_2O)] = (111 mole H atoms) / (liter). However, in a mole of anything, there are 6.02 × 10^23 particles, (Avogadro's number). Therefore, the total number of H atoms per liter of water is [(111 mole H atoms) / (liter)] × 6.02 × 10^23 (atoms/mole) = 6.68 × 10^25 (H atoms/liter) since you have 1 deuterium atom per 7000H atoms. Therefore, the number of deuterium atoms per liter of water is 6.68 × 10^25 (H atoms/liter) × [(1 atom deuterium) / (7000 H atoms)] = 9.55 × 10^21 [D atoms/liter)] Since each deuterium atom yields 5MeV or 5 x 10^6 ev of energy, when it fuses to produce helium, the amount of energy per liter of water is 9.55 × 10^21 (D atoms/liter) × 5 × 10^6 (ev/D atom) = 47.75 × 10^27 (ev/liter). Since, 1.60 × 10^-19 J = one ev = 1.60 × 10^-22 kJ, 47.75 × 10^27 (ev/liter) = 47.75 × 10^27 (ev/liter) × 1.60 × 10^-22 kJ/ev = 7.64 × 10^6 kJ/liter of energy per liter of water. To compute the energy in one liter of gasoline, remember that the density of gasoline equals mass of gasoline divided by its volume. Therefore, the mass of one liter of gasoline is (density × volume) = .703 g/cm^3 × 1000 cm^3 = 703 g. Thus, there are 703 g/liter of gasoline. The molecular weight of gasoline is 142 grams per mole. Therefore, the number of moles of gasoline per liter is (703 g/l) / (142 g/mole) = 6.16 moles/liter for gasoline. The heat of combustion of gasoline is 5100 kJ/mole. Therefore, the energy per liter of gasoline is 6.16 moles/liter × 5100 kJ/mole = 31,400 kJ/liter = 3.14 × 10^4 kJ/liter. To find out how may liters of gasoline is equivalent to water in terms of energy, divide the energy of 1 liter H_2 O by the energy of 1 liter gasoline. For water, it is 7.64 × 10^6 kJ/l. For gasoline, it is 3.14 × 10^4 kJ/l. (7.64 × 10^6 kJ/lwater) / (3.14 × 10^4 kJ/l gasoline) = 243 [(liters gasoline) / (liter water)].

Question:

Hydrogen peroxide solution for hair bleaching is usually prepared by mixing 5.0 g of hydrogen peroxide (H_2O_2, molecular weight = 34 g/mole) per 100 ml of solution. What is themolarityof this solution?

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Solution:

Before employing the definition molarity= [(number of moles)/{volume (liters)}] we must convert 5.0 g of H_2O_2 to the corresponding number of moles. To do this, we use the formula moles = mass/molecular weight. Then, moles = (mass/molecular weight) = [5.0 g / (34 g/mole)]= 0.15 moles H_2O_2 Converting 100 ml to liters, 100 ml = 100 ml × (1 liter/1000 ml) = 0.10 liter. Themolarityis then molarity= [(number of mole) / {volume (liters)}] = (0.15 moles / 0.10 liters) = 1.5 moles/liter = 1.5 molar = 1.5 M.

Question:

Synthesize CH_2BrCH_2Br from CH_3CH_2OH.

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Solution:

You are given the alcohol, ethanol, and are asked to synthesize CH_2BrCH_2Br , 1,2 dibromoethane. The fact that it is a dibromo compound suggests that the bromine added across a double bond. Therefore, if you could convert the alcohol to an olefin, namely, ethylene, you could add Br_2 in the solvent CCI_4 , and via an addition reaction, obtain the desired product. To obtain ethylene, remember that the alcohol can be dehydrated to the corresponding olefin by the addition of acid. This well-documented mechanism is as follows: When you add the acid, you protonate the alcohol. Water is expelled and you obtain a carbonium ion, CH_3CH_2+, The carbonium expels a proton (H+), so that ethylene is obtained. Now that you have CH_2 = CH_2 , add Br_2 and CCI_4 and you obtain CH_2BrCH_2Br , your desired product. The Br_2 is added across the double bond in an addition reaction.

Question:

What causes the characteristic sounds of a beating heart?

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Solution:

The heart produces two sounds per heartbeat which can be heard by using a stethoscope, which is an instrument which magnifies sounds and transmits them to the ear. The sound is sometimes de-scribed as "lub, dub, lub, dub..." The first sound, lub, is low- pitched and of long duration. It is associated with the closure of the atrio-ventricular valves (the tricuspid and bicuspid) at the beginning of ventricular systole. The second sound, dub, is higher-pitched, louder and shorter in duration. It is associated with the closure of the semilunar valves at the beginning of diastole. In the heart, the valves are cushioned by the blood, and it is difficult to understand why the valves create sound. It is not due to the slapping together of the "flaps" of the valves. The cause is believed to be the vibrations of the walls of the heart and the vessels around the heart. For example, contraction of the ventricles causes blood to flow against the A-V valves, clos-ing them. The elasticity of the valves causes the blood to surge backwards and bounce back toward the ventricles. This sets the ventricular walls into vibrations, which are carried to the chest wall (where it contacts the heart), creating the sound waves audible in a stethoscope. A very small portion of the first sound is also caused by the contraction of the heart muscle itself. As any muscle contracts, a small amount of noise can be heard as the internal filaments rearrange. The second heart sound is caused by the vibrations of the walls of the aorta, the pulmonary artery, and the ventricles. The closing of the semilunar valves causes the blood in the arteries to rever-berate between the arterial walls and valves. When these vibra-tions of the walls come into contact with the chest wall, an audible sound is created. When the heartbeat sounds normal, it is an indication that the valves are functioning properly. When the valves are damaged, a hissing or sloshing sound can be heard. This is caused by the blood leaking back through the damaged valve. This condition is known as a heart murmur. It is a common result of rheumatic fever, syphilis, and other diseases.

Question:

200mlof oxygen is collected over water at 25\textdegreeC and 750 torr. If the oxygen thus obtained is dried at a constant temperature of 25\textdegreeC and 750torr, what volume will it occupy? What volume will be occupied by the water vapor removed from the oxygen if it is maintained at 25\textdegreeC and 750 torr? (The equilibrium vapor pressure of water at 25\textdegreeC is 28.3torr.)

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Solution:

This problem will be approached by determining the partial pressures of oxygen and of water vapor in the initial mixture and then using these partial press-ures to determine the separate volumes of each component. The partial pressures exerted by the oxygen and the water vapor must add up to the total pressure. Since the partial pressure of water vapor is equal to its equi-librium vapor pressure, we can determine the partial pressure of oxygen: total pressure = partial pressure of O_2 + partial press-ure of water 750torr= partial pressure of O_2 + 28.3torr or,partial pressure of O_2 = 750torr- 28.3torr = 721.7torr. By definition, this partial pressure is the press-ure the oxygen would exert if it filled the entire 200mlvolume. We must determine the volume this amount of oxygen would fill at 750torr. To do this we use the relationship P_1V_1 = P_2V_2, which is valid for constant temperature and constant number of moles of gas. Let P_1 = 721.7torr, V_1 = 200ml, and P_2 = 750torr. Then, V_2 = [P_1 / P_2] × V_1 = [721.7torr/ 750torr] × 200ml= 192.45ml. Hence, the dried oxygen would occupy a volume of 192.45ml. We follow a similar procedure in calculating the volume that the water vapor would occupy at 750torr. The vapor pressure of water, 28.3 torr, is the pressure the water vapor would exert if it filled the entire 200ml volume. Again, applying the relationship P_1V_1 = P_2V_2, remembering that the only reason we can do so is that we have constant temperature and constant number of moles of gas, we can calculate the volume that the water vapor occupies under a pressure of 750torr. Let P_1 = 28.3torr, V_1 = 200ml, and P_2 = 750torr. Then the volume at 750torris V_2 = [P_1 / P_2] × V_1 =[28.3torr/ 750torr] × 200ml= 7.55ml. As a check, we ascertain that the individual volumes at 750torradd up to the total volume at 750torr: Volume of O_2 + volume of water = 192.45ml+ 7.55ml = 200 ml.

Question:

For the following exothermic reaction at 25\textdegreeC, C_6 H_6(l) + 7(1/2) O_2(g) \rightarrow 6CO_2 (g) + 3H_2O(l), the change in energy, ∆E, is - 780 Kcal/mole. Find the change in enthalpy, ∆H.

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Solution:

The solution is based on the following equation for the enthalpy change ∆H = ∆E + ∆nRT, where ∆n is the number of moles of gaseous products minus the number of moles of gaseous reactants, R is the gas constant (1.987 cal/deg-mole), and T is the absolute tem-perature. Since there are 6 moles of gaseous products (6CO_2(g)) and 7(1/2) moles of gaseous reactants (7(1/2) O_2 (g)), ∆n = 6 - 7(1/2) = - 1.5. The absolute temperature is T = 25\textdegree + 273 = 298\textdegreeK. Hence, ∆H= ∆E + ∆nRT = - 780 Kcal + (- 1.5 mole) (1.987 cal/deg-mole) × (298\textdegreeK) = - 780 Kcal - 888 cal = - 780 Kcal - 0.888 Kcal \cong - 781 Kcal.

Question:

What is the function of the leaves? Describe the structures of a typical dicot leaf.

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Solution:

The leaf is a specialized nutritive organ of the plant. It's function is to carry out photosynthesis, a process requiring a continuous supply of carbon dioxide, water, and radiant energy. The leaf of a typical dicot consists of a stalk, called the petiole, attached to the stem, and a broad blade, which may be simple or compound. The petiole, like a stem in cross-section, contains vascular bundles. Within the blade, the vascular bundles fork repeatedly and form a network of veins. A cross-section of the leaf shows that it is composed of several types of cells. The outer cells, lining both the top and bottom surfaces of the leaf, make up the epidermis which secretes a protective, waterproof cutin covering. Distributed throughout the epidermal surface are many small, specialized pores, the stomata, each surrounded by two guard cells. There are many more stomata on the lower surface than the upper surface of the leaf in most species. The chloroplast- containing guard cells requlate the size of the stomata by changing their turgor pressure and thus their shape. The stomata allow oxygen to diffuse out and carbon dioxide to diffuse into the leaf. Most of the space between the upper and lower epidermal layers is filled with thin-walled cells, called mesophyll, Most of the space between the upper and lower epidermal layers is filled with thin-walled cells, called mesophyll, which are full of chloroplasts. The mesophyll layer near the upper epidermis is usually made of cylindrical palisade cells, closely packed together with their long axes perpendicular to the epidermal surface. The rest of the mesophyll cells are very loosely packed together, with large air spaces between them. These loosely - packed cells form the spongy layer of the leaf. The cells of the palisade layer, containing abundant chloroplasts, are chiefly responsible for the photosynthetic functioning of the leaf, while the air spaces between the mesophyll of the spongy layer hold moisture and gases. These air spaces are continuous with the stomata where exchange of gases takes place.

Question:

Calculate (a) the root-mean-square speed of the molecules of nitrogen under standard conditions, and (b) the kinetic energy of translation of one of these molecules when it is moving with the most probable speed in amaxwelliandistri-bution.

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Solution:

(a) Each molecule of an ideal gas may have a velocity with rectangular componentsv_x,v_y,v_z. Hence, the square of the velocity of one molecule is v^2 = (v_x^2 + v_y^2 + v_z^2) If we calculate this for every gas molecule, add the re-sults, and divide by the number of molecules in the box, N, we find (^N\sum_i_=1 v_i^2)/N =v^2=v^2_x+v^2_y+v^2_z where the bar over a quantity indicates an average value. (The summation symbol \sum indicates that we calculate v^2 for each gas molecule and add the results.) Since no direc-tion of motion is referred for the gas molecule, v^2_x=v^2_y=v^2_z andv^2=3v^2_x we call \surd(v^2) the root mean square speed,v_rms. Now, the average kinetic energy of one molecule is related to the temperature T of the gas by (1/2)mv^2= (3/2)kT(1) where m is the mass of one gas molecule, and k is Boltzmann's constant. Then the kinetic energy of 1 mole of molecules is found by multiplying (1) by N_0, Avogadro's number, or (1/2)N_0mv^2_rms = (3/2)N_0kT The gas constant, R, is, however R = kN_0 and mN_0 = M the mass of 1 mole of gas. Hence (1/2)mv^2= (3/2)RTandv^2= 3RT/M If we consider a kg mole of the gas, we have v^2= 3 × 8.31 × 103(joule/\textdegree K\bulletmole K\bulletmole ) × 273\textdegreeK × (1 mole/28kgm) ) × 273\textdegreeK × (1 mole/28kgm) = 2.43 × 10^5 m^2/sec^2 = 2.43 × 10^5 m^2/sec^2 ; v_rms= \surd(v^2) = \surd(2.43 × 10^5 m^2/sec^2) = 492 m/sec. (b)E_k= (1/2)mv^2_m , wherev_mis the most probable speed. The mass of a nitrogen molecule can be calculated from its molecular weight and Avogadro's number: m= M/N_0 = 28 kg/kmole× [(1kmole)/(6.02 × 10^26 molecules)] = 4.64 × 10^-26 kg/molecule. v_m= \surd(8/3\pi)v_rms= .921v_rms HenceE_k= (1/2)mv^2_m = (1/2)m \bullet (.921v_rms)^2 = (1/2)(4.64 × 10^-26 kg)(.921)^2(2.43 × 10^5 m^2/s^2) = 3.76 × 10^-21 Joule.

Question:

The best vacuum that can be produced corresponds to a pressure of about 10^-10 dyne cm^-2 at 300\textdegree K. How many molecules remain in 1 cm^3?

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Solution:

We can use the ideal gas equation to calculate N, the number of molecules in the given volume: N =pV/kT = (10^-10 dyne/cm^2 × 1 cm^3)/(1.38 × 10^-16 dyne-cm/\textdegreeK × 300\textdegree K) \approx 2,500 There is still a large number of molecules left.

Question:

Suppose 100 ml of oxygen were collected over water in the laboratory at a pressure of 700torrand a temperature of 20\textdegreeC. What would the volume of the dry oxygen gas be at STP?

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Solution:

STP means Standard Temperature and Pressure, which is 0\textdegreeC and 760torr. In this problem oxygen is gathered over water, therefore, Dalton's law of partial pressure (each of the gases in a gaseous mixture behaves independently of the other gases and exerts its own pressure, the total pressure of the mixture being the sum of the partial pressures exerted by each gas present; that isP_total= P_1 + P_2 + P_3 \textbullet\textbullet\textbulletP_n) is used to calculate the original pressure of the oxygen. There is both water vapor and oxygen gas present. After you obtain the original pressure of the oxygen, you can use the combined gas law to calculate the final volume of the oxygen. The combined gas law stated that for a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. It can be written as follows: [P_1V_1] T_1 = [P_2V_2] / T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the final temperature, V_2 is the final volume, and T_2 is the final absolute temperature. You are given the temperature in \textdegreeC, so it must be converted to the absolute temperature by adding 273. P_total= 700torr= P_(O)2 + P_(H)2O P_(O)2= 700 - 17.5 = 682.5torr. (b) Converting the temperature to the absolute scale. T_1 = 20 + 273 = 293\textdegreeK T_2 =0 + 273 = 273\textdegreeK (c) Using the combined law [P_1V_1] / T_1 = [P_2V_2] / T_2 P_1 = 682.5torrP_2 = 760torr V_1 = 100 mlV_2 = ? T_1 = 293\textdegreeKT_2 = 273\textdegreeK [(682.5torr) (.100 ml)] / [293\textdegreeK ] = [(.760torr) V_2] / [273\textdegreeK] V_2 = [(682.5torr)(100 ml)(273\textdegreeK)] / [(760torr) (293\textdegreeK)] = 83.7 ml.

Question:

One of the most outstanding features of any assembler lan-guage is its close association with the main memory of a computer . Discuss the uses of explicit addresses and state the importance of checking the format of an instruction be-fore coding it.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0192.htm

Solution:

The process of explicit addressing generally in-volves replacing a symbolin a program by its displacement, its length, and a substitute base register, not necessarily in this order. Going from left to right, the displacementindicates what portion, (how many bytes) of a symbol is to beignored in the calculations. The length tells how many of the remaining bytesare to be used in the calculations, and the substitute base register givesthe address of the symbol relative to the start of the program in main memory. The address of this symbol usually has to be loaded into a registerbefore it is used in an explicit address. The re-gister is referred to asa substitute base register because it is generally not the main base registerof the program. To illustrate this we will add two packed decimal fields, PKDAYS and PKHRS. We will assume PKDAYS to be 8 bytes long andPKHRS to be 6 bytes long. Our intentions are to add the last 3 bytes ofPKHRS to the last 4 bytes of PKDAYS: AK 3(4, 14), 2(3, 7) The above instruction accomplishes our task. Operand 1 is PKDAYS and operand 2 is PKHRS. In each operand, the number outside the parentheses is the displacement D, the one fol-lowing it is the length, L and the last is the substitute base register, B. Therefore, the general format is D( L,B). One should be especially careful when coding the displacement. Note that we are operating on the last 4 bytes of PKDAYS and on the last 3 bytes of PKHRS, but we have used a displacement of 3 and 2 respectively . This is so because the first byte of the symbol is not included in D. As you can see, explicit addressing enables us to op-erate on desired portions of a field. This method of cod-ing is especially useful when table handling is to be done. For example, let us place zeros in a 16 byte table . Each entry in the table will be 4 bytes long, therefore, our zeroing symbol , ZTAB, will be 4 bytes in length. We will name our table TABZERO. Example L 9, = F, 16' STORING LENGTH OF TABLE LA 6, TABZERO LOAD ADDRESS OF TABLE ZERO ZAP 0 (4, 6), ZTAB ZERO 4 BYTES OF TABLE S 9, = F '4' 4 BYTES JUST SET TO ZERO, SO ADJUST REG. 9 TO AMOUNT LEFT C 9, = F '0' SEE IF ZEROING IS COMPLETED BE OUT A 6, = F '4' GO TO NEXT 4 BYTES OF TABLE IF NOT FINISHED B ZERO CONTINUE ZEROING As you can see, through explicit addressing, we are able to operate on a table with great ease. The general operational rules for string handling and mathematics also apply to ex-plicit coding. Thus, we should be careful of such things as overflows. When coding explicit operands you must remember that different instructions have different formats, and so, what can be coded in some instructions cannot be coded in others. Each explicit value is critical to the resulting object code, so the formats must be checked to make sure each explicit value means the right thing.

Question:

Write a program segment in C to display the following message: Hi from C C is a unique and exciting language

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0421.htm

Solution:

To display the above message the program segment is as follows: main ( ) { printf ("Hi from C\textbackslashn"); printf ("C is a unique and exciting language"); } The \textbackslashn in the firstprintf( ) statement is an escape sequence that causes the output line to be advanced. If it is not in-cluded then the program segment becomes, main ( ) { printf ("Hi from C"); printf ("C is a unique and exciting language); } and will output both messages on the same line as follows: Hi from CC is a unique and exciting language

Question:

Determine ∆G\textdegree for the reaction 4 NH_3 (g) + 5O_2 (g) \rightarrow 4NO (g) + 6H_2 O (l) ∆G\textdegree_f of NH_3 (g) = - 4.0 Kcal/mole ∆G\textdegree_f of NO (g) = 20.7 Kcal/mole ∆G\textdegree_f of H_2 O (l) = - 56.7 Kcal/mole

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0530.htm

Solution:

The change in free energy (∆G\textdegree) may be found by subtracting the sum of free energies (∆G\textdegree_f) of the reactants from the free energies of the products. The free energy of formation of pure elements is always 0. When more than 1 mole of a compound is either reacted or formed, the ∆G\textdegree_f of that compound must be multiplied by thestoichiometriccoefficient for the specific compound, ∆G\textdegree = [4 moles × ∆G\textdegree_f of NO (g) + 6 moles × ∆G\textdegree_f of H_2 O (l)] - (4 moles × ∆G\textdegree_fof NH_3 (g) + 5 moles × ∆G\textdegree_f of O_2) G\textdegree = [4 moles × 20.7 kcal/mole + 6 moles × (-56.7 kcal/mole)] - (4 moles × (-4.0 kcal/mole) + 5 moles × 0 kcal/mole) = - 241.4 kcal.

Question:

How many moles are there in one atom?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0045.htm

Solution:

A mole of atoms is defined as containing Avogadro's Number of atoms. Avogadro's number is 6.02 × 10^23. Therefore, the number of moles in one atom is equal to 1 atom divided by 6.02 × 10^23 atoms/mole. No. of moles = [(1 atom)/(6.02 × 10^23 atoms/mole)] = 1.66 × 10^-24 moles.

Question:

Explain the mechanism of the genetic determination of sex in man .

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0665.htm

Solution:

The sex chromosomes are an exception to the general rule that the members of a pair of chromosomes are identical in size and shape and carry allelic pairs. The sex chromosomes are not homologous chromosomes . In man, the cells of females contain two identicalsex chromosomes or X chromosomes. In males there is only one X chromosome and a smaller Y chromosome with which the X pairs during meiotic synapsis. Men have 22 pairs of ordinary chromosomes (autosomes), plus one X and one Y chromosome, and women have 22 pairs ofautosomesplus two X chromosomes. Thus it is the presence of the Y chromosome which determines that an individual will be male. Although the mechanism is quite complex, we know that the presence of the Y chromosome stimulates thegonadal medulla , or sex-organ forming portion of the egg; to develop into male gonads , or sex-organs. In theabscenceof the Y chromosome, and in the presence of two X chromo-somes, the medulla develops into female gametes . [Note, that a full complement of two X chromosomes are needed for normal female development.] In man, since the male has one X and one Y chromosome, two types of sperm, or male gametes, are produced during spermatogenesis (theprocess of sperm formation, which includes meiosis) One half of the sperm population contains an X chromosome and the other half contains a Y chromosome. Each egg, or female gamete, contains a single X chromosome . This is because a female has only X chromo-somes, and meiosis produces only gametes with X chromo-somes. Fertilization of the X-bearing egg by an X-bearing sperm results in an XX, or female offspring . The ferti-lization of an X-bearing egg byaY-bearing sperm results in an XY, or male offspring. Since there are approximately equal numbers of X- and Y- bearing sperm, the numbers of boys and girls born in a population are nearly equal.

Question:

A light emitting diode (LED) used to display a number in decimal form has the following 7-segment configuration. Segment a lights up when input a is at logic level 1 and, segment b lights up when input b is at logic level 1, etc. Design a decoder which will accept a 3 bit number and de-code it to a seven bit number to be accepted by the LED. The LED will then display the decimal form of the three bit number.

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Solution:

To design this decoder we can begin by setting up a table of the input possibilities. Let the 3 bit input number be represented by ABC, C being the least significant digit. INPUT EQUIVALENT DECIMAL SEGMENTS NEED TO LIGHT UP ABC NUMBER a b c d e f g 000 0 1 1 1 1 1 1 0 001 1 0 1 1 0 0 0 0 010 2 1 1 0 1 1 0 1 011 3 1 1 1 1 0 0 1 100 4 0 1 1 0 0 1 1 101 5 1 0 1 1 0 1 1 110 6 1 0 1 1 1 1 1 111 7 1 1 1 0 0 0 0 Note: 3 bits can give you a maximum of 8 possibilities. Next we find which numbers light up a given segment, in other words, which numbers are common to a given segment. From the table we find: a = \sum(0, 2, 3, 5, 6, 7) or a = \sum (000, 010, 011, 101, 110, 111) b = \sum(0, 1, 2, 3, 4, 7) or b = \sum (000, 001, 010, 011, 100, 111) c = \sum(0, 1, 2, 3, 4, 5, 6, 7) or c = \sum (000, 001, 011, 100, 101, 110, 111) d = \sum(0, 2, 3, 5, 6) or d = \sum (000, 010, 011, 101, 110) e = \sum(0, 2, 6) or e = \sum (000, 010, 010) f = \sum(0, 4, 5, 6) or f = \sum (000, 100, 101, 110) g = \sum(2, 3, 4, 5, 6) or g = \sum (010, 011, 100, 101, 110) Next we minimize the logic for each segment a through g using K-maps: Now the logic diagram of figure 1 is synthesized using the simplified segment equations a through g.

Question:

As we have seen in the previous problem, character strings in Pascal have fixed length. However, we treat a string as having a varying length if we ignore all trailing blanks. Consider, for example, the following implementation: CONST STRSIZE = 80; TYPE STRING = RECORD WORD: PACKED ARRAY [1. .STRSIZE] of CHAR; length: 1..strsize END; where the field length represents the position of the last non trailing-blank character. For example, the string 'HELLObbb... 'haslength 5 and 75 trailing blanks, and the string 'GOODbbbBYEbbbb....' has length and 70 trailing blanks . Assuming the above defini-tions, write PROCEDURE SUBSTRING (S1:STRING;var S2:string;p,n: integer); which copies the first n characters, starting with posi-tion p in the string S1 into string S2. For example, if S1='ABCDEFGHIJbbbb.....', then after substring (S1, S2, 4,3 )S2 becomes 'DEFbbb....'.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0407.htm

Solution:

PROCEDURE SUBSTRING (S1: string;varS2: string, p, n : integer); VAR I, J: integer; BEGIN IF (P+N-1)>strsizethen Write1n ('ERROR-SUBSTRING is out of RANGE') {checkto see if desired substring is within the first one} ELSE BEGIN J: = 1; {initial position of the second string} FOR I: = P to P + N - 1 DO {start copying with position p} BEGIN S2.W0RD[ J]: = S1.W0RD[I] J: = J +1; {increment position count for 2string} END;{all required characters copied, pad the rest of the string S2 with blanks) S2.length: = J - 1; FOR I:=J tostrsizeDO S2.W0RD[I]: '' END END;

Question:

An ideal gas is contained within a volume of 2ft^3, when the pressure is 137 atmosphere and the temperature is 27\textdegree C. What volume would be occupied by this gas if it were allowed to expand to atmospheric pressure at a temperature of 50\textdegree C?

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0505.htm

Solution:

We may use the ideal gas law to analyze the behavior of the ideal gas. PV =nRT where P is the pressure of a container (of volume V) of gas at an absolute temperature T. n is the number of moles of the gas and R is the universal gas constant. Since we are dealing with a fixed mass of gas, we may write PV/T =nR= constant. Alternatively, (P_1V_1)/T_1 = (P_2V_2)/T_2 where (V_1,P_1,T_1) AND (V_2,P_2,T_2) are the conditions which describe the behavior of the gas before and after it expands, respectively. Since T_1 = 273\textdegree + 27\textdegree = 300\textdegreeK and T^2 = 273\textdegree + 50\textdegree = 323\textdegreeK then V_2 = V_1 (P_1/P_2)(T_2/T_1) = [2 × (137/1) × (323/300)ft^3] = 295ft^3.

Question:

What pressure is required to compress 5 liters of gas at 1 atm. pressure to 1 liter at a constant temperature?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0028.htm

Solution:

In solving this problem, one uses Boyle's Law: The volume of a given mass of gas at constant temperature varies inversely with the pressure. This means that, for a given gas, the pressure and the volume are proportional; at a constant temperature, and their product equals a constant. P × V = K where P is the pressure, V is the volume and K is a constant. From this one can propose the following equation P_1 V_1 = P_2 V_2 , where P_1 is the original pressure, V_1 is the original volume, P_2 is the new pressure and V_2 is the new volume. In this problem, one is asked to find the new pressure and is given the original pressure and volume and the new volume. P_1 V_1 = P_2 V_2P_1 = 1 atm. V_1 = 5 liters 1atm× 5 liters = P_2 × 1 literP_2 = ? [(1atm× 5 liters)/(1 liter)] = P_2V_2 = 1liter 5 atm. = P_2 .

Question:

A chemist decides to find the vapor pressure of water by the gas saturation method. 100 liters of N_2 gas is passed through 65.44 g of water. After passage of the gas, 63.13 g remained. The temperature of the H_2O (water) is 25\textdegreeC.Find the vapor pressure of water at this temperature.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0253.htm

Solution:

In the gas saturation method, a dry,unreactivegas such as nitrogen or air is bubbled through a specific amount of liquid maintained at constant temperature. After the gas has been bubbled away, the loss in weight of the liquid is determined. The weight loss is the number of grams of liquid in the vapor state. There exists an equation that relates the volume, pressure, weight loss, and molecular weight of the liquid. P = (g / MV)RT, where P = vapor pressure, g = grams of vapor, M = molecular weight of liquid, R = universal gas constant [.0821 (liter- atm )] / (mole \textdegreeK), V = volume, and T = temperature in Kelvin (Celsius plus 273\textdegree) The pressure will be expressed in mm, so that you use the conversion factor of (760 mm) / (atm)g = (65.44) - (63.13) = 2.31atm grams or the weight of the liquid in the vapor state. The molecular weight of water = (18.02 g / mole). Thus, P = (gRT)(MV) = [(2.31 grams) (.0821) {(literatm) / (mole \textdegreeK)} (298\textdegreeK) (760 mm / 1atm)] / [(18.02 g / mole)(200 liters)] = 23.8 mm = vapor pressure of H_2O.

Question:

What is wrong in the following array declaration? int my _ array (200);

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G17-0427.htm

Solution:

Arrays in C are created based on the four data types (int, char, float or double). The basic format of an array declaration is as follows: [number _ of _ subscripts]; The items appearing above i.e. array's __ type and variable _ name and number _ of _ subscripts have to be provided to declare an ar-ray. < > brackets are used only for identity purpose, but they are restricted in C (used only in defining directories). If we check the above statement, int my _ array (200); we note that individual items are as follows: int : array's _ type my _ array: variable _ name 200: number _ of _ subscripts. But the standard notation of enclosing number _ of _ sub-scripts in square brackets i.e. [], instead of which ( ) has been used, is missing. Therefore the correct declaration is: int my _ array [200];

Question:

The world's total reserve of coal and natural gas is estimated to be equivalent to 3.0 × 10^19 Kcal. The world's present consumption of these fuels is 4.2 × 102 7ergs/year. At this rate of consumption, how long will the fuel reserves last?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E29-0914.htm

Solution:

This problem is an example of the conversion between calories and ergs (1 cal = 4.2 × 10^7 ergs). The total reserve is 3.0 × 10^19 Kcal = 3.0 × 10^19 Kcal × 10^3 cal/Kcal = 3.0 × 10^22 cal, or,3.0 × 10^22 cal × 4.2 × 10^7 ergs/cal = 1.26 × 10^30 ergs. Dividing the total reserve by the rate of consumption gives (1.26 × 10^30 ergs) / (4.2 × 10^27 ergs/year) = 300 years. Hence, at the present rate of consumption, the world's supply of gas and coal will last 300 years.

Question:

For the gas-phase decomposition of acetaldehyde, CH_3CHO\rightarrow CH_4 + CO, the second-order rate constant changes from 0.105 m^-1 sec^-1 at 759\textdegreeK to 0.343 m^-1 sec^-1 at 791\textdegreeK. Calculate the activation energy that this corresponds to. What rate constant would you predict for 836\textdegreeK?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0471.htm

Solution:

The rate constant is related to the activation energy by the Arrhenius equation, which is stated k =Ae^-E/RT where k is the rate constant, A is a constant characteristic to the reaction, e is the base of natural logarithms, E is the activation energy, R is the gas constant (8.314 J/mole \textdegreeK) natural log: lnk=lnA- E/RT One is given k, R and T in the problem for two trials. For 759\textdegreeK:k = 0.105 m^-1 sec^-1 a)ln(0.105) =lnA - E/(8.314 J/mole \textdegreeK) (759\textdegreeK) For 791\textdegreeK:k = 0.343 m^-1 sec^-1 b)ln(0.343) =lnA - E/(8.314 J/mole \textdegreeK) (791\textdegreeK) Subtract equation b from equation a to solve for E. ln(0.105) =lnA- E/(8.314 J/mole\textdegreeK) (759\textdegreeK) -[ln(0.343) =lnA- E/(8.314 J/mole\textdegreeK)(791\textdegreeK)] C)ln(0.105) -ln(0..343) = - E/(8.314 J/mole\textdegreeK) (759\textdegreeK) + E/(8.314 J/mole\textdegreeK)(791\textdegreeK) -2.25 - (- 1.07) = - E/6310.33 J/mole + E/6576.37 J/mole (6310.33 J/mole)(6576.37 J/mole)(- 1.18) = E/6310.33 J/mole + E/6576.37 J/mole(6310.33J/mole)(6576.37 J/mole) - 4.8969 × 10^7 J^2/mole^2 = - 6576.37 E J/mole + 6310.33 E J/mole - 4.8969 × 10^7 J^2/mole^2 = - 2.6604 × 10^2 JE/mole E = (- 4.8969 × 10^7 J^2/mole^2 ) /(- 2.6604 × 10^2 J/mole ) E = 1.8407 × 10^5 J/mole One can solve for k at 836\textdegreeK by replacing the values of this third trial for the values found at 791\textdegreeK. ln(0.105) -lnk= -[(1.8407 × 10^5 J/mole) /{(8.314 J/mole\textdegreeK)(759\textdegreeK)}] + [(1.8407 × 10^5 J/mole) / {(8.314 J/mole\textdegreeK)(836\textdegreeK)}] - 2.25 -lnk= - 2.917 × 10^1 + 2.648 × 10^1 -lnk= - 4.4 × 10^-1 k = 1.55 m^-1 sec^-1.

Question:

Hydrogen peroxide, H_2 O_2, can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, H_2 (g) + O_2 (g) \rightarrow H_2 O_2 (l) . The second method involves oxidation of water: 2H_2 O (l) + O_2 (g) \rightarrow 2H_2 O_2 (l). Find the free energy of formation, ∆G\textdegree, for both processes and predict which process is more efficient for the com-mercial preparation of hydrogen peroxide.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0529.htm

Solution:

The standard free energy change for a reaction is equal to the difference between the standard free energy of formation of the products and the standard free energy of formation of the reactants, ∆G\textdegree = ∆G\textdegree _Products- ∆G\textdegree _reactants . The reaction with the more negative free energy change will be the more efficient, since it proceeds spontaneously as written. We will require the following standard free energies of formation: ∆G\textdegree_(H)2 (O)2 (l) = - 27.2 Kcal/mole ∆G\textdegree_(H)2 O (l)= 56.7 Kcal/mole ∆G\textdegree_(H)2 (g)= ∆G\textdegree_(O)2 (g) = O Kcal/mole For the first process H_2 (g) + O_2 (g) \rightarrow H_2 O_2 (l), the standard free energy change is ∆G\textdegree = ∆G\textdegree _products - ∆G\textdegree _reactants = ∆G\textdegree(H)2 (O)2(l) - [∆G\textdegree(H)2 (g)+ ∆G\textdegree(O)2 (g)] = - 27.2 Kcal/mole - (O Kcal/mole + O Kcal/mole) = - 27.2 Kcal/mole For the second process, 2H_2 O (l) + O_2 (g) \rightarrow 2H_2 O_2 (l), the standard free energy change is the standard free energy change is ∆G\textdegree = ∆G\textdegree _products - ∆G\textdegree _reactants = 2∆G\textdegree(H)2 (O)2 (l)- [2∆G\textdegree(H)2 O (l)+ ∆G\textdegree_(O)2 (g)] = 2 (-27.2 kcal/mole) - [2 × (-56.7) kcal/mole + O kcal/mole] = 59.0 Kcal/2 moles H_2 O_2 produced. = 29.5 Kcal/mole. Since the first process proceeds spontaneously as written (negative ∆G\textdegree) and the second process requires energy to proceed as written (positive ∆G\textdegree), the first method for preparing H_2 O_2 is more efficient than the second.

Question:

A motion, starting from rest, has the acceleration - time A motion, starting from rest, has the acceleration - time graph shown in figure (a) Draw the v - t graph and calculate the net displacement.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0067.htm

Solution:

Between t = 0 and t = 2 sec, a = 2 m/sec^2. Thus \Deltav = a\Deltat = 4 Between t = 0 and t = 2 sec, a = 2 m/sec^2. Thus \Deltav = a\Deltat = 4 m/sec. Thus at t = 2, v = 4 m/sec. Between t = 2 and t = 6, a = 3 m/sec^2; thus \Deltav = 3 × (6 - 2) = 12 m/sec. At t = 6, 4 + 12 =, 16 m/sec and so forth. Having found the velocities at various times and plotting the points as in figure (b), we can connect them with straight lines, since, as we can see from the acceleration - time graph, all accelerations are constant (therefore velocity is a linear function of t). Since displacement equals the product of velocity and time, the net displacement can be found by calculat-ing the area under the v - t curve until t = 10 sec. In figure (c), we break the area under the v-t curve into triangles and trapezoid. The total area under the curve is equal to the sum of the areas of these figures: d= area = (1/2)(4 × 2) + (1/2)(4 + 16) × 4 + 16 × 2 + (1/2)(16 + 8) × 2 = 4 + 40 + 32 + 24 = 100 m

Question:

Write a program to find the prime numbers between 1 and 50.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0411.htm

Solution:

The classical algorithm for enumerating prime numbers is the sieve of Eratosthenes. Suppose that the prime numbers less than 10 are to be found . The prime numbers less than 10 can be found by writing down the numbers from 2 to 10: 2345678910 Then the lowest number is to be removed claiming that it is prime. The multiples of 2 are also to be removed. After the first step, 2 is a prime, and the sieve contains odd numbers only. 3579. After the second step, 3 is a prime, and only 5 and 7 remain in the sieve. The process terminates when the sieve is empty. Declare CONST maximum = 100000 VAR Sieve: PACKED ARRAY [2. .maximum] ofboolean; Initially, each component of sieve is to be set to 'true\textasteriskcentered indicating that alltbe numbers are present. As the numbers are removed, the corresponding components are set to 'false'. The program consists of two nested loops, one to find the lowest number still in the sieve, and the other to remove its multiples . The termination condition for the outer loop is that there are no numbers left in the sieve, and it can be expedited by maintaining a count of the numbers currently in the sieve. The program is given below. PROGRAM prime numbers (input, output); CONST first prime = 2; maximum = 100000; VAR sieve : PACKED ARRAY[first prime. .maximum] ofboolean; Left in, range, factor, multiple: 0..maximum; BEGIN read (range); FOR factor: = first primeTorange DO sieve [factor]:=true; Left in:=range - first prime +1; factor :=first prime - 1; REPEAT factor : =factor + 1; IF sieve [factor] THEN {factor isprime} BEGIN write1n (factor); multiple: = 1; WHILE factor \textasteriskcentered multiple \leq range DO BEGIN IF sieve [factor \textasteriskcentered multiple] THEN {remove multiple} BEGIN sieve[factor \textasteriskcentered multiple]:= false; leftin:= left in - 1 END; Multiple: = multiple + 1 END {while} END UNTIL left in = 0 END. {primenumbers} Input: 50 Output: 241 3 543 7 1147 13 17 19 23 29 31 37

Question:

A coil having resistance and inductance is connected in series with an ac ammeter across a 100 volt dc line. The meter reads 1.1 am-peres. The combination is then connected across a 110 volt ac 60 cycle line and the meter reads .55 ampere. What are the resistance, the impedance, the reactance, and the inductance of the coil?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0767.htm

Solution:

Note that with dc by ohm's law, I = (E/R)R = (E/I) R = (110v/1.1A) = 100 ohms (Resistance) On ac, however the reactance of the inductance must also be taken into account (it is zero for dc current). Ohm's law still applies with the impedance taking the place of resistance. Therefore (E/I) = Z = (110/.55) = 200 ohms (Impedance) ButẒ = \surd(R^2 + X^2) where X is the impedance of the inductance. = 100^2 + X^2 = (200\Omega)^2 therefore X^2 =(40000 - 10000)\Omega^2 = 30000 \Omega^2 AndX = \surd(30000\Omega^2) = 173\Omega (Reactance) X = 2\pinL where n is the number of times per second the current alternates. For dc current, n = 0, and the inductance has no effect on the circuit. (The effect of the inductance on the behavior of the circuit increases appreciably as the frequency of the ac current increases). Therefore L = (X/2\pin) = {173\Omega/[2\pi(60)s^-1]} = .459 henry (inductance)

Question:

What is the function of the contractile vacuole?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0270.htm

Solution:

Unicellular and simplemulticellularorganisms lack special excretory structures for the eli-mination of nitrogenous wastes. In these organisms, wastes are simply excreted across the general cell membranes. Someprotozoansdo, however, have a special excretory organelle called the contractile vacuole. It appears that this organelle eliminates water from the cell but not nitrogenous wastes. These organelles are more common in fresh waterprotozoansthan in marine forms. This is because, in fresh water forms, the concentration of solutes is greater in the cytoplasm than in the sur-rounding water, and water passively flows into the cell through the cell membrane. The fresh water protozoan is said to live in hypotonic environment. This inflow of water would cause a fatal bloating if the water were not removed by the contractile vacuole. The vacuole swells and shrinks in a steady cycle, slowly ballooning as water collects in it, then rapidly contracting as it expels its contents, and then slowly ballooning again. The exact process of how the cell pumps water out of the vacuole is still unclear, but it is believed that the process is an energy consuming one.

Question:

What are van der Waals forces? What is their significance in biological systems?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0016.htm

Solution:

Van der Waals forces are the weak attractive forces that molecules of non-polar compounds have for one another. These are the forces that allow non-polar compounds to liquefy and/or solidify. These forces are based on the existence of momentary dipoles within molecules of non- polar compounds. A dipole is the separation of opposite charges (positive and negative). A non-polar compound's average distribution of charge is symmetrical, so there is no net dipole. But, electrons are not static, they are constantly moving about. Thus, at any instant in time a small dipole will probably exist. This momentary dipole will affect the distri-bution of charge in nearby non-polar molecules, inducing charges in them. This induction happens because the negative end of the temporary dipole will repel electrons and the positive end attracts electrons. Thus, the neighboring non-polar molecules will have oppositely oriented dipoles: These momentary, induced dipoles are constantly changing, short range forces. But, their net result is attraction between molecules. The attraction due to van der Waals forces steadily increases when two non-bonded atoms are brought closer together reaching its maximum when they are just touching. Every atom has a van der Waals radius. The atoms are said to be touching when the distance between their nuclei is equal to the sum of their van der Waals radii. If the two atoms are then forced closer together, van der Walls attraction is replaced by van der Waals repulsion (the repulsion of the positively charged nuclei). The atoms then try to restore the state in which the distance be-tween their two nuclei equals the sum of their van der Waals radii. Both attractive and repulsive van der Waals forces play important roles in many biological systems. It is these forces, acting between non- polar chains of phospholipids, which serve as the cement holding together the membranes of living cells.

Question:

Show that if k is the radius of gyration of a body, then Show that if k is the radius of gyration of a body, then k = \surd(I/m) where I is the body's rotational inertia about a given axis.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0181.htm

Solution:

The radius of gyration of a rigid body is defined as the radial distance from a given axis at which the entire mass of the body can be concentrated, without altering the object's moment of inertia about the axis. The definition implies 2 equivalent configurations (see figure), either of which may be used to calculate the body's moment of inertia about axis Z. From the discussion above, the moment of inertia of the body of mass m about axis Z in figure (a) is equivalent to the moment of inertia, about Z; of a mass particle m at a radial distance k from Z. (See figure (b)). Hence mk^2 = I ork = \surd(I/m)

Question:

What trends in the evolution of the life cycle are evident from the algae to flowering plants?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0208.htm

Solution:

In the evolution from the algae to the flowering plants there is, first, a change from a domi-nating population that is mostly haploid to one that is almost entirely diploid. Inmulticellularfilamentous algae such aUlothrix, only one cell during the life cycle, namely the zygote, is diploid; all the rest are haploid. In mosses, the haploid phase is more conspicu-ous and long-lived than the diploid phase, but the latter has evolved a complex, multicellularplant body. The relative importance of the two phases is reversed in the ferns: the diploid phase is the obvious, larger plant, and the haploid gametophyte, though still an independent plant, is small and inconspicuous. Thegymnosperrasand angiosperms show progressive reductions of the haploid generation. In angiosperms, the male gametophyte consists only of three cells and the female gametophyte of eight cells. The evolutionary advantage of a long-lived, dominant sporophyteand a short-lived, inconspicu-ous gametophyte generation is that a diploid individual can survive despite the presence of a deleterious recessive gene. A haploid individual would be more susceptible to the effects of such genes, since there would be no way to counterbalance the lethality of such an allele. Besides the reduction of the gametophyte, there is a gradual reduction in the dependence of fertilization on the presence of moisture. Aquatic plants such as algae have motile sperm that swim to the eggs. Bryophytes and lower vascular plants, which develop on the ground near water, also require a moist habitat. The seed plants have replaced sperm motility by pollination. Pollen may be very light and carried by the wind over large distances. The angiosperms have further evolved showy floral parts and nectar glands to attract insects, birds and other animals which serve to carry pollen on their feet or fur. Pollination has thus two evolutionary advantages: it allows sperm transport and fertilization to occur without moisture, and it enables long distance dispersal of sperm cells. Greater embryonic protection is a third evolutionary trend in the life cycle of plants. The green algaeChlamydomonashas its zygote protected only within a thick wall. Bryophytes and lower vascular plants have their embryos developing within themulticellulararchegoniumwhich draws nutrients from the gametophyte. Gymnosperms and angiosperms have evolved a seed structure for main-taining and protecting the embryo independent of the gametophyte. The seed has contributed immensely to the success of the seed plants. The stored food nourishes the embryo and the tough outer coat protects it from heat, cold, desiccation, drying, and parasites. Seeds also provide an effective means for the dispersal of the species. They may be wind-borne, water-borne, or carried by animals. To secure even greater protection, the angi-osperms have evolved fruit which encloses one or more seeds. Fruits may either decay or be eaten by animals, liberating the seeds.

Question:

What are the characteristics of ecological succession?

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/Users/wenhuchen/Documents/Crawler/Biology/F30-0778.htm

Solution:

Succession is a fairly orderly process of changes of communities in aregion. It involves replace-ment, in the course of time, of the dominant specieswithin a given area by other species. Communities succeed each otherin an orderly sequence in which each successive stage is thought to bedependent on the one which preceded it. Each community in the successionis called aseralstage. The first stage of succession is usually the colonization of barren spacesuch as a sandy beach or a rock, by simple pioneer species such asgrasses. The pioneer species are able to grow and breed rapidly. They haveadopted the strategy of finding and utilizing empty space. Then, graduallythe pioneers are replaced by more complex and bulky species untilfinally the community is characterized by climax species which representthe final stage of succession. Climax commu-nities are the most stableand will only if some outside agent such as new species or a geographicchange dis-places them. Most succession can be classified into eitherprimary or secondary successions. Primary successions proceed by pioneeringnew uninhabited sites. Secondary successions occur on disturbedland where a climax community that existed before had been destroyedby such occurrences as fire, severe wind storms, flooding and landslides. Although succession in different places and at different times is not identical(the species involved are often completely different) some ecologistshave nevertheless formulated generalizations that hold true for mostcases where bothautotrophsandheterotrophsare involved: 1) The species composition changes continuously during the succession, but the change is usually more rapid in the earlier stages than inthe later ones. 2) The total number of species represented increases initiallyand then becomes fairly stabilized in the older stages. This is particularlytrue of theheterotrophs, whose variety is usually much greater inthe later stages of the succession. 3) Both the total biomass in the ecosystemand the amount of nonliving organic matter increase during the successionuntil a more stable stage is reached. 4) The food webs becomemore complex, and the relations between species in them becomebetter defined. 5) Although the amount of new organic matter synthesizedby the producers remains approximately the same, except at thebeginning of suc-cession, the percentage utilized at the varioustrophic levelsrises. In summary, the trend of most successions is toward a more complexecosystem in which less energy is wasted and hence a greater biomass can be supported.

Question:

The normal boiling point of benzene is 80.10\textdegreeC. When 1 mole of a solute is dissolved in 1000 g of benzene, the boiling point of the resulting solution is 82.73\textdegreeC. When 1.2 g of elemental sulfur is dissolved in 50 g of benzene, the boiling point of the solution is 80.36\textdegreeC. What is the molecular weight of sulfur?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0249.htm

Solution:

In order to determine the molecular weight of sulfur, we must determine how many moles of sulfur corres-pond to 1.2 g. When 1 mole of solute is added to 1000 g of benzene, the normal boiling point is raised by 2.63\textdegreeC, from 80.10\textdegreeC to 82.73\textdegreeC. This corresponds to a concentration of 1molal(1 mole solute / 1000 g solvent = 1molal). The ratio of sulfur to benzene in the sulfur-benzene solution is (1.2 g sulfur / 50g)benzene = 0.024. Thus, adding 1.2 g of sulfur to 50 g of benzene is equivalent to adding 24 g of sulfur to 1000 g of benzene (because 24 g sulfur / 1000 g benzene is also 0.024). The observed boiling point elevation in the sulfur- benzene solution is 0.26\textdegreeC (from 80.10\textdegreeC to 80.36\textdegreeC).This is the boiling point elevation one would obtain by adding 24 g of sulfur to 1000 g of benzene. But this is about one-tenth the rise one would obtain by adding 1 mole of sulfur (the rise would then be 2.63\textdegreeC) , hence 24 g of sulfur must correspond to about one-tenth of a mole. Thus, the apparent molecular weight of sulfur is (24 g / 0.1 mole) = 240 (g / mole).

Question:

The density of diamond is 3.5 gm/cm^3. What is the order of magnitude of the average velocity and average energy of a carbon atom in diamond at absolute zero?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0987.htm

Solution:

Density is defined as \varphi = M/V where M is the mass in a volume V. Solving for M M =\varphiV The number of carbon atoms, n, in this mass is M divided by the atomic mass of carbon n = [M/(12amu)] = [\varphiV/(12amu)] Hence the volume occupied by 1 carbon atom is (V/n) = [(12amu)/\varphi] (V/n) = [[(12amu) (1.66 × 10^-27 kg/amu)}/(3.5 kg/m^3)] (V/n) = [(19.8 × 10^-27 m3)/3.5] (V/n) = 5.7 × 10^-27 m^3 Assuming that each carbon atom occupies a cubical volume, the length of one side of this cube, l will be such that l^3 = 5.7 × 10^-27 m^3 l = 1.8 × 10^-10 m The mass of a carbon atom, m, is m = (12amu)(1.66 × 10^-27 kg/amu) m \approx 2 × 10^-26 kg We may now use Heisenberg's uncertainty relationship to find the aver-age velocity of the carbon atom. Since the atom is assumed to be in a cubical volume of length l, the uncertainty in its position is \Deltax= l Hence, using the uncertainty relationship (where\Deltapis the uncertainty in momentumhand is Planck's constant divided by 2\pi) \Deltap\Deltax\geqh \Deltap\geq (h/l) = [(6.63 × 10^-34 J \textbullet s)/{(2\pi) (1.8 × 10^-10 m)}] \Deltap\geq[(6.63 × 10^-34 N \textbullet m \textbullet s)/(11.3 × 10^-10 m)] \Deltap\geq .586 × 10^-24 N \textbullet s \Deltap\geq 5.86 × 10^-25 N \textbullet s The minimum uncertainty in p is then \Deltap_min= 5.86 × 10^-25 N \textbullets But\Deltap_min=m\Deltav_min' since m is constant. Hence \Deltav_min= (\Deltap_min/m) = [(5.86 × 10^-25 N \textbullets)/(2 × 10^-26 kg)] \Deltav_min= 2.93 × 10^1 m/s \Deltav_min= 29.3 m/s This quantity gives us an approximate value for the velocity of the carbon atom. The average energy is then E = (1/2)mv^2 = (1/2) (2 × 10^-26 kg) (29.3 m/s)^2 = (10^-26 kg) (858.5 m^2/s^2) = 8.59 × 10^-24 Joules The zero point energy is the least energy a system can have and occurs at absolute zero. It is E_0 = (1/2)hf and f = (2E_0/h). Assuming that the magnitude of E_0 is the E calculated above, f = [{(2) (8.59 × 10^-24 Joules)} / (6.63 × 10^-34 Joules-sec)] f = [(17.18 × 10^-24) / (6.63 × 10^-34)] s^-1 f = 2.6 × 10^10 s^-1

Question:

What is the radius of the cyclotron orbit in a field of magnetic induction of 1weber/m^2 for an electron traveling with velocity 10^6 m/s in a direction normal to B^\ding{217}? (This radius is termed thegyroradius).

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0716.htm

Solution:

Before starting this problem, we must recognize what type of motion the electron is executing. Since the electron is a charged particle traveling perpendicular to a uniform magnetic field (a field having the same value over all space), the particle will travel in a circle. To find the exact radius of this circular motion, we relate the magnetic force acting on the electron to the electron's acceleration via Newton's Second Law, F^\ding{217} = ma^\ding{217}. The magnetic force for a particle of charge q traveling perpendicular to a magnetic field with velocity v^\ding{217} is F =qvB(1) Using Newton's Second Law, qvB = ma(2) Then because the motion is circular, we know that the acceleration the electron experiences is centri-petal (that is, it always points to the center of thecircle) and has the value (v^2 /R) , where v is the speed of the electron, and R is the radius of its orbit. Substituting this into equation (2), we find qvB = (mv^2 /R)(3) Solving for R, we obtain R = mv /qB(4) Now, the charge of an electron is q = 1.6 × 10^-19 Coulombs, and its mass is m = 9.11 x 10^-31 kilograms. Substituting this information and the values of v and B given in the statement of the problem into (4) R = [{(9.11 × 10^-31 kg) (10^6 m/s)}/{(1.6 × 10^-19 C) (1 w/m^2 )}] R = 5.7 × 10^-6 m

Question:

The pendulum in figure 1 consists of a homogeneous prismatic bar of mass m and length l, which hangs at rest on its frictionless suspension O. A spherical pellet of mass m travels with a horizontal velocity v_1 and strikes it in the middle. Immediately after the impact, the pellet moves with velocity v_2 to the right and the pendulum begins to rotate about O with initial angular velocity \omega. Solve for v_2 and \omega in terms of v_1 for the cases in which the collision is: 1.) completely elastic, and 2.) completely inelastic.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1072.htm

Solution:

Since not only the normal force, but also the reaction at O, becomes large during the impact, the principle of impulsive forces is useless for the determination of v_2 and \omega. However, here we can apply the principle of impul-sive moments about O. The angular momentum about O of the system of two bodies before the impulse originates from the pellet alone. It is normal to the plane of motion and has the magnitude. H_(0)1 = m(l / 2)v_1 .(1) The total angular momentum immediately after the impulse has the same direction and the magnitude H_(0)2 = m(l / 2)v_2 + (ml^2 / 3)\omega.(2) The second term in equation (2) is the angular momen-tum of the bar, I\omega, where I is the moment of inertia of the bar about O, which is equal to ml^2 / 3. Since no external forces that have a static moment about O become large during the impulse, we have H(0)1= H(0)2or substituting the values from (1) and (2), m(l / 2)v_2 + (ml^2 / 3)\omega = m(l / 2)v_1 .(3)- If the impulse is elastic the kinetic energy of the system is conserved. Thus we have (m/2)^v2 _2 + (1/2) (ml^2 / 3)\omega^2 = (m/2)v^2 _1.(4) Hence, by (3) and (4) we have the following equations for the determination of v_2 and \omega: v_2 + (2 / 3)l\omega= v_1,v2_2 + (1 / 3)l^2\omega^2 = v2_1 . In addition, we have the requirement that (l / 2)\omega \geq v_2. The solution yields \omega = (12 / 7)(v_1 / l) and v_2 = \rule{1em}{1pt}(1 / 7) v_1. Thus, both bodies move after the impulse, and in such a way that the speed of the pellet relative to the pendulum if v^1_2 = (l / 2)\omega \rule{1em}{1pt} v_2 = v_1 to the left. The velocity of the pellet relative to the pendulum is reversed. If the pellet remains imbedded in the pendulum, the impulse is completely inelastic. Then, instead of (4) , we have the condition v_2 = (l / 2)\omega.(5) and, from (3) and (4) it follows that \omega = (6 / 7)(v_1 / l) and v_2 = (3 / 7)v_1.

Question:

H_2(g) + 2AgCl (s) + 2H_2O \rightarrow 2Ag(s) + 2H_3O^+(Ag) + 2Cl^-(Ag). It has been determined that at 25\textdegreeC the standard free energy of forma-tion of AgCl(s) is -109.7 kJ/mole, that of H_2O(l) is -237.2 kJ / mole, and that of (H_3O^+ + Cl^-) (ag) is - 368.4 KJ / mole. Now assuming the reaction is run at 25\textdegreec and 1 atm of pressure in a cell in which the H_2(g) activity is unity and H_3O+(ag) and Cl^-(ag) activities are .01,what will be the cell voltage?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0603.htm

Solution:

The answer to this question requires the use of the Nernst equation and the relationship between overall free energy change (∆G) and the cell voltage E. The Nernst equation implies that the voltage of a cell depends on the concentrations or activity of the species involved. It states E = E\textdegree - (RT / nF) 1n Q , where E is the potential for a reaction at nonstandard conditions, E\textdegree = the standard potential, n = number of faradays of electricity, F = the value of a faraday, (for conversion to kilo joules per mole is 96.49 kJ V^-1 equiv^-1 ), T = temperature in kelvin (celsius plus 273\textdegree), R = universal gas constant, (8.31 J mol^-1 deg^-1), and Q is the mass- action expression. The relationship between the overall free energy and cell voltage is ∆G\textdegree = -nFE , where ∆G is the free energy. If all the species are in their standard states, then you can say ∆G\textdegree = -nFE\textdegree . The sum of ∆G\textdegree must be the free energy contribution from each component. Therefore, ∆G\textdegree = 2∆G\textdegreeH3O+ Cl-^ -2∆G\textdegreeAgCl-2∆G\textdegree_H2O = 2(-368.4) - 2(-109.7) - 2(-237.2)kJ = -43.0 kJ per mole of reaction. E\textdegree = (- ∆G\textdegree) / nF = -[(-43.0k J) / {2(96.49kJ / v)}] = .223v. Since the activities of solid Ag, of solid AgCl and of H_2O are unity, they can be eliminated from the mass-action expression: {[H_3O^+]^2 [Cl^-]^2 } / [H_2] . Thus, E = E\textdegree -[RT / nF] ln({[H_3O^+]^2 [Cl^-]^2 } / [H_2]) . Therefore, E = .223 -[{(8.31)(298)(2.303)} / {2(96490)}] log [{(.0100)^2(.0100)^2} / (1.00)] E = .223 + .236 = .459v.

Question:

For the first order decomposition of azomethane at 600\textdegreeK, it takes30 minutes for the original concentration to decrease tohalf its value. After 60.0 minutes have elapsed, what percentageof the azomethane originally present remains?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0461.htm

Solution:

The decomposition is first order which means rate = k [A], where k isa constant and [A] is the con-centration of azomethane. The half-life is independentof time, i.e., it is a constant and is defined as the time necessaryfor half the particular reactant present to decompose. Thus, if youknew the half-life of azomethane, you could calculate the percentage ofazomethane that remains after a certain period of time. The half-life of azomethaneis 30 minutes. Thus, after 30 minutes, you have half of the originalmaterial left. After 60 minutes half of this original material is left or 1/4 of the original material remains.

Question:

Distinguish between copulation, insemination and fertilization.

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Solution:

Copulation is the act of physical joining of two sex partners. In copulation, accessory sex organs play an important role and synchronized hormonal control is often extensively involved. Insemination is the actual process of depositing semen into the vagina. This occurs during copulation or sexual intercourse, but could also be accomplished by means of artificial injection. When copulation and insemination are complete, fertilization can occur. Fertilization is the union of the male and female gametes to produce the zygote. It should be noted that successful copulation and insemination do not guarantee that fertilization will occur. This may be due simply to the fact that the female has not ovulated and thus provided a mature ovum at the time of insemination. However, it is possible that an ovum may be present, and yet not be fertilized. For fertilization to occur, sperm must not only survive and reach the ovum, they must also succeed in penetrating it. In addition, if one or both partners are sterile, fertilization cannot take place. Sterility refers to any number of specific conditions which result in the failure to produce viable gametes. For example, sterile males may produce viable sperm, but in such low amounts that sufficient numbers do not survive in the female to insure fertilization. Sterile females may produce viable eggs but have a blockage preventing their passage from the ovaries to the oviducts or uterus. Some may produce sperm antibodies in their cervical secretions which resist sperm passage.

Question:

Calculate the kinetic energy of an automobile weighing 2000 pounds and moving at a speed of 50 miles per hour. Record your answer in (a) Joules and (b) ergs.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E29-0915.htm

Solution:

Kinetic energy is equal to 1/2 mV^2, where m is the mass of the object and V is its velocity. One is given that the mass of the car is 2000 lbs and that its velocity is 50 miles/hr. Solving for the kinetic energy in lbs miles^2/hr^2: kinetic energy = (1/2) (2000 lb) (50 mile/hr)^2 = (1/2) (2000 lb) (2500 mile^2/hr^2) = 2.5 × 10^6 lb mile^2/hr^2 (a) 1 erg = 1 g cm^2/sec^2. Thus, the units of the kinetic energy found must be converted. Pounds must be converted to grams, miles^2 to cm^2 and hours^2 to sec^2. This is done by the use of conversion factors. There are 454 g in 1 lb, thus to convert pounds to grams one must multiply the number pounds by 454 g/lb. There are 1.0 × 10^5 cm in 0.6214 mi, squaring these two quantities, one finds that there are 1.0 × 10^10 cm^2 in .386 mi^2. The conversion factor from mi^2 to cm^2 is then 1.0 × 10^10 cm^2/ .386 mi^2 . There are 3600 seconds in an hour, therefore there are (3600 sec)^2 in 1 hour^2 or 1.296 × 10^7 sec^2 per 1 hr^2. The appropriate conversion factor from hr^2 to sec^2 is (1 hr^2) / (1.296 × 10^7 sec^2 ) Solving for ergs: [2.5 × 10^6 (lb mi^2 / hr^2)] [454 (g/lb)] [(1.0 × 10^10 cm^2 ) / (.386 mi^2)] [(1 hr^2) / (1.296 × 10^7 sec^2 )] = 2.27 ×10^12 (g cm^2 / sec^2 ) = 2.27 × 10^12 ergs. (b) 1 Joule = 10^7 ergs, thus the number of Joules is equal to the number of ergs divided by 10^7. The conversion factor used when converting ergs to Joules is 1 Joule/10^7 ergs. no. of Joules = 2.27 × 10^12 ergs × [(1 Joule) / (10^7 ergs)] = 2.27 × 10^5 Joules.

Question:

A camera photographs a 120-W sodium lamp 100 m away, thefilm being exposed for 1/100s. The aperture of the camerahas a diameter of 4cm, How many photons enter the cameraduring the exposure? The wavelength of sodium lightis 5893 \AA, and all the energy of the lamp is assumed to beemitted as light.

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Solution:

The energy possessed by one photon of sodium light is E =hf, whereh is Planck's constant and f is the photon frequency. But c =\lambdaf where\lambda is the wavelength of light and c is the speed of light. Therefore, hf= h(c/\lambda) = 6.625 × 10^-34 J\textbullets × [(2.998 × 10^8 m\textbullets^-1)/(5893 × 10^-10 m)] = 3.37 × 10^-19 J. The lamp consumes 120 J of energy per second. Therefore the number of photonsemitted per second is n = [(120 J \textbullet s^-1)/(3.37 × 10^-19 J)] = 3.55 × 10^20 s^-1. These photons are emitted in all directions and 100 m from the lamp are distributedevenly over a spherical surface of that radius. The number N enteringthe aperture of the camera per second is thus n multiplied by the fractionof this surface occupied by the aperture. That is, N =n[\pir^2/4\piR^2] whereR is the distance from the lens to the lamp, and r is the aperture radius. Then N = 3.55 × 10^20 s^-1 × [{\pi × (0.02)^2m^2}/{4\pi × (100)^2m^2}] = 3.55 × 10^12 s-1. . Thus in 1/100 s the number of photons entering the camera is 3.55 × 10^10.

Question:

A chemistfinds that an unknown compound contains 50.05 % S and 49.95 % O by weight. Calculate its simplest formula.

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Solution:

To calculate a compound's simplest formula, you need the relative number of moles of atoms in the compound. The percentages by weight of the elements allow for this calculation. If you had 100 g of the unknown compound, it would consist of 50.05 g of S and 49.95 g of O. A mole is de-fined as weight in grams/atomic weight. Therefore, the number of moles of sulfur (s) = (50.05 g) / (32.06 g/mole)= 1.561 moles . The number of moles of oxygen = (49.66) / (15.999 g/mole) = 3.122 moles. You see that the ratio of S to O is 1.561 / 3.122 = 1 / 2 ; 1 ; 2. Therefore, the simplest formula can be expressed as SO_2.

Question:

Why are organ transplants generally unsuccessful?

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Solution:

Under normal circumstances, if a piece of skin, or some organ such as the heart or kidney, is trans-planted from one person to another, it initially appears to be accepted. But within two weeks, the graft (trans planted tissue) is invaded by white blood cells and the tissue dies. This graft rejection is an immunological reaction, also known as a primary set reaction. Second grafts from the same donor are rejected within a few days. This accelerated rejection is termed the second set reaction and can be explained on the basis of theclonalselection theory. Althoughhumoral(circulating) antibodies may be involved in the rejection of transplanted tissues and organs, the active agents are intact small lymphocytes.Immunoglobulinson the surface of these lymphocytes form a complex with the surface antigens on the foreign tissue cells. These lymphocytes specifically destroy these cells by secreting toxic substances or by stimulating increasedphagocyticactivity of macrophages. This immunological response is a reflection of the ability of the body's cells (self) to recognize and destroy anything that is foreign (unself). The rejection of tissue from another member of a different species is termed a heterograft reaction.Heterograftsare usually unsuccess-ful. When the donor is a member of the same species, the transplantation, called a homograft, is also usually un-successful. But when the donor and recipient are either identical twins or members of the same purebred strain, the transplantation is usually successful. These trans plants, calledisografts, are successful because the donor and recipient are genetically very similar. The genes that determine the antigens responsible for the trans-plantation immunological response are termed histocompati-bility genes. The greater the number ofhistocompatibility genes that the donor and recipient, have in common, the greater the probability that the transplant will be successful. That is whyautografts, or tissue transplants from one location to another on the same individual, are usually successful. The success ofhomograftsdepend on how well the donor's and recipient's tissues are matched, that is, how well the donor and recipient resemble each other genetically. Sometimes certain tissues of an individual's own body may be rejected, resulting in disease. The immuno-logical mechanism for distinguishing between self and non-self becomes disrupted and the body begins to destroy itself. These autoimmune diseases include rheumatoid arthritis, rheumatic fever and multiple sclerosis.

Question:

An electric kettle contains 2 liters of water which it heats from 20\textdegreeC to boiling point in 5 min. The supply voltage is 200 V and a kWh (kilowatt-hour) unit costs 2 cents. Calculate (a) the power consumed (assume that heat losses are negligible),(b) the cost of using the kettle under these conditions six times,(c) the re-sistance of the heating element, and (d) the current in the element.

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Solution:

The heat gained by the water in being raised to the boiling point is given by the expression H = mc(t_2 - t_1) where c is the specific heat of water (the amount of heat required to raise the temperature of the substance 1\textdegreeC), m is the mass of the water being heated and (t_2 - t_1) is the temperature difference before and after heating. (a) H = 2 × 10^3 cm^3 × 1 g \bullet cm^-3 × 4.18J \bullet g^-1 \bullet C deg^-1 × (100 - 20)C deg = 6.69 × 10^5 J . Since heat losses are neglected, the conservation of energy requires that the heat energy generated be equal to the electrical energy consumed by the kettle. Thus the electric energy E = 6.69 × 10^5 J. The power is the energy consumed per second, which is thus P = (H/\tau) = [(6.69 × 10^5 J)/(5 × 60 s)] = 2.23 x 10^3 J \bullet s^-1 = 2.23 kW.[for IW = 1 J \bullet s^-1 .] (b)The kettle uses 2.23 kW for 5 min each time the water is boiled. When it is used six times, 2.23 kW is used for 30 min = (1/2) hr. The cost is thus 2.23 kW × (1/2) hr × 2 cents \textbullet kWh^-1 = 2.23 cents. (c) The power P consumed is 2.23 kW and the supply voltage V is 200 V. But P = V^2/R, where R is the resistance of the kettle's heating element. R = (V^2 /P) = [(200^2 V^2 )/(2.23 × 10^3 W)] = 17.9\Omega. \Omega (d)But one may also write the power as P = IV, where I is the current through the heating element. I = (P/V) = [(2.23 × 10^3 W)/(200 V)] = 11.2 A .

Question:

Identify the minerals most essential to the body and give the function of each.

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Solution:

About fifteen minerals are known to be essen-tial in the human diet. A few of these are required in minute quantities and are referred to as trace elements. The essential minerals include sodium, chlorine, potassium, magnesium, phosphorus, calcium, iron, copper, manganese and iodine. Sodium chloride is a major salt in blood and other body fluids. Sodium and chloride ions play an important role in maintaining osmotic balance and acid-base balance in the body fluid. Secretions of the digestive tract such as the hydrochloric acid of the stomach and the pancreatic and intestinal juices, contain a large amount of sodium and chloride ions. Potassium and magnesium are required for muscle contraction and for the functioning of many enzymes. Calcium and phosphorus form the major components of bones and teeth, and are thus essential for maintenance of locomotion and support of the body. Phosphorus is also im-portant in the synthesis of DNA, RNA, NAD+ (NADH) , NADP+ (NADPH) and ATP. Moreover,phosphorylation(the addition of phosphate) of sugars and glycerol must take place before they can becatabolizedto produce biologically useful energy. Trace elements such as iodine, iron, copper and zinc generally function as metal or non-organic components of specific enzyme systems. Iodine is a constituent ofthyroxine, the hormone secreted by the thyroid gland. Hemoglobin and thecytochromescontain iron. Small amounts of copper are necessary in the diet to facilitate proper utilization of iron for normal growth and as a component of certain enzymes. Traces of manganese, molybdenum, zinc and cobalt are required for normal growth and as activators of certain enzymes. Fluorine, present in trace amounts in drinking water, is effective in preventing dental decay.

Question:

What are the functions of the xylem and the phloem?

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Solution:

In vascular plants, the vital function of the transportation of food, water and minerals is performed by the vascular system, composed of the xylem and the phloem. The xylem is chiefly concerned with the conduction of water and mineral salts from the roots to the above-ground portion of the plant, where they are used for photosynthesis and other metabolic purposes. The xylem also serves as a means of support in larger vascular plants. It contains several specialized cell types, the two major kinds being the tracheids and vessel elements. Both are elongated cells with thickened cellulose walls heavily impregnated with lignin; both contain no living protoplasm, and hence are dead cells. The tracheids (see Figure 1) have tapering ends and lateral pores, or pits, which al-low water to flow between adjacent cells. Although structurally similar, vessel elements are a more efficient system for conduction than tracheids. At maturity, the end of the vessels are broken down to form a perforation area, so that each individual cell is continuous with the cell above and the cell below. Vessel cells are thought to have evolved from tracheids, having increasing degrees of perforation with evolutionary advancement, (see Figure 2) Indeed, vessel elements are characteristic of the flowering plants and do not occur at all in most of the less advanced gymnosperms. Vessel elements have lateral pits but movement of materials is chiefly through their ends. Carbohydrates are manufactured primarily in the leaves of a plant. They are transported to the other parts of the plant by the phloem, which runs parallel to the xylem throughout the plant body. The phloem consists of elongated, tube-like cells, called sieve-tube elements, with specialized pores at each end (See Figure 3). Unlike the tracheids and vessels, mature sieve-tube elements are living and have a protoplasm, but no nuclei. The sieve-tube-elements are associated with small, narrow cells having dense contents and prominent nuclei, knowns as the companion cells. These cells probably function as the `nuclei' for the sieve - tube elements and make it possible for them to continue to function. Besides conduction, the phloem serves as a supporting tissue, due to the presence of strong fibers in the walls of its cells.

Question:

A Martian performs an experiment to determine the Martian g with the local type of Atwood's machine (see figure). He hangs two equal weights of mass 0.02 slug over a frictionless pulley and adds a rider of mass 0.002 slug to one side. When the heavier side has descended 2 ft the rider is removed and the system travels 4 ft in the next 3.5s. What value does he obtain for g?

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Solution:

Initially, there are equal masses on both sides of the Atwood's machine (see figure). The apparatus is then in equilibrium. When we add the rider mass (m), the system is no longer in equilibrium, and each mass accelerates, as shown In the figure. After the heavier side drops 2 ft., the system Is moving with a velocity v^\ding{217}, and then we remove the rider. By Newton's First Law, the left side must continue to move with a velocity v\ding{217} since the system is no longer accelerating. As will be shown below, the acceleration of the system is related to the Martian value of g. If we know the velocity attained by the left hand mass at the instant the rider is removed, we can find a numerical value for g. To find the acceleration of the system, we apply Newton's Second Law to each separate mass shown in the figure. Therefore, taking the acceleration of the left hand mass as positive downward, we obtain (M + m)g - T = (M + m)a_1 where a_1 is the acceleration of the left hand mass. Similarly, taking the acceleration (a_2) of the right hand mass as positive upward, we find T - Mg = Ma_2 But a_1 = a_2, since the 2 masses are connected by a string, and, therefore accelerate as a unit, whence (M + m)g - T = a(M + m)(1) T - Mg = Ma(2) To find a, we eliminate T by adding (1) and (2) mg = (2M+m)a whencea = mg/2M + m(3) Since a is constant, we know that a is related to the position (y) and velocity (v) of the left hand mass by v^2 = v2_0 + 2a y - y_0 where y_0 and v_0 are the initial position and velocity of this mass. Since the left hand mass is initially at rest, v_0 =0. v^2 = 2a(y - y_0) Using (3)v^2 = [{(2mg) / (m + 2M)}] {y - y_0} Hence[{(m + 2 M)v^2} / {2m(y - y_0)}] = g and, to find g, we must know v. Note that y - y_0 is the distance traversed during acceleration. But, we observe that the left side must move with constant velocity after the rider is removed (m = 0) since, by (3), a = 0. Since this mass moves 4 ft in 3.5s after the rider is removed, and its velocity is constant, we may write v = (4 ft) / (3.5 s) = (8/7)ft/s Therefore,g = [{(.042 sl)(64/49 ft^2/s^2)} / {(2)(.002 sl)(2 ft)}] g = 6.86 ft/s^2

Question:

If 2.02 g of hydrogen gas occupies 22.4 liters, calculate the density of hydrogen gas in grams per milliliter.

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Solution:

Density = (weight / volume) = g/ml. One is given the weight as 2.02 g but the volume is given in liters. Therefore, before calculating the density in g/ml,, one must convert liters to milliliters. 1 liter = 1,000 ml; therefore, 22.4 l = 22,400 ml. Solving for the density: density = (2.02 g / 22,400 ml) = 9.0 × 10 ^\Elzbar5g/ml.

Question:

A Rowland ring made of iron has a mean circumferential length of 50 cm and a cross-sectional area of 4 cm^2 . It is wound with 450 turns of wire which carry a current of 1.2 A. The relative permeability of iron under these conditions is 550. What is the magnetic flux through the ring? What would be the flux through the ring if a gap of 2 cm were to be cut in its length, assuming that the flux did not spread from the gap?

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Solution:

Assume that the iron the Rowland ring is operated in the linear region. Then the magnetic induction through a Rowland ring (see fig. A) is given by the formula B = \muH Applying Ampere's circuital law about the circumference (of length l) of the Rowland ring, \int H^\ding{217} \bullet dl^\ding{217} = NI_C H^\ding{217} is constant along the circumference and is parallel to dl^\ding{217}, an infinitesimal element of length. N is the number of windings of current- carrying wire around the ring. The H \int dl = Hl = NI_C H = (NI_C /l) Hence B = \mu (NI_C /l) = K_m \mu_0 (NI_C /l) Where the relative permeability of iron K_m = \mu/\mu_0 . The flux through the ring is \varphi = \int B^\ding{217} \bullet dA^\ding{217} = BA = K_m \mu_0 A (NI_C /l)(1) where A is the cross-sectional area of the ring Since we are given A = 4 cm^2 , N = 450, l = 50 cm = 0.5 m, I_C = 1.2 A and K_m = 550, then \varphi = 550 × 4\pi × 10^-7 N \bullet A^-2 × (4 × 10^-4 m^2 ) × [(450 × 1.2 A)/(0.5 m)] = 2.99 × 10^-4 Wb. Equation (1) may be written as \varphi = [(NI_C )/(l/\muA)] = ( M/R_1 ) where M is the magnetomotive force, and R the reluctance of the ring. This is the magnetic analogue of Ohm's law, with \varphi the analogue of current, M the analogue of emf and R the analogue of resistance (see figure A). When a gap is cut in the ring (see fig. B), the flux may be obtained by use of the magnetic circuit relation \varphi = [M/(R_1 + R_2)], where R_1 and R_2 are the reluctance of ring and gap, respectively (see fig. B), Thus \varphi = [(NI_C )/[(l_1 /\mu_1 A_1 ) + (l_2 / \mu_2 A_2 )] l_1 is the length of the ring minus the gap (= 0.5 cm - .02m = 0.48 m), \mu_1 the permeability of iron ( = 550 \mu_0) , A_1 the cross sectional area of the ring, l_2 the length of the gap (= 0.02 m), \mu_2 = \mu_0 and A_2 = A_1 . Thus \varphi = (450 × 1.2 A) / [{0.48 m/(550 × 4\pi × 10^-7 N \bullet A^-2 × 4 × 10^-4 m^2)} + {0.02 m/(4\pi × 10^-7 N \bullet A^-2 × 4 × 10^-4 m^2)}] = [(450 × 1.2 × 4\pi × 10^-7 × 4 × 10^-4 )/[(0.48/550) + 0.02)] = 1.30 × 10^-5 Wb.

Question:

Use the laws of Boolean algebra to simplify the following Boolean expression: [(A+B) + (A+C) + (A+D)] \bullet [A \bullet \textasciitilde B] \textbullet (1) + is the inclusive OR, \bullet is AND and N is NOT. Show the switching and gate representations of the resulting expression.

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Solution:

One suggested plan of attack is to eliminate redundancy in the expression. Since A appears four times, we try to simplify the terms using A. The law that allows each step is given on the right: [(A+B) + (A+C) + (A+D)] \textbullet [A \textbullet \backsimB] = [(B+A) + (A+C) + (A+D)] \textbullet [A \bullet \backsimB] Comm. of + = [B + (A+A) + C + (A+D) ] \bullet [A \bullet \backsimB] Assoc. of + = [B + A + C + (A+D) ] \textbullet [A \textbullet \textasciitildeB] Idempotent law = [B + C + A + (A+D) ] \bullet [A \textbullet \textasciitildeB] Comm. of + = [B + C + (A+A) + D] \textbullet [A \bullet \textasciitildeB] Assoc. of + = [B + C + A + D] \textbullet [A \textbullet \textasciitilde B] Idempotent law(1) Working on B and \textasciitilde B, (1)= [B + C + A + D] \textbullet [\textasciitilde B \textbullet A] Commutativity of \bullet = [(B + C + A + D) \textbullet \textasciitilde B] \bullet A Associativity of \textbullet = [(B \bullet \textasciitilde B) + (C \textbullet \textasciitilde B) + (A \bullet \backsim B) + (D \bullet \backsim B) \bullet A Distributive law of \bullet over + = [0 + (C \textbullet \textasciitilde B) + (A \textbullet \textasciitilde B) \textbullet (D \bullet \backsim B) \bullet A Law of complement = [(C \textbullet \textasciitilde B) + (A \textbullet \backsim B) + (D \textbullet \textasciitilde B)] \bullet A Identity of \textbullet = [(C + A + D) \textbullet \textasciitilde B) \textbullet A Dist. law of \textbullet over +(2) The switching representation of (2) would be as shown in Fig. 1 Note that the circuit conducts if and only if both A andare closed, regardless of the settings of C and D. The gate representation of (2) would be as shown in Fig. 2. The reduction of statement (1) to statement (2) was tedious but straightforward. It allows an eight-switch circuit to be replaced by a five-switch circuit, or a seven-gate circuit to be replaced by a three-gate circuit. However, the reduction method depended on observation and a "feel" for algebraic manipulation. Fortunately, minimization algorithms exist which do not depend on such subjective methods and can, in fact, be implemented on a computer.

Question:

Write the micro-operations for the indirect cycle of the computer whose registers and control signals are specified as follows: M - memory PC - program counter MAR - memory address register MBR - memory buffer register OPR - op-code register F } - cycle identification registers R } I - indirect register M - memory The cycle code is shown in Figure 1. Cycle ID Registers FR Cyclecode Cycledescription 00 C_0 Fetch cycle 01 C_1 Indirect cycle 10 C_2 Execute cycle 11 C_3 Interrupt cycle Fig. 1 Each cycle consists of four steps T_0, T_1, T_2, T_3.

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Solution:

The difference between a direct memory reference instruction and an indirect memory reference in-struction is explained with the use of Figure 2. In a direct memory-reference instruction the address por-tion of the instruction format (bits 4 through 15) con-tains the address of the operand; hence, when the fetch cycle is completed, bits 4 through 15 of the MBR contain the address of the operand. The operand is not taken from memory until the execution cycle. In an indirect memory-reference location the address por-tion of the instruction format contains the location of the address of the operand; hence, after the fetch cycle the address of the operand must still be found. Thus, an extra cycle called the indirect cycle is used during an indirect memory reference instruction to fetch the ad-dress of the operand. The indirect cycle is designated by the control signal Instruction format : 013415 I OP-CODE ADDRESS Direct address instruction : c_1 (F = 0, R = 1).A step is specified when the control signal expression to the left of the colon is true (Logical 1). C_1 t_0 :MAR \leftarrow [MBR]_4-15(Step 1) Note: [] means "contents of" In the first step of the indirect cycle the address por-tion of the fetched instruction is transferred to the MAR; thus, the computer is preparing to fetch the address of the operand. c_1t_1 :MBR \leftarrow [M](Step 2) After the second step, the address of the operand has been fetched. .The only thing that remains to be done is to go to the execute cycle. c_1t_2 :(Step 3) c_1t_3 :F \leftarrow 1, R \leftarrow 0(Step 4) The execute cycle control signal cannot be set until the last step. If it were set on step three, control would go to the fourth step of the execute cycle rather than the first step of the execute cycle. (In other words, the computer would skip three steps in the execute cycle.)

Question:

When a small drop of stearic acid is placed on the surface of water, the liquid spreads out on the water to form a single layer of molecules which are all standing on end, as shown in the figure. In a stearic acid molecule, one end is polar. This occurs in the 0-H bond, since the electron the two atoms share does not orbit an equal time around the individual atoms. The electron spends more time in orbit around the oxygen atom, causing it to be slightly negative and the hydrogen atom to be slightly positive. The charge of the two atoms as a whole remains neutral. Since water molecules (H_2 O) contain 0-H bonds, they are also polarized. The stearic acid and water molecules therefore attract each other causing the acid molecules to stand on end. Estimate the length of the stearic acid molecule if the volume of the drop was 1.56 × 10^-10 m^3 and the area of the stearic acid film is 6.25×10^-2 m^2. Also compute the approximate size of and atom, assuming that the carbon, oxygen, and hydrogen atoms are the same size. The chain is 20 atoms long.

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Solution:

The volume V is related to the area, A, of the liquid and L, the length of the stearic acid molecule, as follows: V = LA The length of the molecule is, therefore, L = V/A = [(1.56 × 10^-10m^3)/(6.25 × 10^-9 m^2)] = 2.5 × 10^-9 m Since the chain is 20 atoms long, (see figure), each atom has a diameter d that is approximately d = L/20 = [(2.5 × 10^-9 m)/(2.0 × 10^1)] = 1.3 × 10^-10 m This method is a very simple way to estimate the length of linear molecules. The technique was suggested independently in 1890 by both W.C. Rontgen and Lord Rayleigh.

Question:

Why are genetic ratios more reliable when there are large numbers of offspring? Discuss the above with a reference to one inherited character in human beings.

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Solution:

Genetic ratios are often misunderstood. When we say that a certain type of cross yields a three-to-one ratio, this does not mean that for every four offspring, there will always be three of one type and one of another. A ratio is worked out on the basis of mathematical pro-bability and will be approached when large numbers are considered. In tossing a coin there is an equal probability of obtaining either a head or a tail, and the ratio of heads to tails would be 1:1. This is because each toss or event is independent of, and therefore not influenced by, the results of any preceding or subsequent tosses. The 50% chance of obtaining either side of the coin is the pro-bability within which each individual event operates, and does not change, regardless of the number of time the event occurs. However, when the event takes place a large number of times, the results do tend to average out to the expected probability, and the actual ratio approaches the anticipated ratio of 1/2. The same is true for certain genetic events. For instance, the cross between a heterozygous brown-eyed man (Bb) and a blue-eyed woman (bb) would be depicted as the following: p Each offspring produced has a 1/2 chance of being brown eyed and a 1/2 chance of being blue eyed. Thus, the expected ratios are; 1 Bb : 1 bb; or phenotypically, 1 brown : 1 blue. We might expect then, a one-to-one ratio in eye color among their offspring. But suppose their first child had blue eyes. This does not automatically mean that their second child must have brown eyes. That child has, like the first child, an equal probability of being either brown or blue-eyed; either outcome is equally probable regardless of what eye color the first child has. This is because the separation of alleles in gamete formation is, like a coin toss, a purely random event, not effected by preceding events. So, while we know that in this case the egg can only be carrying the gene for blue eyes, (because the female in this case is homozygous for blue eyes) . The gene carried by the sperm that will fertilize the egg has an equal chance of being brown or blue, and we cannot predict which it will be because of the randomness of gene segregation. Therefore, though we might expect a one-to-one ratio for brown-eyed and blue-eyed offspring, the actual ratio may deviate from expected entirely by chance. However - and this is the important point - if we were to tabulate the eye color of thousands of children from many families of parents having the same genotypic combinations as this couple, we would indeed find that close to one half of them will have blue eyes and the other half will have brown eyes. As was the case for the coin toss, the larger our sampling population, the closer our ratio approaches one-to-one, the expected ratio. Our actual ratio approaches the probable ratio, and our results become more reliable.

Question:

Construct a flowchart describing the problem-solving procedure. Include a description of debugging the program.

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Solution:

The novice is apt to imagine that bugs crawl into the program only during the process of converting the written program to a form acceptable to the computer. In fact, errors can arise even during the phase when one is formulating a mathematical solution to the problem. Hence, the process of debugging a program should properly start by studying, once again, the algorithm for solving the problem and then proceed to checking the written program. Finally, the input medium (cards, C.R.T., etc.) should be checked for typographical errors. A typical flowchart of the problem-solving procedure might be as follows:

Question:

In the simple case of the field due to a single point charge q, check the method for obtaining E from \varphi See the figure.

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Solution:

Choose the points A and B to be on the same radius, in which case the component of E in the direction AB^\ding{217} is the total field. Let A be at a distance R from q and B at a distance (R + \sigma), where \sigma is very small. In general, work = \intF^\ding{217} \textbullet dr^\ding{217} where the integral is evaluated over the path of travel of the object which we are working on. For this problem, F^\ding{217} and dr^\ding{217} are parallel because we are moving q along a radius. Hence work =(\sigma + R)\int_R F dr Now, because \sigma is very small, F will be essentially constant along the path of travel, and we may replace it by its average value, F^\ding{217}. Therefore work \approxF\sigma(1) (This relation becomes more exact as \sigma gets smaller.) Also note that, by definition of the potential difference, between R and R + \sigma, we obtain work = q∆v Substituting this n (1). We find q∆V =F\sigma ButF= qE Hence ∆V =E̅̅̅\sigma orE = ∆V / \sigma(2) As \sigma \rightarrow0, we will obtain E, the exact value of the electric field intensity. The potential at A is by definition \varphi^A _e = q / R The potential at B is \varphi^B _e = q / R + \sigma Using equation (2) ,E= Using equation (2) ,E= [(q / R) \rule{1em}{1pt} {q / (R + \sigma)}] / \sigma = 1 / \sigma [{(q(R + \sigma)) / (R(R + \sigma))} \rule{1em}{1pt} {qR / ((R + \sigma)R)}] = 1/ \sigma [(qR+q\sigma-qR) / {R(R + \sigma)}] = q / [R(R +\sigma)] = q / [R^2{1 + (\sigma / R)}] Taking the limit as \sigma \rightarrow 0, we find E = q / R^2 for the value of E at a distance R from q.

Question:

Working-Storage entries describing two heading lines are shown in fig. 2. Description of report items, including data- names and starting columns to line up with the headings, are given in the table of fig. 1. Use these descriptions of edited items to write a record description for TRANS-LINE.

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Solution:

In writing the record description of fig. 3, defined in WORKING STORAGE SECTION and the data table, the VALUE clause is used to specify the appearance of spaces. Observe that the solution is following the data table step by step. The editing characters used in the PICTURE of each data item are described in fig. 4. Data-Name Starting Column Description C-NUMB 9 6 digits; insert 0 between sets of two C-NAME 26 20 characters; insert blank bet. 12th. and 13th. C-CODE 55 1 digit C-AMT 64 6 digits, 2 to right of decimal point; all digits replaced with blanks if zero. C-NEW-BAL 79 6 digits, 2 to right of decimal point; comma in appropriate position, fixed dollar sign, sign if value negative, item all asterisks if zero. C-DUE 96 5 digits, 2 to right of decimal point; dollar sign just before first nonzero digit. Blank if zero. C-DATE 110 3 sets of two-digit numbers; insert blanks between sets. Fig. 1. 0 B Z \textbullet $ \textasteriskcentered Inserts a zero at this position Inserts a blank at this posi-tion Replaces leading zeros with blanks up to the amount ofZ's used. Inserts a decimal point in this position Inserts dollar sign at this position if just one, or at farthest right position for more than one, suppressing leading zeros. Replace leading zeros with as-terisks up to the amount of asterisks data PICTURE printed 5432 AB26 00265 00023 265 125 0026 00015 0003 0002 9099 XXB9(2) Z 9 (5) Z Z Z Z Z 9.99 $9(4) $$$$ $$9(5) $\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered 50432 AB 26 0265 23 2.65 $125 $26 $015 $\textasteriskcentered\textasteriskcentered\textasteriskcentered3 \textasteriskcentered\textasteriskcentered02 fig. 4

Question:

Suppose someone told you that plants bend toward the light so that their leaves will be better exposed for photosynthesis. Tell how you could demonstrate by experiments that this may not be a scientific explanation of the growth toward the light.

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0229.htm

Solution:

If plants bend toward the light in order to have their leaves exposed to sunlight, one would assume then, that the signal for stem-bending comes from the leaves. In other words, there will be a mechanism present in the leaves which allows them to receive the stimulus of sunlight and consequently direct the stem to elongate and bend in the direction of the light. We can test the validity of this hypothesis by defoliating the plant. If our hypothesis is correct, removing the leaves would also remove the receptors of the stimulus by light and the effectors of the signal to bend, and no bending of the stem should occur. We know, however, that as long as the apex is intact, a defoliated stem will bend toward the light. The coleoptile tip of the oat seedling, for instance, is a non-leaf-bearing structure which has been used in classical demonstrations of phototropism (the growth response to light). Since a stem will bend toward the sun even if it has no leaves, one cannot really say that plants bend toward the sun purposefully in order to better expose their leaves for photosynthesis. Our experiment has shown that the leaves themselves do not hold the controlling key to bending. We can advance a step further and test the role of the tip of the stem in phototropism. If the apex is involved in the bending of the stem toward light, covering the tip from the light or removing it totally should prevent phototropism. It has, in fact, been observed in experiments that both operations do prevent phototropism as well as inhibit growth. Replacing the tip on the decapitated stem actually brings about resumed growth and bending of the stem. Thus it is clear that the apical meristem of the stem rather than the leaves of the plant is responsible for the phenomenon of phototropism.

Question:

Write a subroutine in FORTRAN to compute the square root of the complex number w = u + iv (where i = \surd-1 and where u and v are any real numbers) according to the formula w1/2 = + \surdw = [{(u2 + v2)1^/2 + u} / 2]1^/2 \pm i [{(u2 - v2)1^/2 - u} / 2]1^/2 Note that the imaginary part of w1^/2 is positive if v > 0, but negative if v < 0.

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Solution:

It is convenient to begin the calculation by first computing the modulus of w, which is defined as │w│ = R = (u2 + v2)1^/2 . Once R is evaluated, it is simple to calculate the real part of w1^/2, denoted by the variable REAL, as REAL= ((R + u)/2)1^/2. Then one calculates the imaginary part of w1^/2, denoted by AMAG, as AMAG = ((R - U)/2)1/2 . If v < 0, one must then change the sign of AMAG from positive to negative. This completes the calculation of the real and imaginary parts of w. The logic of these calculations is present-ed in the flowchart. COMPUTER PROGRAM SOLUTION [Note all FORTRAN statements begin in column 7.] SUBROUTINE C0MSQ (U,V,REAL,AMAG) CSQRT IS A LIBRARY FUNCTION FOR TAKING THE CSQUARE ROOT OF A REAL NUMBER R = SQRT (U \textasteriskcentered U + V \textasteriskcentered V) REAL = SQRT ((R + U)/2.) AMAG = SQRT ((R - U)/2.) IF (V.LT.O.O) AM&G = -AMA CABOVE STATEMENT SWITCHES SIGN OF CIMAGINARY PART IF V IS LESS THAN ZERO RETURN END

Question:

A high-powered rifle whose mass is 5 kg fires a 15-g bullet with a muzzle velocity of 3 × 10^4 cm/sec. What is the recoil velocity of the rifle?

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0304.htm

Solution:

The momentum of the system after the gun has fired must equal the momentum before the gun went off. Originally, the momentum of the bullet and rifle is zero since they are at rest. Using the conservation of momentum equation: (m_1 + m_2)v_0 = m_1v_1 + m_2v_2 = 0 m_1v_1 = - m_2v2 v_1 = -[(m_2v_2)/(m_1)] = - [ {(15g) × (3 × 10^4_ cm/sec)}/ {5 ×10^3g } ] = - 90 cm/sec. This is a sizable recoil velocity and if the rifle is not held firmly against the shoulder, the shooter will receive a substantial "kick". However, if he does hold the rifle firmly against his shoulder, the shooter's body as a whole absorbs the momentum. That is, we must use for m_1 the mass of the rifle plus the mass of the shooter. If his mass is 100 kg, then the recoil veloci-ty (now of the rifle plus shooter) is v_1 = - [ {(15g) × (3 × 10^4cm/sec)} / {5 × 10^3g + 10^5g} ] \cong - 4.5 cm/sec This magnitude of recoil is quite tolerable.

Question:

For the following voltaic cell, write the half-reactions, designating which is oxidation and which is reduction. Write the cell reaction and calculate the voltage of the cell made from standard electrodes. The cell is Co; Co^+2 ││Ni^+2 ; Ni .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0587.htm

Solution:

The cell reaction is the algebraic sum of the reactions that take place at the electrodes. Every cell has 2 electrodes an anode and a cathode. Oxidation, which is the loss of electrons, occurs at the anode. Reduction, which is the gain of electrons, takes place at the cathode. The cell is always written as solid; ion in solution ││ ion in solu-tion; solid (anode)(cathode) Oxidation and reduction are the half reactions that take place in the cell. For this cell, they are Co \rightarrow Co^+2 + 2e^- (oxidation at anode) Ni^+2 + 2e^- \rightarrow Ni (reduction at cathode) Sum: Ni^2+ + Co \rightarrow Ni + Co^+2 (Cell reaction). Since Co is losing electrons, it provides the oxidation reaction and Ni^+2, gaining these electrons, takes part in the reduction re-action. The voltage of a cell is the sum of the oxidation and reduction potentials in units of volts and is designated by E\textdegree (under standard conditions) . E\textdegree_cell = E\textdegreeoxidation+ E\textdegree_reduction. The voltages of half cell reactions are usually given as the reduction potentials. The oxidation potential is opposite in sign to the re-duction potential. The potentials can be obtained from a table of standard reduction potentials. Ni^+2 + 2e^- \rightarrow Ni , the potential is -.25v, E\textdegree_red = -.25v. For Co^+2 + 2e^- \rightarrow Co, the potential is -.277 v. Since Co \rightarrow Co+2+ 2e^- is the oxidation reaction, E\textdegree_ox equals the negative of -.277v, or E\textdegreeox= .277v. Substituting these values into the equation E\textdegree = E\textdegreeoxid+ E\textdegree_red , one obtains E\textdegree = +.277 + (-.25) = .027 v . Since E\textdegree is positive, the reaction proceeds spontaneously and can be used to supply current.

Question:

Classify the bonds in the following as ionic, ionic - covalent, covalent - ionic and covalent: I_2,LiCl,MgTe, Cl_2O, H_2S.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0623.htm

Solution:

The bonds in these molecules can be classified by considering the electronegativities of the atoms in-volved . Theelectronegativityof any element is the tendency of that element to attract electrons. The greater theelectronegativity, the stronger the attraction. When 2 atoms join, the type of bond formed can be determined by calculating the difference inelectronegativityof the two atoms involved. When the difference is greater than the bond is classified as ionic, if less than 1.7, then the bond is covalent. An ionic bond involves the complete transfer of electrons from 1 atom to another, while in a covalent bond, the electrons are shared between like or similar atoms. Covalent-ionic indicates a mixed character which is more covalent in nature. Ionic - covalent bonds have more of an ionic character. Consider the cases in point. For I_2, there is no difference in electronegativity because the 2 atoms involved in the bond are the same, and hence contains a covalent bond with equal sharing of the electrons. InLiCl, theelectronagetivitiesare Li = 1.0 andCl= 3.0. (The electronegativity values are obtained from anelectronegativitytable). The electronegative difference is 2.0. Therefore, the bond is ionic. One proceeds in a similar manner for the remainder of the compounds. Compounds and Electronegatives Of the Elements Difference in Electronegatives Types of Bond Mg - Te 1.22.1 2.1 - 1.2 = 0.9 Covalent-ionic Cl_2 - O 3.03.5 3.5 - 3.0 = 0.5 Covalent-ionic H_2 - S 2.12.5 2.5 - 2.1 = 0.4 Covalent-ionic

Question:

Water flows at the rate of 300 ft^3/min through an inclined pipe as shown in the figure. At A, where the diameter is 12 in., the pressure is 15 lb/in^2. What is the pressure at B, where the diameter is 6.0 in. and the center of the pipe is 2.0 ft lower than at A?

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Solution:

The mass of liquid entering the tube at point A, in a time ∆t, should be equal to the mass leaving it at point B in ∆t. Let the velocities and the densities at A and B be v_1,v_2 and \rho_1,\rho_2,respectively. In a time ∆t, the liquid entering at A moves a distancev_1∆t, hence the volume of liquid entering is Av_1∆t . Therefore, the mass of this fluid is \rho_1v_1A_1∆t. Similarly, the mass of liquid leaving at B is \rho_2v_2A_2∆t. Dividing both terms by ∆t, we get the continuity equation for the flow \rho_1v_1A_1 = \rho_2v_2A_2 . For all practical purposes, we can assume liquids to be incompressible, therefore the density of liquid during flow remains constant \rho1= \rho_2 and we have v_1A1= v_2A_2 For this problemA_1v1= [(300 ft^3/min)/(60 sec/min)] = 5.0 ft^3/sec Hencev_1 = (A_1v_1)/(A_1) = (A_1v_1)/(\pir^2_1) = (5.0 ft^3/sec)/(3.14 × {1/2 ft}^2) = 6.4 ft/sec v_2= [(A_1)/(A_2-)]v_1 = [\pi(1/2 ft)^2]/[\pi(1/4 ft)^2] v_1 = 4v_1 v_2= 25.6 ft/soc. The pressure P_1 at point A is P_1= (15 lb/in.^2)(144 in.^2/ft^2) = 2200 lb/ft^2 The weight density of water D = 62.4 lb/ft^3, and therefore \rho = 1.94slugs/ft^3. The Bernoulli equation for the pressures at A and B is P_1 - P_2 = \rhogh + (1/2) \rho(v^2_2 - v^2_1) where h is the difference between the heights of the centers of the cross- sections at the two ends of the pipe. The pressure at point B is P_2 = P_1 + \rhog(h_1 - h_2) + \rho/2 (v^2_1 - v^2_2) = 2200 lb/ft^2 + (62.4 lb/ft^3)(2.0 ft) + [(1.94 slug/ft^3)/2][(6.4)^2 - (25.6)^2] ft^2/sec^2 = 2200 lb/ft^2 + 125 lb/ft^2 - 596 lb/ft^2 = 1729 lb/ft^2 = 12 lb/in.^2

Question:

Electrons of energies 10.20eV, 12.09eV, and 13.06eV colliding with a hydrogen atom, can cause radiation to be emitted from the latter. Calculate in each case the principal quantum number of the orbit to which the electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.

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Solution:

Bohr theory relates the wavelength of the radiation emitted by a hydrogen atom to the principal quantum numbers of the energy levels involved by the equation (1/\lambda) = R [(1/n^2_1) - (1/n^2_2)] or f = (c/\lambda) =Rc[(1/n^2_1) - (1/n^2_2)], where c is the speed of light, f is the frequency of the radiation and R is theRydbergconstant. The energy of the quantum of radiation emitted is E =hf=Rch[(1/n^2_1) - (1/n^2_2)]. (Here, h isPlahcK'sconstant.) This is the energy emitted in the form of a quantum of radiation when the electron drops from the orbit characterized by the quantum number n_2 to that character-ized by the quantum number n_1 If the electron is raised from the orbit with quantum number n_1 to the orbit with quantum number n_2, it must absorb an equal amount of energy. Before being struck by an electron, a hydrogen atom will be in the ground state with n_1 = 1. It is, in any case, true that if n_1 had a value higher than 1, then E, the absorbed energy, could not be greater than 3.40eV, the limiting energy for n_1 =2 . Thus, if E has the values 10.20eV, 12.09 eV, and 13.06eV, the atom absorbing these energies from the incoming electrons, then E =Rch[(1/n^2_1) - (1/n^2_2)]. E = (1.097 × 10^7 m^-1) (2.997 × 10^8 m \textbullet s^-1) (6.625 × 10^-34 J \textbullet s) [(1/n^2_1) - (1/n^2_2)] E = 2.178 × 10^-18 [(1/n^2_1) - (1/n^2_2)] J Since the given energies are calculated in terms of electron volts, we transform 2.178 × 10^-18 J intoeVnoting that 1 J = 6.242 × 10^18eV whence E = (2.178 × 10^-18 J) (6.242 × 10^18eV/J) × [(1/n^2_1) - (1/n^2_2)] E = (13.60eV) [(1/n^2_1) - (1/n^2_2)] Assuming that n_1 =1, and using the given data, we obtain [1 - (1/n^2_2)] = (10.20/13.60) = (3/4)or n_2 = 2 [1 - (1/n^2_2)] = (12.09/13.60) = (8/9)or n_2 = 3 [1 - (1/n^2_2)] = (13.06/13.60) = (24/25)or n_2 = 5 We interpret these results to mean that an incident electron, of energy 10.20eV, will cause the hydrogen atom electron to jump from n = 1 to n = 2, and similarly for the other 2 cases. In dropping back to the ground state, the wavelength emitted in the three cases will be given from the original equation quoted. Thus (1/\lambda) = R [1 - (1/n^2_2)] (1/\lambda) = R [1 - (1/4)] = (3R)/4or \lambda = 4/(3R) (1/\lambda) = R [1 - (1/9)] = (8R)/9or \lambda = 9/(8R) (1/\lambda) = R [1 - (1/25)] = (24R)/25or \lambda = 25/(24R) Whence \lambda = [4/(3 × 1.097 × 10^7 m^-1)] = 121.57 × 10^-9 m \lambda = [9/(8 × 1.097 × 10^7 m^-1)] = 102.57 × 10^-9 m \lambda = [25/(24 × 1.097 × 10^7 m^-1)] = 94.97 × 10^-9 m

Question:

A + 2B + C = AB + BC. You collect the following data: Concentration(moles/liter) [A] [B] [C] -dA/dt (moles/liter - sec) 1 1.00 1.00 2.00 1.00 2 2.00 1.00 2.00 2.00 3 2.00 2.00 2.00 8.00 4 2.00 2.00 4.00 8.00 (a)Determine the experimental rate law expression. (b) Find the specific reaction rate constant.(c) Calculate the rate of reaction, if the [A], [B], and [c] are, respectively,1.0, 2.0, and 3.0: (d) Speculate on the rate-controlling step, i.e., the slow step.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0450.htm

Solution:

First write a general rate law for the reaction, and then write a specific law based on the given data. The specific rate constant can be found from a given set of data. The rate with given concentrations is obtainable from the specific rate law. From this, a rate-controlling step can be guessed at. Proceed as follows: (a)To write the rate law, note that, in general, the rate = r = k[A]x[B]^y [C]^z, where the brackets indicate concentrations of A, B, and C, k = specific rate constant and the exponents (x, y, and z) are the orders ofthe re-action with respect to each reactant. The order of a chemical reaction is the number of particles needed to form the activated complex or transition state. To find the actual law, you must determine these exponents. To do this, consult the data. You are told the actual rate of the re-action, since you are given - dA/dt , which is an indication of rate. Thus, to find the order, see what happens when one of the concentrations is changed. In going from (1) to (2) , the concentration of A is doubled from 1.00 to 2.00 . Notice - dA/dt also doubles, from 1.00 to 2.00, which means the re-action is first order, x = 1, with respect to A. In going from (2) to (3), [B] goes from 1.00 to 2.00 (doubles). But notice, - dA/dt quadruples going from 2.00 to 8.00. Thus, the reaction is second order, y = 2, with respect to B. In going from (3) to (4) , [C] is doubled going from 2.00 to 4.00, while - dA/dt remains constant at 8.00. Thus, the reaction is zero order, z = 0, with respect to C, i.e., rate doesn't depend on [C]. You then have, r = k[A][B]^2[C]^0 orr = k[A][B]^2 for the experimental rate law expression. (b)To find k, substitute any of the four sets of data in the experimental rate law expression. rate = (-dA/dt) = k[A][B]^2 For example, let us take the data from line (1). You have 1.00 mole/liter- sec = k[1.00 mole/liter] [1.00 mole/liter]^2 , or k = 1.00. Thus, the value of the specific rate constant k is 1.00 liter^2/mole2- sec. (c)To find the rate of the reaction at the concen-trations [A] = 1, [B] = 2, and [c] =3, substitute these values into the rate law, r = k [A][B]^2, Thus, r = (1)[1][2]^2 = 4.0 moles/liter-sec. (d)You can use previous information gained in this problem to speculate about what the rate-controlling step must be. From part (a) you know the rate does not depend on [C]. Thus, C cannot be involved in the slow step. The rate of the reaction is influenced by [A] and [B], however. The rate-controlling step is probably the combination of B_2 and A,i.e., B_2 + Aslow\rightarrow AB_2.

Question:

Explain how the stack area of a computer's memory oper-ates.

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Solution:

The stack is similar to a plate dispenser in a cafeteria. Figure 1 shows a plate dispenser with yel-low, blue, and red plates. Notice that plates can only be put in or taken from the top of the stack. If someone wants the yellow plate, the red and then the blue plate must be removed before the yellow plate. Notice that the last plate in is always the first plate out. This "last in-first out" (LIFO) rule is characteristic of all computer stacks. Most computers have a specific section in memory which is used as a stack and a specific register called the stack pointer. The stack pointer (SP) contains the ad-dress of the first "empty\textquotedblright memory location in the stack. Figure 2 shows the stack with contents A, B, C, D and E at memory locations 25, 26, 27, 28 and 29 respectively. To remove A the POP instruction is used and the SP is incremented. With A removed the SP still contains the first empty position. To insert F a PUSH instruction is used;\textasteriskcentered the SP is decremented, and F goes into the stack at location 25. When using the stack, care must be taken to ensure that the stack is not in "over fill." If too many Push instructions are used, the stack will grow upward and interfere with a main program. The programmer must always keep track of what is at the top of the stack and how many items are in the stack, otherwise errors will be made. Also, care must be taken to see that not too many items are removed from a stack, since it is im-possible to "pop" an. "empty" stack.

Question:

The electric field in the space between the plates of a discharge tube is 3.25 × 10^4 newtons / coul. What is the force of the electric field on a proton in this field? Compare this force with the weight of the proton, if the mass of the proton is 1.67 × 10^\rule{1em}{1pt}27 kg and its charge is 1.60 × 10^\rule{1em}{1pt}19 coul

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Solution:

E^\ding{217}, the electric field strength, is defined as the force per unit charge q, or E^\ding{217} = F^\ding{217} / q The magnitude of F^\ding{217} is F = Eq = 3.25 × 10^4 nt/coul× 1.60 × 10^\rule{1em}{1pt}19 coul = 5.20 × 10^\rule{1em}{1pt}15 nt The weight of the proton is given by the force of gravitational attraction on m. By Newton's Second Law, F_g = ma where a is the acceleration of m and F_g is the force of gravitational attraction on m (the weight of m). Be-cause the acceleration of m is due to the earth's gravity, we may write F_g = mg = (1.67 × 10^\rule{1em}{1pt}27 kg)(9.80 m/s^2) = 1.64 × 10^\rule{1em}{1pt}26 nt To compare the force on the proton in the electric field to the weight on a proton (also a force), we take their ratio. Hence F / W = (5.20 × 10^\rule{1em}{1pt}15 nt ) / (1.64 × 10^\rule{1em}{1pt}26 nt) = 3.17 × 1011 This emphasizes the fact that the force caused by an electric field may be much greater than that due to a gravitational field.

Question:

The net dipole moment of water is 1.84 debyes and the bond angle is 104.45\textdegree. What moment can be assigned to each O-H bond?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0644.htm

Solution:

To answer this problem, consider the definition of a dipole, how it is expressed quantitatively, and the geometry of a water molecule. A dipole consists of a positive and negative charge separated by some distance. Quantitatively, it is ex-pressed as dipole moment, which is the charge times the distance between the positive and negative centers. The units are debyes. The geometry of a water molecule is not linear. It is represented by FIG. a. The oxygen atom shares its electrons with the hydrogen atoms; covalent bonding. Because the oxygen atom is electronegative, which means it has a tendency to attract electrons, the shared electrons will spend more time on the oxygen atom than on the hydrogen atoms. Thus, a polar covalent bond exists. The oxygen tends to develop a negative charge because the electrons spend more time on the oxygen atom. Since a water molecule is neutral (overall), the charge on the hydrogen atoms become positively charged (Fig. b). Therefore, a dipole exists. The net dipole is represented by Fig. c. As given, the dipole moment is 1.84 D. One is asked to compute the moment of each OH bond. One can represent the net moment and the OH moment as in Fig. d. The net dipole moment is the longer line and bisects the bond angle of 104.45\textdegree. The shorter lines represent the components of the net dipole, which comes from each bond. This can be simplified and drawn as Fig. e. This is the representation for one OH bond. The net dipole must be divided by two (1.84/2 = 0.92), since the total number O-H bonds is 2. The bisected angle of 104.45\textdegree yields 52.22\textdegree. To find X, the unknown OH dipole, use trigonometry. The cos 52.22\textdegree is = adjacent leg / hypotenuse = .92 / X, since a right triangle is present, cos 52.22\textdegree = 0.6127. Thus, equating 0.6127 = 0.92 / X. Therefore, X = .92 / .6127 = 1.50 Debyes.

Question:

What do each of the embryonic germ layers give rise to?

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Solution:

Ectoderm gives rise to the epidermis of the skin, including the skin glands, hair, nails, and enamel of teeth. In addition, the epithelial lining of the mouth, nasal cavity, sinuses, sense organs, and the anal canal are ectodermal in origin. Nervous tissue, including the brain, spinal cord, and nerves, are all derived from embryonic ectoderm. Mesoderm gives rise to muscle tissue, cartilage, bone, and the notochord which in man is replaced in the embryo by vertebrae. It also provides the foundation for the organs of circulation (bone marrow, blood, lymphoid tissue, blood ves-sels), excretion (kidneys, ureters) and reproduction (go-nads, genital ducts). The mesoderm produces by far the greatest amount of tissue in the vertebrate body. Endoderm gives rise to the epithelium of the digestive tract, the tonsils, the parathyroid and thymus glands, the larynx, trachea, lungs, the bladder and the urethra with its associated glands.

Question:

Pentobarbital, sometimes prescribed for the treatment of chronic insomnia under the trade name Nembutal, has the following structure: It is toxic in doses of 0.5 g and lethal in doses of 1.5 g. Barbituric acid, denoted by H Bar, and which may be used in the synthesis of pentobarbital, has the following structure: H Bar dissociates according to the following equation: H Bar \rightleftarrows H^+ + Bar^-. It is a weak acid, having a pK_a of only 4.01. Approximately what percent of H Bar molecules are dissociated in a 0.10 N solution?

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Solution:

We will solve this problem by determining the concentrations of H^+ and Bar- formed from the dissociation of H Bar. If we let x denote the equilibrium concentration of H+, then, since one mole of Bar- is formed per mole of H^+ formed, the equilibrium concentration of Bar- is also x. Furthermore, the equilibrium concentration of H Bar is 0.10 - x. Then K_a = [H^+] [Bar-] / [HBar] = x^2/(0.10 - x). Note that we have neglected the contribution to [H^+] from the dissociation of water. Assuming that x is much smaller than 0.10 (a justifiable assumption, since H bar is so weak an acid), the 0.10 - x \allequal 0.10 and we have K_a = x^2/(0.10 - x) \allequal x^2 / 0.10 orx^2 = 0.10 × K_a 0 x = (0.10 × K_a)^1/2. The value of K_a is determined from the pK_a. By definition, pK_a = - log K_a. Hence K_a = 10^-pKa = 10^-4.01, and x = (0.10 x K_a)^1/2 = (0.10 ×10^-4.01)^1/2 = (10^-5.01)^1/2 = 10^-2.50 \allequal 3.2 × 10^-3. The equilibrium concentrations of H^+, Bar-, and H Bar are then [H^+] = x = 3.2 × 10^-3 M, [Bar-] = x = 3.2 ×10^-3 M, and [H Bar] = 0.10 - x = 0.10 - 3.2 ×10^-3 \allequal 0.10 M. The percent dissociation of H Bar is given by % dissociation = {[H^+] / [HBar]}× 100 % = {[Bar-] / [HBar]}x 100 % = [(3.2 × 10^-3 M) / (0.10 M)] × 100 % \allequal 3%.

Question:

Refer to the accompanying figure and determine the number of unit particles (atoms) in a unit cell of each of the three type lattices. Unit cells in cubic lattices: (a) simple cubic lattice; (b) body- Unit cells in cubic lattices: (a) simple cubic lattice; (b) body- centered lattice; (c) face-centered lattice. centered lattice; (c) face-centered lattice.

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Solution:

The following rules may be used to determine the number of unit particles (atoms) associated with each type of cubic unit cell: 1) An atom at a corner contributes 1/8 of its volume to each of eight adjacent cubes. 2) An atom on an edge of a cube contributes 1/4 of its volume to each of 4 adjacent cubes. 3) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. 4) An atom completely within a cube contributes all of its volume to the unit cell. The number of unit particles in each of the three types of lattices can be found as follows: Simple cubic: In the simple cubic lattice all eight units are of the same variety - they are all at corners, thus rule 1 is used. Total atoms in cell = 8 corners × 1/8 atom/corner= 1 atom Therefore the area of unit cube is equivalent to 1 atom. Body-centered: In the body-centered lattice there are nine atoms contributing to the unit cube. There are 8 corners and one atom completely-within the cube, thus, rules 1 and 4 are used. Total atoms in cell = 8 corners × 1/8 atom/corner + 1 atom in center = 2 atoms. Face-centered: In the face-centered lattice there are 14 atoms contributing to the unit cube. There are 8 corners and 6 faces, thus, rules 1 and 3 will be used. Total atoms in a unit cell = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/corner = 4 atoms.

Question:

What is the energy equivalent in MeV of a particle whose mass is 1 amu?

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Solution:

The energy equivalent is given by the Einstein relation E = mc^2. In M.K.S. units, a mass of 1 amu is 1.66 × 10^-27 kg. Hence its equivalent energy is E = mc^2 = (1.66 × 10^-27 kg)(3 × 10^8 m/s)^2 = 1.49 × 10^-10J Since 1 eV = 1.6 × 10^-19 J, this energy can be expressed in eV E = (1.49 × 10^-10J)/(1.6 × 10^-19J/ev) =9.31 × 10^8 eV = 931 MeV Thus the energy equivalent of 1 amu is 931 MeV.

Question:

CH_3COC_4H_9; C_2H_5OH; CH_3NH_2; (C_2H_5)_2O; CH_3COOC_3H_7; CH_3COOH.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0747.htm

Solution:

All of these compounds are derivatives of hydrocarbons, i.e. compounds that contain only hydrogen and carbon atoms. The derivative of a hydrocarbon is the substitution of elements or radicals for one or more hydrogen atoms. These derivatives may be classified into functional groups, which, to a great extent, determine the properties of the compound. This is the key to naming these compounds, identify the functional groups and name it accordingly. First one must describe these functional groups. Alcohols: They have the general formula R\rule{1em}{1pt}OH, where R is any alkyl group. Amines: They are derivatives of ammonia (NH_3), where one or more H's have been replaced by an alkyl group. Ethers: General formula is R\rule{1em}{1pt}O\rule{1em}{1pt}R', where R may be the same as R'. C_2H_5OH fits into the alcohol classification group. It is named ethanol or ethyl alcohol. CH_3NH_2 fits the description of an amine and is named methylamine. (C_2H_5)_2O can be written as H_5C_2\rule{1em}{1pt}O\rule{1em}{1pt}C_2H_5, which fits the description of an ether. It is named ethyl ether. CH_3COOC_3H_7 indicates that an ester is present, once it

Question:

Given a solution of 1.00M HC_2H_3O_2 , what is the concentration of all solute species? What is the percentage of acid that dissociated? AssumeK_diss= 1.8 × 10^-5 .

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Solution:

Determine the equilibrium equation, write anequilibriamconstant expression, and substitute the concentrations of the species into the expression. By definition, an acid is a substance which donates protons (H^+). Since water is the only other species in the solution, it must act as a base; it receives protons. HC_2H_3O_2 + H_2O \rightleftarrows H_3O^+ + C_2H_3O_2^- . There is a constant calledK_dissociation_, which measures the extent of the HC_2H_3O_2 donation of protons. For the general reaction, HA + H_2O \rightarrow H_3O+ + A^- ,K_diss= {[A^-][H_3O^+]} / [HA] . For the reaction in this problem, you have {[C_2H_3O_2^-][H_3O^+]} / [HC_2H_3O_2] =K_diss= 1.8 × 10^-5 . You are asked to find these concentrations. Let x = the moles per liter of HC_2H_3O_2 that dissociated. From the chemical equation, you see that x must also be the concentration of H_3O^+ and C_2H_3O_2^- , since for each mole HC_2H_3O_2 of that dissociates, one mole of H_3O^+ and one mole of C_2H_3O_2^- is produced. With this in mind, you can represent HC_2H_3O_2 as 1-x, since you started with 1M of HC_2H_3O_2 and x moles per liter dissociate. You have left 1-x moles per liter. Now substituting, you obtain {[H_3O^+][ C_2H_3O_2^-]} / [HC_2H_3O_2] = 1.8 × 10^-5 = [(x \textbullet x) / (1 - x)] . Solving for x, you obtain x = .0042. Therefore, the concentrations of the species are [HC_2H_3O_2] = 1 - .0042 = .9958M [H_3O^+] = x = .0042M [C_2H_3O_2^-] = x = .0042M The percentage of acid that dissociated is, thus, (.0042/1) × 100 = .42% .

Question:

Write the Born-Haber cycle for the formation of crystalline sodium fluoride (Na^+F^-), starting with solid Na and gaseous F. Then, using the thermochemical data supplied below, de-termine its heat of formation: (1) Na(s)\ding{217} Na(g)∆H = +26.0 Kcal : sublimation (2) F_2 (g)\ding{217} 2F(g)∆H = +36.6 Kcal : dissociation (3) Na(g) \ding{217} Na^+(g) + e-∆H = +120.0 Kcal : ionization (4) F(g) + e^-\ding{217} F^-(g)∆H = -83.5 Kcal : electron addition (5) Na^+(g) + F^-(g)\ding{217} Na^+, F^-(s)∆H = -216.7 Kcal: lattice formation.

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Solution:

The Born-Haber cycle shows that the heat of the reaction is the sum of the ∆H's for the sequence of the reactions used in the formation of crystalline sodium fluoride. The Born-Haber cycle relates reactions which produce the gaseous atoms from solids, those that produce the gaseous ions, and those that indicate the union of these ions to form the crystalline product. Thus, one can write the Born-Haber cycle for this reaction as (The numbers above the arrows correspond to the numbered reactions in the problem.) One is given the energies, i.e., the thermodynamic data, for these processes. If one numbers the reactions in this problem (1) to (5), the reaction becomes (1) +(1/2)(2) + (3) + (4) + (5). The 1/2 value for (2) is derived from the fact that one mole of NaF is required, but, in (2), 2 moles of F(g) atoms are generated. Thus, take 1/2 this number. Therefore, heat of reaction is: ∆H= ∆H (1) + 1/2∆H (2) + ∆H(3) + ∆H(4) + ∆H(5) = 26 + 18.3 + 120 - 83.5 - 216.7 = - 135.9 Kcal / mole.

Question:

For a uniform sphere with moments of inertia I_1 = I_2 = I_3, use the Euler equations to find the equation of motion of the sphere.

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Solution:

The Euler equation are I_1(d\omega_1/dt) + (I_3 - I_2)\omega_3\omega_2 = N_1 I_2(d\omega_2/dt) + (I_1 - I_3)\omega_1\omega_3 = N_2 I_3(d\omega_3/dt) + (I_2 - I_1)\omega_2\omega_1 = N_3 where the subscript 1 refers to the first principal axis of the sphere, N and \omega are the net external torque and angular velocity of the sphere. Noting that I_1 = I_2 = I_3, we obtain I_1(d\omega_1/dt) = N_1 I_2(d\omega_2/dt) = N_2 I_3(d\omega_3/dt) = N_3 Defining I_1 = I_2 = I_3 = I, we nay write (d\omega_1/dt) = (N_1/I), (d\omega_2/dt) = (N_2/I), (d\omega_3/dt) = (N_3/I)(1) In free motion N = 0, and (1) tells us that \omega = const. The result \omega = const is a special feature of the free rotating sphere.

Question:

Describe the mechanism that prevents food from entering the trachea or nasal pasages during swallowing.

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Solution:

Swallowing is a complex reflex initiated when the tongue forces a bolus of food into the rear of the mouth. Here, pressure-sensitive cells are stimulated to send impulses to the swallowing control center in the medulla, which coordinates the remainder of the process. Swallowing is an all-or-none response, and once initiated, it cannot be stopped. The swallowing reflex is a sequen-tial process in which several responses occur at timed in-tervals determined by nerve connections in the control center. As the bolus is pushed into the pharynx, the soft palate rises and lodges against the back wall of the pharynx, preventing the entrance of food into the nasal cavity. Breathing is temporarily stopped and the glottis, the opening between the vocal cords, is closed as the larynx, the upper region of the trachea, rises. The rear portion of the tongue pushes the bolus down the throat over the epiglottis and the epiglottis is simultaneously bent backwards over the glottis. It is primarily the closure of the glottis, not the folding of the epiglottis, that pre-vents food from entering the trachea. At the top of the esophagus is the hypopharyngeal sphincter. This sphincter is normally closed but opens as food reaches it. The sphincter immediately closes after the food passes it, preventing the food from reversing its direction. Peristalsis, a series of rhythmic wave-like muscular con-tractions, pushes the food through the esophagus into the opening of the stomach.

Question:

A research worker isolated an unknown compound from one of his reaction products. This compound was a volatile liquid with a boiling point of 65\textdegreeC. A.134 g sample of the liquid was vaporized and the gas was collected. The temperature of the collecting chamber was kept at 98\textdegreeC in order that liquefaction would not occur. After all the sample was vaporized, the volume of the gas was measured as .0532 liters at 98\textdegreeC and 737 mm. Calculate the molecular weight.

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Solution:

We are dealing with the variables m = mass, V = volume, P = pressure, and T = temperature. Whenever this occurs, it indicates that one should employ the ideal gas equation, PV = nRT, where n = moles and R = universal gas constant. An ideal gas is one in which the gas molecules take up no space and do not attract one another. Real gases do not completely meet these conditions, but at values in the neighborhood of standard conditions ( 0\textdegreeC and 1 atm) , the difference in the real and ideal values is small enough such that the ideal gas law is accurate enough to carry out the calculations. One further bit of information is required. Since M.W. (Molecular Weight) does not appear anywhere in the equation, a relation must be found between M.W. and one of the 4 variables. The relation is n = moles = mass/M.W. Substituting this into the ideal gas equation we obtain PV = (m/M.W.) RT Solving for the molecular weight, we obtain: MW = (mRT/PV) m = mass of the sample = .134 g R = gas constant = .082 liter atm/mole - \textdegreeK T = 98\textdegreeC = 273 + 98 = 371\textdegreeK V = 0.0532 l P = 737 mm. For the formula, P must be given in atmospheres. To do this, we multiply 737 mm by the conversion factor of [(1 atm)/(760 mm)].Thus, 737 mm = 737 mm \bullet [(1 atm)/(760 mm)] = [(737)/(760)] atm. MW = [{(0.134) (0.082) (371)}/{(737/760) (0.0532)} = 79.018 g/mole.

Question:

The total speed of a projectile at its greatest height v_1, is \surd(6/7) of its total speed when it is at half its greatest height, v_2. Show that the angle of projection is 30\textdegree.

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Solution:

When a particle is projected as shown in the figure i the component of the velocity in the x- direction stays at all times the same, v_x = v-_0 cos \texttheta_0, since there is no acceleration in that direction, owing to the fact that there is no horizontal component of force acting on the projectile. In the y-direction, the upward velocity is initially v_0 sin \texttheta_0 and gradually decreases, due to the acceleration g acting downward. At its greatest height h, the upward velocity is reduced to zero. The kinematic relation for constant acceleration which does not involve time is used to find the greatest height of the trajectory. It is v^2_f = v^2_i + 2as(1) In this case v_f = 0, v_i, the initial velocity, is v_0 sin \texttheta_0, a = - g and s = h. Then 0 = (v_0 sin \texttheta_0)^2 - 2gh orh = [(v_0 sin \texttheta_0)^2]/2g. The total velocity at the highest point is thus the x-component only. That is, v_1 = v_0 cos \texttheta_0. At half the greatest height, h/2 = (v_0 sin \texttheta_0)^2/4g, the velocity in the y-direction, v_y, is obtained from the equation (1) with v_f = v_y, v_i = v_0 sin \texttheta_0, a = -g, and s = h/2. v^2_y2 = (v_0 sin \texttheta)^2 - 2g (h/2) = (v_0 sin \texttheta)^2 - (1/2) (v_0 sin \texttheta)^2 = (1/2) (v_0 sin \texttheta)^2.(2) In addition, there is also the ever-present x- component of the velocity v_0 cos \texttheta_0. Hence the total velocity at this point is obtained by the Pythagorean theorem, v^2_2 = v^2_x + v^2_y2 = (v_0 cos \texttheta_0)^2 + (1/2) (v_0 sin \texttheta_0)^2 = (v_0 cos \texttheta_0)^2 + (1/2)v^2_0(1 - cos^2 \texttheta_0) = (1/2)v^2_0 + (1/2) (v_0 cos \texttheta_0) 2.(3) Here we used the trigonometric identity sin^2 \texttheta + cos_2 \texttheta = 1. However, we are given that v_1 = \surd(6/7) v_2or(v^2_1)/(v^2_2) = 6/7 . Therefore,[(v_0 cos \texttheta_0)^2]/[(1/2)v^2_0 + (1/2)(v_0 cos \texttheta_0)^2] = 6/7 ; or7(v_0 cos \texttheta_0)^2 = 3v^2_0 + 3(v_0 cos \texttheta_0)^2, or4 cos2\texttheta_0 = 3. One can therefore say that cos \texttheta_0 = \surd3/2or\texttheta_0 = 30\textdegree.

Question:

A saturated solution of CaF_2 contains .00168g of CaF_2 per 100g of water. Determine theK_sp.

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Solution:

When a salt is added to water, it dissociates into ions. Once a saturated solution is obtained, the addition of more salt re-sults in precipitation of the solid. In such a case, an equilibrium is established between the solid phase and the ions in solution. This equilibrium is expressed as the solubility product constant orK_sp. To determine theK_sp, you must set up an equation that expresses this equilibrium. You must also determine themolarityof the solution.Molarityrefers to the number of moles per liter of solution. The equilibrium reaction is CaF_2 (s) \rightleftarrows Ca^2+ (aq) + 2F-(aq). The products have their oxidation states expressed because they are ions in solution. TheK_spfor the reaction is defined as K_sp= [Ca^2+] [F-]^2 . Because this is a heterogeneous equilibrium, there is no need to in-clude CaF_2 (s) in the expression forK_sp. You include only those substances with variable molar concentrations. The fluoride ion con-centration is squared because 2 moles of fluoride ion are generated in the reaction. In general, then, the exponent of the concentration will equal the number of moles of that ion that is generated. Now that theK_spis defined, determine the molarityof the solution. Sincemolarityis defined as moles of solute per liter of solution, you have M =molarity= [(.00168g) / {78g/mole (CaF_2)}] / [100 ml / (1000ml/1)] Solving, you obtain M = .214 × 10-^2 M. You will notice that the 100 grams of water do not appear in this calculation. The density of water is 1g/ml. Since Density = Mass/Volume, 100g H_2 O has a volume of 100ml. Now that you know themolarity, go back to CaF_2 (s) = Ca^2+ (aq) + 2F-(aq). If CaF_2 (s) has M = .214 × 10-^2 , then by looking at the coefficients, you fina in solution 2.14 × 10-^3 M Ca^2+ and 4.28 × 10-^3 M F- . These results are obtained because for every mole/liter of CaF_2 in solution, 1 mole/liter of Ca^2+ and 2 moles/liter of F- is obtained, Therefore, recalling K_sp= [Ca^2+] [F-]^2 , you have K_sp= (.214 × 10-^2 ) (.428 × 10-^2 )^2 = 3.92 × 10-^8 .

Question:

Draw the structure of the isomer of trinitrobenzene that would be easiest to synthesize from benzene, nitric acid, and sulfuric acid.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0773.htm

Solution:

The nitration of benzene proceeds quickly if a mixture of concentrated nitric and sulfuric acids is used. For each mole of nitric acid dissolved in sulfuric acid, four moles of particles are formed: 2H_2SO_4 + HNO_3 \rightarrow H_3O+ + 2HSO-_4 + NO_2+ The nitronium ion (NO_2+) is the reagent responsible for the nitration of aromatic compounds. When benzene derivatives undergo substitution, a "directing effect" on the position of the new substituent is observed. The first phase of the nitration reaction is the attack of the nitronium ion on the electrons of the benzene ring (seen in figure A). The loss of a hydrogen atom, or proton, to HSO-_4 (a proton acceptor) leads to the restoration of the original \pi - bonding system. The attack on the benzene ring is generally accomplished by an electron-deficient species such as a positive ion, or the positive end of a polar molecule. The electron withdraw-ing nitro group makes it more difficult for an electrophilic species to add onto the benzene ring; vigorous conditions are required for further nitration of nitrobenzene. The nitro group will "direct" the substitution of additional nitronium ions in a meta orientation. With this in mind, the structure of the isomer of trinitrobenzene is at once known. Each nitronium ion is di-rected to the meta position of the other ion. Thus, the structure is that of figure B and has the formula 1,3,5-trinitrobenzene.

Question:

A wire of diameter 1 mm carrying a heavy current has a temperature of 1200\textdegreeK when equilibrium has been achieved. We assume that the heat lost from the wire is purely by radiation. The temperature of the sur-roundings is 300\textdegreeK, the resistivity of the wire at this temperature is 5 × 10^-8 \Omega \bullet m, and the temperature coefficient of resistance of the wire is 4 × 10^-3 per C deg. What is the magnitude of the current in the wire?

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Solution:

Since the head is being lost by radiation only, the energy lost per second by a 1 - m length of the wire (i.e. the intensity of radiation emitted by the wire)is given by Stefan's law W =A\sigma(T^4 - T^4 _0 ) , where A is the surface area of the length of wire and \sigma is Stefan's constant. The wire is assumed to radiate as a blackbody (i.e. it emits all of the radiation produced by the excited atoms of the wire). T_0 is the wire temperature and T is the ambient temperature. But this energy is supplied by the current flowing. Thus, if R is resistance of 1 m of the wire, then the energy supplied per second is Power = I^2 R = W =A\sigma(T^4 - T^4 _0 ). But R = (\rhol/A') where A' is the cross-sectional area, l is the length and \rho is the resistivity of the wire at temperature T. It is related to T by the following relation: \rho = \rho_0 [1 + \alpha (T - T_0)], where \rho_0 is the resistivity at 300\textdegreeK, and \alpha is the temperature coefficient of resistance. Hence R = [{\rho_0 [1 + \alpha (T - T_0)] l}/(A')] But, for 1 meter of wire, P = W and I^2 = [{A\sigma(T^4 - T^4 _0)} / ({\rho_0 [1 + \alpha (T - T_0)] l} / {A'})] = [{AA'\sigma(T^4 - T^4 _0)} / {\rho_0 [1 + \alpha (T - T_0)] l}] = [{1m×2\pi×(1/2)×10^-3m×\pi[(1/2)×10^-3 ]^2 m^2 ×5.67×10^-8 W\bulletm^-2 (K deg)^-4 × (1200^4 - 300^4 ) (K deg)} / {5×10^-8 \Omega\bulletm[1 + 4 ×10^-3 (K deg)^-1 (1200 - 300) K deg] × 1m}] = 1258 A^2 . \thereforeI = 35.5A.

Question:

Construct a tree which represents the Polish prefix expres-sion + XA - B12 \div C4. Remember that in Polish notation, all oper-ators have the same precedence; only their position is sig-nificant to the order of operations.

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Solution:

Trees are useful to display the way in which the computer translates the algebraic form of an expression (the way in which you generally enter various formulae) into Polish notation during the process of compiling a program. As a general rule, a tree will have operators (+, -, ×, \div , \uparrow) in its nonterminal nodes and operands (constants or variables) in its terminal nodes. The root node is the exception; it may be thought of as representing the middle operator of an expression. In Polish prefix notation, however, the middle operator actually comes at the beginning of the string. To begin constructing the tree, we number each element to correspond with the number of the node it will occupy. Hence, + × A - B 12 \div C 4 1 2 3 4 5 6 7 8 9 The first element (+) will be the root node. We see that the next element is (x). Remembering that trees are always constructed left to right, this element becomes the left subnode of the root. Since it is also an operator, we will neglect the right subnode for a moment in order to find the two operands which appear on either side of the (x). (See Figure 1.) The following figures should illustrate the growth of the tree in a clearer fashion: At this point, we need to find the operands which should appear on either side of the (-). B and 12, since they are the next two elements, will be those operators. Fig. 2 shows that the left side of the tree is complete: Finally, we notice that the next operator is (\div). The two operands, C and 4, follow directly, so we have the com-pleted tree in Figure 3.

Question:

What is a resting potential? Describe the chemical mechanism responsible for the resting potential. How can a resting potential be detected?

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Solution:

There is a difference in electrical potential between the inside and the outside of all living cells. For example, the potential difference across the plasma membrane of the neuron is measured to be about 60 milli-volts, the inside being negative with respect to the outside. This potential difference is called the resting potential. The chemical basis for the resting potential is as follows (refer to Figure). By active transport, an energy- requiring process that transfers substances across the plasma membrane against their concentration gradients, the concentration of potassium (K^+) ions is kept higher inside the cell than outside. At the same time there is a lower concentration of sodium (Na^+) ions in the cell interior than the exterior. Moreover, in the resting state the permeability of the plasma membrane is different to K+ and Na^+ ions. The membrane is more permeable to K^+ ions than to Na^+ ions. Hence, K^+ ions can move across the membrane by simple diffusion to the outside more easily than Na^+ ions can move in to replace them. Because more positive charges (K^+) leave the inside of the cell than are replenished (by Na^+), there is a net negative charge on the inside and a net positive charge on the outside. An electrical potential is established across the membrane. This potential is the resting membrane potential. We can measure the resting potential by placing one electrode, insulated except at the tip, inside the cell and a second electrode on the outside surface and con-necting the two with a suitable recording device such as a sensitive galvanometer. The reading on the galvanometer should be approximately 60 millivolts if the cell tested is a neuron. (Different types of cells, such as skeletal and cardiac muscle cells, vary in their values of resting potential.) Note, however, that if both electrodes are placed on the outside surface of the cell, no potential dif-ference between them is registered because all points on the outside are at equal potential. The same is true if both electrodes are placed on the inside surface of the plasma membrane.

Question:

Calculate the energy, in ergs and eV, of the photon which has just sufficient energy to disintegrate a deuteron. What are its frequency and wavelength?

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Solution:

We can find the minimum photon energy needed to disintegrate the deuteron by assuming that the products of the photon-deuteron collision (a proton and neutron) are at rest. Hence, the conserva-tion of total relativistic energy yields hv + M_Dc^2 = (m_p + m_n)c^2(1) where h is Planck's constant, v is the photon's frequency, and c is the speed of light. Note that the initial energy (left side of (1)) is that of the incident photon, plus the rest mass energy of the target deuteron. The final energy [right side of (1)] is the rest mass energy of the reaction products. hv = (m_p + m_n - m_D)c^2 Since m_p \approx m_n = 1.6747 × 10^-27 kg M_D = 3.3441 ×10^-27 kg hv = [(3.3494 - 3.3441) × 10^-27 kg] × (9 × 10^16 m^2/s^2) hv = 0.0053 × 9 × 10^-11 Joules hv = 4.77 × 10-1 3Joules(2) But1 Joule = 6.242 × 101 8eV hv = (4.77 × 10-1 3J) (6.242 × 101 8J/eV) hv = 2.977 × 10^6 eV From(2) , we obtain v = [(4.77 × 10-1 3J)/(6.63 × 10-3 4J\textbullets)] = 7.2 × 10^20 s^-1 Because the wavelength of the photon is we find \lambda = (c/v) we find \lambda = [(3 × 10^8 m/s) / (7.2 × 10^20 s^-1)] = 4.2 × 10^-13 m Note that the above analysis gives only a lower bound on hv. The reason for this can be seen by examining the figure. In order to conserve momentum, the react-ion products must be in motion, contrary to what we assumed in equation (1). Hence, in reality, hv is greater than the value we calculated, since some of the photon's energy goes into the kinetic energy of the reaction products.

Question:

Describe the procedures involved in adding and subtracting hexadecimalnumbers.

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Solution:

The process of adding and subtracting hexadecimal (Hex) digits is almostidentical to that of adding and subtracting Arabic digits. Actually, thereare a few ways in which these calculations may be per-formed. The mostefficient one is the method used in the decimal cal-culations. The operationswill be done from right to left. In using this method, each Hex symbolwill be translated to a decimal digit before the calculations, then, uponcompletion of the calculations, each re-sulting decimal digit will be retranslatedto a corresponding Hex symbol. Addition: In decimal calculations, as we go from one column to the next, wecarry units of tens. For example: col2col1col\o 128 88 216 In Hex addition, instead of carrying unitsof ten, we carry units of sixteen. Example 1 col3col2col1col\o 6E88 +1D3 Forcol\o: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum 888 333 11\rightarrow B Note in this column the decimal sum is less than 16, and no units of sixteenare carried. Forcol1: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum 888 D1313 21\rightarrow 16 + 5 Here the decimal sum is greater than 16. After factoring out the units of sixteen(here only one), the remaining digit should be placed in this column. Forcol2: Hex \rightarrow Decimal \rightarrow Decimal Sum \rightarrow Hex Sum E1414 111 15 + 1(fromcol1) 16 \rightarrow 16 + \o One unit of sixteen can be factored out, leaving zero as the remaining digit forcol2. Forcol3: Hex\rightarrowDecimal\rightarrow Decimal Sum \rightarrow Hex Sum 666 6 + 1(fromcol2) 7 \rightarrow7 Thus, the final result is: 6 E 88 + 1 D 3 7 0 5 B Subtraction: In the subtraction procedure the units of sixteen are borrowed insteadof being carried. Example 2: col2col1col\o 35E -2B8 Forcol\o Hex \rightarrow Decimal \rightarrow Decimal Difference \rightarrow Hex Difference E1414 88- 8 6 \rightarrow 6 Forcol1: Hex \rightarrow Decimal \rightarrow Decimal Difference \rightarrow Hex Difference 55(5) + 16 (Borrowed unit fromcol2) B1111 10\rightarrowA The difference incol2 is zero since one unit was borrowed from 3. The finalresult is: 2+16 35E -2B8 A6

Question:

.0324 Faradays (F) liberated .651g of Calcium. What is the atomic weight of Calcium?

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Solution:

A faraday may be defined as the number of coulombs or charges of electricity that liberate one gram-equivalent weight of an element in solution. Note, the equivalent weight of a substance is the amount of substance which 1 mole of electrons can oxidize or reduce into a neutral species. Calcium has an equivalent weight of (1 / 2) its atomic weight, since its oxidation state is +2. In other words, one mole of electrons will neutralize only half of the calcium ions present, since each ion needs two electrons. Thus, if you can find the gram-equivalent weight of calcium, you can find its atomic weight by multiplying it by two. However, you can determine the number of gram-equivalents given the number of faradays of electricity used. You are told that .0324F deposits .651g. By definition, 1F deposits the gram-equivalent weight. You have, then, the following proportion: (.651g Ca) / (.0324 )F = (g-eq.wt. of Ca) / (1F) = 20.1. Hence, 20.1g is the gram-equivalent of calcium. Thus, the atomic weight is found to be 2 × 20.1 = 40.2.

Question:

On a given diet yielding 3000 Kcal per day, a 70 kg man excretes27.0g of urea daily. What percentage of his daily energy requirement is met by protein? Assume that gram of protein yields 4 Kcal and 0.16g of nitrogen as urea.

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Solution:

If 1.0 g of protein yields 0.16 g of urea, and 27.0 g of urea is excreted daily, then the man eats 27. 0 g/0.16 g or 168.75 gof protein a day. Each gram of protein contributes 4 Kcal towards the energy re-quirement. Therefore, 168.75 gof protein produce 4 × 168.75or 675 Kcal.675 Kcal is (675/3000) × 100 or 22.5% of the daily energy requirement.

Question:

Given the function f(x,y,z) below, write f(x,y,z) as a product ofmaxterms. f(x,y,z) = (z +x) (y +z) (x + y + z) (x+y)

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G03-0055.htm

Solution:

If a switching function is specified in Product of Sum form, then it may be expanded to canonical PS form by repeatedly using the theorem given below. THEOREM: (a + b) (a +b) = a Note: in the original function f(x,y,z), the first, second, and the fourth terms can be changed using the theorem defined above. These changes will be; (z +x) = (x+ y + z) (x+y+ z) (y +z) = (x + y +z) (x+ y +z) (x+y) = (x+y+ z) (x+y+z) then the function f(x,y,z) becomes; f(x,y,z) = (x+ y + z) (x+y+z) (x + y +z) \bullet (x+ y +z) (x+y+ z) (x+y+z) \bullet (x + y + z) this is the desired final form of the function, now it can be translated into MAXTERM code by assigning 1's to false variables, and 0's to true variables. This procedure of coding MAXTERMS is illustrated below using the function f(x,y,z); f(x,y,z) = (x+ y + z) (x+y+z) (x + y +z) 100111001 (x+ y +z) (x+y+ z) (x+y+z) 101110111 (x + y + z) 000 In decimal arithmetic each code group represents numbers such as 4,7,1,5,6,0, with the redundant 7 being used only once, and the solution can be written as; f(x,y,z) = M_4 M_7 M_1 M_5 M_6 M_0 or f(x,y,z) = \prod M(0,1,4,5,6,7)

Question:

Using Vein der Waal's equation, calculate the pressure exerted by 1 mole of carbon dioxide at 0\textdegreeC in a volume of (a) 1.00 liter,(b) 0.05 liter.

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Solution:

Real gases are more compressible than ideal gases because their molecules attract each other. The intermolecular attraction is provided for by adding to the observed pressure a term of n^2 a/V^2 in the ideal gas law (PV = nRT) where n is the number of moles, V is the volume, and a is a van der Waal's constant. At very high pressures, real gases occupy larger volumes than ideal gases; this effect is provided for by subtracting an excluded volume nb from the observed volume V to give the actual volume in which the molecules move (b is a constant). Thus, van der Waal's full equation is: [p + (n^2 a/V^2 )] (V - nb) = nRT To solve this problem one must 1) convert the tem-perature from units of \textdegreeC to \textdegreeK by adding 273\textdegree to the \textdegreeC temperature. This is done because T is expressed in the absolute. (P = pressure and R = gas constant.) 2)Find the values for the constants (a) and (b) (these can be found in any reference text). For CO_2; a = 3.592 liter^2 atm/mole^2; b = 0.04267 liter/mole. 3)Sub-stitute the known values into the van der Waal's equation and solve for pressure. Known values: n = 1R = 0.082 [(liter-atm)/(mole-\textdegreeK)]T = 273\textdegreeK (a)V = 1.00 liter.Substituting, [P + {(1)^2 (3.592)}/(1.00)^2 ] [1.00 - (1) (0.04267)] = (1) (0.082) (273) P = [{(0.082) ( 273\textdegreeK)}/(1.00 - 0.04267)] - [(3.592)/(1.00)^2 ] = 23.38 - 3.592 = 19.79 atm (b)V = 0.05 liter.Substitute [P + {(1)^2 (3.592)}/(0.05)^2 ] [0.05 - (1) 0.04267] = (1) (0.082) (273 K) P = [{(0.082) ( 273\textdegreeK)}/(0.05 - 0.04267)] - [(3.592)/(0.05)^2 ] = 3054 - 1437 = 1617 atm.

Question:

Memory cell, Registers, Decoders, Multiplexers.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G05-0092.htm

Solution:

The memory cell is the basic unit of memory. It can store 1 bit of information (i.e. it can have value "1" in it or value "0" in it). The cell may be either a core memory or a flip-flop. Registers: A group of flip-flops constitutes a Register. A 4 bit Register is made up of 4 flip-flops and holds 4 bits of information. Registers can be used in various ways; for example, as counters, shift registers, Incre-ment/Decrement counters with parallel load, etc. These are temporary storage devices used in the internal storage of intermediate data in the CPU. Decoders: Decoders are devices by which we can divide n number of input lines to it into 2^n output lines; see Figure 1. The decoder is used in conjunction with a truth table, Figure 2. From the truth table we can find the decoder's output for any combination of the inputs; enable and bits A_1, A_2, A_3. INPUTS OUTPUT Enable A_1 A_2 A_3 C1 C2 C3 C4 C5 C6 C7 C8 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 0 0 1 Fig. 2. Truth Table for 3 line to 8 line decoder Multiplexer: This is just the reverse of the Decoder. Given 2^n input lines there will be only one output line, see Figure 3. Here, which input lines proceed to output depends on the bit combination of the n select lines. The truth table for the 4 line to 1 line multiplexer, shown in Figure 3, is given in Figure 4. INPUTS OUTPUT S_0 S_1 X 0 0 A 0 1 B 1 0 C 1 1 D Fig. 4

Question:

In the change from unicellular organisms to multi-cellular organisms, a number of different events occur. Describe these changes as they are evidenced by the motile-colony series of green algae.

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Solution:

The motile-colony series, or the volvocine series, of Chlorophyta shows a gradual progression from the unicellular condition of Chlamydomonas to elaborate colonial organization. Gonium is an example of the simplest colonial stage. The cells of a Gonium colony are morphologically similar to the unicellular Chlamydomonas. Each cell has a single large cup-shaped chloroplast, containing a pyrenoid which functions in starch synthesis, and an eye spot or stigma which functions in light detection. There are two flagella at the anterior end of each cell, and near the base of the flagella two small contractile vacuoles, which dis-charge rhythmically, expelling excess water from the cell. The Gonium cells are embedded in a mucilaginous matrix. There is coordination of activity between the cells, since the flagella of all the cells beat together, enabling the colony to swim as a unit. Pandorina is at a more advanced stage of develop-ment than Gonium. The colony has anterior and posterior halves, and there is regional differentiation. The colony swims in an oriented fashion, with a definite anterior half. The stigmas are larger in the anterior cells. In addition, the vegetative cells of the colony are so dependent on one another that they cannot live apart from the colony, and the colony cannot survive if broken up. The genus Pleodorina is further advanced with a considerable division of labor among the cells. The anterior cells are purely vegetative, and the posterior cells are larger and function in both sexual and asexual reproduction. Each Gonium colony consists of 4, 16 or 32 cells and the Pandorina colony contains 8, 16, or 32 cells, and the Pleodorina colony has 32 to 128 cells. Volvox is the most advanced of the volvocine series. The colony consists of 500 to 50,000 cells. All the anterior cells are exclusively vegetative, and the cells in the posterior half of the colony are much larger than the vegetative cells, and are specialized for asexual reproduction. This is the production of male gametes, or of non-motile egg cells. The changes represented by the volvocine series are as follows: (1) increasing size of colonies, (2) increasing coordination of activity among the cells, (3) increasing interdependence among the cells, so that vegetative cells cannot survive apart from the colony, and (4) increasing division of labor, particularly between vegetative and reproductive cells.

Question:

The two adrenal glands lie very close to the kidneys. Each adrenal gland in mammals is actually a double gland, composed of an inner corelike medulla and an outer cortex. Each of these is functionally unrelated. Outline the function of the adrenal medulla.

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Solution:

The adrenal medulla secretes two hormones, adrenalin (epinephrine) and noradrenalin (norepinephrine, NE), whose functions are very similar but not identical. The adrenal medulla is derived embryologically from neural tis-sue. It has been likened to an overgrown sympathetic gan-glion whose cell bodies do not send out nerve fibers, but release their active substances directly into the blood, thereby fulfilling the criteria for an endocrine gland. In controlling epinephrine secretion, the adrenal medulla behaves just like any sympathetic ganglion, and is dependent upon stimulation by sympathetic preganglionic fibers, as shown below: Epinephrine promotes several responses, all of which are helpful in coping with emergencies: the blood pressure rises, the heart rate increases, the glucose content of the blood rises because of glycogen breakdown, the spleen contracts and squeezes out a reserve supply of blood, the clotting time of blood is decreased, the pupils dilate, the blood flow to skeletal muscle increases, the blood supply to intestinal smooth muscle decreases, and hairs become erect. These adrenal functions, which mobilize the resources of the body in emergencies, have been called the fight-or-flight response. Norepinephrine stimulates reactions similar to those produced by epineph-rine, but is less effective in the conversion of glycogen into glucose. The significance of the adrenal medulla may seem questionable since the complete removal of the gland causes few noticeable changes in the animal; the animal can still exhibit the fight-or-flight response. This occurs because the sympathetic nervous system compliments the adrenal medulla in stimulating the fight-or-flight response, and the absence of the hormonal control will be compensated for by the nervous system.

Question:

The resistance of a copper wire 2500 cm long and 0.90 cm in diameter is 0.67 ohm at 20\textdegreeC. What is the resistivity of copper at this temperature?

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Solution:

The resistivity of a conductor is directly proportional to the cross-sectional area A and its resistance, and inversely proportional to its length l. Therefore, knowing the resistance, we have \rho = R (A/l) = [(0.67 ohm)/(2500 cm)] [{\pi (0.090 cm)^2 } / (4)] = 1.7 × 10^-6 ohm-cm Resistivity is a characteristic of a material as a whole, rather than of a particular piece of it. For a given temperature, it is constant for the particular medium, as opposed to resistance which depends on the dimensions of the piece.

Question:

A horizontal shelf moves vertically with simple harmonic motion, the period of which is 1 s and the amplitude of which is 30 cm. A light particle is laid on the shelf when it is at its lowest position. Determine the point at which the particle leaves the shelf and the height to which it rises from that position, g being taken as \pi^2 m\bullets^-2.

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Solution:

The period of a simple harmonic motion is given by the expression T = 1/f = 2\pi/2\pif = 2\pi/\omega Therefore, the angular frequency in this case is \omega = 2\pi/T = 2\pi/l s = 2\pi rad\bullets^-1. The only forces acting on the particle are its weight mg^\ding{217} downward and the normal force N\ding{217}exerted by the shelf upward. (See the figure.) At \ding{217} any time, according to Newton's second law, N - mg = ma, where a is the upward acceleration of shelf and particle. If a = -g, (i.e. the shelf accelerates downward with magnitude g) N becomes zero and, if a becomes more negative, the shelf is retarded at a greater rate than the particle; therefore the particle moves away from the shelf. Since the shelf undergoes simple harmonic motion, its displacement may be described by x = A cos (\omegat + \delta), where A is the amplitude of the oscillation, and \delta is a phase factor dependent on the initial conditions (i.e. where the shelf is at t = 0). The acceleration a = d^2x/dt^2 = -\omega^2 A cos (\omegat + \delta) = -\omega^2x. Hence, the displacement x at which the particle leaves the shelf is given by -g = -\omega^2xor x = g/\omega^2 = (\pi^2 m\bullets^-2)/(4\pir2rad^2\bullets^-2) = 1/4 m. (for g = \pi^2 m\bullets^-2 and \omega = 2\pi rad\bulletsec^-1). The particle thus leaves the shelf when it is 1/4 m above the mean position. At that point the common velocity of shelf and particle is given by the formula relating to velocity to displacement for simple harmonic motion: v = \pm \omega\surd(A^2 - x^2) Since the shelf is rising, v is positive and v = 2\pi rad\bulletsec^-1 ×\surd[(0.3m)^2 - (0.25m)^2] = 2\pi\surd(0.0275) m\bullets^-1. We now have a new problem concerning a particle thrown upward from a platform with an initial speed v. If the platform level is taken as the reference level for measuring potential energy, then the law of conservation of energy requires that at each moment of time that the particle is in motion, the sum of its kinetic and potential energies must remain constant. At the time the particle leaves the platform, its potential energy is zero and E_T = PE + KE = 0 + 1/2 mv^2. At its maximum height h, v = 0 and E_T = PE + KE = mgh + 0 \therefore 1/2 mv^2 = mghor h = v^2/2g = (4\pi^2 ×0.0275 m^2\bullets^-2)/(2 ×\pi^2 m\bullets^-2) = 0.055 m = 5.5 cm.

Question:

Two springs are joined and connected to a mass m which rides on a frictionless surface (see figure (a)). What is the frequency v of the oscillation that will result if the mass is displaced a small distance x?

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Solution:

We must show that the configuration of figure (a), is equivalent to that of mass m, connected to a single spring having force constant k'. In figure (b), a mass m is shown connected to such a spring. A segment of the spring is divided into two sections of length x_1 and x_2. The tension in these two sections must be equal, as must be true of the tension in any other arbitrary sections into which the spring may be divided. If this were not the case, the spring would buckle. In general, when the spring is stretched or compressed by a force F, the changes in length ∆x_1 and ∆x_2 of sections x_1 and x_2 will not be equal since the magnitudes of such distortions will be proportional to the lengths of the segments involved. We can see this by symmetry. If a spring is stretched to twice its usual length, for example, each arbitrary section must stretch to twice its usual length. If we choose the sections to be of unequal length, the amount that each section stretches will be different. Thus: F = -k_1 ∆x_1 = -k_2 ∆x_2 since we know that the force that the spring exerts on the mass equals the tension within each section. The force constants of sections x_1 and x_2 are represent-ed by k_1 and k_2. It is not unreasonable to assume that each arbitrary section of the spring can be considered to have a dif-ferent force constant as long as: F= -k_1 ∆x_1 = -k_2 ∆x_2 = -k_3 ∆x_3 = . . . -k_n ∆x_n = -k'x where x = ∆x_1 + ∆x_2 + ∆x_3 + . . . + ∆x_n is the total displacement of the mass. Thus we see that in the case of the two connected springs above, the tension in each must be equal: F = -k_1 ∆x_1 = -k_2 ∆x_2 = -k'x = -k'(∆x_1 + ∆x_2) ∆x_1 = -(F/k_1) ∆x_2 =-(F/k_2) Thus: k' = -[F/(∆x_1 + ∆x_2)] = -[F/{-(F/k_1) - (F/k_1)}] = [1/{(1/k_1) + (1/k_2)}] = 1/[(k_1 +k_2)/(k_1k_2)] = (k_1k_2)/(k_1 +k_2) Since the frequency of an oscillator having force constant k' is: v = [1/(2\pi)]\surd(k'/m) then:v = [1/(2\pi)]\surd[(k_1k_2)/{(k_1 + k_2)m}]

Question:

The Eiffel Tower is 984 feet high. Express this height in meters, in kilometers, in centimeters, and in millimeters.

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Solution:

A meter is equivalent to 39.370 inches. In this problem, the height of the tower in feet must be converted to inches and then the inches can be converted to meters. There are 12 inches in 1 foot. Therefore, feet can be converted to inches by using the factor 12 inches/1 foot. 984 feet × 12 inches/1 foot = 118 × 10^2 inches. Once the height is found in inches, this can be converted to meters-by the factor 1 meter/39.370 inches. 11808 inches × 1 meter/39.370 inches = 300 m. Therefore, the height in meters is 300 m. There are 1,000 meters in one kilometer. Meters can be converted to kilometers by using the factor 1 km/1000 m. 300 m × 1 km/1000 m = .300 km. As such, there are .300 kilometers in 300m. There are 100 centimeters in 1 meter, thus meters can be converted to centimeters by multiplying by the factor 100 cm/1 m. 300 m × 100 cm/1 m = 300 × 102cm. There are 30,000 centimeters in 300 m. There are 1,000 millimeters in 1 meter; therefore, meters can be converted to millimeters by the factor 1000 mm./1 m. 300 m × 1,000 mm/1 m = 300 × 10^3 mm. There are 300,000 millimeters in 300 meters.

Question:

Draw the inter-connections between the control section (CPU) and the memory section of the computer and state the purpose of each inter-connection. What takes place when the CPU transfers data to or from memory?

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Solution:

Figure 1 shows the inter-connections between the CPU and memory section of the computer. The address bus transfers the address of the location to be read from, or written into, the memory section The read/write line designates whether data is to be read from memory or written into memory. The output data bus is used to transmit data to be written into memory. The input data bus is used to transmit data read from memory. Many times a single data bus is used for more than one purpose. However, more control lines are needed to designate the purpose of the bus. Such multifunction buses are said to be "time shared" or "multiplexed." The CPU transfers data to the memory section as follows: 1.The CPU sends the address to the memory section via the address bus. 2.The CPU sends the data to the memory section via the output data bus. 3.The CPU signals a Write operation to the memory sec-tion via the Read/Write line. The CPU transfers data from memory as follows: 1.The CPU sends the address to the memory section via the address bus. 2.The CPU signals a Read operation to the memory sec-tion via the Read/Write line. 3.The CPU waits, then the data is sent out to the CPU via the input data bus.

Question:

A bullet weighing 4 g is fired at a speed of 600 m/s into a ballistic pendulum of weight 1 kg and thickness 25 cm. The bullet passes through the pendulum and emerges with a speed of 100 m/s. Calculate the constant retarding force acting on the bullet in its passage through the block, and the height to which the pendulum rises.

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Solution:

The ballistic pendulum is used to measure bullet speeds. The three stages of the motion of the system are as shown in the diagram. In the first stage, the bullet of mass m approaches with velocity V^\ding{217} the ballistic pendulum of mass M. In the second, the bullet, having passed through the pendulum, is moving off with velocity v_1^\ding{217}, leaving the pendulum just starting to move with velocity v_2^\ding{217} Since momentum must be conserved, mV^\ding{217} = mv_1^\ding{217} + Mv_2^\ding{217}. \thereforev_2 = [mg(V - v_1)] / [Mg] = (4 × 10-3Kg (600 - 100) m/s)] / (1 kg) = 2 m/s. In the third stage, the pendulum, which has acquired kinetic energy (1/2) Mv2_2, swings through a certain angle such that the bob loses all its kinetic energy but gains an equivalent quantity of potential energy in rising a height h, in accordance with the principle of conservation of energy. But = Mgh = (1/2)MV^2 _2.Therefore h = \surd(V22/2g) = 0.45 m = 45 cm. The loss of kinetic energy as the bullet passes through the pendulum is (1/2) mV^2 - (1/2) mv2_1 - (1/2) Mv22 = [(4 × 10^-3 kg) / (2)] (600^2 - 100^2) m^2/s2 - [(1kg) / 2] × 4 m^2/s^2 = 698 J. By the principle of conservation of energy, this quan-tity of energy must represent the work done against the retarding force, F^\ding{217}, as the bullet pushes its way through the pendulum. Thus, W = F^\ding{217} \bullet s^\ding{217} = F × 0.25 m = 698 J or F = 4 m^-1 × 698 J = 2792 N.

Question:

The space between the plates of a parallel-plate capacitor is filled with dielectric of coefficient 2.5 and strength \textasciigrave5 ×10^6V \bullet m^\rule{1em}{1pt}1. The plates are 2mm apart. What is the maximum voltage which can be applied between the plates? What area of plates will give a capacitance of 10^\rule{1em}{1pt}3\muF, and at maximum voltage what are the free and bound charges per unit area of the plate and dielectric surface?

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Solution:

The dielectric strength of the dielectric is the largest electric field which it can withstand before becoming a conductor. We must relate the voltage between the capacitor plates to E_max (dielectric strength). This can be done by realizing that voltage differences are defined V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}^b \int_a E^\ding{217} \bullet dl^\ding{217}(1) where V_b,V_a refer to the potentials at z = b and z = a, respectively (see figure), and E^\ding{217} is the electric field. The integral in (1) is a line integral, and it is to be evaluated over an arbitrary path connect-ing a and b, dl^\ding{217} being a small element of path length. Looking at the figure, we see that E^\ding{217} is composed of contributions from two sources-the conducting plates (E^\ding{217}_c)and the dielectric material (E^\ding{217}_D). Note that the latter is in opposition to E^\ding{217}_C . Since E^\ding{217} is uniform in direction and magnitude for a parallel-plate capacitor,(1) becomes V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}E^\ding{217} \bulletb\int_a dl^\ding{217}(2) The easiest path to evaluate this line integral along is straight line path from z = a to z = b. In this case. dl^\ding{217} = dl k˄ Since E^\ding{217} = \rule{1em}{1pt}E k˄ (2) becomes V_b \rule{1em}{1pt} V_a = \rule{1em}{1pt}(\rule{1em}{1pt}Ek˄) \bullet ^b\int_a (dl k˄) V_b \rule{1em}{1pt} V_a = E(b-a) But b \rule{1em}{1pt} a = d (see figure) and we define Vb\rule{1em}{1pt} V_a = ∆V as the potential difference between the plates. Therefore, ∆V = Ed(3) The maximum voltage which can be applied between the plates is ∆V_max = E_max d = [5 × 106(V / m)](2 × 10^\rule{1em}{1pt}3 m) ∆V_max = 10^4V To answer the second part of the problem, we must relate the capacitance of the capacitor to its geometry. We do this by beginning with the definition of capacitance (C) C = Q / ∆V(4) where Q is the charge on 1 capacitor plate. If the free charge density on the plate of cross-sectional area A is \sigma_f, (see figure) then Q = \sigma_fA(5) Using (5) and (3) in (4) C = (\sigma_fA) / Ed(6) Since we don't know E in this portion of the problem, we must eliminate it from (6), This may be done by using Gauss' Law for a dielectric \oint_s E^\ding{217}. ds^\ding{217} = (Q / \epsilon)net(7) where \epsilon is a constant, andQ_net is the total free charge enclosed by the surface s over which the integral is evaluated. Evaluating (7) over the Gaussian surface indicated in the figure, \oint_a E^\ding{217} \bullet ds^\ding{217} = \int_(s)T E^\ding{217}(s_T)\bullet ds^\ding{217} + \int_(S)B E^\ding{217}(s_E)\bulletda^\ding{217}= Q_net / \epsilon(8) ^\ding{217} where S_T and S_B are the top and bottom of the rectangular surface. Now, ds^\ding{217} is a vector element of surface area. For S_T, ds^\ding{217} = ds k˄ and for S_B, ds^\ding{217} = \rule{1em}{1pt}ds k˄. Then, using the fact that E^\ding{217} = \rule{1em}{1pt}Ek˄ everywhere, (8) becomes \oint_S E^\ding{217} \bullet ds^\ding{217} = \int_(s)T {\rule{1em}{1pt}E^\ding{217}(s_T)k˄}\bullet(ds^\ding{217}k˄) + \int_(S)B {\rule{1em}{1pt}E(s_B)k˄}\bullet(-ds k˄) = Q_net / \epsilon \oint_s E^\ding{217} \bullet ds^\ding{217} = \rule{1em}{1pt}\int_(s)T E(s_T)(ds) + \int E(s_B)(ds) = Q_net / \epsilon Now, E(S_T) is the value of E at S_T, which lies in the top conduct-ing plate. By definition of a conductor, then,E(S_T) = 0, Therefore \oint_s E^\ding{217} \bullet ds^\ding{217} = \int_(s)B E(s_B)(ds) = Q_net / \epsilon(9) But E(s_B) = E, the electric field between the capacitor plates. This is constant, and (9) becomes E \int_(s)B ds =EA = Q_net / \epsilon since the area of s_B is A. Hence E = Q_net / \epsilonA. Now, Q_net includes the free charge found within the Gaussian surface. Therefore (see figure) Q_net / \epsilon = \sigma_fA and, E = \sigma_f / \epsilon(10) Using (10) in (6) C = [(\sigma_fA) / {(\sigma_f / \epsilon)d}] = \epsilonA / d But K, the dielectric constant) is defined by K = \epsilon / \epsilon_0 Hence,\epsilon = \epsilon_0Kand C = k\epsilon_0A / d(11) Solving for A A = Cd / K\epsilon_0(12) Substituting the given data in (12) A ={(10^\rule{1em}{1pt}3 \muF)(2 × 10^\rule{1em}{1pt}3m)} / {(2.5)(8.85 × 10^\rule{1em}{1pt}12 c^2/N \bullet m^2)} A = .09m^2 The free charge density can be found using (10) \sigma_f = \epsilonE = \epsilon_0KE or \sigma_f = (5 × 10^6 V/m)(2.5)( 8.85 × 10^\rule{1em}{1pt}12 C^2 / N \bullet m^2) \sigma_f = 110.6 × 10^\rule{1em}{1pt}6 C / m^2 From the figure, note that the net field E^\ding{217} is the superposition of the field due to the bound charges of the dielectric (E^\ding{217}_D) and the field due to the free charge on the capacitor plates (E^\ding{217}_C). Then E^\ding{217}= E^\ding{217}_C + E^\ding{217}_D(13) ^\ding{217} But by (10) and the figure, E^\ding{217} = \rule{1em}{1pt} (\sigma_f / \epsilon)(k˄) . Since the bound charge accumulates only on the surface of the dielectric, we may consider the dielectric slab to be equivalent to a parallel plate capacitor with surface density \sigma_b and an air dielectric. E^\ding{217}_D = (\sigma_b / \epsilon_0 )(k˄) Similarly, E^\ding{217}_C = \rule{1em}{1pt} (\sigma_f / \epsilon_0 )(k˄) Hence, from (13) \rule{1em}{1pt}(\sigma_f / \epsilon) (k˄) = \rule{1em}{1pt}(\sigma_f / \epsilon_0 )(k˄) + (\sigma_b / \epsilon_0 )(k˄) or \rule{1em}{1pt}\sigma_f = \rule{1em}{1pt}\sigma_f(\epsilon / \epsilon_0) + \sigma_b (\epsilon / \epsilon_0) \sigma_f [(\epsilon / \epsilon_0) \rule{1em}{1pt} 1] = \sigma_b(\epsilon / \epsilon_0) Since K = \epsilon / \epsilon_0 \sigma_b = \sigma_f [(k \rule{1em}{1pt} 1) / (k)] = [(1.5) / (2.5)] \sigma_f = [3 / 5] \sigmaf Noting that \sigma_f = 110.6 × 10\rule{1em}{1pt}6c/m^2 \sigma_b = (331.8 / 5)×10\rule{1em}{1pt}6c/m^2 \sigmab=66.4 ×10\rule{1em}{1pt}5c/m^2

Question:

A woman who is Rh-negative marries a man who is Rh- positive. What possible problems could arise if the couple wants more than one child?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0371.htm

Solution:

Besides the A and B agglutinogens, there is another type of blood antigen, designated the Rh factor. There are at least eight different types of Rh antigens which may be present on human erythrocytes. People are classified into two groups with respect to the Rh antigen: Rh-positive individuals have an Rh antigen, while Rh- negative individuals have such a weakly antigenic factor that it can be disregarded. The A\rule{1em}{1pt}B\rule{1em}{1pt}O blood type system is an example of a rare case in which the body normally synthesizes antibodies in large amounts against an antigen to which it has not yet been exposed. The Rh system follows the rule rather than the exception. An individual's plasma will not con-tain an antibodies against the Rh antigen unless it has been sensitized to the antigen by previous exposure. Thus, while Rh-positive individuals have the Rh antigen, Rh-negative individuals have neither an effective antigen nor an antibody against the antigen. If Rh-positive blood is given to an Rh negative individual, the Rh-negative individual will gradually synthesize antibodies against the Rh antigen introduced. Agglutination of the cells occurs, plugging up the small blood vessels, inactivating the erythrocytes as oxygen carriers, and eventually leading to death of the recipient. In our particular problem, the couple's first child might inherit the Rh-negative characteristic. There is thus no particular problem since the mother is also Rh-negative. But if the fetus is Rh-positive, trouble may occur. If some blood from the Rh-positive fetus enters the maternal circulation, its antigens stimulate the mother's plasma cells to synthesize Rh antibodies. Normally, the mother cannot produce antibodies rapidly enough and in large enough amounts to affect the first child before birth. But if the mother, who is Rh-sensitized by the first pregnancy, has subsequent pregnancies, she will have produced sufficient antibodies which can diffuse into the blood of an Rh-positive fetus and react with its erythrocytes. This agglutination of the blood of the fetus with the antibodies produced in its mother is a disease known as erythroblastosis fetalis. Only about 3% of the second or later Rh-positive fetuses born to Rh-negative mothers show signs of erythroblastosis fetalis. The blood of the mother may not diffuse into the fetal circulation, thus preventing agglutination. In addition, only one of the types of Rh antigens is responsible for erythroblastosis; other forms will not cause any trouble. Also, the longer the time interval between pregnancies, the less intense is the anti-Rh immunal response in the mother. One method of preventing erythroblastosis is to in-ject Rh antibodies into a Rh-negative mother immediately after the birth of her first Rh positive infant. The antibodies kill any fetal red blood cells that have en tered her circulation and thus prevent stimulation of maternal antibody production. The injected antibodies eventually are destroyed arid the mother is now unsensi-tized and able to bear a second Rh-positive infant.

Question:

Discuss various types of computer systems in terms of the level of expense, sophistication, and capabilities to carry out intended commercial and scientific data processing tasks.

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Solution:

Traditionally, the computer systems from lowest to highest level of capabilities have been classified as: Microcom-puters, Minicomputers, and Mainframes. It used to be that micros such as INTEL 8085 had 8-bit registers, minis such as PDP-11 had 16-bit registers, and mainframes such as IBM 370 had 32-bit registers. Note that the amount of addressable Random Access Memory (RAM) is dependent upon the register length. Accordingly, micros had small, minis had larger, and mainframes had much larger memories. Although the above classification still serves a useful purpose for quick sizing and comparisons, the rapid developments in VLSI technologies for fourth generation computers have blurred these distinctions. There have also been many grades and shades in between this family of computers. Generally, the mi-crocomputers have caught up with mainframes in terms of process-ing speeds and memory capacities, but the mainframes still re-tain the edge in terms of secondary storage, communication, and I/O peripherals. As of this time (1989-90), a plausible (by no means complete or precise) classification is as follows: Types and sub-types Description and example Home computers 8-16 bits registers and memory words. - Simple Commodore VIC-20 and 64 (64K RAM) -Complex Commodore AMIGA with sound and graphics Personal computers 16-32 bits registers and memory words. -Low-end IBM PC/XT, AppleIIe, 0p6 AT&T 7300, etc.m -High-end IBM PC/AT, Apple. SE, and com-patibles. -Super high-end IBM PS/2 80, Apple MACIIx, AT&T 6386, and NeXT. Personal Supercomputers Either employs 64-bit vector type processors, or a number of 16/32-bit high-end processors. A personal computer that employs INTEL 860 with 64-bit vector register would be an example of the former. IntelSugarCubethat uses 8 16-bit processors that are connected according to hypercube topology is an example the of latter. Work Stations These are generally UNIX machines that can support multiple users and have large memories and special constructions for various types of tasks such as high-resolution graph-ics symbolic processing for artifi-cial intelligence applications. The typical examples are Sun 3 and Sun series, Texas Instrument Explorer II, andSymbolicsseries. Minicomputers DEC PDP-11 and HP series of comput-ers that use 16-bit registers are typical examples. Superminicomputers DEC VAX 11 and 85XX series that use 32-bit registers, and AT&T 3B20 machines are typical examples. Mainframes IBM 370 series (formerly), and 4000 series recently are typical ex-amples . Supercomputers Cray 1, 2, X-MP, and Y-MP that use 64-bit vector CPUs are examples. Parallel Computers These are multiple processor systems that use 8 to 1024 processors that are connected according to a regular topology and cooperatively solve a large-scale problem with parallelism.

Question:

A slender homogeneous bar, shown in Figure 1, is 4 ft long and weighs 1.2 lb. A spring is attached to it as shown, 3 ft from the lower end which is hinged. When the bar is in a vertical position the spring is compressed 0.12 ft. The bar is released from rest. What should the spring con- stant be if the bar comes to rest when it just reaches the horizontal position, Figure 2?

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Solution:

The system is conservative since only spring and gravity forces are acting, and if the hinge is assumed to be frictionless. We therefore use the principle of conser-vation of mechanical energy, i.e., (KE)_1 + (PE)_1=(KE)_2 + (PE)_2 (Bar in vertical position 1)(Bar in horizontal position, 2.) Further, since the bar is at rest in both positions, the above equation reduces to (PE)_1 = (PE)_2(1) The potential energy of the system is the sum of that of the spring and the gravitational potential energy. Potential energy of a spring is (1/2) K\sigma^2 where K is the spring con-stant and \sigma its extension or compression. In position 1, \sigma_1 = 0.12 ft. Thus the free length of the spring is 1.2 + 0.12 = 1.32 ft. In position 2, the length of the spring is (4.2^2 + 3^2)^1/2 = 5.16 ft. Thus \sigma_2 = 5.16 \rule{1em}{1pt} 1.32 = 3.84 ft. The gravitational potential energy of a mass m is mgh where h is the height of its center of gravity above the datum. Let us assume the datum to be the horizontal plane passing through the hinge at O. Thus the total potential energies in the two positions are: (PE)_1 = (1/2)K (0.12)^2 + (1.2) (2) = 0.0072 K + 2.4 ft-lb (PE)_2 = (1/2)K (3.84)^2 + (1.2) (0) = 7.378 K ft-lb. Substituting these values into equation (1) yields 0.0072 K + 2.4 = 7.378 K orK = 0.3256 lb / ft. Note that this is not the equilibrium position. The bar will oscillate. The bottom position will be an amplitude extreme, where the bar is momentarily at rest. To consider the equilibrium position we need to use force equilibrium.

Question:

What are spinal nerves and what are their functions?

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Solution:

Spinal nerves belong to the peripheral nervous system (PNS). They arise as pairs at regular intervals from the spinal cord, branch, and run to various parts of the body to innervate them. In human beings, there are 31 symmetrical pairs of spinal nerves. The size of each spinal nerve is related to the size of the body area it innervates. All the spinal nerves are mixed nerves, in the sense that all have both motor and sensory components in roughly equal amounts. They contain fibers of the somatic nervous system, which are both afferent (conducting impulses towards the nervous system) and efferent (conducting impulses to effector organs). They also contain fibers of the auto-nomic nervous system, which are uniquely efferent and are separable into parasympathetic and sympathetic nervous systems. Because of this nature, the spinal nerves function to convey messages from the external environment to the central nervous system, and from the central nervous system to various effectors of the body. In other words, the spinal nerves, as part of the peripheral nervous system, serve as a link between the central nervous system and the effector organs. Somatic fibers of the spinal nerves innervate skeletal muscles of the body, and are under voluntary regulations. We can bend our arm or leg at will. The auto-nomic nerves innervate the smooth and cardiac muscles and glands of the body, and cannot be voluntarily controlled. We cannot speed up our stomach contractions or heart beat at will. The autonomic nerve fibers leave the spinal cord and run for a certain distance with the somatic fibers in the spinal nerves. Then the two types of fibers diverge and run to their respective body areas which they innervate.

Question:

The lens system of a certain portrait camera may be taken as equivalent to a thin converging lens of focal length 10.0 in. How far behind the lens should the film be located to receive the image of a person seated 50.0 in. from the lens? How large will the image be in comparison with the object?

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Solution:

The general equation for thin lenses states 1/p + 1/q = 1/f where p is the distance of the object, in this case the person, in front of the lens, and q is the distance between the image formed and the lens. The focal length is repre-sented by f. All distances are measured with respect to the optical center of the lens. This center is defined is the point through which light can pass without being bent. Since the lens is a converging one, it is convex. This means that it is thicker at the center than at the edges. Substitute values in the above equation noting that q is positive when it is on the opposite side of the lens in relation to the object. 1/q = 1/10 - 1/50 = (50 - 10) / {(10)(50)} = 4/50 q = 50/4 = 12.5 in. To find the magnification M ], the equation M = q/p can be used Substituting, M = 12.5 in. / 50.0 in. = 0.250 . The image will be one-fourth as large as the object.

Question:

Trace the sequence of events between nerve action potential and contraction and relaxation of a muscle fiber.

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/Users/wenhuchen/Documents/Crawler/Biology/F19-0466.htm

Solution:

The cell membranes of muscle fibers are excitable membranes capable of generating and propagating action potentials by mechanisms very similar to those found in nerve cells. An action potential in the muscle cell membrane results from nervous stimulation and provides the signal for the initiation of contractile activity within the muscle cell by a mechanism known as excitation-contraction coupling. An action potential is initiated and propagated in a motor axon which innervates the muscle fiber. This potential is the result of synaptic events on the cell body and dendrites of the motor neuron in the central nervous system, (see Figure 1) The action potential in the motor axon causes the release of the neurotransmitter known as acetylcholine from the axon terminals into the synaptic space between the nerve and muscle, (see Figure 2) This junction between nerve and skeletal muscles is known as the neuromuscular junction. The nerve cells which form junctions with skeletal muscles are known as motor neurons, and the cell bodies of these neurons are located in the brain and spinal cord. Once released, acetylcholine binds to receptor sites on the muscle membrane which lie directly under the terminal portion of the axon. This region of the muscle membrane is known as the motor end plate. Acetylcholine increases the permeability of the motor end plate to sodium and potassium ions, causing a depolarization of the end plate called an end-plate potential (EPP). The EPP depolarizes the muscle membrane to its threshold potential, generating a muscle action potential which is propagated over the surface of the muscle membrane. The molecules of acetylcholine re-leased from the motor-neuron have a lifetime of only about 5 milliseconds before they are destroyed by the acetylcholin-esterase, an enzyme on the muscle cell membrane near the receptors for acetylcholine. Once acetylcholine is de-stroyed, the muscle membrane permeability to sodium and po-tassium ions returns to its initial state, and the depolar-ized end plate returns to its resting potential. The muscle action potential depolarizes the T tubules at the A-I junction of the sarcomeres. This depolarization leads influx of calcium from the extracellular fluid and to the release of calcium ions from the terminal cisternae of the sarcoplasmic reticulum surrounding the myofibrils. Cal-cium ions bind to troponin in the thin actin myofilaments. This calcium-troponin complex causes tropomyosin to shift its position, releasing the inhibition that prevented actin from combining with myosin. Actin then combines with myosin. This binding activates the myosin ATPase, which splits ATP, releasing energy which is used to produce the movement of the cross bridge of the myosin molecule. ATP then binds to the myosin bridge, breaking the bond between the actin and myosin, thus allowing the cross bridge to dissociate from actin. As long as the concent-ration of calcium ions is high enough to counteract the inhibitory action of the troponin-tropomyosin system, cycles of cross-bridge contraction and relaxation will continue. The concentration of calcium ions falls as they are moved back into the sarcoplasmic reticulum by an energy- requiring process which splits ATP. Removal of calcium ions restores the inhibitory action of troponin- tropomyosin, and in the presence of ATP, actin and myosin remain in the dissociated, relaxed state.

Question:

Draw and explain a system dynamics diagram of a market model.

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Solution:

System dynamics analyzes the forces operating in a system in order to determine their influence on the stability or growth of the system. Here, stability is determined by the rates of changes of different variables. Consider the following system dynamics diagram of market model. This diagram depicts a modified exponential growth model in which a variable increases in value at first at an increas-ing rate and then later at a decreasing rate until it reaches an upper limit. This corresponds to a typical marketing situation. At first, as the product is intro-duced into the market, sales will increase rapidly. Later as the saturation point is approached, the increase in sales will gradually taper off. Next, consider the various symbols used in the System Dynamics model. The number of products sold can be viewed as a level in a reservoir of products. Levels are represented by boxes. Rates are represented by boxes with valves attached. The constants of the system are indicated by the symbol \cyrchar\CYROTLD. The differential equation corresponding to the system diagram is: ẋ = k(X - x)(4) with initial condition x(o) = 0 The solution is x = X(1 - e^-kt) which is known as a modified exponential curve. A computer simulation model would solve (4) for various values of k, the sales rate constant, and X, the market limit.

Question:

80 standard characters,and E-MESSAGE: 52 standard characters.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G11-0268.htm

Solution:

The DATA DIVISION is shown in fig. 1. The statement BLOCK CONTAINS 5 RECORDS in the INDEXED file description means that the record BAS-REC will be stored in the indexed file in blocks of 5. This allows more efficient searching for the unique keys requested in the program. The computer handles all the details of record storage and retrieval.

Question:

What is the difference betweenelectronegativityand electron affinity? Use specific examples.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E17-0625.htm

Solution:

Electron affinity represents the energy released when an electron is added to an isolated neutral gaseous atom.Electronegativityis the tendency of an element to attract electrons, not in the gaseous state, but from a nucleus joined to it by a covalent bond. The electron affinity is a precise quantitative term while theelectro-negativityis a more qualitative concept.

Question:

Two hollow spherical shells are mounted concentrically, but are insulated from one another. The inner shell has a charge Q and the outer shell is grounded. What is the electric intensity and potential in the space between them? When the outer shell is not grounded, why does a charge outside the system experience a force when the inner shell is charged?

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0614.htm

Solution:

In the region between the shells, because of the symmetry of the arrangement, the electric in-tensity E^\ding{217} must have the same magnitude at all points a distance r from the common center, and must thus be everywhere radial in direction. Hence, applying Gauss's law to a region bounded by a spherical surface of radius r (see the figure), we have \int E^\ding{217} \bullet ds^\ding{217} = Q / \epsilon_0(1) where Q is the total charge enclosed by the Gaussian surface, and dS is an element of surface area of the Gaussian surface. E^\ding{217} is parallel to dS^\ding{217}, thus (1) becomes E \intds = 4\pir^2E = Q / \epsilon_0orE = Q / 4\pi\epsilonr^2 . But E = \rule{1em}{1pt} (dV / dr) by definition of E as a potential gradient. \therefore dV = \rule{1em}{1pt}(Q / 4\pi\epsilon_0r^2)drorV =(Q / 4\pi\epsilon_0r) + C, where C is a constant of integration. But at the outer spherical shell of radiusr = b, V = 0. \therefore 0 = (Q / 4\pi\epsilon_0b) + CorC = \rule{1em}{1pt}(Q / 4\pi\epsilon_0b). \thereforeV = Q / 4\pi\epsilon_0 [(1 / r) \rule{1em}{1pt} (1 / b)]. Lines of force come from the inner shell and all end on the outer shell. The same number of lines of force end on the outer shell as start on the inner shell; hence the charge induced on the inside of the outer shell is equal and opposite to that on the inner shell. Thus a charge \rule{1em}{1pt}Q is induced on the inside of the outer shell. But this shell was initially un-charged. Hence a charge + Q must be left on the out-side of the outer shell. If the outer shell is grounded, electrons flow from the earth to neutralize this positive charge. In the absence of grounding, the positive charge remains on the outside of the shell and produces a field of force around the system which affects any other charge in the vicinity.

Question:

What cell components typically found in aeucaryoticcell are missing from a bacterial (procaryotic) cell?

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Solution:

Cells are classified as eukaryotic or prokaryotic. The former means "truenucleus" while the latter means "before the nucleus". Prokaryotes include all bacteria including thecyanobacteria(blue-green algae), whereas eukaryotes include theprotistan, fungal, animal and plant cells. The most obvious difference between bacteria andeucaryotesis the absence of the nuclear membrane in bacteria. The DNA of the bacterial cell is generally confined to a specific area but this region is not enclosed by a membrane. In addition, proteins, such as thehistonesofeucaryotic cells , are not found in association with bacterial DNA. Another distinctive feature is that during replication of the bacterial nuclear region, the mitotic spindle apparatus is not seen. Bacterial cells also lack Golgiapparati, mitochondria, and endoplasmic reticulum. This means that theribosomesfound in the cytoplasmic region are free, i.e. they are not bound to endoplasmic reticulum . Membranous systems are not completely absent in bacteria. The plasma membrane folds inward at various points to formmesosomes. These membranes may be involved with the origin of other intra-cellular structures , and the compartmentalization and integration of biochemical systems . For example, although the electron transport system is located in the plasma membrane of bacteria, most of the respiratory enzymes are located in themesosomes. The plasma membrane andmesosomesare the bacterial counterpart of theeucaryoticcells' mitochondria, serving to compartmentalize the respiratory enzymes. Those bacteria that contain chlorophyll do not contain plastids to "house" the chlorophyll. Instead, the chlorophyll is associated with membranous vesicles arising from mesosomes .

Question:

Discuss some properties and functions of (a) carbohydrates (b) lipids, (c) proteins and (d) nucleic acids.

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Solution:

(a) Carbohydrates are made up of carbon, oxygen and hydrogen, and have the general formula (CH_2O)_n. Carbohydrates can be classified as monosaccharides, disaccharides, oligosaccharides or polysac-charides. The monosaccharides ("simple sugars") are further categorized according to the number of carbons in the molecule. Trioses contain 3 carbons: pentoses contain 5 carbons (i.e. ribose, dexyribose): and hexoses contain 6 carbons (i.e. glucose, fructose, galactose). The hexoses, which exist as straight chains or rings, are important building blocks for disaccharides and the more complex carbohydrates. Disaccharides, important in nutrition, are chemical combinations of two monosaccharides: Lactose = glucose + galactose Sucrose (table sugar) = glucose + fructose Maltose = glucose + glucose As can be seen by the double arrows, the reverse of this reaction, hydrolysis, is also possible. Oligosaccharide means "few sugars" and is arbitrarily defined as com-pounds which upon hydrolysis yield 2 - 10 monosaccharides. Polysaccharides are complex carbohydrates made up of many monosaccharides bonded by glycosidic linkages. These long chains are formed by dehydration synthesis. They also can be broke down into monosaccharide units by hydrolysis. There are many complex polysaccharides that are of great biological significance. Their primary functions include both storage and structural properties. Examples of these are: starches (principal storage product of animals), and cellulose (major supporting material in plants). These are all polymers of glucose. (b) Lipids are also composed principally of carbon, hydrogen and oxygen. However, they can also contain other elements, particularly phosphorus and nitrogen, and typically contain a much smaller proportion of oxygen than do carbohydrates. Lipids are insoluble in water. There are many known lipids, the most common include the fats, the phospholipids and the steroids. Fats are composed of two different types of compounds: glycerol (an alcohol) and fatty acids (organic compounds with a carboxyl group . Each molecule of fat contains one glycerol molecule and three fatty acids joined together by dehydration reactions. There are basically two groups of fats: the saturated (those fats which have the maximum possible number of hydrogen atoms and therefore have no carbon to carbon double or triple bonds) and the unsaturated (those which have at least one carbon to carbon double or triple bond). Unsaturated fats are an important part of our diet due to their storage capabilities. Phospho-lipids are composed of glycerol, fatty acids, phoshoric acid and usually a nitrogenous compound. They are important components of many cellular membranes. Steroids are classified as lipids because their solubility characteristics are similar to those of fats and phospholipids - they are all insoluble in water, but soluble in ether. However, steroids differ structurally from fats and phospholipids - they are not based on the bonding together of fatty acids and an alcohol. Instead, they are composed of four interlocking rings of carbon atoms with various side groups attached to the rings: They are very important biologically. Steriods include the sex hormones, various other hormones and some vitamins. In addition they are also important structural elements in living cells, especially plants. c) Proteins are much more complex than either carbohydrates or lipids. They are made up of the four essential elements: carbon, hydrogen, oxygen and nitrogen bonded together to form compounds called amino acids. Some amino acids contain sulfur. The amino acid is represented as: where R is a chain which can bevery simple (as in the amino acid glycine where R=H) or it can be very complex (as in tryptophan where R contains two ring structures). Proteins are long and complex polymers of varying combinations of the twenty amino acids which are formed by condensation reactions between the -COO^- and -NH_3^+ groups of the amino acid building blocks. The bonds formed by these reactions are called peptide bonds. A dipeptide is a molecule with two amino acids joined together by one peptide bond. A tripeptide refers to three linked amino acids. Oligopeptides, which includes tripeptides, is the term for a short chain of amino acids. Polypeptides are the polymer of amino acids and may contain 1000 amino acids. Finally, a protein is one or more polypeptide chains coiled or folded into complex three- dimensional configurations. Often a metal ion or organic molecule is an integral part of the protein structure. Proteins are found in every part of the cell and are an integral part in both the structure and function of living things. They play an important part in many of the chemical reactions that occur within cells because the enzymes that catalyze these reactions are proteins themselves. Proteins also function as the structural and binding materials of organisms. Hair, fingernails, muscle, cartilage, tendons and ligaments are all structures which contain large amounts of proteins. d) Nucleic acids, as their name implies, are found primarily in the nucleus. There are two types of nucleic acid: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA and RNA molecules are very long chains composed of repeating subunits called nucleotides. A nucleotide is composed of any one of the following five nitrogenous bases: adenine, guanine, cytosine, thymine (only in DNA) or uracil (only in RNA), a five carbon sugar (ribose in RNA, deoxyribose in DNA) and a phosphate group. Both DNA and RNA are composed of many nucleotides linked together and are called poly-nucleotides or nucleic acids. In a polynucleotide, any two nucleotides are linked together by a dehydration re-action between the phosphate group of one nucleotide and the sugar group of another. The bases are attached to the sugars. Nucleic acids primarily function in heredity and governing the synthesis of many different kinds of proteins and other substances present in organisms. Chromosomes and genes are predominantly composed of DNA. Some DNA is also found in the mitochondria and the chloroplasts. Large quanitites of RNA are present in the nucleoli, the cytoplasm, and the ribosomes of most cells.

Question:

What are the chemical and physical processes involved in transmission at the synapse?

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Solution:

The nervous system is composed of discrete units, the neurons, yet it behaves like a continuous system of transmission of impulses. For this to occur, there have to be functional connections between neurons. These con-nections are known as synapses. A synapse is an anatomical-ly specialized junction between two neurons, lying adjacent to each other, where the electric activity in one neuron (the presynaptic neuron) influences the excitability of the second (the postsynaptic neuron). At the synapse, the electric impulse is transformed into a chemical form of transmission. Chemical transmission at the synapse involves the processes of neurosecretion and chemoreception. The arrival of a nerve impulse at the axon terminal stimulates the release of a specific chemical substance, which has been synthesized in the cell body and stored in the tip of the axon, into the narrow synaptic space between the ad-jacent neurons. This process constitutes neurosecretion. The chemical secreted, known as neurotransmitter, can cause local depolarization of the membrane of the postsynaptic region and thus transmit the excitation to the adjacent neuron. Chemoreception is the process in which the neurotransmitter becomes attached to specific molecular sites on the membrane of the dendrite (postsynaptic region), producing a change in the properties of the cell membrane so that a new impulse is established. The chemical transmitter, say acetylcholine, passes from the presynaptic axon to the postsynaptic dendrite by simple diffusion across the narrow space, called the synaptic cleft, separating the two neurons involved in the synapse. The synaptic clefts have been measured under the electron microscope to be about 200 \AA in width. Diffusion is rapid enough to account for the speed of trans-mission observed at the synapse. After the neurotransmitter has exerted its effect on the postsynaptic membrane, it is promptly destroyed by an enzyme called cholinesterase. This destruction is of critical importance. If the acetyl-choline were not destroyed, it would continue its stimulatory action indefinitely and all control would be lost.

Question:

What is a cone? Describe its structure.

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Solution:

Cones are the typical reproductive structures of the gymnosperms. A cone or strobilus is a spiral aggregation of modified leaves called cone scales (see figure problem 8\rule{1em}{1pt}3). These cone scales are thesporophylls, and each scale bears on its surface the sporangia. In most gymnosperms, themicrosporangiaandmegasporangiaare borne on separate cones. In the conifers, both male and female cones are borne on the same plants. It would thus appear that self-fertilization occurs. However, the male cones are borne on the lower branches of the conifers, while the female cones are borne on the higher branches. Thus, self-fertilization is essentially prevented. The ovulate, or female cones, are larger. In these cones, the megasporangiaare enclosed by enveloping integuments to form structures called ovules. Each scale bears two ovules; each ovule is composed of the outer integuments and the innernucellus (megasporangium) and has a small opening at one end known as the micropyle, through which the pollen grains will enter. It is from the tissue of thenucellus, that the megaspore mother cell will be produced. It is of interest to note that the megaspore, which gives rise to the megagametophyte, is never released from themegasporangium. It remains embedded in the sporangium, which is enclosed to form the ovule. It is the ovule, which when mature, constitutes the naked seed of the gymnosperms. Male or staminate cones are smaller and bear themicrosporangia on their scales. The number ofmicro-sporangiaproduced by each microsporophyll varies among gymnosperms; in conifers, two is the usual number, while cycads have manymicrosporangiascattered on the lower surface of each scale. Each microsporangium will produce the microspore mother cell, which will give rise to numerousmicrogametesor pollen grains.

Question:

Given ∆H = 2800 kJ for the production of one mole of C_6H_12O_6 from CO_2 and H_2O, calculate what wavelength of light would be needed per molecule of CO_2 to achieve this photosynthesis in a one-step process.

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Solution:

To solve this problem: (1) set up the balanced equation for the production of 1 mole of C_6H_12O_6from CO_2 and H_2O. (2) From thestoichiometryof this equation determine the number of moles of CO_2 necessary to make 1 mole of C_6H_12O_6. (3) Calculate the number of molecules of CO_2 necess-ary to make 1 mole of C_6H_12O_6 using Avogadro's number (6.022 × 10^23 molecules / mole). (4) calculate the energy necessary per molecule of CO_2 by dividing ∆H by the total number of CO_2 molecules. (5) Use the expression E =hѵwhere E is the energy of light, h is Planck's constant, and ѵ is the frequency of light. Set the energy per molecule of CO_2 equal tohѵand solve for the frequency, ѵ. (6) Use the expression c =\lambdaѵ, (where c is the speed of light, ѵ is the frequency of light (from step 5) and \lambda is the wavelength of light) to find \lambda. To begin, the balanced equation for the production of 1 mole of C_6H_12O_6 is 6CO_2 + 6H_2O \rightarrow C_6H_12O_6 + 6O_2 Note that this process is photosynthesis. From thestoichiometryof this equation, 1 mole of C_6H_12O_6 is produced for every 6molesof CO_2 consumed. And, therefore, the number of molecules of CO_2 necessary to make 1 mole of C_6H_12O_6 is (6 moles CO_2) (6.02 × 10^23 molecules/mole) = 3.61 × 10^24 molecules. The energy necessary per molecule of CO_2 is (2,800,000 J) / (3.61 × 10^24 molecules CO_2) = 7.75 × 10-^19 J/molecule. Steps 5 and 6 are combined to give the expression E = h (c/\lambda) where h = 6.63 × 10-^34 J-s, and c = 3.00 × 10^10 cm/s. Setting this equal to 7.75 × 10-^19 J and solving for \lambda gives \lambda = (hc/ E) = {(6.63 × 10-^34 J-s) (3.00 × 10^10 cm/s)} / {7.75 × 10-^19 J} = 2.57 × 10-^5 cm = 2.57 × 10-^5 cm × [(10^8 \AA) / cm] = 2570 \AA.

Question:

BothHClandNaOHare strong electrolytes. What is the pH of the solution formed by adding 40 ml of 0.10 MNaOHto 10 ml of 0.45 MHCl?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0346.htm

Solution:

We will solve this problem by considering the number of moles of H_3O^+ and of OH^- formed by the complete dissociation ofHClandNaOH, respectively. If we assume thatHClandNaOHdissociate completely, then the concentration of H_3O^+ in theHClsolution is equal to the initial concentration ofHCl, or [H_3O^+] = 0.45 M, and the concentration of OH^- in theNaOHsolution is equal to the initial concentration ofNaOH, or [OH^-] = 0.10 M. Since moles = concentration × volume, the number of moles of H_3O^+ in 10 ml of the acid solution is moles H_3O^+ = [H_3O^+] × volume = 0.45 M × 10 ml = 0.45 M × 0.01 liter = .0045 mole = 4.5 × 10^-3 mole. Similarly, the number of moles of OH^- in 40 ml of the basic solution is moles OH^- = [OH^-] × volume = 0.10 M × 40 ml = 0.10 M × 0.04 liter = 0.004 mole = 4 × 10^-3 mole H_3O^+ and OH^- neutralize each other according to the reaction H_3O^+ + OH^- \rightarrow 2H_2O. Ignoring the dissociation of water, we can assume that this reaction is complete. Hence, the 4 × 10^-3 moles of OH^- will be neutralized by 4 × 10^-3 moles of H_3O^+, leaving 4.5 ×10^-3 - 4.0 ×10^-3 = 0.5 ×10^-3 = 5 ×10^-4 mole of H_3O^+ remaining. Thus, when the two solutions are mixed, no OH^- remains and 5 ×10^-4 mole of H_3O^+ remains. Since the final volume of the solution is 40 ml + 10 ml = 50 ml = 0.05 liter = 5 ×10^-2 liter, the concentration of H_3O^+ is [H_3O^+] = 5 ×10^-4 mole/5 × 10^-2 liter = 10^-2 mole/liter. The pH is defined as pH = - log [H_3O^+], hence pH = - log [H_3O^+] = - log (10^-2) = - (- 2) = 2.

Question:

If, in tracing evolutionary relationships, anatomic evidence pointedone way and biochemical evidence the other, which doyou think would be the more reliable? Why?

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Solution:

It has been stressed again and again that evolutioncan notoccur withouta change in the genotype. Biochemical properties are controlled to agreater extent by genes than are anatomic ones. Anatomic characteristicsare more susceptible to modification by the external environmentin which the organism possessing them lives. Plants of the samespecies growing in two quite different habitats may demonstrate strikinglydifferent characteris-tics; this is because the two groups of plants areexposed to different environmental forces and achieve a different developmentalpotential even though the genes present in them are very similar. When we compare two organisms or groups of organisms in seekingan evolutionary relationship, we are essentially looking for a geneticrelationship between them. Since biochemical characteristics are moregreatly controlled by genes than are anatomic characteristics, a biochemicalsimilarity between the two organisms should be more reliable thanan anatomic similarity in indicating a genetic linkage and thus an evolutionaryrelationship. Therefore, biochemical evidence is more useful inthe tracing of evolutionary relationships between organisms.

Question:

If the year 1975 (year 1) is started with a 2 year old truck, what decision should be made at the start of this year and the start of the next four years in order to maximize the total return.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0460.htm

Solution:

This is a problem in dynamic optimization. A series of decisions must be made over time so as to maxi-mize some chosen quantity. Each decision by itself may be non-optimal; the optimality of the sequence of decisions is what matters. As an example, suppose X wishes to travel from city A to city F and has the choice of routes ABCF and ADEF. It is possible for AB to cost less than AD and yet ADEF may be cheaper overall (DE is less than BC and EF is less than CF). Going back to the initial problem, let 1)r_i (t) = revenue in period i from a truck that was made in year (i - t) and is t years old at the start of period i. For example r_3(3) = 8. 2)ui(t) = upkeep in period i on a truck that was made in year (i - t) and is t years old at the start of period i. For example, a truck made in 1975 requires an up-keep of 2 in 1978, i.e., u_5 (3) = 2. 3)c_i(t) = cost to replace a truck that was made in year (i - t) and is t years old at the start of period i. 4)IT = age of the given truck = 2. 5)fi(t) = optimal return for periods i, i+1,...., 5 when the i-th period is started with a truck that is t years old. 6)X_i (t) = decision to make at the start of period i that will yield f_i(t). The only two possible decisions are to keep the old truck or to purchase a new one. Suppose the current truck was 2 years old at the start of 1975 (year 1), and the estimates in table I are available. Suppose we decide to keep the current truck until 1979. Then the total revenue is 10 + 8 + 8 + 6 + 4 = 36. Total upkeep is3 + 3 + 4 + 4 + 5 = 19. Thus total return is 17. On the other hand, suppose we decide to replace the 1973 truck with a truck made in 1975 at the start of 1975. Then, total revenue is 14 + 16 + 16 + 14 + 12 = 72 and the upkeep costs are TABLE I:Truck made... 1973 (year minus 1) Age Revenue Upkeep Replacement 2 10 3 25 3 8 3 26 4 8 4 27 5 6 4 28 6 4 5 29 Truck made in 1975 (year 1) Age Revenue Upkeep Replacement 0 14 1 20 1 16 1 22 2 16 2 24 3 14 2 25 4 12 3 26 Truck made in 1976 (year 2) Age Revenue Upkeep Replacement 0 16 1 20 1 14 1 22 2 14 2 22 3 12 2 25 Truck made in 1977 (year 3) Age Revenue Upkeep Replacement 0 18 1 20 1 16 1 22 2 16 2 24 Truck made in 1978 (year 4) Age Revenue Upkeep Replacement 0 18 1 21 1 16 1 22 Truck made in 1979 (year 5) Age Revenue Upkeep Replacement 0 20 1 21 1 + 1 + 2 + 2 + 3 = 9 Replacement costs are 25 and hence total return is 38. Of course we could keep the truck for a year and then sell it. The total return in this case works out to 37. Clearly there is a large number of possible choices. The functional equation for period i is fi(t) = max│purchase:ri(0) - u_i (0) + Ci(t)_ + fi+1(1)│ │keep:ri(t) - u_i (t) + fi+1(t + 1)│ for i = 1,2,...,5 and t = 1,2,....(i - 1), (i+ IT - 1) We now apply the functional equation working back-wards. That is, assume that the fifth year is reached with a truck that is 1, 2, 3, 4, or 6 years old. Thus, for i = 5 and t = 1 f5(1) = max│Purchase: r5(0) - u_5 (0) + C5(1)+ f6(1) │ │keep:r_5 (1) - u_5 (1) + f6+(2)│ Note that f_6 (j) = 0 for all j. Then f5(1) = max│20 - 1 - 24 + 0│ │16 - 1 + 0│ = 15 Thus, we keep the 1978 truck in 1979, i.e., x_5 (1) is keep. At t = 2 f5(2) = max│Purchase: r5(0) - u_5 (0) + C5(2)│ │keep:r5(2) - u_5 (2)│ = max│20 - 1 - 24│ │16 - 2│ = 14 x_5(2) = keep t = 3 f5(3) = max│Purchase: 20 - 1 - 25│ │Keep:12 - 2│ = 10 x5(3) = keep t= 4 f5(4) = max│Purchase:20 - 1 - 26│ │Keep:12 - 3│ = 9 x5(4) = keep t = 6 f5(6) = max│Purchase:20 - 1 - 29│ │Keep:4 - 5│ = -1 X5(6) = keep. Now, assume that the start of the fourth period is reached with a 1, 2, 3, or 5 year old truck. The optimal decision to make in order to maximize the return from the last two periods is given by the functional equation: f4(t) = max│P: r_4(0) - u_4(0) - C_4(t) + f_5(1)│ │K: r_4(t) - u_4(t) + f_5(t + 1)│ t = 1,2,3,4,5 Thus, f_4 (1) = max│P: 18 - 1 - 22 + 15│ │K: 16 - 1 + 14│ = 29 x_4(1) = Keep f_4 (2) = max│P: 18 - 1 - 24 + 15│ │K: 14 - 2 + 10│ = 22 x_4(2) = Keep f_4 (3) = max│P: 18 - 1 - 25 + 15│ │K: 14 - 2 + 9│ = 21 x_4(3) = Keep f_4 (5) = max│P: 18 - 1 - 28 + 15│ │K: 6 - 4 + (-1)│ = 4 x_4(5) = purchase Moving back to the start of the third period and consider-ing the optimal policy over the last three periods when the third period is started with a 1, 2 or 4 year old truck, we have f_3(t) = max│P: r_3(0) - u_3(0) - C_3(t) + f_4 (1)│ │K: r_3(t) - u_3(t) + f_4(t + 1)│ t = 1,2,4 f_3(1) = max│P: 18 - 1 - 22 + 29│ │K: 14 - 1 + 22│ = 35 x_3(1) = Keep f_3(2) = max│P: 18 - 1 - 24 + 29│ │K: 16 - 2 + 21│ = 35 x_3(2) = Keep f_3(4) = max│P: 18 - 1 - 27 + 29│ │K: 8 - 4 + 4│ = 19 x_3(4) = purchase Now move backwards from period 2 with t = 1, 3 f_2(1) = max│P: 16 - 1 - 22 + 35│ │K: 16 - 1 + 35│ = 50 x_2(1) = Keep f_2(3) = max│P: 16 - 1 - 26 + 35│ │K: 8 - 3 + 19│ = 24 x_2(3) = purchase or keep Finally, period 1 is reached where we have a 2-year old truck. To arrive at a decision whether to keep it or replace it in order to maximize the total return over the next five periods, we have f_1(2) = max│P: 14 - 1 - 25 + 50│ │K: 10 - 3 + 24│ = 38 x_1(2) = purchase Thus, we purchase a truck at the start of 1975. We keep this truck in 1976, 1977, 1978 and 1979 in order to maxi-mize total returns over the next four years.

Question:

An automobile moves with a constant speed of 50 mi/hr around a track of 1 mi diameter. What is the angular velocity and the period of the motion?

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Solution:

For circular motion, the angular velocity \omega, the radius r, and the linear velocity v obey the re-lation: v\ding{217}= \omega\ding{217}× r\ding{217}as shown in the figure. \ding{217} \ding{217} \ding{217} Since \omega and r are perpendicular to each other, this reduces to v = \omegar. \omega = v/r = (50 mi/hr)/(0.5 mi) = 100 rad/hr The period \cyrchar\cyrt is the time duration of one complete cycle of motion around the circular path. In linear motion, x = vt. This equation can be applied to circular motion with linear velocity v replaced by \omega and linear distance x replaced by \texttheta expressed in radians. In one cycle of motion, the automobile travels 2\pi radians. Therefore\omegat = \textthetat = \texttheta/\omega\cyrchar\cyrt = 2\pi/\omega \cyrchar\cyrt = 2\pi/\omega= 2\pi/(100 rad/hr) = 0.063 hr = 3.8 min.

Question:

Discuss the current theory of the evolution ofautotrophs fromprimitiveheterotrophs.

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Solution:

The first organisms on earth were almost certainlyheterotrophs, which, in an atmosphere lacking free oxygen, presumably obtained energy bythe fermentation of certain organic substances in the sea. Because of theirdependence upon the external environment for food, they could surviveonly as long as the supply of organic molecules in the sea lasted. Because food in the sea was necessarily limited, before the supply was exhausted, some of the then existingheterotrophsmust have evolved the mechanismsto becomeautotrophs. They would then be able to make theirown organic molecules either by chemosynthesis or photosynthesis. An explanation of how anautotrophmay have evolved from one of theprimitiveheterotrophswas offered by N.H. Horowitz in 1945. Horowitz postulatedthat the evolution was a result of successive gene mutations in certainheterotrophswhich resulted in the enzymes needed to synthesize complexsubstances from simple ones. According to this hypothesis, the primitiveheterotrophswere able to survive only if a certain specific organic substancex, limited in supply, was present in the surrounding sea. If a mutationoccurred for a new enzyme enabling theheterotrophto synthesizesubstance x from some other more abundantly available organicform, the strain ofheterotrophswith this mutation would be able to survivewhen the supply of x in the sea was exhausted. A second mutation thatestablished an enzyme catalyzing another reaction by which x could bemade from inorganic substances would again have survival value. Finally, when all the organic molecules in the sea were exhausted, a mutationfor the synthesis of x from some inorganic substances, present in largequantities, marked an important evolutionary advantage. In this way, successive mutations over time led to the production of astrain of organisms able to synthesize all of its metabolic requirements fromsimple inorganic compounds. Such organisms, called theautotrophs, areexemplified by the many photosynthetic algae and chemosynthetic bacteriathat are believed to be among the earliest inhabitants of the earth.

Question:

Compute the speed of sound in the steel rails of a rail road track. The weight density of steel is 490 lb/ft^3, and Young's modulus for steel is 29 x 10^6 lb/in^2.

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0828.htm

Solution:

For an elastic medium, the speed of longitudinal waves is given by v = \surd(Y/\rho) = \surd[(Yg)/D] Where Y is Young's modulus, \rho is the density of the medium and \rho = D/g where D is the weight density of the medium. v = \surd[{(29 × 10^6 lb/n^2) (32 ft/s^2) }/{ 490 lb/ft^3 }] In order to keep all length dimensions consistent, we must change 29 × 10^6 lb/in^2 to lb/ft^2, Hence, 29 × 10^6 lb/in^2 = [(29 × 10^6 1b) / (1/44 ft^2)] 29 × 10^6 lb/in^2 = 29 × 144 x 10^6 lb/ft^2 Therefore, v = \surd[{(29 × 10^6 144 lb/ft^2) (32 ft/sec^2)}/{ 490 lb/ft^3 }] = 1.6 × 10^4 ft/sec

Question:

Write a Basic program to simulate the game ofAwari.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0568.htm

Solution:

Awariis an African game played with seven sticks and 36 beans, laid out as shown: Y 6 5 4 3 2 1 Y's Home Away From Home 000 000 000 000 000 000 X's Home 000 000 000 000 000 000 1 2 3 4 5 6 X In the given diagram X and Y are the two players. Suppose X has the first move. A move is made by taking all the beans from any non-empty pit on X's side and sowing them in a counterclockwise direction, one in each pit. A turn consists of one or two moves. If the last bean of X's move is sown in his own home he can take another move. Also, if the lastbeansown in a move lands in an empty pit and the pit opposite is not empty, then all the beans in that pit (together with the last bean sown) are captured and moved to the player's home. When either side is empty, the game is finished and the player with the most beans is the winner. Consider the following sample part game: Y 333333 00 333333 X X starts sowing from 4 333333 01 333344 Y starts sowing from 2: 344403 01 333344 X starts from 1 344403 01 044144 (He now captures the 4 stones on Y (3) and his X (4) to obtain 344003 06 044044 Y plays from 4: 040003 56 044044 X starts from 3 to obtain 040003 57 040155 Since he finished at home he can play again. He plays from 5: 040114 58 040006 and so on... The program follows: 10PRINT "AWARI\textquotedblright 15DATA 0 20DIM B (13), G (13), F (50): READ N 30E = 0 40FOR I = 0 TO 12 50B (I) = 3 60NEXT I 70C = 0 80F (N) =0 90B(13) = 0 100B (6) = 0 110GOSUB 540 120PRINT "YOUR MOVE"; 130GOSUB 310 140IF E = 0 THEN 210 145IF M = H THEN GOSUB 300 146IF E = 0 THEN 210 150PRINT "MY MOVE IS"; 160GOSUB 820 170IF E = 0 THEN 210 180IF M = H THEN PRINT " , " 190GOSUB 820 200IF E > 0 THEN 110 210PRINT "GAME OVER" 220D = B (6) - B (13) 230IF D < 0 THEN PRINT "I WIN BY"; -D; "POINTS" 240GOTO 30 250N = N + 1 260IF D = 0 THEN PRINT "DRAWN GAME" 270GOTO 30 280PRINT "YOU WIN BY"; D ;"POINTS" 290GOTO 30 300PRINT "AGAIN"; 310INPUT M 320IF M < 7 THEN IF M > 0 THEN M = M - 1 330GOTO 360 340PRINT "ILLEGAL MOVE" 350GOTO 300 360IF B (M) = 0 THEN 340 370H = 6 380GOSUB 400 390GOTO 540 400K = M 410GOSUB 680 420E = 0 430IF K > 6 THEN K = K - 7 440C = C + 1 450IF C < 9 THEN F (N) = F (N)\textasteriskcentered6 + K 460FOR I = 0 TO 5 470IF B (I) <> 0 THEN 500 480NEXT I 490RETURN 500FOR I = 7 TO 12 510IF B(I) <> 0 THEN E = 1 520RETURN 530GOTO 480 540FOR I = 12 TO 7 STEP - 1 550GOSUB 650 560NEXT I 570PRINT: I = 13 580GOSUB 650 590PRINT ""; 600PRINT B (6) 610PRINT ""; 620FOR I = 0 TO 5: GOSUB 650 630NEXT I 640RETURN 650IF B (I) < 10 THEN PRINT ""; 660PRINT B (I); 670RETURN 680P = B (M) 690B (M) =0 700FOR P = P TO 1 STEP - 1 710M = M + 1 720IF M > 13 THEN M = M - 14 730B (M) = B (M) + 1 740NEXT P 750IF B (M) = 1 THEN IF M <> 6 THEN IF M <> 13 THEN IF B (12 - M) <> 0 THEN 780 770RETURN 780B (H) = B(H) + B(12 - M) + 1 790B (M) = 0 800B (12 - M) = 0 810RETURN 820D = - 99 830H = 13 840FOR I = 0 TO 13 850G (I) = B (I) 860NEXT I 870FOR J = 7 TO 12 880IF B(J) = 0 THEN 1110 885G = 0: M = J: GOSUB 680 886FOR I = 0 TO 5: IF B (I) = 0 THEN 970 890L = B (I) + I 900R = 0 910IF L > 13 THEN L = L - 14 920R = 1 930GOTO 910 940IF B (L) = 0 THEN IF L <> 6 THEN IF L <> 13 THEN R = B(12 - L) + R 960IF R > Q THEN Q = R 970NEXT I 980Q = B(13) - B (6) - Q 990IF C > 8 THEN 1050 1000K = J 1010IF K > 6 THEN K = K - 7 1020FOR I = 0 TO N - 1 1030IF F (N)\textasteriskcentered6 + K = INT (F(I)/6 \uparrow (7 - C) +.1) THEN Q = Q - 2 1040NEXT I 1050FOR I = 0 TO 13 1060B (I) = G (I) 1070NEXT I 1080IF Q > = D THEN A = J 1090D = Q 1100NEXT J 1110M = A 1120PRINT CHR$ (42 + M); 1130GOTO 400 1140FOR I = 0 TO N - 1 1150PRINT B (I) 1160NEXT I 1170END

Question:

Use the uncertainty principle to estimate the kinetic energy of a neutron in a nucleus of radius R = 2 × 10^-13 cm.

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0985.htm

Solution:

We write the uncertainty principle in the form \Deltap\Deltax= h Then \Deltax\cong 2R for the neutron can be found anywhere in the nucleus. The\Deltaxgiven expresses this lack of certainty as to where in the nucleus it can be found. As a result of the uncertainty principle, the neutron can acquire a momentum P =\Deltap = (h/2R) The kinetic energy,T = (1/2) mv^2 = [1/2{(mv)^2/m}] = (p^2/2m) = [h^2/(8mR^2)] T = [{(6.63 × 10^-27 erg-sec)^2}/{8(1.67 × 10^-24 gm)(2 × 10^-13 cm)^2}] = 8.22 × 10^-5 [(erg^2-sec^2)/(gm-cm^2)] = 8.22 × 10^-5 [(erg^2) / {(gm-cm/sec^2) -cm}] = 8.22 ×10^-5 (erg^2/dyne-cm) = 8.22 ×10^-5 erg

Question:

What are the roles of coenzyme A in metabolism ?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0074.htm

Solution:

Coenzyme A is a large, complex organic molecule involved in a number of key reactions in the cell. The functions of coenzyme A are several. In glycolysis, glucose is converted to pyruvate. To enter the Krebs acid cycle (also known as the tricarboxylic acid, or TCA cycle and the citric acid cycle) , pyruvate must be con-verted to another form; such a conversion first involves decarboxylation of pyruvate (loss of carbon dioxide), then a reaction with coenzyme A, and finally, a dehydrogenation re-action, to yield acetyl coenzyme A. The reactions are rep-resented as follows: The acetate group is bonded to the coenzyme A molecule via a high energy bond. The energy of this bond in acetyl coenzyme A is utilized to transfer the acetyl group to oxaloacetic acid, the first reaction of the TCA cycle. Coenzyme A is also involved in a reaction farther along in the cycle. An intermediate (derived from \alpha- Ketoglutarate) combines with coenzyme A to yield succinyl coenzyme A. The bond connecting coenzyme A to succinate is also of high energy. The energy in this bond is converted to an energy-rich phosphate bond,\simP, through the addition of an inorganic \sim phosphate. The phosphate group is subsequently transfered to GDP to produce GTP and free succinate. The reactions are diagrammed below: Coenzyme A is also utilized in fatty acid oxidation. The metabolism of a fatty acid initially involves its activation by ATP and coenzyme A to fatty acyl coenzyme A. Through a repeating series of reactions, fatty acyl coenyzme A is converted to acetyl coenzyme A, which, in turn, goes into the TCA cycle As a general rule, coenzyme A is involved in reactions which utilize its high-energy bond in order to more effectively facilitate metabolism. The sequence of reactions, wherein Acyl CoA is shortened by two carbon atoms, repeats itself until the Acyl CoA is shortened to Acetyl CoA.

Question:

If I_2 and Br_2 are added to a solution containing I^- and Br^-, what reaction will occur if the concentration of each species is 1m? Some half-reactions and their standard reduction potentials. Increasing strength as oxidixing agents Half - reaction Increasing strength as reducing agents Standard reduction potential, V 2e^- + F_2(g) \rightarrow 2F^- +2.87 2e^- + Cl_2(g) \rightarrow 2Cl^- +1.36 4e^- + 4H_3O^+ +O_2(g) \rightarrow 6H_2O +1.23 2e^- + Br_2(g) \rightarrow 2Br^- +1.09 e^- + Ag^+ \rightarrow Ag(s) +0.80 2e^- + I_2 \rightarrow 2I^- +.054 2e^- + Cu^2+ \rightarrow Cu(s) +0.34 2e^- + 2H_3O^+ \rightarrow H_2(g) + 2H_2O Zero 2e^- + Fe^2+ \rightarrow Fe(s) -0.44 2e^- + Zn^2+ \rightarrow Zn(s) -0.76 3e^- + Al^3+ \rightarrow Al(s) -1.66 2e^- + Mg^2+ \rightarrow Mg(s) -2.37 e^- + Na^+ \rightarrow Na(s) -2.71 e^- + Li^+ \rightarrow Li(s) -3.05

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Solution:

When I_2 and Br_2 are added to the solution, a dynamic equilibrium is attained between the non-charged element and its ion form. These reactions, redox reactions, are written as: 2e^- + I_2 \rightarrow 2I^-+ 0.54V 2e^- + Br_2 \rightarrow 2Br^-+ 1.09V The voltages listed indicate the electric potential between two electrodes. In other words, it is a measurement of the work done in moving a unit charge (an electron) from one electrode to the other. For any reaction to occur, the half reactions must be written such that the overall vol-tage is positive. It is written positively to indicate that the reaction tends to go spontaneously. Any redox reaction for which the overall potential is positive has the tendency to take place as written. Therefore one must pick the larger voltage in the half reaction to be positive and the other negative. Thus, 2e^- + Br_2 \rightarrow 2Br^-+ 1.09V 2I^- + I_2+2e^- -0.54V 2I^- + Br_2 \rightarrow I_2 + 2Br^- The half-reactions are written so that when added together, the electrons cancel out. Because the reaction, 2I^- + Br_2 \rightarrow I_2 + 2Br^- , has a positive voltage, it will proceed as written.

Question:

A girl is a hemophiliac, a) What are the possible genotypes and phenotypes of her parents?b) Assuming that her mother is normal, what were this girl's chances of being born with the disease?c) Several cases of hemophilia in girls have been reported within a small region in England where there is much close intermarriage. Explain this high frequency of hemophilia in girls.

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Solution:

a) Hemophilia is a sex-linked trait and its gene is carried on the X chromosome. It is also a recessive trait, and so will not be expressed in the presence of a normal X chromosome. It will be expressed, however, in the presence of a normal Y chromosome, because that chromosome does not carry any genes that would mask its expression. This means that in order for a boy to be a hemophiliac, he need only have one parent who carries the gene. A girl, however, would have to have both parents carry the gene, since she receives one X chromosome from each parent. Let ^+X represent the chromosome carrying the gene for hemophilia, and X and Y represent the normal sex chromosomes. The father of a hemophiliac girl, in order to be able to give his daughter the gene for hemophilia, must be ^+X Y, and therefore a hemophiliac himself. The mother could be either ^+X^+X, (a hemophiliac), or ^+XX, (a carrier), in order to pass the hemophilia gene to her daughter. b) If the mother is normal, her genotype must be ^+XX. Let us look at the cross that produced the hemop-hiliac girl. F^1 + X Y + X + X^+X + XY X + XX XY The offspring will be as follows: 1/4 female carrier (X^+X) 1/2 normal1/4 normal male (^+X^+X) 1/4 hemophiliac male (^+X Y) 1/2 hemop-hiliac1/4 hemophiliac female (^+X^+X) 1/2 hemop-hiliac Since half the possible offspring are hemophiliac, the girl had a 50% chance of being born with the disease. c) While within the family above, a female has the same chance as a male of being a hemophiliac, this is not the case in the population as a whole. Because of the fact that it requires that a father be hemophiliac and a mother be at least a carrier in order to produce a hemo-philiac girl, there is usually only a rare case now and then of a girl being born with hemophilia. The hemophilia gene has a lower frequency than the normal gene in the population, and the chances that both parents will have this gene are very low. However, within the particular region in question, there is much intermarriage. If a gene is carried by a family member, its frequency among other family members will be much higher than in the population as a whole. If an individual carries a particular gene, in this case the gene for hemophilia, and marries within the family, there is a good chance that their spouse will also carry the gene. There is thus an increased probability that it will be carried by offspring in the homozygous state, in this case resulting in hemophiliac girls.

Question:

In a double slit interference experiment the distance between the slits is 0.05 cm and the screen is 2 meters from the slits. The light is yellow light from a sodium lamp and it has a wavelength of 5.89 × 10^-5 cm. What is the distance between the fringes?

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Solution:

To find the distance between fringes in the double slit experiment, we must first derive the formulas for the location of the maxima and minima of the fringe pattern. Let us examine this experiment in more detail. Light is incident on the 2 slits from the left. (See figure) MP and AP represent 2 rays of light, one from each slit, arriving at P. Typically, L >> d, and we may con-sider MP to be equal to BP. Assuming that the light rays emerging from the slits are in phase, the two light rays arriving at P will be out of phase because light from A must travel the extra distance AB when compared with light from M. If this path difference (AB = d sin \textphi ) is equal to an even number of half wavelengths, P will be a maximum point. If AB equals an odd number of half wavelengths, P will be a minimum point. Hence, For a maximum sin \textphi = (2n) \lambda/2d(n = 0, 1, 2,...) For a minimum sin \textphi = [(2n + 1)] \lambda/2d(n = 0, 1, 2,...) Therefore, the angular location of adjacent maxima on the screen (say the nth and (n + 1) th maxima,) is sin (\textphi_n+1) = [{2(n + 1)\lambda} / {2d}] = [(n + 1)\lambda/d](1) sin (\textphi_n) = [{(2n)\lambda} / {2d}] = n\lambda/d But, if \textphi is small, sin (\textphi_n+1) \approx tan (\textphi_n+1) sin (\textphi_n) \approx tan (\textphi_n). Hence, using (1) and the figure, (Y_n+1)/L = [{(n + 1)\lambda}/d] Y_n/L = n\lambda/dhence, Y_n+1 - Y_n = [{(n + 1)\lambdaL}/d] - (n\lambdaL)/d Y_n+1 - Y_n = \lambdaL/d. This is the screen separation of 2 adjacent maxima. If\lambda = 5.89 × 10^-5 cm L = 200 cm d = 0.05 cm. Y_n+1 - Y_n = [{(5.89 ×10-5× 200)cm^2} / {0.05 cm}] = .233 cm.

Question:

A rectangular coll 30 cm long and 10 cm wide is mounte in a uniform field of flux density 8.0 × 10^-4 nt/amp -m. There is a current of 20 amp in the coil, which has 15 turns. When the plane of the coil makes an angle of 40\textdegree with the direction of the field, what is the torque tending to rotate the coil?

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Solution:

The torque on a circuit in a field of magnetic induction, B^\ding{217}, is T^\ding{217} = \mu^\ding{217} × B^\ding{217}(1) where \mu^\ding{217} is the magnetic moment of the circuit. (This is the property of the circuit which causes the torque to be exerted.) The magnitude of the magnetic moment is \mu = NIA(2) where N is the number of turns in the circuit, I is the current in the circuit, and A is the area it encloses. The direction of \mu^\ding{217} is given by the right hand rule: wrap the fingers of Your right hand around the circuit in the direction of the current, and the direction in which your thumb points will then be the sense of \mu^\ding{217}. Since we ^\ding{217} only want the magnitude of T^\ding{217}, we write T = \muB sin \texttheta(3) where T, \mu, B are the magnitudes of T^\ding{217}, \mu^\ding{217}, B^\ding{217}, and \texttheta is the angle between the directions of \mu^\ding{217} and B^\ding{217} (see figure). Sub-stituting (2) into (3), we obtain T = NiAB sin \texttheta(4) However, the data is given in terms of flux density, not in terms of B. But flux density is actually equal to B because Flux density = (\cyrchar\CYRF/A) = (BA/A) = B where A is the area enclosed by the circuit, and \cyrchar\CYRF is the flux cutting through the circuit. We still cannot proceed yet, because we do not have \texttheta. The question gives us the angle between the plane of the coil and the direction of B^\ding{217}. (In the figure this is \alpha.) The angle we need, \texttheta, is 90\textdegree - a = 50\textdegree. Inserting the given data in (4), we find T = [8 × 10^-4 (nt/A \bullet m)] (15) (20A) (.3m) (.1m) (.77) T = .0055 nt \bullet m

Question:

1) Two linearly polarized waves are in phase, but have different amplitudes. At x = 0 E^\ding{217}_1 = \^{\i} A_1 cos 2\piѵt + \^{\j} B_1cos 2\piѵt, E^\ding{217}_2 = \^{\i} A_2 cos 2\piѵt + \^{\j} B_2cos 2\piѵt. Show that E^\ding{217}_T = E^\ding{217}_1 + E^\ding{217}_2 is also linearly polarized, and find its polarization direction. 2) Two circularly polarized waves (a right and a left) can be added to form a linearly polarized wave: At x = 0, E^\ding{217}_1 = \^{\i} E_o cos 2\piѵt + \^{\j} Eosin 2\piѵt, E^\ding{217}_2 = \^{\i} E_o cos (2\piѵt + \alpha) - \^{\j} E_osin (2\piѵt + \alpha) . Show that E^\ding{217}_T is linearly polarized, and find its polariza-tion direction. 3) An elliptically polarized wave is written (at z = 0) as E^\ding{217}_T = \^{\i} A_1 sin 2\piѵt + \^{\j} Bcos 2\piѵt, Show that this can be decomposed into a linearly and a circularly polarized wave.

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Solution:

1) E_1 = (\^{\i}A_1 + \^{\j}B_1) cos 2\piѵt, E_2 = (\^{\i}A_2 + \^{\j}B_2) cos 2\piѵt. Now, add the components of E_1 and E_2 together to find the expression for E_T. Then, E_T = {\^{\i}(A_1 + A_2) + \^{\j}(B_1 + B_2)} cos 2\piѵt. The point here is that the quantity in { } just specifies a vector, and the time variation is the same for all parts of it. So the vector lies along a line and changes size, but not direction. Therefore, E^\ding{217}_T is linearly polarized. In the case of circular polarization, this is not true, since the x and y components vary differently in time, one being large when the other is small. 2) Left: E_1 = \^{\i}E_o cos 2\piѵt + \^{\j}E_o sin 2\pivt, Right: E_2 = \^{\i}E_o cos (2\piѵt + \alpha) - \^{\j}E_o sin (2\pivt + \alpha). E_T = \^{\i}E_0{cos 2\piѵt + cos (2\piѵt + \alpha)} + \^{\j}E_0{ sin (2\piѵt - sin(2\piѵt + \alpha)} cos(2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2) + (\alpha / 2)] Using one of the double angle formulas, we find cos (2\piѵt + \alpha) = cos[2\piѵt + (\alpha / 2)] cos (\alpha/2) - sin [2\pivt + (\alpha/2)] sin (\alpha/2) = cos [2\piѵt + (\alpha / 2) cos (\alpha / 2) - [sin 2\piѵtcos(\alpha / 2) + cos 2\piѵt sin(\alpha / 2)] sin(\alpha / 2) where we have made use of the formula sin (x + y) = sin x cos y + cos x sin y. Then, cos (2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2)] cos (\alpha / 2) - sin 2\piѵtcos (\alpha / 2) sin(\alpha / 2) - cos 2\piѵtsin^2 (\alpha / 2). Therefore, cos 2\piѵt + cos(2\piѵt + \alpha) = cos 2\piѵt + cos[2\piѵt +(\alpha / 2)] cos (\alpha / 2) - sin 2\piѵtcos (\alpha / 2) sin(\alpha/ 2) - cos 2\piѵt sin^2 (\alpha / 2). Substituting 1 - cos^2(\alpha / 2) for sin^2(\alpha / 2) in the preceding expres-sion yields the following result: cos(2\piѵt) + cos(2\piѵt + \alpha) = cos(2\piѵt) + cos[2\piѵt + (\alpha / 2)] cos(\alpha / 2) -sin(2\piѵt) cos(\alpha / 2) sin(\alpha / 2) + cos(2\piѵt) cos^2(\alpha / 2) - cos(2\piѵt). The cos(2\piѵt) terms cancel one another, and so, cos (2\piѵt) + cos (2\piѵt + \alpha) = cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) + cos(\alpha / 2)[cos(2\piѵt)cos (\alpha / 2) - sin(2\piѵt) sin(\alpha / 2)]. By applying the double angle formula cos (x + y) = cos x cos y -sin x sin y, we have the result that cos (2\piѵt) + cos (2\piѵt + \alpha) = 2 cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) Similarly, sin (2\piѵt) - sin (2\piѵt + \alpha) = 2 sin(\alpha / 2) cos [2\piѵt + (\alpha / 2)]. Therefore, E_T = 2\^{\i}E_0 cos [2\piѵt + (\alpha / 2)] cos(\alpha / 2) + 2\^{\j}E_0 sin (\alpha / 2) cos[2\piѵt + (\alpha / 2)] = 2E_0 [(\^{\i}cos (\alpha / 2) + \^{\j} sin(\alpha / 2)] cos [2\piѵt + (\alpha / 2)] . Since the two components are in phase, we can add them in a vector diagrams:ET is a vector of magnitude 2A cos { 2\piѵt + (\varphi_1 / 2) } at an angle of \varphi_(1/2) to the y axis. Notice that we can see some easy limiting cases in the first two lines: If \alpha = 0 the y components cancel out, while the x components add to E_T = \^{\i}2E_0 cos (2\piѵt )(\alpha = 2\pi just multiplies everything by -1). If \alpha = (\pi / 2), we get E_T =(2E_0) [\^{\i} cos (\pi / 4)+ \^{\j}sin (\pi / 4)] cos [2\piѵt + (\alpha / 4)] = \surd(2E_0)_ (\^{\i} + \^{\j}) (cos 2\piѵt cos (\alpha / 4)-sin 2\piѵt sin(\alpha / 4)] = E_0(\^{\i} + \^{\j}) (cos 2\piѵt - sin 2\piѵt) linearly polarized at 45 degrees 3) E = \^{\i}A sin 2\piѵt + \^{\j}B cos 2\piѵt. First take out a circularly polarized part : E_o = \^{\i}A sin 2\piѵt + \^{\j}A cos 2\piѵt . What is left is E_lin =\^{\j}(B - A) cos 2\piѵt , a linearly polarized wave. To show that the original wave is elliptically polar-ized, draw the vector at various times, as shown in the figure.

Question:

The figure shows a box being dragged along a horizontal surface by a constant force P making a constant angle \texttheta with the direction of motion. The other forces on the box are its weight W^\ding{217} = mg^\ding{217}, the nor-mal upward force N^\ding{217} exerted by the surface, and the friction force f^\ding{217}. What is the work done by each force when the box moves a distance s along the surface to the right?

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Solution:

The component of P^\ding{217} in the direction of motion is P cos \texttheta. The work of the force P^\ding{217} is by definition W_p = \int P^\ding{217} \textbullet ds^\ding{217} = P cos \texttheta \int ds = P(cos \texttheta)(s) for P^\ding{217} is a constant force, ds^\ding{217} is a vector in the direction of horizontal motion. The forces w^\ding{217} and N^\ding{217} are both at right angles to the displacement. Hence W_w = \intw^f \textbullet ds^\ding{217} = ws cos 90\textdegree = 0 (W^\ding{217} is constant. Therefore it may be taken out of the integral). W_N = \intw^\ding{217} \textbullet ds^\ding{217} = Ns cos 90\textdegree = 0 (N^\ding{217} is constant. Therefore it was taken out of the integral). The friction force f^\ding{217} is opposite to the displacement, so the work of the friction force is W_F = \intf^\ding{217} \textbullet ds^\ding{217} = fs cos 180\textdegree = -fs Since work is a scalar quantity, the total work W of all forces on the body is the algebraic (not the vector) sum of the individual works. W = W_p + W_w + W_N + W_f = (P cos \texttheta) \bullet s + 0 + 0 - f \bullet s = (P cos \texttheta - f)s. But (P cos \texttheta - f) is the resultant force on the body. The sum of the forces in the vertical direction, acting on the body is zero, for the object moves only in the horizontal direction. Hence the total work of all forces is equal to the work of the resultant force. Suppose that w = 100 lb, P = 50 lb, f = 15 lb, \texttheta = 37\textdegree , and s = 20 ft. Then W_p = (P cos \texttheta) \textbullet s = 50 × 0.8 × 20 = 800 ft.lb, W_f= - fs = -15 × 20 = -300 ft.lb, W= W_p + W_w + W_N + W_f = 800 ft\bulletlb + 0 + 0 - 300 ft\bulletlb = 500 ft\bulletlb

Question:

The quadrilateral ABCD is a square of side 1 ft which can rotate about the fixed point 0, which is the midpoint of the diagonals. Forces 2, 3, 2, and 1 lb act along sides AB, BC, CD, and DA, respectively. Find the magnitude and line of action of a single force which would produce the same effect as these four forces.

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Solution:

At first sight this may not appear to be a problem in equilibrium, but the easiest method of solution is obtained when it is changed into one. Add a fifth force E\ding{217}, the force necessary to produce equilibrium. This force is just sufficient to negate the translational and rotational effects of the resultant force. E\ding{217} (see diagram) acts at the angle \texttheta to AB, at a distance of x from A. The resultant required in the problem must be equal and opposite to E\ding{217}, since the single force equivalent to the four given forces must, with E\ding{217}, produce equilibrium. Resolve the five forces parallel to AB and at right angles to AB. Since the forces are in equilibrium, 2 lb - 2 lb - E cos \texttheta = 0 and1 lb - 3 lb + E sin \texttheta= 0. E cos \texttheta = 0andE sin \texttheta = 2 lb. \texttheta = 90\textdegreeandE = 2 lb. Taking moments about 0 and using the condition for equilibrium, one obtains -2lb × (1/2)ft - 3lb × (1/2)ft - 2lb × (1/2)ft - lb × (1/2)ft + E[x - (1/2)ft] = 0 E[x - (1/2)ft] = 4 ft \textbullet lb x - (1/2)ft = (4ft \bullet lb)/(2lb) = 2 ftorx = 2(1/2)ft. Thus, the equilibrant has a magnitude of 2 lb and acts at right angles to AB in a direction away from 0, along a line passing through a point at a distance of 2(1/2) ft from A. The resultant required has thus the same magnitude and position but acts at right angles to AB toward 0.

Question:

What must be the width a of a rectangular guide such that the energy of electromagnetic radiation whose free-space wavelength is 3.0cm travels down the guide (a) at 95% of the speed of light? (b) At 50% of the speed of light?

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Solution:

The group speed is given by: v_gr v_gr = 0.95c = c\surd = 0.95c = c\surd [1 - (\lambda / 2a)^2] . Solving for a yields a = 4.8cm; repeating forv_gr= 0.50c yields a = 1.7cm. If \lambda = 2a, thenv_gr= 0 and energy cannot travel down the guide. For the radiation considered in this example \lambda = 3.0cm, so that the guide must have a width a of at least (1/2) × 3.0cm = 1.5 cm if it is to transmit this wave.The guide whose width we calculated in (a) above can transmit radiations whose free-space wavelength is 2 × 4.8cm = 9.6 cm or less.

Question:

How do higher plants get rid of their waste products?

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Solution:

Since green plants undergo both photosyn-thesis and respiration, the products of one process may become the raw materials of the other and vice versa. Thus oxygen, the product of photosynthesis, is utilized in cellular respiration while the products of respiration, carbon dioxide and water, are used in photosynthesis. These products may also diffuse out through the pores of the leaves, depending on the dominance of either process at a particular time. For example, during the night, when respiration predominates, carbon dioxide and water vapor escape from the pores of the leaf surfaces. The amount of nitrogenous wastes in plants is small compared to that in animals, and can be eliminated by diffusion either as ammonia gas through the pores of the leaves or as nitrogen-containing salts through the membrane of the root cells into the soil. Other wastes such as oxalic acid accumulate in the cells of the leaves and are eliminated from the plant when the leaves are shed.

Question:

The solubility products of Fe(OH)_2 and Fe(OH)_3 are 10-^17 and 10-^38 , respectively. If the concentrations of Fe^2+ and Fe^3+ are each 10-^5 M, at what pH will each hydroxide just begin to precipitate?

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Solution:

To solve this problem, set up the solubility product equa-tions for Fe(OH)_2 and Fe(OH)_3 . Once this is done, substitute the values of Fe^2+ and Fe^3+ concentrations and solve for the OH- ion concentrations. Since pH +pOH= 14, then pH = 14 -pOH, wherepOH= -log[OH-]. One has, therefore, [Fe^2+] [OH-]^2 = 10-^17 and [Fe^3+] [OH-]^3= 10-^33 . For precipitation to occur in these solutions [Fe^2+] [OH-]^2 \geq 10-^17 and [Fe^3+] [OH-]^3 \geq 10-^33 . When the product of the concentrations is equal to zero the concentrations are just sufficient for preci-pitation to occur. Solving for the pH at these concentrations: For Fe(OH-)_2 , [OH-]^2 =(10-^17 ) / [Fe^2+] = (10-^17 ) / (10-^5 ) = 10-^12, [OH-]^2 =(10-^17 ) / [Fe^2+] = (10-^17 ) / (10-^5 ) = 10-^12, so that [OH-] = 10-^6 therefore,pOH= 6 and thus pH = 8 . For Fe(OH)_3 , [OH-]^3 = (10-^38 ) / [Fe^2+] = (10-^38 ) / ((10-^5 ) = 10-^33 , so that [OH-] = 10-^11 . Consequently, pOH= 11andpH = 3 .

Question:

Find a general formula to give information about the number of passes through the loop in both the best case and the worst case for the selection sort. Express in terms of N elements.

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Solution:

Take the worst case first. For the integers 5, 4, 3, 2, 1, the selection sort will take the following ac-tions: 1. Find the smallest entry and interchange it with the element in the first position. 1, 4, 3, 2, 5 2. Sort the remaining N - 1 elements in positions 2 through N of the array 1, 2, 3, 4, 5 If you refer back to the program, notice that on the first pass through the outer loop, the inner loop is executed for K from 2 to N; on the second pass, from 3 to N, and so on. The total number of passes through the inner loop is (N - 1) + (N - 2) + (N - 3) + ... 3 + 2 + 1, which comes to N (N - 1) / 2 passes through the inner loop. Unfortunately, the same number of passes is required in the best case to check the list. The best way to understand the formula above is to write the series in the order given and in reverse: (N - 1) + (N - 2) + (N - 3) + ... 3 + 2 + 1 1 + 2 + 3 + ... + (N - 3) + (N - 2) + (N - 1) Adding term by term, we get twice the sum of the series, which is N + N + N + ... + N + N + N Since there are N-l terms in the above series, twice the sum of the series may be written as N (N - 1). Dividing by 2 gives us the sum of the series, or N (N -1) / 2.

Question:

Explain how a "Skinner box" can be used in operant conditioning.

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Solution:

In operant conditioning, a particular behavior is rewarded or punishedwhen it occurs, thereby increasing or decreasing the probability thatit will occur again. The Skinner box (named after B. F. Skinner, a behavioralpsychologist) is a principal means of studying this type of learning. The typical Skinner box is a small cage with a bar and food dish. When a rat or other small animal is placed in the box, its exploratorybehavior might cause it to press the bar accidentally. If the rat isrewarded by being given a food pellet, it is more probable that it will pressthe bar again. This is called positive reinforce-ment. It is critical that thereward be given within a second or so after the response in order for conditioningto be affective. The reward is given whenever the rat presses thebar. The rat eventually learns to associate bar-pressing with food. Repetition produces a degree of learning beyond which there is no improvement. One can eliminate the bar pressing behavior by eliminating the reinforcement. This is called extinction, which is the decay of the original learnedbehavior in the absence of the reward.Therateof extinction dependson the extent and schedule of positive reinforcement. In-creased repetitionmakes a response more resistant to extinction. Also, reinforcementat temporal intervals (giving a reward for every fourth bar press) is more difficult to extinguish than continuous reinforcement (rewardfor every bar press). It is also possible to condition the rat not to pressthe bar through the use of punishment. By electrically shocking the ratcon-currently with bar pressing behavior, the rat associates the two and learnsnot to press the bar. Operant conditioning is sometimes called trial-and- error learning whenit occurs in nature. Young chicks will initially peck at any small object; eventually they peck only at edible objects. When the chick pecks atinedible objects, it gets no reward, so this behavior is extinguished. It learnsto peck only at edible objects since they act as a positive reinforcement.

Question:

Why isn't a biome map of the earth a true representationof thevegetation formations found on land?

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Solution:

The naturalist travelers of the eighteenth and nineteenth centuries sawstrikingly different kinds of vegetations parceled out into formations. They drew maps showing the extent of each formation. However, drawing amap involves drawing boundaries. Where there are real, distinct boundarieson the ground this is an easy task. Such boundaries include thosebetween sea and land, along deserts or beside mountain ranges. But often there are no real boundaries on the ground. In fact, most of the formationboundaries of the earth are not distinct. Real vegetation types usuallygrade one into another so that it is impossible to tell where one formationends and another begins. The apparent distinctness of the vegetation formations, when viewedfrom afar, is largely an optical illusion as the eye picks out bands whereindividual species are concentrated. A vegetation map-maker could plotthe position of a tree line well enough, if he had a few reports from placeswith distinct tree lines which he could plot and link up. However, thegradual transitions, such as between the, temperate woods and tropicalrain forests would be much more difficult to plot. So mapmakers drawboundaries using what seems to be the middle of the transitions. When the map is finished, the area is parceled out by blocks of formations ofplants. A map like this shows the vegetation of the earth to be more neatly set into compartments than it really is.

Question:

Chlorine is found in nature in two isotopic forms, one of atomic mass 35amuand one of atomic mass 37amu. The average atomic mass of chlorine is 35.453amu. What is the percent with which each of these isotopes occurs in nature?

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Solution:

This is a mathematical problem, the solution to which centers on defining the average value properly. Consider a set of N observations or measurements, M_1, M_2, .... , M_N, Let the probability that the observations M_1 is madebep_1, the probability that the observation M_2 is madebeP_2, and so on. Then, the average A is defined as: A = p_1M_1 + p_2M_2 + .... +p_NM_N. Since, by definition, the sum of the probabilities must be one, then: 1 = p_1 + p_2 + ... + P_N. For this problem there are two observations, hence N = 2. Let the first observation be M_1 = 35amuand the second observation be M_2 - 37 amu. The average A = 35.453amu. Hence, A= p_1M_1 + p_2M_2 35.453amu= p_1 × 35amu+ p2× 37amu where1= p_1 + p_2. One must determine p_1 and p_2 (the probabilities of occurrence of isotope 35 and 37, respectively). From the second of these relationships, 1 = p_1 + p_2 one obtains p_2 = 1 - p_1. Then A= p_1M_1 + p_2M_2 = p_1M_1 + (1 - p_1)M_2 = p_1M_1 + M_2 - p_1M_2 = M_2 + (M_1 - M_2)p_1 or, p_1 = (A - M_2)/(M_1 - M_2) = (35.453amu- 37amu)/(35amu- 37amu) = 0.2265 Then, p_2 = 1 - p_1 = 1 - 0.7735 = 0.2265. Thus, the isotope of mass 35amuoccurs with probability of 0.7735 or 0.7735 × 100 % = 77.35 % and the isotope of mass 37amuoccurs with a probability of 0.2265 or 0.2265 × 100 % = 22.65 %. In reality, the two isotopes do not have integral atomic masses, and the percent occurrences calculated above are not exactly correct.

Question:

What is the critical angle of incidence for a ray of light passing from glass into water. Assume n_glass = 1. 50 and n_water =1.33

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Solution:

When light passes from a more optically dense to a less optically dense medium, there is an angle of incidence i, called the critical angle, at which the light ray (ray c of the diagram) will be refracted parallel to the interface of the media. The angle of refraction r will be 90\textdegree in this case. Therefore, by Snell's Law, n_1 sin i = n_2 sin r. Also critical angle of incidence means sin r = 1. \therefore sin i = (n_2/n_1) = (1.33/1.50) = .887 \therefore i_c = 62\textdegree (approx).

Question:

Explain the four important evolutionary events that occurred inthe Paleozoic Era.

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Solution:

The Paleozoic Era was marked by four important evolutionary events. First, in the Ordovician Period, the ancestral backbone-bearing animalsor vertebrates appeared. These early vertebrates were the ostracoderms, small, jawless, armored, fresh-water, bottom-dwelling fisheswithout fins. Theostracodermsthen gave rise in the later periods to agreat variety of finned fish. In the period following the Ordovician, the Silurian, two events of great biologic importance marked the second and thirdmajor evolutionary advancement of the Paleozoic Era. The first land plantsappeared, signifying the beginning of the invasion of the land by plants. These plants most closely resembled the ferns. At about the same time, the first air-breathing animals arose. These were thearachnids- landdwellers resembling to some extent modern scorpions. Finally, in the Carboniferous Period, the first reptiles, called the stem reptiles, appeared. This was an important evolutionary event because the stem reptiles were thecommon primal ancestors of a great variety of animals that still exist today. The modern reptiles, birds, and mammals are all derivatives of the stem reptiles.

Question:

What are tropisms?Where in a plant cantropistic responses occur?

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Solution:

Most animals perceive external stimuli via specialized sense organs and respond with the aid of an elaborate nervous system. Plants have neither sense organs nor nervous systems, but react to stimuli by means of tropisms. A tropism can be defined as a growth movement effected by an actively growing plant in response to a stimulus coming from a given direction. This results in the differential growth or elongation of the plant toward or away from the stimulus. Tropisms are named after the kind of stimulus eliciting them. Phototropism is a response to light; geotropism, a response to gravity; chemotropism, a response to a certain chemical; and thigmotropism, a response to contact or touch. The mechanisms of tropisms are now under much investigation. At this point, they are believed to involve certain plant hormones which are produced primarily inmeristemicregions (regions showing rapid cell division) of the plant. In appropriately low concentrations, these hormones stimulate growth and differentiation. These plant hormones are along the phloem and exert their effects on parts somewhat removed from the site of production. But because they can stimulate only those cells that are capable of rapid division,tropisticresponses can occur only in those parts of the plant which are actively growing or elongating, such as the apical part of the stem and the tip of the root.

Question:

What volume of hydrogen at STP is produced as sulfuric acid acts on 120 g. of metallic calcium. Equation for the re-action is Ca + H_2 SO_4 \rightarrow CaSO_4 + H_2

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0206.htm

Solution:

This problem may be solved by using either the mole method or the proportion method. Mole method: In using the mole method, one looks at this equation and sees that for every mole of calcium acted on by H_2 SO_4 one mole of hydrogen is produced. This means that to find how muchhyrdogenis produced, one must first find out how much calcium is present. There will be the same number of moles of hydrogen produced as there are calcium reacted. After one knows how many moles of hydrogen are produced, one can calculate the volume. To calculate the number of moles of calcium present, one must divide the amount present by the molecular weight (molecular weight of Ca = 40). number of moles = [(number of grams present) / (molecularweigth)] number of moles of Ca = [(120 g)/(40 g/mole)] = 3.00 moles Therefore, 3.00molesof hydrogen gas are produced. At STP (Standard Temperature and Pressure), the volume of one mole of any gas occupies 22.4 liters. Thus, when 3 moles of gas are generated, as in this problem it occupies 3 × 22.4 liters = 67.2 liters. Proportion method: In the proportion method, the molecular weights (multiplied by the proper coefficients) are placed below the formula in the equation and the amounts of substances (given and unknown) are placed above. In this case, because one is trying to find the volume of hydrogen and not its weight, the volume of one mole will be placed below the equation as shown. 120 gX( Vol.) Ca+H2 SO_4\rightarrowCaSO_4+H2 40.0g22.4 liters [(120 g) / (40.0 g)] = [(X) / (22.4 liters)] X = unknown volume of H_2 produced. Solving for X: X = [{(22.4 liters) (120 q)} / {40.0}] = 67.2 liters.

Question:

What is a hermaphrodite? How and why is self-fertilization in hermaphrodites prevented?

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Solution:

Any organism that has both male and female gonads is said to be hermaphroditic. The majority of flowering plants are hermaphroditic in that they bear both the pistil (the female sex organ) and the stamen (the male sex organ).Hermaphroditismis rarer in animals than in plants. It occurs chiefly in sessile forms, notably the sponges and some mollusks. It is also found in the earthworms, the parasitic tapeworms and flukes. Hermaph-roditism due to abnormal embryonic sexual differentiation resulting in individuals born with both a testis and an ovary has also been documented in man. Some hermaphroditic animals, the parasitic tapeworms, for example, are capable of self-fertilization. Most hermaphrodites, however, do not reproduce by self-ferti-lization; instead, two animals, such as two earthworms, copulate and each inseminates the other. Self-fertiliza-tion is prevented by a variety of methods. In some animal species, self- fertilization is prevented by the development of the testes and ovaries at different times. In others, such as the oysters, the different gonads produce gametes at different times, so that self-fertilization cannot occur. The advantage accompanying cross-fertilization is that genetic diversity is enhanced. Self-fertilization preserves the same genetic composition and with mutation occurring at a slow rate genetic variation is limited. Hermaphrodites undergoing cross-fertilization produce offspring showing the same genetic diversity as those resulted from individuals having a distinct sex. Thus, by ensuring gene exchange between individuals, cross--fertilization enhances genetic variation in the population. Only when a population carries a sufficient degree of genetic variation, can physical and environmental forces act to ultimately improve the characteristics of the individuals. And, as we shall see in a later chapter, the survival andpropagativeability of a species depends to a large extent on its genetic diversity.

Question:

Three forces acting on a particle and keeping it in equilibrium must be coplanar and concurrent. Show that the vectors representing the forces, when added in order, form a closed triangle; and further show that the magnitude of any force divided by the sine of the angle between the lines of action of the other two is a constant quantity.

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Solution:

Let the three forces be P\ding{217}, Q\ding{217}, and S\ding{217}, at angles \alpha, \beta, and \Upsilon to one another as shown in figure(a). In order that the three forces shall be in equilibrium, the resultant R\ding{217} of P\ding{217} and Q\ding{217} must be equal and opposite to S\ding{217}. The vectors P\ding{217}, Q\ding{217}, and S\ding{217} are concurrent and, since the vector R\ding{217} is in the same plane as P\ding{217} and Q\ding{217}, they are coplanar. But the resultant of P\ding{217} and Q\ding{217} is obtained by vector addition, as in figure (b). That is, R\ding{217} is the third side of the triangle formed by placing the tail of Q\ding{217} at the head of P\ding{217}. The force S\ding{217} is equal and opposite to R\ding{217} and thus will occupy the same space as R\ding{217}, the third side of the triangle, but will be opposite in di-rection to R\ding{217}. Thus P\ding{217} + Q\ding{217} + S\ding{217} taken in order, form a closed triangle and their sum is of necessity zero. Applying the law of sines to the triangle of figure (b)P/(sin \texttheta) = Q/(sin \varphi) = S/(sin \Psi) \thereforeP/[sin(180 - \alpha)] =Q/[sin(180 - \beta)] = S/[sin(180 - \Upsilon)] \thereforeP/[sin \alpha] =Q/[sin \beta] = S/[sin \Upsilon] = const.

Question:

It can be shown that the set of operations S = {+, \bullet, \textasciitilde} is functionally complete. That is, every Boolean function can be represent-ed by a form f(x_1,....,x_n)in variables x_1,...,x_n and operations + ,\bullet, \textasciitilde . Equivalently, the set of gates S^I = {OR, AND, NOT} is function-ally complete. Show that: (a) S_1 = {+, \backsim} (b) S_0 = {\bullet, \backsim} (c) S3= {\uparrow} where x_1\uparrowx_2 = \backsim(x_1 \bullet x_2)(1) (d) S4= {\downarrow} where x_1\downarrowx_2 = \backsim(x_1 + x_2)(2) Are functionally complete.

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Solution:

(a) It is sufficient to show that we can construct the \textbullet operation from S_1, since we know that S = { +, \textbullet , \backsim } is functionally complete. DeMorgan's law suggests itself here: \textasciitilde(A \bullet B) = \textasciitilde A + \textasciitilde B9B Applying the \textasciitilde operator to both sides and using the involution law for the left side, we get: A \textbullet B = \textasciitilde [\textasciitilde (A \bullet B)] = \backsim [\backsim A + \backsim B] Hence, S_1 = [+, \backsim ] is functionally complete. The gate representation of A \textbullet B = \textasciitilde [ \backsim A + \textasciitilde B] is shown in figure 1 This means that every gating function can be implemented by a network of OR and NOT gates, by substituting the network of fig. 1 for any AND gate in the original OR-AND-NOT network. (b) It is sufficient to show that we can construct the + operation from S_2. But DeMorgan's law states that: \textasciitilde(A + B) = \textasciitilde A \textbullet \textasciitilde B Applying the \textasciitilde operator to both sides and using the involution law for the left side, we get: A + B = \textasciitilde[\backsim(A + B) ] = \textasciitilde(\textasciitilde A \textbullet \backsim B] Hence, S_2 = {\bullet, \backsim } is functionally complete. The gate representation of A + B = \textasciitilde[\textasciitilde A \bullet \textasciitilde B] is shown in fig. 2 . This means that every gating function can be implemented by a network of AND and NOT gates. (c) The \uparrow operator is called the Shaeffer stroke function and its associated gate is a NAND(for NOT-AND) gate. It is sufficient to show that we can construct S_2 = { \textbullet , \backsim} from the stroke function alone, since it has already been shown that S_2. is functionally complete. The idempotent law for \textbullet allows substitution of x_1 \textbullet x_1 for x_1 so that: \backsim x_1 = \backsim (x_1 \bullet x_1) But from the definition of the stroke function, equation (1), we have: \backsim (x_1 \bullet x_1) = x_1 \uparrow x_1 Therefore, \textasciitilde x_1 = x_1 \uparrow x_1 \bullet(3) To obtain \textbullet from the stroke function, recall that the involution law allows us to write: x_1 \bullet x_2 = \backsim [\backsim(x_1 \bullet x_2)] But from the definition of the stroke function, equation (1), we have: \textasciitilde[\textasciitilde(x_1 \textbullet x_2)] = \textasciitilde[x1\uparrow x_2] Using the result obtained equation (3), we can write: \backsim[x_1 \uparrow x_2] = (x_1 \uparrow x_2) \uparrow (x_1 \uparrow x_2) Therefore,x_1- \bullet x_2 = (x_1 \uparrow x_2) \uparrow (x_1 \uparrowx_2)(4) and S_3 is functionally complete. The gate representation of equation (3) is shown in fig. 3 (3) is shown in fig. 3 . The gate representation of equation (4) is shown in fig. 4 . This means that every gating function can be implemented by a network of NAND gates only. (d)The \downarrow operator is called the NOR (from NOT-OR) function and its associated gate is a NOR gate. It is sufficient to show that we can construct the S_1 = { + , \textasciitilde} from the NOR function alone. Using the idempotent law for + and the definition of the NOR function, equation (2), we can write: \textasciitildex1= \textasciitilde(x_1 +x_1) = x_1 \downarrow x_1(5) To obtain + from the NOR function, we apply the involution law and use the definition of the NOR function and equation (5): x_1 + x_2 = \textasciitilde[\textasciitilde(x_1 + x_2)] = \textasciitilde[x_1 \downarrow x_2] = [x_1 \downarrow x_2] \downarrow[x_1 x_2](6) Hence, S_4 = {\downarrow} is functionally complete. The gate representation of equation (5) is shown in fig. 5 The gate representation of equation (6) is shown in fig. 6 Every gating function can be implemented by a network of NOR gates only: Though, for simplicity, only two variables were used in showing the functional completeness of S_1 - S_4, the results can be shown to be true for any number of variables by a simple induction argument.

Question:

Compare cardiac muscle to skeletal and smooth muscle.

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/Users/wenhuchen/Documents/Crawler/Biology/F15-0382.htm

Solution:

Cardiac muscle is the tissue of which the heart is composed. Cardiac muscle shows some characteristics of both skeletal and smooth muscle. Like skeletal muscle, it is stria-ted; it has myofibrils composed of thickand thinmyofilaments, and contains numerous nuclei per cell. The slidingfilament mech-anism of contraction is found in cardiac muscle. Cardiac muscle resembles smooth muscle in that it is innervated by the autonomicnervous system. The cells of cardiac muscle are very tightly compressedagainst each other and are so intricatelyinterdigitatedthat previouslyno junctions were thought to exist between cells. They do, however, exist, and are visible under the light micro-scope as dark-colored discs, calledintercalcalateddiscs. It is believed that these discs may help totransfer the electrical impulses generated by the S-A node between musclecells due to their low resistance to the flow of current. The metabolism of cardiac muscle is designed for endurance rather thanspeed or strength. A continuous supply of oxygen and ATP must be providedin order for the heart muscle to main-tain its contractile machinery. Cardiac cells deprived of oxygen for as little as 30 seconds ceaseto contract, and heart failure ensues.

Question:

The pineal gland is a lobe on the upper portion of the forebrain, and has long intrigued investigators by its glandular appearance. Is this lobe an endocrine gland? If yes, what is its function(s)?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0547.htm

Solution:

It was only recently demonstrated that the pineal gland has an endocrine function. In some primitive vertebrates, the pineal responds to light by generating nerve impulses and by secreting a hormone called melatonin. Melatonin lightens the skin by concentrating the pigment granules in melanophores, which are specialized pigment- containing cells. In frogs, the action of melatonin is opposed by melanocyte-stimulating hormone (MSH). MSH is secreted by the pituitary gland, and serves to disperse the pigment granules, causing a darkening of the skin. In mammals, the pineal secretes melatonin but does not generate impulses because light-sensitive cells are not present. Although the pineal does secrete melatonin in mammals, a problem arises when we realize that mammals have no melanophores (the sites of the hormonal action). Thus, the function of the pineal and its hormone in mammals has been a source of puzzlement for a long time. Recently, a theory has been proposed that the pineal gland functions as a neuroendocrine transducer. It was theorized that the pineal receives neural input from the visual receptors, and transduces this data into chemical messages (hormones). Apparently, impulses from the eyes reach the pineal via sympathetic nervous pathways from the cervical ganglia. The pineal responds by secreting more or less melatonin. In mammals, melatonin no longer affects skin pigmentation, but appears to have become a regulator of the anterior pituitary. Melatonin seems to in-hibit reproductive activities of rodents and birds by inhib-iting the release of gonadotrophic hormones (LH and FSH) . The diurnal rhythm of the light/dark cycles seems to mediate this function of melatonin. Thus, in these animals, repro-duction coincides with the optimal season, as light of day is interpreted by the brain via stimulation of the photore-ceptors. The function of the pineal gland in man remains highly speculative, though it may play a role in circadian rhythms.

Question:

Two hikers set off in an eastward direction. Hiker 1 travels 3 km while hiker 2 travels 6 times the distance covered by hiker 1. What is the displacement of hiker 2?

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Solution:

From the information given the displacement vector is directed east. The magnitude of the displace-ment vector for hiker 2 is 6 times the magnitude of the displacement vector for hiker 1. Therefore, its magnitude is 6 × (3 km) = 18 km

Question:

Plot the graph of a sine curve with point at 15\textdegree intervals.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0234.htm

Solution:

The TAB function is useful here. By specifying the appropriate coordinates, it causes the printer to place an X in the correct column. We use column 40 to correspond to the X-axis so that there will be room to plot negative values. In the program, C = \pi/180. It converts from degrees to radians. The PRINT statement, however, represents the points in 150 steps, as given by the FOR ... NEXT construct. The program is given below: 1\O REM A PLOT OF THE SINE CURVE 2\O LET C = 3.14159/18\O. \O 3\O FOR A = \O TO 36\O STEP 15 4\O LET R = C\textasteriskcenteredA 5\O PRINT A; "DEG"; TAB(4\O. \O + 25\textasteriskcenteredSIN(R)); "X" 6\O NEXT A 7\O END -10+1 \O DEG X X X X X X X X X X X X X X X X X X X X X X X X X 15 DEG 3\O DEG 45 DEG 6\O DEG 75 DEG 9\O DEG 1\O5 DEG 12\O DEG 135 DEG 15\O DEG 165 DEG 18\O DEG 195 DEG 21\O DEG 225 DEG 24\O DEG 255 DEG 27\O DEG 285 DEG 3\O\O DEG 315 DEG 33\O DEG 345 DEG 36\O DEG

Question:

Discuss the structure of the chloroplast and the possible relation of its structure to the process of photosynthesis. What is a quantosome?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0212.htm

Solution:

Chloroplasts are small organelles found in plant cells that serve as the sites of photosynthesis. There are some 20 to 100 chloroplasts in a plant cell; these can grow and divide to form daughter chloroplasts. A chloroplast, like a mitochondrion, has a double - layered outer membrane. Lying within each chloroplast are many smaller stacks called grana (see figures). Under the electron microscope, the grana are seen as part of an elaborate system of membranes organized in parallel pairs. The pairs of membranes are joined at their ends to form a closed disk or thylakoid. The grana themselves are stacks of these thylakoids. Each granum is composed struc-turally of layers of protein molecules alternating with layers of chlorophyll, carotenes and other pigments, and special types of lipids containing galactose or sulfur but only one fatty acid. These surface - active lipids are believed to be absorbed between the layers and serve in stabilizing the lamellae composed of the alternate layers of protein and pigments. This lamellar structure of the grana is important in permitting the transfer of energy captured from the sun from one molecule to the adjacent one during the light phase of the photosynthetic process. This is evidenced by the fact that chlorophyll extracted from the chloroplast and isolated in a test tube can no longer carry out any photosynthetic reaction owing to the destruc-tion of the lamellar structure. The semifluid matrix within the chloroplast and surrounding the grana is called the stroma. The stroma contains the enzymes responsible for the dark reactions of photosynthesis. Electron microscopy has revealed the presence of repeating unit structures on the surface of the lamellae within the chloroplast. These structures, given the name quantosomes, are found to each contain some 230 molecules of chlorophyll. Botanists now believe that quantosomes are the functional photosynthetic units.

Question:

What average force is necessary to stop a bullet of mass 20 gm and speed 250 m/sec as it penetrates wood to a distance of 12 cm?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0269.htm

Solution:

As it travels through the block, the bullet experiences an average force, F_avg^\ding{217}, which retards its motion. By the work-energy theorem, the work done by the net force on an object equals the change in kinetic energy of the object. Hence \int F^\ding{217} \textbullet ds^\ding{217} = (1/2)mv^2_f - (1/2)mv^2_0 But we only know F^\ding{217} as an average value. Hence \int F^\ding{217} \textbullet ds^\ding{217} \approx F_avg^\ding{217} \textbullet ∆s^\ding{217} = (1/2)mv^2_f - (1/2)mv^2_0 By definition F_avg^\ding{217} \textbullet ∆s^\ding{217} = \vert F_avg^\ding{217}\vert\vert∆s^\ding{217}\vert cos \texttheta = (1/2)mv^2_f - (1/2)mv^2_0- where \texttheta is the angle between F_avg^\ding{217} and∆s, 180\textdegree in this problem. Whence -\vert F_avg^\ding{217}\vert\vert∆s^\ding{217}\vert = (1/2)mv^2_f - (1/2)mv^2_0- \vert F_avg^\ding{217}\vert = [(1/2)mv^2_o - (1/2)mv^2_f] / [\vert∆s^\ding{217}\vert] Hence,\vert F_avg^\ding{217}\vert = [(1/2)(0.2 kg)(250m/s)^2 - 0] / [.12 m] = 5.2 × 10^3 nt This force is nearly 30,000 times the weight of the bullet. The initial kinetic energy, (1/2) mv^2 = 620 joules, is largely wasted in heat and in work done in deforming the bullet.

Question:

Develop a FORTRAN program to useLagrangian interpolationto evaluate F(x) = x3 at x = 3, given the table below: i 1 2 3 4 x_i 1 2 4 7 f_i 1 8 64 343 f_i \equiv f(x_i)

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Solution:

The Lagrange polynomials of degree m are defined as P_j(x) =A_j^m^+1\prod(k=1)k\not =j(x -x_k) whereA_j= ^m+1\prod(k=1)k\not =j[1/(x_j-x_k)] Here m = 3 and we approximate f(x) by 4\sum_i_=1f(x_i)P_i (x). The evaluations of P_j(x) are shown below, for x = 3. P1(x) \equiv P(x,1) = {(x-2)(x-4)(x-7)} / {(x1-x2)(x1-x3)(x1-x4)} = {(x-2)(x-4)(x-7)}/ (-18) P2(x) \equiv P(x,2) = {(x-1)(x-4)(x-7)} / {(x2-x1)(x2-x3)(x2-x4)} = {(x-1)(x-4)(x-7)} / 10 P3(x) \equiv P(x,3) = {(x-1)(x-2)(x-7)} / {(x3-x1)(x3-x2)(x3-x4)} = {(x-1)(x-2)(x-7)} / 18 P4(x) = P(x,4) = {(x-1)(x-2)(x-4)} / {(x4-x1)(x4-x2)(x4-x3)} = {(x-1)(x-2)(x-4)} / 90 ThusF(3) \approx 4\sum_i_=1 f(x_i) P(3,I) = 1(-2/9) + 8(4/5) + 64(4/9) + 343(-2/90) = 26.9999980 which is very close to 33 = 27. The program and sample output are given below: CPROGRAM 3.3 CPROGRAM FOR LAGRANGIAN INTERPOLATION CUNEVENLY SPACED PIVOTAL POINTS DIMENSIONX(50), P(50), F(50) PUNCH 994 READ 999, N, (X(I), F(I), I = 1,N) 1READ 998, XO DO 10 J = 1,N P(J) = 1 DO 10 I = 1,N IF (I - J) 9, 10, 9 9P(J) = P(J) \textasteriskcentered (XO - X(I))/(X(J) - X(I)) 10CONTINUE FO = O DO 20 I = 1,N 20FO = FO +P(I) \textasteriskcentered F(I) PUNCH 997 PUNCH 996, (I,X(I),F(I), P(I), I = 1,N) PUNCH 995, XO, FO GO TO 1 994FORMAT (//, 19X, 24H RESULTS FROM PROGRAM 3.3) 995FORMAT (19X, 6H AT X = F5.2, 21H THE VALUE OF F(X) IS F12.7) 996FORMAT (I10, 3F14.7) 997FORMAT (//, 9X, 1H1, 8X,4HX(I), 10X, 4HF(I), 9X, 6HP(X,I)) 998FORMAT (8F10.0) 999FORMAT (I5/8F10.0) END [SAMPLE]RESULTSFROMPROGRAM 3.3 X(I) F(I) P(X,I) 1 1.0000000 1.0000000 -0.2222222 2 2.0000000 8.0000000 0.8000000 3 4.0000000 64.0000000 0.4444444 4 7.0000000 343.0000000 -0.0222222 AT X = 3.00 THE VALUE OF F(X) IS 26.9999980

Question:

What are the mole fractions of solute and solvent in a solution prepared by dissolving 98 g H_2SO_4 (M.W. 98) in 162 g H_20 (M.W. 18)?

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Solution:

The mole fraction of solute is defined as the moles of solute divided by the sum of the number of moles of the solute and the number of moles of the solvent. Mole fraction of solute = [(moles of solute)/( moles of solute + moles of solvent)] The mole fraction of the solvent is defined similarly, Mole fraction of solvent = [(moles of solvent)/(moles of solute + moles of solvent)] Here, the solute is H_2SO_4 and the solvent is H_2O. One is given the amount of H_2SO_4and H_2O in grams. Therefore, these quantities must be converted to moles. This can be done by dividing the number of grams available by the molecular weight. No. of moles = [(no. of grams)/MW] No. of moles of H_2SO_4 = [98 g/(98 g/mole)] = 1 mole No. of moles of H_2O = [162 g/(18 g/mole)] = 9 mole Now that the number of moles of both solvent and solute are known, the mole fraction can be found. Mole fraction of H_2SO_4 = [(1mole)/(1 mole + 9 moles)] = 0.1 Mole fraction of H_2O = [(9mole)/(1 mole + 9 moles)] = 0.9.

Question:

What would be the surface tension of a liquid (density 0.876 g / cm^3) which rises 2.0 cm in a capillary where water at 20\textdegreeC rises 4.5 cm?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0271.htm

Solution:

Surface tension refers specifically to the force within a liquid that acts parallel to the surface and tends to stretch the surface out. The equation that relates surface tension to the density of a liquid is \Elzpgamma = (1 / 2 rh\rhog where \Elzpgamma is the surface tension, r is the radius of the tube, h is the height the liquid rises, \rho is the density of the liquid and g is the downward acceleration of gravity. This equation can be rewritten rg = (2\Elzpgamma) / (h\rho) Because r and g are constant for a given capillary tube (2\Elzpgamma H_2O)/(h_H(2)O \rho_H(2)O) = (2\Elzpgamma liq)/(h_liq \rho_liq) One can find \Elzpgamma_H(2)O and \rho_H(2)O from standard tables. \Elzpgamma_H(2)O = (72.62 dyn / cm), \rho_H(2)O = (1 g / cm^3). Solving for liq: [2(72.62 dyn / cm) / (4.5 cm)(1g / cm^3)] = (2\Elzpgamma_liq) / [(2.0cm)(0.876 g / cm^3)] \Elzpgamma_liq = [2(72.62 dyn / cm)(2.0cm)(0.876 g / cm^3)] / [2(4.5 cm)(1g / cm^3)] \Elzpgamma_liq = 28.27 (dyn / cm)

Question:

A 0.10 M solution ofNaOHis prepared. What species of ions are present atequlibrium, and what will be their equilibrium concentrations?

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Solution:

Two processes are occurring simultaneously, the dissociation of NaOH, and theautoionizationof H_2O. NaOHdissociates according to the equation NaOH\rightleftarrows Na^+ + OH^-. For every mole ofNaOHthat dissociates, one mole of Na^+ and one mole of OH^- are produced. The initial concentration ofNaOHis 0.10 M. Thus, if we assume thatNaOHdis-sociates completely[Na^+] = 0.10 M and [OH^-] = 0.10 M. Waterautoionizesaccording to the equation H_2O + H_2O \rightleftarrows H_3O^+ + OH^-. The water constant for this process is K_W = 10^-14 moles^2/liter^2 = [H_3O+][OH^-]. Hence, [H_3O^+] = [(10^-14 moles^2/liter^2)/(OH^-)] Since [OH^-] was determined to be 0.10 M = 10^-1 M = 10^-1 moles/liter, by substitution, we obtain [H_3O^+] = [(10^-14 moles^2/liter^2)/(OH^-)] = [(10^-14 moles^2/liter^2)/ (10^-1 moles/liter)] = 10^-13 moles/liter = 10^-13 M. Hence, at equilibrium, H_3O^+, OH^-, and Na^+ are present in the concentrations [H_3O^+] = 10^-13 M, [OH^-] = 0.10 M, [Na^+] = 0.10 M.

Question:

A snowball at - 15\textdegreeC is thrown at a large tree trunk. If we disregard all frictional heat loss, at what speed must the snowball travel so as to melt on impact? The specific heat of fusion of ice is 79.7 cal/gm.

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Solution:

To solve this problem, one must know that the translational energy or kinetic energy of the snowball is completely converted into heat energy. The kinetic energy is given by the equation k. e. = 1/2 mv2, where m = mass and v = velocity. The total heat energy for this particular problem is Q =m∆Tc+m∆H_fus_ , where m = mass, ∆T = change in temperature, c = heat capacity (for H_2O, 0.5 cal/\textdegreeC gm) , and ∆H_fus= 79.9 cal/gm. These two equations are set equal to each other. The kinetic energy must be large enough so that the snowball can increase its temperature to 0\textdegreeC and then be completely converted to liquid water at 0\textdegreeC. Thus, m∆Tc+m∆H_fus= (1/2) mv^2 ∆Tc+ ∆H_fus= (1/2) v^2 (15\textdegreeC) (0.5 cal/\textdegreeC gm) + 79.7 cal/gm= (1/2) v^2 v^2 = 174 cal/gm = (174 cal/mg) (4.18 J/cal) = 727 J/gm = 727 kgm^2 / s^2 / gm = 7.27 × 10^5 m^2/s^2 [Note: 1 kg = 1000 gm; 727 × 1000 m^2/s^2 = 7.27 × 10^5 m^2/s^2] Thus, v = \surd(7.27 × 10^5 m^2/s^2) = 853 m/s.

Question:

Differentiate betweenisolecithal,telolecithal, and centrolecithal eggs. Which organisms are characteristicof each ?

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Solution:

The eggs of different animals vary greatly in the amount and distribution of the yolk they contain. Some, such as those of many annelids , mollusks, and echinoderms, are small and contain only a little yolk . If the yolk is evenly distributed they are calledisolecithaleggs. Other eggs are larger and have a moderate supply of yolk. Of this sort are amphibian eggs and those of primitive fish. Still others, such as those of reptiles and birds, have a tremendous supply of yolk. Since, in these types , the yolk is concentrated toward one pole of the egg (known as the vegetal pole), these eggs are termedtelolecithal. Arthropod eggs usually have the yolk massed toward the center, and hence are called centrolecithal .

Question:

A skier is filmed by a motion-picture photographer who notices him traveling down a ski run. The skier travels 36 ft during the fourth second of the filming and 48 ft during the sixth second. What distance did he cover in the eight seconds of filming? Assume that the acceleration is uniform throughout.

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Solution:

The fact that the acceleration is uniform gives us a big advantage since, in this case, the instantaneous acceleration is equivalent to the aver-age acceleration: a\ding{217} =\Deltav\ding{217}/\Deltat= (v_f\ding{217} - v_0\ding{217})/(t_f- t_0) wherev_f\ding{217} and v_0\ding{217} are the velocities at timest_fand t_0 respectively. To solve for a we use the kinematic equation s = v_0t + (1/2)at^2 where s is the distance covered in time t. 36 = v_0(1) + (1/2)a(1) = v_0 + (1/2)a 48 =v_f(1) + (1/2)a(1) =v_f+ (1/2)a. where v_0 andv_fare the velocities at the beginning of the fourth and sixth seconds respectively, and both time intervals are one second long. (1/2)a = 36 - v_0 = 48 -v_f v_f= v_0 + 12 Since there is a two second interval between the times when the skier has velocities v_0 andv_f: a= (v_f- v_0)/\Deltat = {(v_0 + 12) - v_0}/2 = 12/2 = 6 ft/sec^2 Knowing the acceleration, we can now solve .for the skier's velocity v_0 at the beginning of the 4th second: 36 = v_0(1) + {(1/2)(6)} (1) v_0 = 36 - 3 = 33 ft/sec Now, we may solve for v_0', the velocity at the beginning of the filming v_0 = v_0' + at,v_0' = v_0 - at v_0' = 33 - (6) (3) = 15 ft/sec Thus the distance covered in the eight seconds of filming is: s= v_0' t + (1/2)at^2 = (15 ft/sec) (8 sec) + {(1/2)(6 ft/sec^2)} (8 sec)^2 = 312 ft.

Question:

A certain gas occupies a volume of 100 ml at a temperature of 20\textdegreeC. What will its volume be at 10\textdegreeC, if the pressure remains constant?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0032.htm

Solution:

In a gaseous system, when the volume is changed by increasing the temperature, keeping the pressure constant, Charles' Law can be used to determine the new volume. Charles' Law states that, at a constant pressure, the volume of a given mass of gas is directly proportional to the absolute temperature. Charles' Law may also be written (V_1 /T_1 ) = (V_2 /T_2 ) where V_1 is the volume at the original temperature T_1 and V_2 is the volume at the new temperature T_2. To use Charles1 Law, the temperature must be ex-pressed on the absolute scale. The absolute temperature is calculated by adding 273 to the temperature in degrees Centigrade. In this problem, the centigrade temperatures are given and one must convert them to the absolute scale. T_1 = 20\textdegreeC + 273 = 293\textdegreeK T_2 = 10\textdegreeC + 273 = 283\textdegreeK Using Charles' Law, V_1 = 100 ml T_1 = 293\textdegreeK T_2 = 283\textdegreeK V_2 = ? (V_1 /T_1 ) = (V_2 /T_2 )V_2 = (V_1 T_2 /T_1 ) V_2 = [{(100 ml) (283\textdegreeK)}/(293\textdegreeK)] V_2 = 96.6 ml.

Question:

A mass of 5 kg hangs on a spring which has a spring constant of 2 × 10^3 new tons/m. (a) Calculate the minimum energy of the system, (b) By optical methods the position of the mass can be determined to within 10^-7 m. What is the uncertainty in the velocity of the mass?

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Solution:

(a) The natural frequency of the vibrating mass is given by \omega = \surd(k/m) = \surd[(2 × 10^3nt/m)/5 kg] = 20 sec^-1 The zero point energy is the residual energy the vibrating particle has at absolute zero. It is given by E = (1/2)hf = (1/4\pi) × (6.63 × 10^-34 × 20)J \textbullet s \textbullet s^-1 = 1.05 × 10^-32 J (b) If we equate this to the maximum kinetic energy the vibrating particle has (when the elastic potential energy of the particle is zero) then E = (1/2) mv^2 v^2 = [(2 × 1.05 × 10^-32 J)/5 kg] v = 2.0 × 10^-17 m/sec If the uncertainty in the position is\Deltax= 10^-7 m, then, according to the uncertainty principle, \Deltax\Deltap= (h/2\pi) \Deltap= (h/2\pi\Deltax) where\Deltap=m\Deltavis the uncertainty in the momentum of the particle. Therefore the uncertainty in the velocity is \Deltav= (h/2\pim\Deltax) = [(6.63 × 10^-34 J \textbullet sec)/(2\pi × 5 × 10^-7 m \textbullet kg)] = 2.1 × 10^-28 m/sec This uncertainty is less than the zero point velocity. Suppose that we attempt to measure the velocity by measuring the time required for the mass to move through x = 2 × 10^-7 m. That is, t = (x/v) = [(2 × 10^-7 m)/(2 × 10^-17 sec)] = 10^10 sec(about 300 years) Such a measurement is liable to be interrupted before it is completed.

Question:

Calculate ∆G\textdegree andK_pat 25\textdegreeC for CO (g) + H_2 O (g) \rightarrow CO_2 (g) + H_2 (g), CO (g) + H_2 O (g) \rightarrow CO_2 (g) + H_2 (g), ∆G\textdegree_ffor H_2 (g), CO_2 (g), H_2 O (g), and CO (g) are 0, - 94, 2598, - 54.6357, and - 32.8079 Kcal/mole, respectively.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0547.htm

Solution:

The Gibbs free energy of formation (∆G\textdegree_f) of a substance is the Gibbs free energy change, i.e., the amount of free energy available for work, for the re-action in which the substance in its standard state is formed from its elements in their standard states. By use of the following equation, ∆G\textdegree = ∆G\textdegree_f,products- ∆G\textdegree_f,_ reactants the Gibbs free-energy change for a reaction may be calculated from the Gibbs free energies of formation. The equilibrium constant for the reaction may then be calculated using the standard Gibbs free energy. Thus,∆G\textdegree = [∆G\textdegree_f, _CO2 + ∆G\textdegree_f, _H2] - [∆G\textdegree_f,_CO + ∆G\textdegree_f, _H2O] = (- 94.2598 + 0) - (- 32.8079 - 54.6357) = - 6.8162 Kcal/mole = - 6816.2 cal/mole. ∆G\textdegree = - RT InK_p, where R = gas constant, T = absolute temperature (Celsius plus 273\textdegree) and K_p = equilibrium constant, which is to be calculated. As such, one can, via substitution, write ∆G\textdegree = - 6816.2 cal/mole = - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (298\textdegreeK) (2.303) logK_p logK_p= 4.997 K_p= [(P_H2 P_CO2)/(P_CO PH2 O)] = 9.931 × 10^4.

Question:

A ball is thrown upward with an initial speed of 80 ft/sec. How high does it go? What is its speed at the end of 3.0 sec? How high is it at that time?

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Solution:

Since both upward and downward quantities are involved, upward will be called positive. At the highest point the ball stops, and hence at that point v_1 = 0. The only force acting on the ball is the gravitational force which gives a constant acceleration of g =- 32 ft/sec^2. For constant acceleration and unidirectional motion, 2as = v^2_1 - v^2_0 2(-32 ft/sec^2)s_1 = 0 - (80 ft/sec)^2 s_1 = [-(80 ft/sec^2)^2]/[2(-32 ft/sec^2)] \approx 100 ft. s_1 is the highest point the ball reaches. To find the speed of the ball after 3 seconds, v_2 = v_0 + at = 80 ft/sec + (-32 ft/sec^2)(3.0 sec) = 80 ft/sec - 96 ft/sec = -16 ft/sec. After 3 seconds, the speed of the ball is v_2 = 16 ft/sec downward. The height of the ball after 3 seconds can be found from s_2 = v_0t + (1/2)at^2 s_2 = v_0t + (1/2)at^2 = (80 ft/sec)(3.0 sec) + (1/2)(-32 ft/sec^2)(3.0 sec)^2 = 240 ft - 144 ft = 96 ft. As a check, s_2 can also be found by using s_2 =vt= [(v_2 + v_0)/2]t = [(-16 ft/sec + 80 ft/sec)/2] × 3.0 sec = 96 ft wherevis the average velocity. Note that s is the magnitude of the displacement, not the total distance traveled. If the ball returns to the starting point or goes on past it, s will be zero or negative, respectively.

Question:

The accompanying figures show a cathode-ray tube (figure A) and the deflection of an electron in a region of magnetic field (figure B). If deflection in the tube is being produced by the magnetic field alone, predict the effect on the observed deflection by in-creasing (a) mass of particles, (b) velocity of the particles, (c) magnetic field, and (d) charge on the particles.

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Solution:

The magnetic field, charge, velocity, mass and the deflection of a particle are related by the following equation, Hev = [(mv^2 )/(r)] where H is the strength of the magnetic field, e the charge on the particles, r the radius of deflection, v the velocity of the particles, and m the mass of the particles. The derivation of this equation need not be considered in solving this problem. Rewriting the given equation and solving for e/m, the charge to mass ratio, (e/m) = (v/Hr). Since e/m is a constant if (a) the mass is increased, then the radius of deflection must also increase. This is seen more clearly by solving the given equation for r, r = [(vm)/(He)]. Here, if m increases (keeping v, H, and e constant), then r increases. Following this type of reasoning, one can predict: (b) the radius of deflection increases with increasing velocity; (c) the radius of deflection decreases with in-creasing magnetic field strength; and (d) the radius of deflection increases with increasing charge on the particles. Note that the radius of curvature is inversely proportional to the distance the particle is actually deflected from its original path.

Question:

Show that in theBalmerseries that the frequencies of Balmer series that the frequencies of successive lines tend toward a limiting value ofcR/4, where successive lines tend toward a limiting value ofcR/4, where R is theRydbergconstant and c the speed of light. R is theRydbergconstant and c the speed of light.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0700.htm

Solution:

Here, one must try to relate frequency ѵ with c and R. TheRydberg

Question:

Using the given d-t curve calculate the velocity- time curve.

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Solution:

A velocity-time curve is found from a displacement-time curve by plotting the slope of the d - t curve versus time. Our task in this particular case is made easier by the fact that the velocity in each segment of the trip is constant. From A to Bv_AB= d_AB/t_AB = {(3 - 0)m}/{(2 - 0)sec} = 1.5 m/sec From D to Ev_DE= d_DE/t_DE = {(0 - 5)m}/{(14 - 12)sec} = -2.5 m/sec. The corresponding segments on the v - t curve are represented by horizontal lines. The rest of the curve is found similarly. Note that the area under the v - t curve at each time gives the displacement. That is, at t = 6 sec,area= (1.5 × 2 + 0.5 × 4)m = 5 m at t = 14 sec,area= (1.5 × 2 + 0.5 × 4 - 2.5 × 2)m = 0 m

Question:

Write a FORTRAN subroutine to determine the largest element (in absolute value) in theithrow of N × N array called ARRAY.

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Solution:

Let the arguments of the subroutine be denoted by ARRAY, N, I, BIG, J. Here ARRAY stands for a square array whose dimensions are N × N, I stands for the row in which we wish to determine the maximum element, denoted by BIG, is to be found and J stands for the column in which BIG is located. The only trick to the algorithm is to implement a sequential comparison of each element of theithrow to the largest preceding value. The subroutine looks as follows: SUBROUTINE LARGE (ARRAY, N, I, BIG, J) DIMENSION ARRAY (N, N) BIG = ABS (ARRAY (1, 1)) J = 1 DO 9 K = 2, N IF (ABS (ARRAY (I, K)) .LT.BIG) GO TO 9 CSET BIG TO THE NEW VALUE BIG = ABS (ARRAY (I,K)) CUPDATE THE COLUMN OF THE NEW BIG ELEMENT. J = K 9CONTINUE RETURN END

Question:

The pressure of the nitrogen in a constant-volume gas thermometer is 78.0 cm at 0\textdegreeC. What is the temperature of a liquid in which the bulb of the thermometer is immersed when the pressure is seen to be 87.7 cm?

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Solution:

In a thermometer there is one physical property (thermometric property) whose change is used to indicate a change of temperature. The thermometric property of the thermometer, in this case the pressure, is taken as being directly proportional to the Kelvin temperature. Therefore T = cp where c is a constant of proportionality. We can then state that for two temperatures on this scale, the following relationship holds: T_1/T_2 = p_1/p2 Let T_1 be 0\textdegreeC or 273\textdegreeK. Substituting the known values, we get T_2 = T_1p_2/p_1 = (273\textdegreeK) (87.7 cm)/(78.0 cm) = 307\textdegree K Reconverting to the Celsius scale, T_2 = 307\textdegreeK - 273\textdegreeK = 34\textdegreeC.

Question:

Water at 30\textdegreeC has a vapor pressure of 31.82 mm Hg. When 25.0 g of ethylene glycol is added to 1000 g of water, the vapor pressure is lowered to 31.59 mm Hg. Determine the molecular weight of ethylene glycol.

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Solution:

This problem is an application of raoult's law. Let p\textdegree denote the vapor pressure of pure water, p the vapor pressure of the ethylene glycol - water solution, n_W the number of moles of water in the solution, and n_e the number of moles of ethylene glycol. Then from Raoult's law, p = p\textdegree (n_W) / (n_e +n_w). Multiplying by minus one and adding p\textdegree to both sides of this equation gives p\textdegree - p = p\textdegree[(n_W) / (n_e +n_w)] = p\textdegree[1 - {(n_W) / (n_e +n_w)}] = p\textdegree[{(n_e +n_w) / (n_e +n_w)} -{(n_W) / (n_e +n_w)}] or,p\textdegree - p = p\textdegree [n_e(n_e +n_w)] Since the number of moles is equal to the number of grams divided by the molecular weight, this equation becomes p\textdegree - p = p\textdegree(g_e / M_e) /[(g_e / M_e) + (g_w / M_w)] where g_e denotes the number of grams of ethylene glycol, g_w the number of grams of water, M_e the molecular weight of ethylene glycol (which we are trying to determine) , and M_w the molecular weight of water. For this problem, p\textdegree = 31.82 mm, p = 31.59 mm, g_e = 25.0 g. g_w = 1000g and the molecular weight of water, M_w = 18. Substituting into the last equation, we obtain 31.82 mm - 31.59 mm = 31.82 mm [(25.0g/M_e)/(1000g /18) + (25.0 g/M_e)] Or,0.23 mm = 31.82 mm [(25.0 g / M_e) / (1000 g / 18) + (25.0 g / M_e)]. Solving for M_e, (0.23 mm) / (31.82 mm)[(1000 g / 18) + (25.0 g / M_e)] = (25.0 g / M_e) [(0.23 mm) / (31.82 mm)] × [(1000 g / 18) = (25.0 g / M_e)] = [(0.23 mm) / (31.82 mm)] × (25.0 g/M_e) = (1 / M_e) × 25.0 g[1 - {(0.23 mm)/(31.82 mm)}], Or, M_e =[(31.82 mm) / (0.23 mm)] × (1000 g / 18) × 25.0 g[1 -[(0.23 mm) / (31.82 mm)] = 62. Thus, the molecular weight of ethylene glycol is 62.

Question:

Calculate the minimum photon energy (in electron volts) required to create a real electron-positron pair. Express all numbers to four signi-ficant figures.

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Solution:

The rest mass-energy E of an electron or positron is E = m_ec^2 =(9.110 × 10^-31kg) (2.998 × 10^8m/s)^2 = 8.188 × 10^-14 J If the created pair of particles possess no kinetic energy, then by the conservation of energy, the photon energy E_\Upsilon must be equal to the sum of the rest mass-energies of the two particles: E \Upsilon = 2E = 1.638 × 10^-13J Since 1 eV = 1.602 × 10^-19 J, the required energy may be written E_\Upsilon = (1.638 × 10^-13\Delta) / (1.602 × 10^-19\Delta/ eV) = 1.022 × 10^6 eV This energy is the minimum or threshold energy required for creation of the electron-positron pair. Photons with this such energy are from the x-ray or gamma-ray region of the electromagnetic spectrum. Note however, that although we have satisfied the conservation of energy in our analysis, we have not satisfied the conservation ofmomen- tum. The photon has an initial momentum, but the electron and positron have no final momentum, since we have required them to be at rest. Hence, the energy of the incoming photon must be greater than 1.022 × 10^6eV, since, in any practical experiment, some agent must carry away momentum (and energy) at the end of the reaction. This is why pair production cannot occur in a vacuum.

Question:

Describe what is known as the shunting-yard algorithm for converting infix (or algebraic) notation into the Polish postfix notation. Enumerate the five rules of priority that must be applied to the input string in order to explain the algorithm.

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Solution:

The metaphor of the shunting-yard is an appropriate one for this algorithm. We can think of a string of variables and operators as the cars of a train. The shunting-yard is the place where cars are added, removed, and rearranged. In programming terminology, the shunting-yard can be represented by a stack. This stack can be likened to a rail-road track siding, as shown in Figure 1. As the string is moved from right to left, each symbol is evaluated. A symbol will be placed either in the stack or directly into the output string, contingent upon the fol-lowing rules of priority: 1) Operands at the front of the input string are moved directly to the output string. 2) A left parenthesis at the front of the input string is pushed onto the stack. 3) A right parenthesis at the front of the input string causes all operators to be popped up from the stack into the output string until a left parenthesis (if necessary) appears. The parentheses are then discarded from the output string. 4) Operators at the front of the input string cause operators in the stack to be popped up and into the output string until the top item in the stack is one of lower priority than the operator in the front of the input string. Priority is assigned to items according to the following table: ItemsPriority (low + , -high \textasteriskcentered , /higher If the stack is empty, or if the item at the top of the stack is one of lower priority, an operator at the front of the input string is pushed onto the stack. 5) When the input string becomes empty, all items in the stack are popped up and into the output string sequen-tially. Let us now "walk through" this algorithm, showing the movement of the items in the string and stating the rules to be applied at each step.

Question:

How can glycogen (a polymer of glucose) be converted into fats (triacylglycerals) in the body?

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Solution:

A fat, or triacylglycerol molecule, consists of glycerol bonded to three fatty acids. Therefore, in order for glycogen to be converted into fats, it must first be broken down and converted into glycerol or an activated form of glycerol and fatty acids. Glycogen is broken down by the enzyme glycogen phosphorylase, to give glucose 1-phosphate (G1P). G1P is next converted to glucose 6-phosphate (G6P) by the enzyme phosphoglucomutase. G6P is then isomerized to fructose 6-phosphate by an isomerase enzyme. Further reaction of fructose 6-phosphate result in the formation of two triose phosphates: dihydroxy- acetone phosphate and glyceraldehyde 3-phosphate. At this point, the conversion branches. Dihydroxyacetone-P (where P indicates phosphate) is reduced to glycerol 3-P which then gets hydrolyzed by a phosphatase to glycerol. Glyceraldehyde 3-P is converted to pyruvate by a series of reactions. Pyruvate now undergoes oxidative decarboxylation in the mitochondria and is activated by Coenzyme A to become acetyl-CoA. The latter enters the citric acid cycle where it combines with oxaloacetate to form citrate. Citrate, unlike acetyl-CoA, can pass through the mitochond-rial membrane to the cytoplasm, where it is reconverted to acetyl-CoA. Acetyl- CoA is then converted in the cytoplasm into malonyl-CoA, and finally, via a complex sequence of reactions, into fatty acids. Three fatty acids then react with one glycerol molecule to form a triacylglycerol. This sequence of events, discussed only briefly above, is summarized below: Note that pyruvate can only be converted to acetyl-CoA in the Note that pyruvate can only be converted to acetyl-CoA in the mitochondria. Fatty acids, however, are synthesized in the cytoplasm. Since acetyl-CoA cannot directly leave the mitochondria, it must first be converted to citrate, which can leave the mitochondria, and then be reconverted to acetyl-CoA in the cytoplasm. Although the enzymes for each reaction step are not indicated, it is important to realize that each step is catalyzed by an enzyme or complex of enzymes.

Question:

Compare the cost of operating 3 lamps in series and in Compare the cost of operating 3 lamps in series and in parallel on a 115 volt circuit, if each lamp has a resistance of 100ohms. ohms.

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Solution:

We calculate the cost of operation for each configuration (see figures) by calculating the net power expended by each circuit. Assuming both circuits run for the same time, the energy used by each can then be found. The circuit using less energy is the more economical. For the series circuit, the same current I flows through each lamp. Each one then uses power P = I^2 R where R is the lamp resistance. Since the resistance of each lamp is the same, the net power expended in the series circuit is P_net = 3I^2 R(1) Looking at the equivalent circuit in figure (a), we realize that the net resistance of the series con-figuration is R_net = 3R By Ohm's Law, the current in this circuit is I = (V/R_net ) = (V/3R)(2) Using (2) in (1) P_net = [(3V^2 )/(9R^2 )] R = (V^2 / 3R) Using the given data, the series configuration uses power P_net = [(115 V)^2 /(300 \Omega)] = 44.1 Watts For the parallel connection, each lamp has the same voltage V applied across it. Each one uses power P ' = (V^2 /R) where R is the lamp resistance. Since all the resist-ances are equal, the net power expended by the parallel circuit is P'_net = (3V^2 /R) = [{3 (115 V)^2 } / (100 \Omega)] = 396.75 Watts If both circuits operate for a time \cyrchar\cyrt, the energies used are E_series = P_net \cyrchar\cyrt = (44.1 Watts) \cyrchar\cyrt E_parallel = P'_net \cyrchar\cyrt = (396.75 Watts) \cyrchar\cyrt Hence, the series combination is cheaper to run.

Question:

Write a computer program in APL to analyze an electrical circuit using the AND and OR operators of Boolean algebra. Assume four switches in the circuit; two are connected in parallel and two are connected in series.

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Solution:

The circuit in fig. 1 is equivalent to the truth function Y (A,B,C,D) = (AVB) V (C\LambdaD). The complete truth table for Y is: ABAVBCDC\LambdaBY = AVBVC\LambdaD 0000000 0000100 0001000 0001111 0110001 0110101 0111001 0111111 1010001 1010101 1011001 1011111 1110001 1110101 1111001 1111111 The truth-table is equivalent to the circuit. For example, if switches A,B,C are off and D is on, the light will not shine. Similarly, if propositions A,B,C are false (=0) and proposition D is true (=1), the compound proposition (A or B) or (C and D) is false. \nablaBOOLE [1]'ENTER TRUTH VALUES FOR A,B,C AND D' [2]L \leftarrow [] [3]TEST \leftarrow (L[1] V L[2]) V L[3] \Lambda L[4] [4]'THE VALUE OF Y IS:'; TEST [5]\nabla The header of the program is composed of \nabla (del) and BOOLE. The del operator causes the system to change to programming mode while BOOLE is the name of the program. As an example, suppose the values 0,1,1,0, are entered for A, B, C and D respectively. Then the value of Y would be 1. The program may be modified to handle other types of circuits. (FIG. 2) This circuit is equivalent to the truth function Y (A,B,C,D) = (AVB)V(CVD). To test this function replace [3] by [3] TEST \leftarrow (L[1]VL[2])V (L[3]VL[4]) The circuit, (FIG. 3) represents the function (AVB)\Lambda(CAD). The test for this function is [3] TEST \leftarrow (L[1]VL[2]) \Lambda L [3] \LambdaL [4].

Question:

A man is using a uniform ladder of weight W = 75 lb, one end of which is leaning against a smooth vertical wall, the other end resting on the sidewalk. It is prevented from slipping by rubber suction pads rigid-ly attached to the feet of the ladder and stuck firmly to the concrete. If the man of weight w = 150 lb is standing symmetrically three-quarters of the way up the ladder, and if the normal force N\ding{217}exerted by \ding{217} the wall on each leg of the ladder is 43.3 lb, what is the force exerted on the ladder by each suction pad?

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Solution:

The ladder is uniform and thus its weight acts as its center. The man is symmetrically placed on the ladder. Hence, by symmetry, the normal forces exerted by the wall on the two legs of the ladder are equal, as are the forces exerted by the two suction pads. Let the force exerted by either suction pad on the ladder be resolved into component forces F_x and F_y along the sidewalk and normal to it, respectively. The complete force system acting on the ladder is as shown in the figure, the man exerting a force equal to his weight on the ladder. The ladder, of course, exerts an equal and opposite force on him, since he is in equilibrium. The whole system is in equilibrium. It follows from Newtons Second Law that 2F_x = 2N, F_x = N = 43.3 lb and2F_y = W + w,F_y = (W + w)/2 = (75 lb + 150lb)/2 = 112.5 lb The total force exerted by each suction pad on the ladder thus has magnitude F= \surd(F_x^2 + F_y^2) = \surd[(43.3 lb)^2 + (112.5 lb)^2] = 120.6 lb and it acts at an angle \texttheta to the horizontal, where tan \texttheta = (F_y/F_x) = (112.5 lb)/(43.3 lb) = 2.60,\texttheta = 69\textdegree.

Question:

Give examples, in pseudocode or with flowcharts, of the following control constructs: a) DO-WHILE b) D0-F0R c) IF-THEN-ELSE

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G08-0183.htm

Solution:

Control constructs are program segments which interrupt the sequencing of the entire program to perform some specific task. These constructs may consist of decision instructions, loops, or a combination of both. If rendered faithfully, these constructs will make your program efficient and readable. The DO-WHILE construct requires the continued execution of a sequence of statements, as long as a given condition is satisfied. A pseudocode description looks something like this: do while (X) tasks A - Z end do while X is some logical expression. The loop is repeated until X is no longer satisfied. Program control drops out of the loop and on to the next statement. A flowchart of the situation is as shown in Fig. 1 The DO-F0R construct differs from the DO-WHILE construct in that it specifies explicitly the number of tines the loop should be executed. When the desired number of loops has been executed, control passes out of the loop. Here are pseudocode and flowchart descriptions of the DO-FOR: (see Fig. 2) do for I=1 through N tasks A - Z end do for When translated into a high-level language, the loop, will contain some statement that declares the value of N. Exit from the loop is contingent upon N executions. It should be noted that there are two distinct types of DO-FOR loops. The PL/1 implementation of a DO-FOR loop is as shown in Fig. 1(a). If N = 0, the body of the loop (tasks A - Z) will not be performed. The FORTRAN implementation of a DO-FOR loop is described by the flowchart see Fig. 2(b). Note that the body of the loop (tasks A - Z) will always be perforated at least once in a FORTRAN DO-loop, even if N = 0. An unwary student might look at the flowchart given for a DO-FOR loop and think that it describes FORTRAN'S DO-loop, but actually it does not. Note also that DO FOR constructs may be nested inside each other, such as in the following pseudocode: do for I = 1 through N do for J = 1 through M tasks A - Z end do for J end do for I Tasks A - Z will be performed M × N times. Notice two important conventions in this example. First, in both the DO FOR and DO WHILE constructs, only the first and last statements are left-justified; all other statements within the loop are indented. This is purely a stylistic convention, but it serves to clarify the difference between the loop's control statements and its task-performing, or assignment, statements. Second, realize that you may not terminate a nested loop outside the bounds of the "nest". In the example above, because the DO-FOR J loop is the nested loop, you must terminate it before term-inating the DO-FOR I loop. A violation of this rule will cause serious logical errors. The IF-THEN-ELSE construct is a decision-rendering construct. If the given condition is satisfied, then a certain sequence of statements is executed. If the condition is not satisfied, a different sequence is followed. In pseudocode we have If (x) then tasks A - M else tasks N - Z end if then else with x as the given condition. Notice again that the first and last statements are not indented, while the others are. See flowchart Fig. 3. The true condition corresponds to the THEN part of the construct, The true condition corresponds to the THEN part of the construct, while the false condition is interpreted as the ELSE part.

Question:

A flywheel of mass 12 kg and radius of gyration 20 cm is mounted on a light horizontal axle of radius 5 cm which rotates on frictionless bearings. A string wound round the axle has attached to its free end a hanging mass of 4 kg, and the system is allowed to start from rest. If the string leaves the axle after the mass has descended 3 m, what torque must be applied to the fly-wheel to bring it to rest in 5 revs?

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Solution:

Consider the hanging mass. It has two forces acting on it, its weight, mg^\ding{217}, downward and the tension in the string, T^\ding{217} upward. Since the mass descends with acceleration a^\ding{217}, by Newton's second law mg - T = ma But, if the flywheel rotates with angular acceleration \alpha, then the tangential acceleration a = r\alpha,and thusmg - T = mr\alpha(1) The tension T^\ding{217} acts at distance r from the axis of the flywheel of mass M and radius of gyration k. Therefore the torque is \Gamma = Tr. Using the rigid body analogue of Newton's second law, where torque takes the place of force, moment of inertia, I, takes the place of m, and angular acceleration \alpha takes the place of linear acceleration, \Gamma = Tr = l\alpha. Since I = Mk^2, Tr = Mk^2\alpha. Also, upon multiplication of both sides of equation (1) by r, mgr - Tr = mra = mr^2\alpha. \thereforemgr = \alpha(mr^2 + Mk^2). \alpha= mgr/(mr^2 + Mk^2) = (4 kg × 9.8 m\bullet^-2 × 0.05 m) / (4 kg × (0.05)^2m^2 + 12 kg × (0.2)^2 m^2) = [(1.96) / (0.49)]s^-2 = 4 rad\bullets^-2. The flywheel starts from rest and accelerates as long as the string is exerting a couple on it. In that time the mass descends 3 m. But in 1 rev, the string unwraps a length equal to its circumference, 2\pir, and the mass descends by this distance. Thus the angular distance the flywheel turns during the period of acceleration is 3m / (2\pi × 0.05m/rev) = (3/0.05) × (1rev/2\pi) × (2 \pirad/1rev) = (3/0.5) rad = 60 rad The angular speed when the string leaves the axle can be found by using the rigid body analogue of the kinematic equations for constant acceleration. If \omega is the analogue of linear velocity, v, \alpha the analogue of linear acceleration, a, and \texttheta the analogue of linear displacement, s, then the kinematic equation not involving the time variable is \omega^2 = \omega^2_0 + 2\alpha\texttheta. The initial angular velocity \omega_0 = 0 and \omega^2 = 2 × 4 rad\bullets^-2 × 60 rad or\omega= 4 × \surd30 rad\bullets^-1= 21.9 rad\bullets^-1. If a torque \Gamma' is now applied to the wheel, it produces a constant deceleration \alpha', and since the flywheel is being brought to rest, \omega_final = 0 or 0^2 = \omega^2 + 2\alpha' × 10\pi rad. \therefore\alpha' = (-\omega^2)/(20\pi rad) = - [(480 rad^2\bullets^-2)/(20\pi rad)] = - (24/\pi) rad\bullets^-2.. But\Gamma' = l\alpha' = Mk^2\alpha' = - 12 kg × 0.04 m^2 ×(24/\pi) rad\bullets^-2 = - 3.67 N\bulletm. Thus a torque of 3.67 N\bulletm applied against the direction of rotation is necessary. Note that the result is more easily obtained from a consideration of energy of the mass-flywheel system. The mass m descends a height h. In so doing it loses potential energy, which reappears in the form of kinetic energy of the mass and of the flywheel. Thus, if the bottom of the fall is taken as the reference level, then mgh = (1/2)mv^2 + (1/2)I\omega^2 = (1/2)mr^2\omega^2 + (1/2)Mk^2\omega^2 \therefore\omega^2 = (2mgh)/(mr^2 + Mk^2) = 480 rad^2\bullets^-2. In the final stage, the work done by the retarding torque in the 5 rev = 10\pi radians must equal the kinetic energy possessed by the flywheel before the couple is applied. Thus - 10\pi\Gamma = (1/2)I\omega^2 = (1/2)Mk^2\omega^2. \therefore\Gamma = - [(Mk^2\omega^2)/(20 \pi)] = - 3.67 N\bulletm.

Question:

A mass m = 2 kg is hung on a spring of constant k = 3.92 × 10^3N/m. The natural angular frequency of the system is \omega_0 = 44.3 sec^-1. Now we will force the system to vibrate at different frequencies by an alternating force F = 2 cos \omegat newtons That is, the maximum force is two newtons. A static force of two newtons would cause a deflection of about 1/2 mm to the system. What will be the amplitude of vibration for \omega = 15 sec^-1, and 60 sec^-1?

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Solution:

This problem is a case of harmonic motion with a periodic force applied to a vibrating mass. The forces acting on the mass m are the given applied force F = 2 cos \omegat, and the restoring force F_R = -ky due to the spring. y is the vertical displacement of the mass from its rest position. The negative sign indicates that the restoring force acts in such a direction as to oppose a change in vertical displacement. By Newton's second law, -ky + 2 cos \omegat = m[(d^2y)/(dt^2)] or[(d^2y)/(dt^2)] + (k/m)y = 2/m cos \omegat(1) The following function is a solution to the above differential equation, y(t) = A cos \omegat(2) where A is the amplitude of vibration. This can be verified by sub-stitution into equation (1): -\omega^2 A cos \omegat + (k/m)A cos \omegat = 2/m cos \omegat Dividing out the common factor, and solving for A, we find A = (2/m)/[(k/m) - \omega^2] Let k/m = \omega^2 _0 (where \omega_0 is the natural frequency of vibration) then A = (2/m)/(\omega^2 _0 - \omega^2) Since m = 2kg, we can calculate values of \vertA\vert as shown in the table. \omega \vertA\vert 15 sec^-1 5.8 × 10^-4m 44 3.8 × 10^-3 60 6.1 × 10^-4 When the frequency \omega of the applied force is near the natural fre-quency \omega_0 the amplitude becomes much larger (see figure (b)). We can also calculate the average total energy of the system. The total energy H of the system at any moment is H = (1/2) mv^2 + (1/2)ky^2(3) where the first term is the kinetic energy of the object and the second term is the elastic potential energy of the object. Upon differentiating equation (2), substituting it into equation (3), we obtain H = (1/2) mA^2\omega^2 sin^2\omegat + (1/2) kA^2 cos^2\omegat. But k = \omega^2 _0m (by definition of \omega2_0). Therefore H = (1/2) mA^2\omega^2 sin^2\omegat + (1/2) mA^2\omega^2 _0 cos^3\omegat The average value of cos^2\omegat over 1 period is, by definition = 1/2\pi^2\pi\int_0 cos^2\omegat d(\omegat) = 1/2 Similarly, = 1/2 Hence = 1/4 mA^2\omega^2 + 1/4 mA^2\omega^2 _0 or = 1/4 mA^2(\omega^2 +\omega2_0) which shows that the energy of the system becomes much larger as \omega \ding{217} \omega_0. \omega \vertA\vert 15 sec^-1 3.7 × 10^-4 44 2.8 60 1.0 × 10^-3 As the frequency of the applied force nears the resonance frequency it is possible to transfer considerably greater energy to the system.

Question:

A cell possesses two electrodes. Both half-cells are .01M MnO^-_4 ion. One cell is .01MH_3O^+ ion, while the other has a H_3O^+ concentration of.10M. The electrode reaction for the reduction half-cell may be written: MnO^-_4 + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O . The oxidation half-cell is the reverse of this reaction. 1) Write the net equation for the spontaneous cell process taking place; 2) Find ∆E for the reaction; (3) Find the value of the equilibrium constant.

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Solution:

1) The net reaction, in such a situation, is the sum of the balanced half-reactions, i.e., oxidation reaction plus reduction reaction. Since, you are given both reactions, add the equations together to find the overall reaction. You have, with concentrations included, ox:MnO_2 + 2H_2O \rightarrow MnO_4^- (.01M) + 4H^+ (.01M) + 3e^- red:MnO^-_4 (.01M) + 4H (.10M) + 3e^- \rightarrow MnO_2 + H_2O Net Reaction: 4H^+(.10M) \rightarrow 4H^+(.01M) . Notice: all species cancelled out, except H^+ (actually H_3O^+ ). This is the net equation for the spontaneous reaction taking place. 2) To find ∆E for the reaction, use the Nernst equation, which states ∆E = ∆E\textdegree -(.059 / n)log K for a temperature at 25\textdegreec, where ∆E\textdegree = potential for cells under other than standard conditions, ∆E\textdegree = standard cell potential, n = number of electrons transferred and K = equilibrium concentration expression, n = 3, since from either the oxidation or reduction reaction, 3 electrons are being transferred; ∆E\textdegree = 0, since ∆E\textdegree = E\textdegree_prod - E\textdegree_reactants and both the product and reactant are the same species. K is defined as the ratio of products to reactants, each raised to the power of their coefficients in the net equation. Substituting these values into the Nernst equation, you have ∆E = 0 -(.059 / 3)log [(.01)^4 / (.10)^4 ] = 0 - .0197(log 10^-4) = (-.0197)(-4) = 0.079 volt. 3) To find the value of the equilibrium constant, note that there exists a relationship between ∆E\textdegree and the constant at 25\textdegreec. Namely, ∆E\textdegree =(.059 / n)log K. From part 2, you found ∆E\textdegree = 0 . Thus, K = unity (one), since this is the only value that permits log K = 0, which, then, allows ∆E\textdegree = 0, as it must.

Question:

Determine the approximate solubility ofAgClin 0.10MNaCl solution.K_spforAgCl= 1.1 × 10-^10 .

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Solution:

The approximate solubility ofAgClin this solution is equal to the concentration of Ag+ in the solution. This is because the concentration of Ag+ is equal to the amount ofAgClionized. The ionization ofAgClcan be stated AgCl\rightleftarrows Ag+ +Cl+ . This means that one Ag+ ion is formed each time one molecule ofAgClis ionized. The concentration of Ag+ can be obtained by using the solubility product constant (K_sp).+This constant is equal to the pro-duct of the concentration of Ag+ ([Ag+]) times the concentration ofCl-([Cl-]) . K_sp= [Ag+] [Cl-] = 1.1 × 10-^10 . The concentration of theCl- ions is also affected by theCl- ions from the NaCIsolution. BecauseNaCIis a salt, it is 100% ionized in solution NaCI\rightleftarrows Na+ +Cl- . There is oneCl- ion formed for everyNaCIionized. The original concentration ofNaCIis 0.10 molar, and after ionization, the solu-tion is 0.10 molar inCl-ions. The Na+ ions can be disregarded because here we are looking at the solubility ofAgCl, which does not contain any Na+ ions. One can assume that the total concentration ofCl- ions in solution will be approximately 0.10, since the amount con-tributed by theAgClionization is small, and can be neglected. The [Ag+] can now be found. [Ag+] [Cl-] = 1.1× 10-^10 [Ag+ ] (0.10) = 1.1 × 10-^10 [Ag+] = 1.1 × 10-^9 M .

Question:

Why is genetic drift less important in human evolution at present than it was twenty thousand or more years ago?

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/Users/wenhuchen/Documents/Crawler/Biology/F29-0757.htm

Solution:

The communities that man lived in about twenty thousand years ago all had something in common - they all were very small villages. The total world population then was low and people living in these small social groups were relatively distant from people in other places. With no means of transportation and the danger of traveling alone, people tended to stay close to their shelters. As a result, most breeding took place within individual social groups with each probably ranging up to about 100 members. In other words, these people were probably genetically similar for in inbreeding, no genes could be exchanged with other populations that existed. As an illustration, we can consider the western European Neanderthal people who were cut off from the rest of the world population by geographic factors. They were no longer able to exchange genes and consequently their further evolution was independent. Thus, they were sheltered from different genes present in the world's gene pool. These people became more unlike the rest of the world population and more like each other. Because of their reduced variability, when faced with an environmen-tal crisis such as sudden cold or disease, they were not able to adapt successfully and faced extinction. Let us take a look at the genetic effects involved with small inbreeding populations. Evolutionary change is not usually automatic, it occurs only when something disturbs the genetic equilibrium. A condition necessary for a popu-lation to be in equilibrium is that it must be large enough to make it highly unlikely that chance alone could signi-ficantly alter gene frequencies. Any breeding population with more than 10,000 members of reproductive age is probably not significantly affected by random change. This is the condition that exists in the world today. But gene frequencies in small isolated populations of less than 100 reproductive members are highly susceptible to random fluctuations, which can easily lead to loss of an allele from the gene pool and the fixation of another even if the former allele is an adaptively superior one. This is the phenomenon called ge-netic drift. Genetic drift explains why small populations tend to have a high degree ofhomozygosity, while large popu-lations tend to be more variable and diverse.

Question:

A water tank standing on the floor has two small holes vertically above one another punched in one side. The holes are 3.6 cm and 10 cm above the floor. How high does water stand in the tank when the jets from the holes hit the floor at the same point?

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Solution:

For any hole in a tank, the velocity of discharge of the liquid is given by Torricelli's theorem. To derive it, consider a hole a distance d below the surface of the liquid in the tank (see figure 1). Using Bernoulli's equation, P_1 + (1/2) \rhov^2_1 + \rhogy_1 = p_2 + (1/2) \rhov^2_2 + \rhogy_2 take point 1 to be at the hole and point 2 at the surface of the liquid. The pressure at each point is the atmospheric pressure p_a since both are open to the atmosphere. If the hole is small, the level of the liquid in the tank falls slowly and its velocity v_2 can be assumed to be zero. Using the bottom of the tank as the reference level, P_a + (1/2) pv^2_1 + pgy_1 = p_a + 0 + pgy_2 orv^2_1 = 2g(y_2 - y_1) = 2g d Therefore, the velocity of discharge from a hole in a tank is given by v = \surd2gd In this problem, the velocities of efflux are horizontal from both holes. Using Torricelli's theorem, one gets for the upper hole (see figure 2), v_1 = \surd[2g(h - h_1)], and for the lower one, v_2 = \surd[2g(h - h_2)]. Water from the upper hole has a horizontal velocity v_1 and no initial vertical velocity u. In time t_1 applying the formula (y - y_0) = ut_1 + 1/2 gt^21to the vertical motion , one obtains h_1 = 0 + 1/2 gt^2_1 or t_1 = \surd(2h_1/g). In that time the horizontal distance gone is v_1jt_1, which is the distance from the tank at which the jet strikes the floor. Similarly, the distance at which the jet from the lower holes strikes the floor is v_2 t_2, where t_2 = \surd[2h_2/g]. But these distances are equal, and thus v_1t_1 = v_2t_2 or (v_1t_1)^2 = (v_2t_2)^2. Then 2a(h - h_-1) × (2h_1/g) = 2a(h - h_-2) × (2h_2/g) (h - h_-1) × h_1 = (h - h_-2) × h_2 hh_1 - h^2_1 = hh_2 - h^2_2 hh_1 - hh_2 = h^2_1 - h^2_2 h = (h^2_1 - h^2_2)/(h_1 - h_2) = h_1 + h_2 .\therefore h = 13.6 cm.

Question:

The motor driving a grindstone is switched off when the latter has a rotational speed of 240 rev\bulletmin^-1 . After 10 s the speed is 180 rev\bulletmin^-1. If the angular retardation remains constant, how many additional revolutions does it make before coming to rest?

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Solution:

The initial speed\omega_ois 240 rev\bulletmin^-1 and the later speed \omega is 180 rev\bulletmin^-1. Thus since the angular acceleration, \alpha, is constant, we may write \omega = \omega_0 +\alphat. Here, t is the time it takes for the grindstone's angular velocity to go from \omega_0 to \omega. Solving for \alpha, \alpha = (\omega - \omega_0)/t = (180 rev\bulletmin^-1 - 240 rev\bulletmin^-1)/(10 s) Noting that 1 min^-1= 1/60 s^-1, we find \alpha = - (60 rev\bullets^-1)/(60 × 10s) = -.1 rev\bullets^-2 Considering the subsequent slowing-down period, the final speed is zero and the grindstone traverses an angular distance \texttheta. Hence, using the equation \omega^2 = \omega^2_0 + 2\alpha\texttheta with \omega = 0 and \omega_0 = 180 rev\bulletmin^-1 = 3 rev\bullets^-1,we find 0 = 9 rev^2 \bullets^-2 + 2(-.1 rev\bullets^-2)\texttheta or\texttheta = (9 rev)/(.2) = 45 rev.

Question:

Using positive logic, show that the diode-transistor circuit of fig. 1 represents a three-input NAND gate.

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Solution:

The circuit is broken down into two "sub--circuits"; fig. 2(a) and (b). In sub-circuit a, fig. 2(a), if any of the inputs are low(or), its corresponding Diode will conduct, pulling the output X_4 Low. If all the inputs are high (\geq 5V), none of the diodes will conduct, leaving the output X_4 high (5V). The truth table of figure 3 illustrates sub-circuit a's action, that of an AND gate. InputsOutput X_1 X_2 X_3 X_4 L L L L L L H L L H L L L H H L H L L L H L H L H H L L H H H H Fig.3 In sub-circuit b, figure 2(b), the transistor acts as a switch, conducting when X_4 is high ( >.7V), and open when X_4 is low (OV). When X_4 is high the transistor is conduct-ing and the output Y is pulled low. When X_4 is low the transistor is open and Y floats to 5V (high). Thus, sub--circuit b acts as an inverter and the whole circuit, whose truth table is shown in fig. 4, acts as a NAND gate. INPUTS X_4 Y(OUTPUT) L L L L H L L H L H L H L L H L H H L H H L L L H H L H L H H H L L H H H H H L

Question:

By photographic and optical means, an image can be enlarged almost indefinitely. However, most light microscopes offer magnifications of only 1000 to 1500 times the actual size, while electron microscopes offer magnifications of 100,000 times or more. Explain.

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Solution:

All microscopes are characterized by limits of resolution. Resolution refers to the clarity of the image. Objects lying close to one another can not be distinguished (resolved) as separate objects if the distance between them is less than one half the wavelength of the light being used. The average wave-length of visible light is 550 nanometers (or 5500 \AA). Thus, for light microscopes, objects can be distinguished only if they lie farther apart than about 275 nanometers. Objects closer together than 275 nm are not resolved and appear to be one object. Increasing the size of image, or the magnification, will not give meaningful information unless resolution is also increased. Increasing magnification without increasing resolution results in a larger image that is still blurred. Electron microscopes offer resolution of details separated by .1 to .5 nanometers. Electrons, rather than light, are the radiation used in electron micros-copes. Recall that electrons have a wave property in addition to a particle property and may be regarded as a radiation of extremely short wavelength. Since the wavelength of an electron in motion is so much shorter than the wavelength of light, resolution is more than a thousandfold better. Structures such as the plasma membrane, endoplasmic reticulum,ribosomes, microtubules and microfilaments were not visible until the advent of the electron microscope. These structures are all less than 275 nm in width. Theplagmamembrane has a thickness of 7.5 to 10 nm (or 75 to 100 \AA). The ribosome is 15 to 25 nm in diameter (or 150 to 250 \AA) . Microtubules are 20 to 30 nm in diameter, and microfilaments range from 5 to 10 nm. Electron microscopy has also made possible the visualization of the nuclear envelope and the internal membranes of mitochondria and chloroplasts.

Question:

Two true-breeding varieties of unicorns differ with respect to three traits. The first has a straight horn, a green coat, and a straight mane. The other variety has a twisted horn, a blue coat, and a curly mane. When a male of either variety is crossed with a female of the other, all the progeny have straight horns, green coats, and straight manes. These hybrid offspring are then crossed to unicorns with twisted horns, blue coats, and curly manes. This cross produces offspring in the following numbers: straight horn, green coat, straight mane855 twisted horn, blue coat, curly mane855 twisted horn, green coat, straight mane95 straight horn, blue coat, curly mane95 straight horn, blue coat, straight mane50 twisted horn, green coat, curly mane50 50 2000 Construct a linkage map for the three genes involved in this cross.

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Solution:

This is a complicated problem on linkage. Go over the previous problems on recombination. Some new concepts will be introduced here. This cross involves three traits (that is, a three-point cross) each governed by a gene. To begin, we have to determine the dominance or recessiveness of the phenotypes. This can be obtained from the F_1 since all the F_1 descendants have straight horn, green coat, and straight mane, the gene for straight horn (H) is dominant over twisted horn (h); green coat (C) is dominant over blue coat (c); straight mane (M) over curly mane (m) . Since the parents are true-breeding, one of them isand the other; The actual sequence of the genes has yet to be determined. In the cross between two homozygous parents, Note that all the dominant genes are on one chromosome while all the recessive ones are on the other. These are the parental typesHCMand hcm . All other combinations of genes are recombinants. To calculate the recombination frequency we have to know the number of recombinants. There are several ways to do this; one of them begins by translating the F_2 phenotypes to genotypes keeping the F_2 progeny in the same order we get: 1 HCM 855 2 hcm 855 3 hCM 95 4 Hcm 95 5 HcM 50 6 hCm 50 Consider a cross-over between H and C. The number of recombinants is 95 3 + 95 4 + 50 5 + 50 6 = 290. recombinants is 95 3 + 95 4 + 50 5 + 50 6 = 290. The RF between H and C = (290) / (2000) = 0.145. The RF between H and C = (290) / (2000) = 0.145. An RF of 0.145 × 100 = 14.5 map units, meaning H and C are 14.5 An RF of 0.145 × 100 = 14.5 map units, meaning H and C are 14.5 map units apart. map units apart. Consider a recombination between C and M . By similar reasoning, Consider a recombination between C and M . By similar reasoning, the number of recombinants = 50 5 + 50 6 = 100 the number of recombinants = 50 5 + 50 6 = 100 RF = (100) / (2000) RF = (100) / (2000) That is, a map distance of 5 units separates C and M. Finally, consider a cross-over between H and M. The number of recombinants = 95 3 + 95 4 = 190. RF = (190)/(2000) = .095 RF = (190)/(2000) = .095 That is, H and M are 9.5 map units apart. Now that we have calculated a map distance for any two of the three genes, we are ready to plot the linkage map. As our reference points we use H and M, which are 9.5 units apart. We know that H and C are 14.5 units apart. But C can be to the right or left of H. To determine which side of C is on, we look at the distance between C and M , which is 5 units. This means C must be on the right side of H. If C were 14.5 units to the left of H, then C and M would be separated by 24 units.' This is not the case. Therefore, the correct linkage map for the three genes is

Question:

Estimate the upper limit to the frequency of "sound" waves in ordinary matter.

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Solution:

The relation v = v/\lambda ,(1) where ѵ, v, and \lambda are the frequency, velocity and wave-length of the sound wave, respectively, makes it clear that the higher the frequency the shorter is the wavelength. We are therefore asking what is the shortest wavelength sound can have. The answer is that, when the wavelength becomes shorter than the distance apart of the atoms, the concept of a wave breaks down because there is nothing in between the atoms to oscillate (see figure). The atoms are closest together in solids and highly compressed gases. However, when two atoms are about 10^-8 cm apart they repel one another very strongly and it is difficult to force them any nearer. Consequently, in no material are the atoms or molecules found to be closer together than about 10^-8 cm. This is therefore the order of magnitude of the shortest wavelength of sound. The highest value found for the velocity of sound is about 10^6 cm/sec. Using equation (1). Highest frequency = (Largest wave velocity) / (Shortest wavelength) = (10^6 / 10^-8) approximately = 10^14 sec^-1 approximately. This is much larger than the highest frequency yet produced experimentally, which is 2.5 x 10^10 sec^-1. The reader should note the character of these arguments, which are not precise calculations, but rather intelligent guesses. We have made it very plausible that the highest frequency is not very much bigger than 10^14 sec^-1. It is conceivable that, by applying a very high pressure to the right gas or solid, we could force its atoms a little closer than 10^-8 cm and achieve a velocity of sound a little larger than 10^6 cm/sec. In this way we might perhaps realize a frequency of, say, 2 × 10^14 sec^-1. We should be very surprised, though, if we ever pushed the frequency up to 10^15 sec^-1. In fact we know that there is a limit to the pro-cedure, because there is reason to believe that, when the pressure on a substance becomes extremely large, its atoms break up into separate electrons and nuclei.

Question:

The reaction A + 2B \rightarrow products was studied kinetically and the following data was obtained. Experiment [A] (moles/liter) [B] (moles/liter) Rate(mole/liter - sec) 1 2.0 × 10^-3 4.0 × 10^-3 5.7 × 10^-7 2 2.0 × 10^-3 8.0 × 10^-3 11.4 × 10^-7 3 4.0 × 10^-3 4.0 × 10^-3 22.8 × 10^-7 Determine the rate law for this reaction.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0449.htm

Solution:

The rate of reaction is equal to some rate constant, k, multiplied by the concentrations of A and B raised to the appropriate powers. That is, rate = k [A]^m [B]^n where the exponents m and n are to be determined. The presence of the coefficient 2 in the reaction equation A + 2B \rightarrow products in no way affects the form of the rate equation. Comparing experiments 2 and 1 we see that doubling [B] while holding [A] constant doubles the rate of reaction (2 × 5.7 × 10^-7 = 11.4 × 10^-7) . Hence, the rate is directly proportional to [B] and n = 1. Comparing experiments 3 and 1 we see that doubling [A] while holding [B] constant quadruples the rate (4 × 5.7 × 10^-7 = 22.8 × 10^-7). Thus, the rate has a square dependence on [A] (2^2 = 4), that is, it is second order in [A],and m = 2. Substituting m = 2 and n = 1 into the rate expression gives rate = k [A]^2 [B]^1, or rate = k [A]^2 [B] .

Question:

Iron may crystallize in the face-centered cubic system. If the radius of an Fe atom is 1.26 \AA, (a) Determine the length of the unit cell, (b) Calculate the density of Fe if its atomic weight is 55.85.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0223.htm

Solution:

Picture the face of the cube with an iron atom at each, corner and one in the center of each face as shown in the figure. The diagonal BC will be equal to the length L 4 radii of Fe atoms. Hence, BC = 4 × 1.26 \AA = 5.04 \AA Using the Pythagorean Theorem, one knows that the squares of the two sides of a right triangle added together equal the square of the hypotenuse. Thus, AB 2 +AC^2 =BC^2 Because the sides of a square are equal -AB=AC. From this it can be seen that 2AB2=BC^2.ABis defined as the length of the unit cell. Using 2AB^2 =BC^2, it can be seen that AB^2 =BC^2 / 2andAB = \surd[BC^2 / 2] = BC / \surd2 Since we have already found BC, this equation can be solved. (BC = 5.04 \AA.) AB =BC/ \surd2 =5.04 \AA / 1.42 = 3.55 \AA The length of the unit cell is 3.55 \AA. (b) The density is defined as the mass in grams of a unit cell in one unit volume. In this problem we are given the mass of one mole of Fe and we have calculated the length of one side of the unit cube. To find the density one must calculate the number of atoms in one unit cell, the weight of one atom of Fe and the volume of one unit cube first. The following rules may be used to determine the number of unit particles (atoms) associated with the face-centimeter lattice: 1) An atom at a corner contributes 1/8 of its volume to each of 8 adjacent cubes. 2) An atom in the face of a cube contributes 1/2 of its volume to each of 2 adjacent cubes. In this lattice we have 8 corner atoms and 6 face atoms. To calculate the total number of atoms contribut-ing to the unit cube one can use the above rules, Total number of atoms = 8 corners × 1/8 atom/corner + 6 faces × 1/2 atom/face = 4 atoms. The unit cube therefore contains the equivalent of 4 atoms. To find the weight of 1 atom one divides the molecular weight by Avogadro's Number (6.02 × 10^23), because there are Avogadro's Number of atoms contained in a mole of atoms. weight of 1 atom = (55.85 g) / (6.02 × 10^23) = 9.28 × 10^-23 g Since there are 4 atoms in this unit cube this figure is multiplied by 4. weight of 4 atoms = [4×(55.85 g)] / [6.02 × 10^23] = weight of unit cube The volume of the cube is found by cubing the length of a side of the square because the volume is equal to the length × width × height of a rectangular solid. In a cube all three of these quantities are equal. The volume of the cube = (3.55 × 10^-8 cm)^3. (1 \AA = 10^-8 cm) . We now have the quantities necessary to find the density, cc = cubic centimeters = cm^3. density= weight of unit cube / volume of unit cube in cc = {[4×(55.85 g)] / [6.02 × 10^23]} / (3.55 × 10^-8 cm)^3 = 8.30 g/cc The density of Fe is therefore 8.30 grams per cubic centimeter.

Question:

Two possible metabolic fates of \alpha-D-glucose are given below. Compare the standard free energy changes at 25\textdegreeC for these processes. Which process represents the more efficient utilization of \alpha-D-glucose? (a) \alpha-D-glucose (1 M) \rightarrow 2 ethanol (1 M) + 2CO_2 (g) (b) 6O2(g) + \alpha-D-glucose (1 M) 6CO_2(g) + 6 H_2O (l) ∆G\textdegree_fin Kcal/mole are CO_2(g) = - 94.26, H_2O (l) = - 56.69, ethanol (1 M) = - 43.39, \alpha-D-glucose = - 219.22.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E28-0909.htm

Solution:

For the standard free energy changes, ∆G_reaction=\sum ∆G_formationof products-\sum∆G_formationof reactants The free energy of any element in its standard state is taken as zero. Thus, for reaction (a), ∆G\textdegree_f, reaction (a) is: 2∆G\textdegree_f ,CO_2 + 2∆G\textdegree_f , ethanol -∆G\textdegree_f, \alpha-D-glucose. Substituting the known values into this equation yields 2(- 94.26) + 2(- 43.39) - (- 219.22) = - 56.08 Kcal/mole For reaction (b), ∆G\textdegree_f, reaction (b) is 6∆G\textdegree_f, H_2O + 6∆G\textdegree_f, CO_2 -∆G\textdegree_f, \alpha-D-glucose-∆G\textdegree_f, O_2 Substituting in the known values yields 6 (- 56.69) + 6(- 94.26) - (- 219.22) - (0) = - 685.5 Kcal/mole Reaction (b) is more efficient since its standard free energy change of formation is more negative or, in other words, more exothermic and releases more energy per mole of \alpha-D-glucose.

Question:

N_2O_5 \rightarrow 2NO_2 + (1/2)O_2 . The rate expression = {-d[N_2O_5]} / dt = k [N_2O_5]. The following mechanism has been proposed:N_2O_5k\rightleftarrowsNO_2 + NO_3 NO_2 + NO_3k_1\rightarrow NO_2 + O_2 + NO NO + NO_3k_2\rightarrow^2NO_2 Show the rate of O_2 formation is directly proportional to [N_2O_5].

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0475.htm

Solution:

To relate O_2 formation with the concentration of N_2O_5 investigate the mechanism of this reaction. In the mechanism, oxygen (O_2) is formed only from the second reaction, the rate being {d[O_2]} /dt= k_1 [NO_2] [NO_3]. To relate {d[O_2]} /dtto [N_2O_5], express [NO_2] [NO_3] in terms of [N_2O_5]. This can be done from the first reaction, which is at equlibrium. The equilibrium constant of this reaction, k, measures the ratio of products to reactants, each raised to the power of its coefficient in the chemical equation. Thus, k_eq = {[NO_2][NO_3]} / [N_2O_5]or[NO_2][NO_3] = k_eq [N_2O_5] Recall, {d[O_2]} /dt= k_1 [NO_2] [NO_3]. Thus [NO_2] [NO_3] can be replaced byk_eq[N_2O_5] , so that {d[O_2]} /dt= k_1k_eq[N_2O_5], which means O_2 formation is directly proportional to [N_2O_5] .

Question:

Discuss the excretory system of planaria.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0283.htm

Solution:

The excretory system of planaria involves a network of tubules running the length of the body on each side. These highly-branching tubules open to the body surface through a number of tiny pores. Side branches of the tubules end in specialized cells called flame cells (see Figure), also referred to as protonephridia. Each flame cell consists of a hollow, bulb--shaped cavity containing a tuft of long, beating cilia. It is very probable that flame cells function primarily in the regulation of water balance. The presence of better developed flame cells in freshwater species lends support to the osmoregulatory function of these cells. Primarily, water, and some waste materials, move from the tissues into the flame cells. The constant undulating movement of the cilia creates a current that moves the collected liquid through the excretory tubules to the nephridiopores, through which it leaves the body. The motion of the cilia resembles a flickering flame, hence this type of excretory system is often called a flame-cell system. Most metabolic wastes move from the body tissues into the gastrovascular cavity, and from there they are eliminated to the outside through the mouth. Nitrogenous wastes are excreted in the form of ammonia via diffusion across the general body surface to the external aquatic environment.

Question:

You have just received a Basic Assembler Language programyou ran. As you turn to the execution page to check theprintout, you find that every fourth line of the program reads, for example, PSW =... 00004F17B4C4. What is the PSW and what information does it provide? In addition, what informationis provided in the three lines following the PSW statement?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0210.htm

Solution:

PSW is a mnemonic that stands for Program Status Word. The PSW has a length of 64 bits (8 bytes), and in general, contains the value of the location counter, system protection information, and program interrupt status. The system protection information protects the operating system (data needed by the computer) from errors that could be - detrimental to the system. This feature is an imperative one, since quite often programmers unknowingly make critical system errors. The interrupt status indicates the condi-tion of the Central Processing Unit (CPU) at the time of the interruption. When an interruption occurs, the proces-sor (CPU) places specific interrupt information in the PSW. In a simplified manner, we can say that when an interrupt occurs, the PSW stores the address of the next instruction to be executed; we may say the PSW 'looks-ahead'. The ad-dress of this instruction is indicated by the location counter portion of PSW. Thus, it is this part of the PSW that the beginning assembler language programmer should concern himself with. When an interrupt occurs, starting with the third byte of the 8-bytes PSW, the contents of the PSW are printed on the execution page of the program . It is printed in hexa-decimal notation. To find the instruction about to be ex-ecuted when the interruptoccured, the programmer need only concern himself with the rightmost four digits of the PSW. Using these digits, he can find the instruction in the body of the program. As this is the instruction that was to be executed when the interruption occured , it is the instruc-tion immediately preceding it that caused the problem . By closely inspecting this instruction, the cause of the in-terrupt can be found. The three lines following the PSW statement display the contents of the system's registers at the time of inter-ruption. The first two lines contain the 16 general purpose registers; this is indicated by the GPR (GPR 0-7; GPR 8-F) statement at the beginning of the line. The third line holds the Floating Point Registers (FPR). It should be noted that each time an interrupt occurs, the system prints the contents of the PSW. The program will be terminated if the number of allowed program inter-ruptions is exceeded.

Question:

What is the resistance of an electric toaster if it takes a current of 5 amperes when connected to a 120 - volt circuit?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0644.htm

Solution:

This is an application of Ohm's law. Since we wish to find the resistance, we use R = (E/I) . E = 120 volts,I = 5 amperes Therefore R = [(120 volts)/(5 amperes)] = 24 ohms

Question:

A bar with a cross-sectional area of 6 square centimeters and a permeability of 1000 units is placed in a magnetic field of intensity 5 oersteds. What is the total number of lines of magnetic flux in the iron?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0700.htm

Solution:

Let B = flux density \mu = magnetic permeability H = magnetic field intensity B = \muH and total flux \cyrchar\CYRF = B x area Since H = 5 oersteds, B = 1000 × 5 oersteds = 5000 gauss Then the flux \cyrchar\CYRF = a 5000 gauss × 6cm^2 = 30,000 lines of flux.

Question:

750 ml of gas at 300torrpressure and 50\textdegreeC is heated until the volume of gas is 2000 ml at a pressure of 700torr. What is the final temperature of the gas?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E02-0040.htm

Solution:

Here, one is given a gaseous system involving pressure, volume and temperature, where two of the variables are changed in going from the original system to the final system. This indicates that the combined gas law should be used. It can be stated: For a given mass of gas, the volume is inversely proportional to the pressure and directly proportional to the absolute temperature. This gas law can also be stated (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 where P_1 is the original pressure, V_1 is the original volume, T_1 is the original absolute temperature, P_2 is the final pressure, V_2 is the final volume, and T_2 is the final absolute temperature. In this problem, you are given the original pressure, volume and temperature and the final pressure and volume. You are asked to find the final temperature. The temperature in \textdegreeC must be converted to the absolute scale before using the combined law. This can be done by adding 273 to the temperature in \textdegreeC. Converting the temperature: T_1 = 50 + 273 = 323\textdegreeK Using the combined law: (P_1 V_1 )/T_1= (P_2 V_2 )/T_2 P_1 = 300torrP_2 = 700torr V_1 = 750 mlV_2 = 2000 ml. T_1 = 323\textdegreeKT_2 = ? [{(300torr) (750 ml)}/(323\textdegreeK)] = [{(700torr) (2000 ml)}/(T_2 )] T_2 = [{(700torr) (2000 ml) (323\textdegreeK)}/{(300torr) (750 ml)}] T_2 = 2010\textdegreeK Convert T_2 to centigrade by subtracting 273 from it. 2010 - 273 = 1737\textdegreeC.

Question:

When 2.00 lb of brass at 2120F is dropped into 5.00 lb of water at 35 .00F the resulting temperature is 41.2 F. Find the specific heat of the brass. (Neglect the effect of the container.)

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0486.htm

Solution:

The quantity of heatQ_wadded to the water is Q_w= m_wc_w∆T_w where m_w andc_ware respectively the mass and specific heat of the water, and ∆T_wis the increase in temperature of the water. Similarly, the heat QBlost by the brass to the water is Q_B = m_B c_B∆T_B , ∆T_B being the decrease in the temperature of the brass. The heat will flow from the brass to the water until they are at the same temperature. Since the total energy of the system is conserved, the heat leaving the brass equals the heat entering the water, m_Bc_B∆T_B = m_wc_w∆T_w (2.00 lb)(c_B)(2120F - 41.20F) = (5.00 lb)(1.00 Btu/lbF0)(41.20F - 35.00F) c_B= 0.091 Btu/lbF0

Question:

The combination of electric and magnetic fields used in J.J. Thomson's experiment can be used to measure the speed of the electrons. This measurement is possible if the electric and magnetic fields are arranged so that they produce forces acting in opposite directions. The strength of the electric and magnetic fields are then ad-justed so that the resultant force is zero and the beam isundeflected, (a) Show in this case that the electron speed v is given by, v = (E/B) Where E and B are theelecticand magnetic field strengths, respectively, (b) Compute the velocity of an electron beam that isundeflectedin passing through electric and magnetic fields of 3.7 × 10^4 N/C and 0.23 Weber /m^2 respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0719.htm

Solution:

Thomson measured the ratio of the charge q of the electron to its mass m. Electrons were emitted from a hot filament and accelerated by an applied potential difference in a direction perpendicular to an electric field E and a magnetic field B. E and B are also at right angles so that they accelerate the electron along the same direction. The resultant force F on the electron is, F =qE+qv× B For the electron not to be deflected, this force must be equal to zero. The above equation reduces to, qE =qvB Since v and B are perpendicular. Solving this equation for v, v = E/B (b)Since the values of the electric and magnetic fields are 3.7 × 10_4 N/C and 2.3 × 10^1 Weber/m^2 respectively, v = (E/B) = [(3.7 × 10^4 N/C)/(2.3 × 10^-1 Weber /m^2 )] Since, 1 (Weber/m^2 ) = 1 [N/(A \bullet m)] v = 1.6 × 10^5 m/s The speed of the electrons is 1.6 × 10^5 m/s. Electric and magnetic fields arranged in this manner are used as a ve-locity selector for charged particles (ions) in several types of apparatus.

Question:

Show that a particle subjected to two simple harmonic vibrations of the same frequency, at right angles and out of phase, traces an elliptical path which de-generates to two coincident straight lines if the phase difference is \pi. Indicate the relevance of this to a half-wave plate.

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Solution:

Let the vibrations be taking place along the x\rule{1em}{1pt} and y-axes with a phase difference of \textphi between them. Then if x = a sin \omegat, y = b sin (\omegat + \textphi). Since sin (\alpha + \beta) = sin \alpha cos \beta + cos \alpha sin \beta (y/b) = sin \omegat cos \textphi + cos \omegat sin \textphi = (x/a) cos \textphi + \surd{1 - (x^2/a^2)} \textbullet sin \textphi Here we have used the fact that cos \omegat = \surd(1 - sin^2 \omegat) (y^2/b^2) + (x^2/a^2) cos^2 \textphi \rule{1em}{1pt} (2xy/ab) cos \textphi = [1 - (x^2/a^2)] sin^2 \textphi (x^2/a^2) (cos^2 \textphi + sin^2 \textphi) + (y^2/b^2) - (2xy/ab) cos \textphi = sin^2 \textphi or(x^2/a^2) + (y^2/b^2) - (2xy/ab) cos \textphi = sin^2 \textphi This is the general equation of an ellipse where the major and minor axes do not coincide with the x- and y-axes. Thus the particle always has x- and y- coordinates such that the point they define lies on an ellipse. The particle thus follows an elliptical path. If = \textphi = \pi/2, 3\pi/2, 5\pi/2, ..., the equation of the path reduces to (x^2/a^2) + (y^2/b^2) = 1, which is an ellipse with the major and minor axes coincident with the coordinate axes. When \textphi = \pi, the equation of the path becomes (x^2/a^2) + (y^2/b^2) + (2xy/ab) = 0, that is [(x/a) + (y/b)]^2 = 0. This is the equation of two coincident straight lines x/a = \rule{1em}{1pt} y/b, inclined to the negative x-axis at an angle tan^\rule{1em}{1pt}1 (b/a). (See figure.) In the case of a half-wave plate, plane-polarized light striking the plate is split up in two components, O and E, plane-polarized at right angles to one another and initially in phase. These pass through the plate at different speeds and the thickness is such that on emergence the two beams are out of phase by \pi. Any particle affected by the two components will thus be affected by two simple harmonic vibrations at right angles, out of phase by \pi. As can be seen from the above analysis, the particle would trace a straight- line path. This means that the two components are equivalent to a single vibration at an angle tan^\rule{1em}{1pt}1 (b/a) to the slower component, b/a being the ratio of the amplitudes of the components of the incident light on entering the plate. If the plane- polarized light is striking the plate at an angle of 45\textdegree to the two transmission directions, then it is resolved into two equal components so that b = a. The emerging light is thus plane-polarized in a direction making an angle of \rule{1em}{1pt} 45\textdegree with each of the principal directions in the plate.

Question:

Suppose Stan is on the earth and Mavis is flying past the earth in a spaceship with a speed v = (3/4) c, as shown in the figure. Another spaceship is approaching the earth in the opposite direction and Stan measures its velocity of approach v_s = (3/4)c. The negative sign denotes the fact that the spaceship is moving to the left along the x-axis. What is the speed of the other spaceship as measured by Mavis?

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Solution:

The given quantities are v =(3/4)c and v_s = {─(3/4)c In the nonrelativistic case (i.e., where the speeds of the objects are much less than the speed of light, c) the speed of the other spaceship relative to Mavis is, by the law of addition of velocities, equal to the sum of the velocity of the spaceship relative to Stan and the velocity of Stan relative to Mavis. Or v_m =v_s + (─ v) . The negative sign here indicates the velocity of Stan re-lative to Mavis. It is equal but opposite to the velocity of Mavis relative to Stan (v). In the relativistic case, (where, as in this case, the velocities of the objects are comparable to the speed of light ) , the result above must be modified according to the special theory of relativity ,as follows v_m = {v_s + (─v)} / (1 + [{vs(─ v)} / c^2]) = (vs─ v ) / [1 ─ (vv_s / c^2)] = [{─(3/4)c} ─ (3/4)] / [1─ [{(3/4)c} {─(3/4)c}] / c^2] = {─ (3/2)c} / {1 + (9/16)} = ─ (24/25)c . Thus Mavis measures a speed less than the speed of light.

Question:

Write a single APL statement that gives values for the current, according to Ohm's Law, where I = E/R. I is current, E is voltage, and R is resistance. You can assume that E and R are vectors of four elements each. If E \leftarrow 4 8 9 12 and R \leftarrow 2 4 3 4, find I.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G14-0365.htm

Solution:

Since E and R are vectors of equal length, one can write: I \leftarrow E \div R. This will divide the first elements of each vector, then the second, and so on. The result is a four element vector I. E is given by E \leftarrow 4 8 9 12 andR is given by R \leftarrow 2 4 3 4, therefore I \leftarrow E \div r execution will be performed as follows: I \leftarrow 4 \div 28 \div 49 \div 312 \div 4 I \leftarrow2233

Question:

Brass weights are used in weighing an aluminum cylinder whose approximate mass is 89 gm. What error is introduced if the buoyant effect of air (\rho = 0.0013 gm/cm^3) is neglected? \rho_brass = 8.9 gm/cm^3\rhoAl = 2.7 gm/cm^3

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Solution:

If the objects on the pans occupy different volumes, then the volume, and therefore the weight of the air is different on the two objects and must be accounted for. The additional weight on the object with less volume is just equal to the weight of the additional volume of air. V = m/\rho V_B = (89 gm)/(8.9 gm/cm^3) = 10 cm^3 V_Al = (89 gm)/(2.7 gm/cm^3) = 33 cm^3 The difference in volume V of air displaced on the two pans of the balance is V = V_Al - V_B = 33 cm^3 - 10 cm^3 = 23 cm^3 Hence, the mass error m = V\rho = 23 cm^3 × 0.0013 gm/cm^3 = 0.030 gm The error introduced is only a small fraction of the total mass, but in many experiments where accuracy is important an error of 0.030 gm in 89 gm is too great to allow.

Question:

An atom emits a photon of green light \lambda = 5200 \AA in \tau = 2 × 10^-10 sec, Estimate the spread of wavelengths in the photon.

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Solution:

The Heisenberg uncertainty relationship for energy and time is \DeltaE\Deltat\geqh(1) where \DeltaE is the uncertainty in the energy of the system we are observ-ing, and\Deltatis the time interval over which we observe the system. Also,his Planck's constant divided by 2\pi . The energy of a photon is given by E =hv(2) where v is the frequency of the photon. But c =v\lambda(3) where c and \lambda are the speed and wavelength of light, respectively. From (3) and (2) E = (hc/\lambda) Taking the derivative of E with respect to \lambda, (dE/d\lambda) = (-hc/\lambda^2)(4) Hence \DeltaE \approx (-hc/\lambda^2)\Delta\lambda(5) Using (5) in (1) (-hc/\lambda^2)\Delta\lambda\Deltat\geqh \Delta\lambda\geq (h\lambda^2/-hc\Deltat) \Delta\lambda\geq (\lambda^2/-2\pic\Deltat)(6) Therefore, using the given data, \Delta\lambda\geq [{(5200 \AA)^2}/{(-2) (3.14) (3 × 10^8 m/s) (2 × 10^-10 s)}] Since 1 \AA = 10^-10 m \Delta\lambda\geq - [{(5.2 × 10^-7 m)^2}/(37.68 × 10^-2 m)] \Delta\lambda\geq - .717 × 10^-12 m \Delta\lambda\geq - 7.17 × 10^-13 m(7) Hence, the spread in wavelength is at least 7.17 × 10^-13 m. (We may neglect the minus sign in (7), because, looking at (6). it only indicates that as\Deltatincreases,\Delta\lambdadecreases).

Question:

One curie is defined as the amount of material, which is equivalent to one gram of radium, that gives 3.7 × 10^10 nuclear disintegrations per second. The half-life of ^210At is 8.3 hr. How many grains of astatine (At) would equal one curie?

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Solution:

The half life is defined as the time it takes for one half of the amount of a radioactive substance to disintegrate. To find the number of grams of ^210At that would be equal to 1 curie, one should first find the number of disintegrations of ^210At per second. One can then solve for the number of moles necessary and from that solve for the number of grams. Solving: (1) determining the half-life in seconds. This unit is used because curies are measured in disintegrations per second. t(1/2)= 8.3 hr × 60 min/hr × 60 sec/min = 29880 sec (2) Solving for the number of disintegrations of ^210At in 1 curie. Let x = the total number of nuclei of ^210At in 1 curie, x/2 the number of nuclei left after 1 half-life, 29880 sec. The following ratio can be set up: [(^210At) number of nuclei after 1 half life] / [half-life] = 3.7 × 10^10dis/sec. One can solve this ratio for x, the original number of nuclei present. [(x/2) / 29880 sec] = 3.7 x lo10dis/sec [x/2] = 29880 sec × 3.7 × 10^10dis/sec x = 2 × 29880 sec × 3.7 × 10^10dis/sec = 2.21 × 10^15 dis. Thus, 2.21 × 10^15disof ^210At are equal to 1 curie. (3) Solving for the number of moles. There are 6.02 × 10^23 nuclei per mole. Therefore, the number of moles present is equal to the number of disintegrations divided by 6.02 × 10^23 . no. of moles= [(2.21 × 10^15 nuclei) / (6.02 × 10^23 nuclei / mole) = 3.67 ×10^-9 moles (4) One solves for the number of grams necessary by multiplying the number of moles by the molecular weight of ^210At (MW = 210) no. of grams= 210 g/mole × 3.67 × 10^-9 moles = 7.71 × 10^-7 g. This would be the number of grams necessary at t = t_(1/2).

Question:

Given the reaction A + B \rightleftarrows C + D, find the equilibrium constant for this reaction if .7 moles of C are formed when 1 mole of A and 1 mole of B are initially present. For this same equilibrium, find the equilibrium composition when 1 mole each of A, B, C, and D are initial-ly present.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0305.htm

Solution:

In general, an equilibrium constant measures the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the chemical equation. Thus, for this reaction, the equilibrium constant, K, is equal to {[C] [D] } / {[A] [B]} . You are asked to find K in the first part of this problem. Therefore, you must evaluate this expression. To do this, usestoichiometry. From the equation for the chemical reaction, you see that all the quantities are in equimolaramounts; 1:1:1:1 . It is given that .7 moles of C is produced. This means, therefore, that .7 mole of D must also be pro-duced. If the equimolarquantities are to be maintained, then, .7 mole of each A and B must have been consumed. If you started 1 with 1 mole of each, then .3 mole must be left. These mole amounts are the concentra-tions if you assume they are in a given amount of volume. Thus, sub-stituting these values in K = {[C] [D]} / {[A] [B]} K = {[C] [D]} / {[A] [B]} , you obtain K = (.7)^2 / (.3)^2 = 5.44. Therefore, you have calculated the equilibrium constant of this reaction to be 5.44. The second part of this problem asks you to use this same equi-librium for a reaction that starts with 1 mole each of A, B, C, and D. To find its composition, determine the concentration of each species at equilibrium. This can be found from K = {[C] [D]} / {[A] [B]} = 5.44 . To find the composition, assume X moles/Iiter(concentration) react. Thus, at equilibrium, both C and D have a final concentration of 1 + x, since you started with 1 mole/liter and had x moles/liter of each produced. This means, therefore, that A's and B's initial concentration must be reduced by x moles/liter to 1 - x. Substituting, 5.44 = (1 + x)^2 / (1 - x)^2 . Solving, x = .40 moles/liter . Thus, the concentration of both A and B = 1 -.4 = .6M and of both C and D = 1 + .4 = 1.4M .

Question:

A gyroscope consists of a uniform circular disk of mass M = 1 kg and radius R = 0.2 m. The disk spins with an an-gular velocity\omega =400 sec^\rule{1em}{1pt}1. The gyroscope precesses with the axes making an angle of 30\textdegree with the horizontal. The gyroscopewheel is attached to its axis at a point a dis-tance l = 0.3 m from the pivot which supports the whole structure. What is the precessional angular velocity?

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Solution:

First, establish the general formula for regular precession. Refer to the figure, \vert L \vert = I\omegafor a symmetric object spinning around its symmetry axis. The torque \tau from the weight is mgl sin \alpha this is directed perpendicular to the plane containing L^\ding{217} and the vertical axis, ac-cording to the rule for a vector cross product. In a time ∆t the torque effects a change ∆L = \tau∆t = mgl sin \alpha∆t in the angular momentum, also perpendicular to L^\ding{217}. Thus, after ∆t, the angular momentum will have changed its direction; the horizontal component will have moved by an angle ∆\texttheta = \Omega∆t. (\Omega is the precessional angular velocity.) There is no change in the magnitude of L^\ding{217} since its increment ∆L^\ding{217} is perpendicular. Comparing the arc length to radius in the circle described by the tip of L^\ding{217}, the result is ∆\texttheta = \textthetaL / (L sin \alpha); L sin \alpha is the radius of this circle. Thus, \Omega = ∆\texttheta /∆t = (1 / L sin \alpha) (∆L / ∆t). Substituting the above expressions for L and ∆L gives \Omega = 1 / (I\omegasin \alpha)(mgl sin \alpha∆t / ∆t) = (mgl) / (I\omega). For the uniform disk we have I = (1 / 2)mR^2 , so \Omega = 2gl / (R^2\omega). Use g = 9.8 m / s^2, R = 0.2 m, \omega = 400 s^\rule{1em}{1pt}1, l = 0.3 m. All these are mks units. Then, \Omega = 2 \textbullet 9.8 \textbullet 0.3 / [(0.2)^2 \textbullet 400] \textasciitilde 0.368 s^\rule{1em}{1pt}1. Note that the precessional angular velocity does not depend on the mass of the disk nor on the angle that its axes makes with the horizontal.

Question:

Write an APL program to build 10 rows of Pascal's triangle.

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Solution:

Each row ofpascal'striangle is a set of binomial coefficients of (x + y)^n-1, n > 0. Another way of representing these coefficients is (^N_K) for N = 1,2,....., and for K = 0, 1, 2,..., such that (^N_K) = [n!] / [K!(N - K)!] or,= [N(N-1) .... (N-K+1)] / [K!] Our strategy rests on the fact that each number in the triangle is the sum of two numbers right above it. For ex-ample, let us take rows 1 to 5. ROWS N = 11K = 0 N = 211K = 1 N = 3121K = 2 N = 41331K = 3 N = 514641K = 4 To get from row 4 to row 5, we can do the following: ROW 41331 +1331 14641 By taking a row, moving it one place to the right, and then adding it to itself, we can generate the next row. In APL, the symbol used for catenation is the comma. Catenation means "chaining" elements together. This can be done by the statement P \leftarrow (0,P) + (P,0) . This statement will do the following: 01331 +13310 A zerocatenatedwith string P in (0,P) part, moves P one place to the right, while the zero in the second part, (P,0), just adds itself to the string P. Then, the addition is performed, and the new row is created. The program is given below: \nablaPASCAL [1]P \leftarrow 1 [2]P [3]\rightarrow 2 × N \geq\rho\Rho\leftarrow (0,P) + (P,0) \nabla Line [3] contains the symbol \rho, which is called the re-shaping operator. When you enter the value for N (in this case, N = 10), \rho determines the number of times P \leftarrow (0,P) + (P,0) is to be executed. When ten branches have been execut-ed, the comparison N \geq \rho will be false or, in APL, it will have the value zero. This zero will then be multiplied by 2 telling the program to branch to instruction zero, which is the way of indicating the termination of the program.

Question:

What is the physiological purpose of shivering?

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Solution:

Shivering is a means by which the body maintainsitsnormal temperature when the ambient tem-perature is cold. Recall that heat is produced by virtually all chemical reactions within the body. One way by which the basal level of heat production can be increased is to increase the rate of skeletal muscle contractions. The first muscle changes in response to cold are a gradual and general increase in skeletal muscle tone . This soon leads to shivering, which consists of oscillatory, rhythmic muscle tremors occurring at the rate of about 10 to 20 per second. This intensive muscle activity rapidly uses up ATP, thus stimulating more cellular respiration and more heat production. These contractions are so effective that body heat pro-duction may increase several fold within seconds . Because no external work is performed, all the energy liberated by the metabolic machinery becomes internal body heat. Thus shivering is an adaptation to cold. Shivering tends to ruffle the body hair in most animals, creating dead air space that serves as insulation. In man, shivering causes the erection of body hair, but his sparse coat of hair is insufficient to act as an insulatory mechanism.

Question:

Hydras, like most cnidarians, are carnivorous and feed mainly on small crustaceans. Describe the method of ingestion and digestion in these organisms.

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Solution:

Should a small crustacean make contact with the tentacles of the hydra, the nematocysts, which are lodged in the tentacle walls, would discharge. The nematocysts will entangle the prey and paralyze it. The captured organism is pulled toward the mouth by the tentacles. The mouth then opens to receive the prey. The retraction and bending inward of the tentacles and the opening of the mouth is a reflex response. Mucus secretions in the mouth aid in swallowing. Eventually the prey arrives at thegastrovascularcavity.Secretorycells lining the cavity release enzymes that begin the digestion of protein. Eventually, the tissues of the prey animal are reduced to a soupy broth. Flagel-lated cells lining the cavity continually beat their flagella, thus aiding in mixing the digesting food with enzyme molecules. Subsequent to this extracellular phase of digestion comes intracellular digestion. Amoeba-like cells lining the cavity extend pseudopodia which engulf small fragments of what is left of the prey animal. Continued digestion of proteins and the digestion of fats occur within food vacuoles in these amoeba-like cells lining thegastrodermis, Products of digestion are circulated by cellular diffusion. Glycogen and fat are the chief storage products of excess food. Undigested materials are ejected from the mouth upon contraction of the body.

Question:

Write a program to determine the prime factorizations of integers greater than 1.

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Solution:

According to the Fundamental Theorem of Arithmetic any integer can be expressed as the product of primes. Some examples 108 = 2 × 2 × 3 × 3 × 3 = 22 × 33 390 = 2 × 3 × 5 × 13 -210 = -(2 × 3 × 5 × 7) The algorithm for computing the prime factors of a positive integer greater than 1 is based on the following facts: 1) Let p be a prime \leq \surdn. If n is divisible by no such p, then n is a prime. 2) Let p be a divisor of n. Then q = n/p is an integer, and any other prime divisor of n divides q. 3) n has at most one prime factor greater than \surdn . Using these facts we can construct the following computer program. The program assumes that a table of primes for the range 1-1000 was developed earlier and can be obtained from the secondary memory files. Thus, the program can find prime factorizations for all numbers less than (1000)2 = 1,000,000. Execution terminates when the considered integer N is 1 or less. The program looks as follows: PRIME FACTORIZATION CRESERVE MEMORY FOR PRIME DIVISORS (NPR), FOR PRIME FACTORS (NFAC) , AND FOR THE NUMBER OF TIMES EACH PRIME FACTOR OCCURS (NBR). DIMENSION NPR(168), NFAC(8), NBR(8). CREAD IN THE PRIMES LESS THAN 1000. C(THERE ARE 168 OF THEM) READ (5,5)(NPR(I), I = 1, 168) 5FORMAT (20I4) 10READ (5,15) N 15FORMAT (I7) IF (N - 2) 85,75,20 CCOPY N INTO INT FOR TESTING PURPOSES. 20INT = N RN = INT NRT = SQRT(RN) CSTART A COUNTER FOR THE NUMBER OF DIFFERENT PRIMES THAT DIVIDE N. J = 0 DO 45 I = 1, 168 CNF COUNTS THE OCCURRENCES OF EACH PRIME FACTOR NF = 0 CBRANCH OUT OF THE DO LOOP IF CTHE CURRENT PRIME IS GREATER CTHAN THE SQUARE ROOT OF INT. IF (NPR(I) - NRT) 25,25,50 25KINT = INT/NPR(I) IF (INT - KINT\textasteriskcenteredNPR(I)) 35,30,35 CIF NPR(I) DIVIDES INT, STORE THE CQUOTIENT KINT IN INT. THUS, THIS CQUOTIENT IS NOW THE DIVIDEND FOR FURTHER TESTING. 30INT = KINT NF = NF + 1 GO TO 25 35IF (NF) 45,45,40 CINCREMENT J BY 1 IF NF IS POSITIVE CAND IF NPR(I) DOES NOT DIVIDE INT. CTHEN STORE THE PRIME NPR(I), AS THE CJ-TH MEMBER OF NFAC AND STORE CTHE MULTIPLICITY NF IN NBR(J). 40J = J + 1 NFAC(J) = NPR(I) NBR(J) = NF RN = INT NRT = SQRT(RN) 45CONTINUE CN IS PRIME IF AND ONLY IF J = 0 50IF (J) 75,75,55 55IF (INT-1) 65,65,60 60J = J + 1 NFAC (J) = INT NBR(J) = 1 CPRINT PRIME FACTORIZATION 65WRITE (6,70) N, (NFAC(I) ,NBR(I), I = 1,J) 70FORMAT (1HO, I7, 3H = , 8(I5, 1H( ,I2, 1H))) GO TO 10 75WRITE (6,80) N 80FORMAT (1HO, I7, 9H IS PRIME) GO TO 10 85STOP END

Question:

In contrasting a food chain and a food web, which one do you think actually operates in a real community?

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Solution:

Food is the common word used to describe the various nutrients that all living heterotrophic organisms must ingest in order to obtain energy to sus-tain their life processes. Autotrophic photosynthetic organisms, such as the green plants, can manufacture their own food from simple inorganic molecules with the energy from sunlight. Life on earth ultimately depends on food energy which in turn is dependent on the sun. The radiant energy from the sun that reaches the photosynthetic green plants is responsible for the transformation of basic raw materials such as water, carbon dioxide, nitrogenous compounds and minerals into the development of the plants themselves. Plants store nutrients (starches and sugars) within them and in turn are eaten by heterotrophic organisms such as animals. A food chain is most commonly a sequence of organisms that are related to each other as prey and predators. One species is eaten by another, which is eaten in turn by a third, and so on. Each species forms a step or link in one or more food chains. Every food chain begins with the autotrophic organisms (mainly green plants) that serve as the producers of the community. Any organism that does not produce its own food and must therefore depend on another for nutrients is a consumer. Every food chain ends with decomposers, the organisms of decay, which are usually bacteria and fungi that degrade complex organic materials to simple substances which are reusable by the producers. The links between the producers and the decomposers are variable because the producers may die and be acted upon directly by the decomposers, or the producers may be eaten by primary consumers, called the herbivores. The herbivores consume the green plants and in turn may be either acted upon directly by the decomposers or fed upon by the secondary consumers, the carnivores (animal eaters). Some food chains consist of tertiary consumers, or secondary carnivores, which eat the secondary consumers and sometimes also the primary consumers. In almost all ecosystems there are top carnivores: one or more large, specialized animal species that prey on organisms on the lower steps, but are not ordinarily consumed by predators themselves. The larger whales enjoy this status, as do lions, wolves, and man. The decomposers can feed on any dead organism in the food chain. In addition, there are parasite chains in which small or-ganisms live on or within larger ones from where they obtain food. The successive levels in the food chains of a community are referred to as trophic levels. Thus all the producers together constitute the first (or lowest) trophic level and the primary consumers (herbivores) constitute the second trophic level. The herbivore-eating carnivores constitute the third trophic level, and so on. Energy-flow in an ecosystem is actually more complicated than the flow of nutrients that simplified food chains imply. In most real communities, there are many different possible food chains that are tied together by cross linkages. Any one animal usually eats a variety of food and thus may be part of several food chains that intersect. Also, food chains starting from a common plant source may radiate outward as the plant food is eaten by different herbivores, and as the latter are eaten by different carnivores, and so on. Thus, there is actually a complex food web formed as the food chains first radiate outward from the plants and then come together at the top carnivore level of the web (see Figure). For example, the grasshoppers of the grasslands of Canada eat grass and are eaten by robins. Robins eat many kinds of insects as well as grasshoppers. The great horned owl feeds upon the robin as well as other mammals such as the prairie vole. But the prairie vole is also eaten by the coyote. Hence, even though organisms in a community can be classified into the various trophic levels according to the nature of their prey, it is more characteristic to assign organisms to a food web than to a food chain in a real community.

Question:

Explain briefly the background and applications of the COBOL programming language.

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Solution:

On May 28 and 29, 1959, a meeting was called in the Pentagon for the purpose of considering the desirability and feasibility of establishing a common language for pro-gramming of electronic computers in business data processing. As a result, in a short period of time, a new computer lan-guage COBOL (Common Business Oriented Language) was devel-oped. Since that time COBOL has definitely become the most popular and useful language inside the business community. The basic language element in COBOL is the "word." As in the English language, COBOL contains many types of words with which meaningful thoughts are formed. This makes the language intelligible for many people who do not have much programming experience and only use the computer occasion-ally for solving problems. Every COBOL program is divided into four parts: 1.The Identification Division - identifies the program. 2.The Environment Division - specifies the input-output devices to be used. 3.The Data Division - describes the data to be used in the program. 4.The Procedure Division - includes the instructions that the system will follow in solving the problem. Such a division provides greater simplicity of understanding and analysis of a COBOL program. These are only a few of the factors that account for the constantly growing popularity of the COBOL language in the business world.

Question:

A forced oscillator is a system whose behavior can be described by a second-order linear differential equation of the form: \"{y} + A_1ẏ + A_2y (t) = E(t)(1) where A_1, A_2 are positive constants and E(t) is an external forcing input. An automobile suspension system, with the road as a vertical forcing input, is a forced oscillator, for example, as shown in Figure #1. Another example is an RLC circuit connected in series with an electro-motive force generator E(t), as shown in Figure #2. Given the initial conditions y(0) = y0and ẏ(0) = z_0 , write a FORTRAN program that uses the modified Euler method to simulate this system from t = 0 to t = t_f if: Case 1:E(t) = h where h is constant Case 2:E(t) is a pulse of height h and width (t_2 - t_1) . Case 3:E(t) is a sinusoid of amplitude A, period 2\pi/\omega \omega \omega and phase angle \varphi . Case 4:E(t) is a pulse train with height h, width W, period pW and beginning at time t = t_1 .

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Solution:

If equation (1) is rearranged to the form \"{y} = E(t) - A_1ẏ - A_2y It can be seen to be of the form d^2y/dx^2 = 8_3(x,y,y') , so one can use the Euler 3 subroutine developed earlier In this chapter. Case 1:This Is the CONST FUNCTION also developed earlier in the chapter . The parameters need to be read in only once: they are read in the main program and kept in a COMMON block. REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2,H COMMONA1,A2,H READ,N,TFIN,ACCUR,Y0,Z0,A1,A2,H CALL EULER3(N,TFIN,T,Y0,Z0,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,H COMMON A1,A2,H G3 = H - A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 2:This is the PULSE FUNCTION developed earlier. REAL T/0.0/,TFIN,N,ACCUR,Y0,Z0,A1,A2,H,T1,T2 COMMON A1,A2,H,T1,T2 READ, N, TFIN, ACC UR, YO, ZO ,A1 ,A2, H, T1, T2 CALL EULER3 (N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,H,T1,T2 COMMON A1,A2,H,T1,T2 G3 = -A1\textasteriskcenteredW - A2\textasteriskcenteredV IF(T.GE.T1.AND.T.LT.T2)G = H - A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 3:This is the SINE FUNCTION developed earlier. The variable OMEGA is used for\omegaand PHI is used for \varphi . \omega REAL T/0.0/, TFIN,N,ACCUR,YO, ZO,A1,A2 ,A,OMEGA,PHI COMMON A1,A2,A,0MEGA,PHI READ N, TF IN, ACC UR, YO, ZO, A1, A2, A, OMEGA, PHI CALL EULER3(N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,W) REAL G3,T,V,W,A1,A2,A,OMEGA,PHI COMMON A1,A2,OMEGA,PHI G3 = A\textasteriskcenteredSIN(OMEGA\textasteriskcenteredT + PHI) -A1\textasteriskcenteredW - A2\textasteriskcenteredV RETURN END Case 4:This is the TRAIN FUNCTION developed earlier. The parameters for the G3 FUNCTION subprogram (A1,A2) are kept in the COMMON block named G, while the parameters for the TRAIN FUNCTION subprogram (H,T1,W,P) are kept in the COMMON block named TRANE. The subprogram calling sequence is as follows: The main program calls the EULER3 subroutine,EULER3 calls the G3 FUNCTION subprogram, and G3 calls the TRAIN FUNCTION subprogram. CMAIN PROGRAM REAL T/0.0/, TFIN, N, ACCUR, YO, 20, A1, A2, H, Tl, W, P) COMMON /G/A1,A2/TRANE/H,T1,W,P READ,N,TFIN,ACCUR, Y0,Z0,A1,A2,H,T1,W,P CALLEULER3(N,TFIN,T,YO,ZO,ACCUR) STOP END FUNCTION G3(T,V,U) REAL G3,T,V,U,A1 ,A2,TRAIN COMMON /G/A1,A2 G3 = TRAIN(T) -A1\textasteriskcenteredU - A2\textasteriskcenteredV RETURN END FUNCTION TRAIN(T) REAL TRAIN,T,H,T1,W,P COMMON/TRANE/H, T1, W, P TIME = T IF(T.GE. (T1 + P\textasteriskcenteredW)) TIME =AMOD((T - T1)/(P\textasteriskcenteredW))\textasteriskcenteredP\textasteriskcenteredW + T1 IF((TIME.LT.T1).0R. (TIME.GE. (T1 + W)))TRAIN = 0. IF((TIME.GE.T1) .AND. (TIME.LT. (T1 + W)))TRAIN = H RETURN END

Question:

1.3625 g of an unknown metal X reacts with oxygen to form 1.4158 g of the oxide X_2O. What is the atomic mass of X?

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Solution:

This problem will be solved by calculating the mass and, from this, the number of moles of oxygen in the compound. The corresponding number of moles of metal is twice this amount. The molecular weight is the number of moles of metal divided by the weight of metal that was used. The mass of oxygen in the compound is mass O = mass oxide - mass X = 1.4158 g - 1.3625 g = 0.0533 g. The corresponding number of moles is moles O = mass O \div atomic mass O = 0.0533 g \div 15.9994 g/mole = 0.0033 mole . Since, in the oxide, there are two moles of X per mole of O, the number of moles of X present is 2 × 0.0033 mole = 0.0066 mole X. Hence, 1.3625 g of X corresponds to 0.0066 mole X. The atomic mass of X is then: atomic mass X = (mass X) / (moles X) = 1.3625g / 0.0066 mole =206.4394 g/mole.

Question:

Themolalfreezing point constant for a certain liquid is 0.500\textdegreeC. 26.4 g of a solute dissolved in 250 g of this liquid yields a solution which has a freezing point 0.125\textdegree below that of the pure liquid. Calculate the molecular weight of this solute.

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Solution:

A mole of a substance in 1000 g of H_2O gives a definite and known lowering of the freezing point. By determining the freezing point of a solution of known con-centration, one can calculate the molecular weight of the dissolved substance. A general formula may be developed for this kind of calculation: molecular weight = (grams solute) / (kg of solvent) × (k_f) / (∆T_f) wherek_fis the freezing point constant and ∆T_fis change in the freezing point. Here, 26.4 g of a solute is dissolved in 250 g of liquid. One can find the number of grams of solute using the following ratio: (Note: there are 1000 g in 1 kg.) Let X = number of g of solute in 1000 g (26.4 g) / (250 g) = (X / 1000 g) X = [(1000 g × 26.4 g)] / (250g) = 105.6 g One can now solve for the molecular weight molecular weight= (105.6 g / 1kg) × (0.500\textdegreeC / 0.125\textdegree) = 422.4 g

Question:

A 3200-lb car is slowed down uniformly from 60 mph to 15 mph along a level road by a force of 1100 lb. How far does it travel while being slowed down?

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Solution:

A diagram should first be drawn so that our sign conventions are consistent (see diagram). we are given the change in velocity of the car in the positive x direction, the force acting against it in the negative x direction, and the weight of the car. We can calculate the deceleration of the car, and from this we can calculate the time the car is decelerated by its change in velocity. First, we must find the mass of the car, given its weight. W = mg 3200 lb = m 32 ft/s^2 m = 100 lb-s^2/ft[slugs] From this we can calculate the deceleration of the car. Remember that the force acts in the negative × direction. F = ma -1100 lb = 100 lb-s^2 / fta a = -11 ft/s^2 Assuming constant deceleration: a = (∆v/∆t) where ∆v = 15 mph - 60 mph = - 45 mph. A useful conversion factor to remember is 60 mph = 88 ft/s, so that - 45 mph = -66 ft/s. Hence, -11 ft/s^2 = -[(66 ft/s)/∆t] ∆t = 6s Distance is given by the formula 1/2 at^2 + v_it + d_i = d where v_i is initial velocity, and d_i is the initial position (which we will here set equal to 0). Therefore d = 1/2 (-11 ft/s^2)(6s)^2 + 88 ft/s^2 (6s) + 0 d = 330 ft.

Question:

Suppose that the first half of the distance between two points is covered at a speed v_1 =10mi/hr and, that during the 10 second half, the speed is v2= 40 mi/hr. What is the average 2 speed for the entire trip?

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Solution:

The average speed is the total distance traveled divided by the total traveling time. The average speed is not v= {(10 mi/hr) + (40 mi/hr)} / 2 = 25 mi/hr. Let 2x be the total distance traveled and let t_1 and t_2 de-note the times necessary for the two parts of the trip. Then, v= (2x)/(t_1 + t_2) Since only the velocities are known, the average velocity must be expressed in terms of these variables. In order to eliminate unknown variables, we see that t_1 = x/v_1; t2= x/v_2. t_1 + t_2 = x/v_1 + x/v_2 = [x(v_1 + v_2)]/(v_1v_2) Therefore,v= 2x/[{x(v_1 + v_2)}/(v_1v_2)] = (2v_1 + v_2)/(v_1 + v_2) = [2(10 mi/hr)(40 mi/hr)]/(10 mi/hr + 40 mi/hr) = (800/50)mi/hr = 16mi/hr.

Question:

What is the evidence for the theory that theneurilemma sheathplays a role in the regeneration of severed nerves?

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Solution:

When an axon is separated from its cell body by a cut, it soon degenerates . However, the part of the axon still attached to theperikaryon can re-generate. A healthyperikaryonis important in regene-ration. As long as the cell body of the neuron has not been injured, it is capable of making a new axon. Regeneration begins within a few days following the severing of the nerve. Axon regeneration is believed to involve theneurilemma, a cellular sheath composed of Schwann cells which envelopes the axon. The role the neurilemmasheath plays in regeneration is to provide a channel for the axon to grow back to its former position. What happens is that the growing axon enters the old sheath tube and proceeds along it to its final destination in the central nervous system or periphery. In some experiments, theneurilemmasheath is removed and replaced by a conduit, for example, a section of blood vessel or extremely fine plastic tube. The severed axon is able to regenerate normally within the substituted conduit. This result shows that the sheath is not an absolute requirement for the regeneration of an axon in vitro, since a plastic sheath serves the same function as well.

Question:

Insurance companies have worked out probability tables to estimate the chances a person has of living until a certain age. A study of just over a million people was made, and in 1941, a mortality table was published. It listed the number of persons still alive at. age 0,1,2,3, ..., 99. LetL(X) = number of persons living at age X then,P(X) = probability that a person of age X will live for one more year = [L(X+1)] / [L(X)] Develop a program in BASIC that allows the user to input the year of birth, the present age, and some year after the year of birth. The output should list the age, the year, and the chances of that person living until that year.

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Solution:

To make the problem more meaningful, let us present a con-crete example. Suppose we want to calculate the chances of a 35-year old born in 1945 living until the year 2000. Adding the person's age to hisbirthdate gives us the year 1980. This person must live for 20 more years to reach 2000. His chances of making it are [L(35+20)] / [L(35)] = [L(55)] / [L(35)] = P(X) If we look at the data tables given in the program, we find that L(55) = 754191 and L(35) = 906554. So the answer is [754191/ 906554] = .8319316 or, approximately 0.83. The strategy for our computer program is fairly easy. The program reads in the values, figures out the probability, and returns the answer. To terminate the procedure, input a negative value for the year of birth. We list the data (as it is stored in array L) in a DATA statement at the end of the program. The first entry is L(0), the total number of persons included in the study. 1\O DIM M(99,4) 15 FOR J = 0 TO 99 2\O READ M (J,0) 25 IF J = 0 THEN 30 28 LET M(J-1,1) = M(J-1,0) - M(J,0) 3\O NEXT J 32 LET M(99,1) = M(99,0) 33 REM INPUT YEAR OF BIRTH, AGE, AND EXPECTED YEAR 35 PRINT "INPUT YEAR OF BIRTH, AGE, AND EXPECTED YEAR'' 4\O INPUT B,A,X 45 IF B < 0 THEN 99 48 REM N IS THE PRESENT YEAR; M IS THE 49 REM YEARS BETWEEN PRESENT AND EXPECTED. 5\O LET N = B + A 52 LET M = X - N 55 LET P = L (A+M) / L(A) 58 PRINT 6\O PRINT "CHANCES OF LIVING UNTIL YEAR"; X; "ARE"; P 65 GO TO 35 7\O DATA 1023102, 1000000, 994230, 990114, 986767 72 DATA 983817, 981102, 978541, 976124, 973869 74 DATA 971804, 969890, 968038, 966179, 964266 76 DATA 962270, 960201, 958098, 955942, 953743 78 DATA 951483, 949171, 946789, 944337, 941806 8\O DATA 939197, 936492, 933692, 930788, 927763 82 DATA 924609, 921317, 917880, 914282, 910515 84 DATA 906554, 902393, 898007, 893382, 888504 86 DATA 883342, 877883, 872098, 865967, 859464 87 DATA 852554, 845214, 837413, 829114, 820292 88 DATA 810900, 800910, 790282, 778981, 766967 89 DATA 754191, 740631, 726241, 710990, 694843 9\O DATA 677771, 659749, 640761, 620782, 599824 91 DATA 577882, 554975, 531133, 506403, 480850 92 DATA 454548, 427593, 400112, 372240, 344136 93 DATA 315982, 287973, 260322, 233251, 206989 94 DATA 181765, 157799, 135297, 114440, 93378 95 DATA 78221, 63036, 49838, 38593, 29215 96 DATA 21577, 15514, 10833, 7327, 4787 97 DATA 3011, 1818, 1005, 454, 125 99 END

Question:

Two bodies having masses m_1 = 30 gm and m_2 = 40 gm are attached to the ends of a string of negligible mass and suspended from a light frictionless pulley as shown in the diagram. Find the accelerations of the bodies and the tension in the string.

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Solution:

In order to find the acceleration of the masses, we use Newton's Second Law (F = ma) to relate the external forces applied to each mass to the accelera-tion of each mass. Consider the body of mass m_1. Two external forces act on it, the weight m_1g downward and the upward pull T of the string. The resultant force on this body is T -m_1g upward. Using the Second Law T - m_1g = m_1a(1) where the negative sign implies that T and m_1g are in opposite directions, and a is the upward acceleration of this body. Now consider the body of mass m_2. The forces acting on this body are its weight m_2g downward and the tension T upward. The resultant force is m_2g - T downward and m_2g - T = m_2a(2) where a is the downward acceleration of this body. Since the two bodies move together, the accelerations are equal in magnitude but they are opposite in di-rection. To find the acceleration of the string and two masses, we add (1) and (2) and solve for a: m_2g - m_1g = m_1a + m_2a = (m_1 + m_2)a ora= (m_2g - m_1g) / (m_1 + m_2) = {(m_2 - m_1)g} / {m_1 + m_2} [{(.004 kg - .03 kg)(9.8 m/s^2)} / {.04 kg + .03 kg}] = [{(0.1 kg)(9.8 m/s^2)} / {.07 kg}] = (.098 Nt) / (.07 kg) = 1.4 m/s^2(3) We can substitute this value for a in either of the original equations {(1) or (2)} to obtain the value for T, the string tension. Substituting (3) in (1): T - m_1g = m_1a T = m_1g + m_1a = m_1(g + a) = .03 kg (9.8 m/s^2 + 1.4 m/s^2) = .03 kg (11.2 m/s^2) = .336 Nt.

Question:

Name the two basic classes of electronic computers. Give the basic similarities and differences between them.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G01-0006.htm

Solution:

The two main classes of electronic computers are analog and digital, although analog computers are seldom used these days. The analog computers use continuous electrical quan-tities such as voltages, currents, capacitances, etc. to repre-sent the continuous real-world quantities such as temperature, pressure, mass, and so on. For example, 1 + 1 = 2 may be repre-sented as 1 volt + 1 volt = 2 volts using appropriate circuitry. On the other hand, a more complicated mechanical system shown in Figure 1-2a may be represented by the electrical circuit shown in Figure 1 - 2b. Figure 1-2. Representation of A Problem by Analog Computer. The general equation of oscillation for the mechanical system is given by: {MD + F + (1 / KD)} X = 0, where D is called the differential operator, and X is the posi-tion of the weight at any moment. After the switch is closed, the equation of the electrical oscillation will take the form: {LD + R (1 / CD)} I = 0, where D is same as before, and I is the current at any moment. Note that these two equations are identical, except for some of the letters used. Therefore, the behavior of the two systems will be identical. Although analog computers are good in solving problems that involve continuous quantities, the representation is not always easy and the accuracy is in doubt. For example, an analog com-puter calculates the interest on a million-dollar loan, the result could be off by at least a few dollars. The accountant who must reckon every penny may not be satisfied with that kind of result. The digital computer deals with numbers represented inter-nally as a sequence of binary digits (or bits). The architecture and operation of these systems are detailed throughout this chapter. Although a digital computer cannot directly represent and solve the above problem whose nature is continuous, it can be programmed to solve the problem through a mathematical proce-dure. Moreover, the results will be as accurate as the math-ematical formulation allows. In the case of interest calcula-tions on a million-dollar loan, the result will be accurate down to the pennies, except for possible round-off error that is really the property of mathematical solution procedure and not the fault of the digital computer. The digital computers have advantages of programmability with reliable accuracy, since exactly the same result will be obtained the next time the program is executed. The analog com-puter may sometimes have advantage in solving continuous type problems in a direct way to produce results that are not precise but acceptable for all practical purposes. The similarity be-tween the two classes of computers is the fact that both are electromechanical machines. The difference is that the digital computers, sometimes at the expense of extra storage and fre-quency bandwidths (e.g. more complex electronic signals), are more reliable and accurate.

Question:

Explain briefly the background and applications of APL programming .

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Solution:

APL (A Programming Language) is an extremely pow-erful, simple to learn, easy to use computer language, which lends itself to many applications . Widely recognized as a scientific language, APL is equally useful in business and in education. APL is one of the best-suited languages for special applications, such as computer aided instruction (CAI), operations research, and simulation. Being a versatile language, it can be used in either execution or definition modes. In execution mode the lan-guage enables you to operate the terminal as though it were a desk calculator. In definition mode you are able to enter, store, execute, and debug programs. APL is efficient. When compared with other languages: it takes the least amount of programming effort to accomplish a given job. When in execution mode, APL has immediate diagnostics. The system reports errors as soon as they are made and gives their type and location . All these features make APL a strong tool in solving a wide range of complicated problems.

Question:

Calculate the number of liters in one cubic meter.

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Solution:

There are 1,000 milliliters (ml) or cubic centimeters (cc) in one liter. Thus, if one wishes to convert one cubic meter to liters, the cubic meter mustbe converted to cubic centimeters. 1 meter =100 centimeters (1 meter)^3= (100 centimeters)^3 = 1,000,000 centimeters^3 = 1 × 10^6 cubic centimeters Cubic centimeters can be converted to liters by multiplying the number of cubiccentimeters by the factor 1 liter/1,000 cubic centimeters. 1 × 10^6 cubic centimeters × 1 liter/1,000 cubic centimeters = 1,000 liters. There are 1,000 liters in one cubic meter.

Question:

Find the logarithm of 3^2.

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Solution:

Recall that log_b x^y = y log_b x; thus log_10 3^2 = 2 log_10 3 Referring to a table of common logarithms we find: log_10 3 = .4771; hence, = 2 (.4771) = .9542 . Thus, log_10 3^2 = .9542.

Question:

Discuss the basic features of the molluscan body. How do they differ in the several classes of mollusks ?

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Solution:

The phylum Mollusca is the second largest in the animal kingdom. Snails, clams, oysters, slugs, squids, and octopuses are among the best known molluscs. The adult body plan is remarkably different from that of any other group of invertebrates. The soft body consists of three principal parts: (1) a large ventral muscular foot which can be extruded from the shell (if a shell is present) and functions in locomotion; (2) a visceral mass above the foot, containing most of the organs of the body; and (3) a mantle, a heavy fold of tissue that covers the visceral mass. In most species, the mantle contains glands that secrete a shell. The mantle often projects over the edges of the foot and overhangs the sides of the visceral mass, thus enclosing a mantle cavity, in which gills frequently lie. The chitons, members of the class Amphineura, have an ovoid bilaterally symmetrical body with an anterior mouth and posterior anus. They have paired excretory organs, nerves, gonads, and gills. The shell consists of eight dorsal plates. Class Gastropoda is a large class, containing the snails, slugs and their relatives. Most gastropods have a coiled shell; however, in some species, coiling is minimal, and in others, the shell has been lost. The body plan of the adult gastropods is not symmetrical. During develop-ment, two rotations of the body occur, so that the anus comes to lie dorsal to the head in the anterior part of the body. The organs on one side of the body atrophy, so that the adult has one heart, one kidney, one gonad, and one gill. This embryonic twisting, called torsion, occurs in all gastropods, including those with a flat shell, slugs and other species without a shell Class Pelecypoda or Bivalvia contains the bivalves. These mollusks have two hinged shells. Scallops, clams, mussels, and oysters are well known bivalves. These animals have well-developed muscles for opening and clos-ing the shells. These animals also have a muscular siphon for the intake and output of water. The nautilus squid, and octopus are members of the class cephalopoda. Cephalopod means head-foot, and in these molluscs the foot is fused with the head. In squids and octopuses, the foot is divided into ten or eight tentacles, and the shell is greatly reduced or absent. These molluscs have a well-developed nervous system with eyes similar to vertebrate eyes. The mantle is thick and muscular. Giant squids are the largest living invertebrates; they have attained lengths of 55 feet, and weights of two tons.

Question:

Find the net dipole moment inDebyesfor each of the following situations: (a) One + 1 and one - 1 ion se-parated by 2 × 10^-8 cm, and (b) one +2 and one - 2 ion separated by 2 × 10^-8 cm. One charge = 4.80 × 10^-10esuand 1 Debye = 1 × 10^-18esu-cm.

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Solution:

A dipole consists of a positive and a negative charge separated by a distance. Quantitatively, a dipole is described by its dipole moment, which is equal to the charge times the distance between the positive and negative centers. For both situations, substitute the values into this quantitative expression: net dipole moment = charge × distance (a) Net dipole moment = (4.80 × 10^-10esu) (2 × 10-8cm) = 9.6 × 10^-18esu-cm. 1 Debye= 1 × 10^-18esu-cm. [(9.6 × 10^-18esu-cm) / (1 × 10^-18esu-cm / Debye)] = 9.6Debyes (b) Net dipole moment = (2) (4.80 × 10^-10) (2 × 10^-8) = 19.2 × 10^-18esu-cm Converting toDebyes [(19.2 × 10^-10esu-cm) / (1 × 10^-18esu-cm / Debye)] = 19.2 Debye.

Question:

Calculate ∆G\textdegree for the reaction N_2 (g) + 3H_2 (g) = 2NH_3 (g) at 400\textdegreeC = 673\textdegreeK using K_C . The value ofK_pfor this reaction is 1.64 × 10^-4. Interpret the different values of ∆G\textdegree calculated fromK_pand K_C.

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Solution:

Given the equilibrium reaction: (i)aA_(g) +bB_(g) + . . . =cC_(g) +dD_(g) + . . . there are two constants that measure the equilibrium: K_C relates the concentrations. (ii)K_C = {[C]^c [D]^d . . .}/{[A]^a [B]^b . . .} K_p relates the pressures (iii)K_p= [(Pc_C P^d _D . . .)/(Pa_AP^b_B. . .)] P^b _B To derive K_C fromK_psubstitute for each pressure P =cRT, where R is the gas constant; T = absolute temperature; and c is the number of moles of a gas per liter. For the reaction, N_2 (g) + 3H_2 (g) = 2NH_3 (g), K_p = [(P^2 _NH3)/(P_N2 P^3 _H2)] = [{C^2 _NH3 (RT)^2 }/{C_N2 C^3 _H2 (RT)^4} = [(K_C)/(RT)^2], whereK_C = [(C^2 _NH3)/(C_N2 C^3 _H2)] = (RT)^2K_p.Thus, K_C = [{0.0824 (liter-atm/\textdegreeK-mole)} (673\textdegreeK K_C = [{0.0824 (liter-atm/\textdegreeK-mole)} (673\textdegreeK )]^2 (1.64 × 10^-4 ) = 0.500 and∆G\textdegree_C = - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) log (0.500) = 927 cal/mole. ∆G\textdegree_p= - RT InK_p = - (1.987 cal\textdegreeK^-1 mole^-1) (673\textdegreeK) (2.303) log (1.64 × 10^-4) = 11, 657 cal/mole. ∆G\textdegree_C is the change in Gibbs free energy when 1 mole of N_2 in the ideal gas state at the concentration of 1 mole per liter and 3 moles of H_2 in the ideal gas state at 1 mole per liter react to form 2 moles of NH_3 in the ideal gas state at 1 mole per liter. ∆G\textdegree_pis the change in Gibbs free energy when 1 mole of N_2 in the ideal gas state at 1atm, and 3 moles of H_2 in the ideal gas state at 1atm react to form 2 moles of NH_3 in the ideal gas state at 1 atm.

Question:

When lithium is bombarded by 10-MeV deuterons, neutrons are observed to emerge at right angles to the direction of the Incident beam. Calculate the energy of these neutrons and the energy and angle of recoil of the associated beryllium atom. Relevant masses in amu are: _0n^1 = 1.00893, _3Li^7 = 7.01784, _1H^2 = 2.01472, and _4Be^8 = 8.00776

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Solution:

Since the deuteron's kinetic energy is much smaller than its rest mass energy (\approx1876 MeV), the collision is non-relativistic. The equation governing the reaction is _3Li^7 + _1H^2 =_0n1+ _4Be^8 . The sum of the masses of the initial atoms is 9.03256 amu, and of the final atoms 9.01669 amu. The difference in mass, 0.01587 amu, appears, by Einstein's mass-energy relation, as E = (.01587 amu)c^2 amount of kinetic energy. Hence, E = (.01587 amu) (1.66 × 10^-27kg) (9 × 10^16 m^2/s^2) E = 14.77 MeV Since total energy (that is, rest mass energy plus kinetic energy) is conserved in the reaction, the final products have a combined kinetic energy of (10 + 14.77)MeV = 24.77 MeV . When we remember that the momentum of a particle p is related to its kinetic energy E by the relation p = \surd(2mE), then the principle of conservation of momentum, applied In the Initial direction and at right angles to it, gives the following equations, In self-explanatory notation: (see figure) \surd(2m_DE_D)_ = \surd(2m_BE_B) cos \texttheta(1) and\surd(2m_NE_N)_ = \surd(2m_BE_B) sin \texttheta(2) 2m_DE_D + 2m_NEN= 2m_BE_B(3) It was also shown above that E_N + E_B = 24.77 MeV(4) Solving (4) for EN EN= 24.77 MeV - EB(5) Using (5) in (3) 2m_DE_D + 2m_N( 24.77 MeV) -2m_N EB= 2m_BE_B Solving for E_B E_B (m_B + m_N) = m_DE_D+ m_N( 24.77 MeV) orEB= {m_DE_D+ m_N( 24.77 MeV)} / (m_B + m_N) whence E_B = {(2.01472 amu)(10MeV) + (1.00893amu) ( 24.77 MeV)} / (8.00776 amu + 1.00893 amu) E_B = 45.13840 amu .MeV / 9.01669 amu = 5.01 MeV Using this in (5) E_n = 24.77 MeV - 5.01 MeV = = 19.76 MeV Further, we obtain for the angle of recoil (using (1) and (4) tan \texttheta = \surd{(m_NE_N)_ / (m_DE_C)} = \surd[{(1.00893amu) (19.76 MeV)} / {(2.01472 amu) (10MeV)}] = \surd(.9895) = .9948 Hence ,\texttheta \approx 44050' .

Question:

A copper coil has a resistance of 100 ohms at 20\textdegreeC. What is its resistance at 50\textdegreeC if the temperature coefficient of resistivity for copper is 0.0038 ohm per ohm per degree centigrade?

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Solution:

We are given the resistance of a wire at one temperature, and are asked to calculate its resistance at another temperature given the temperature coefficient of resistivity. The new resistance is given by the old re-sistance plus a factor proportional to the old resistance, the temperature change, and the constant coefficient for the substance. R2 0= 100 ohms,t = 50\textdegreeC,\alpha = 0.0038 ohm/ohm/\textdegreeC. UsingR_t= R2 0+ R2 0\alpha(t - 20\textdegreeC), we have R_t = 100 ohms + 100 ohms × 0.0038 ohm/ohm/\textdegreeC (50\textdegreeC - 20\textdegreeC) = 100 ohms + 11.4 ohms = 111 ohms, approximately.

Question:

Ethanol boils at 78.5\textdegreeC. If 10 g of sucrose (C_12H_22O_11) is dissolved in 150 g of ethanol, at what temperature will the solution boil? Assume k_b = (1.20\textdegreeC / M) for the alcohol.

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Solution:

When a non-volatile solute, such as sucrose, is dissolved in a solvent, such as ethanol, it will raise the boiling point of the solvent. The boiling point elevation can be found by using the equation ∆T_b =k_b\eta, where ∆T_b is the boiling point elevation, kbis the elevation constant, and \eta is themolalityof the solution. You want to determine ∆T_b and are given K_b. Thus, you need to determine themolalityof the solution.Molalityis defined as the number of moles of solute per 1 kg of solvent, i.e. (moles solute / 1 kg solvent). Solving formolarityof sucrose: (MW of C_12H_22O_11 = 342.3). Moles of solute= (grams solute) / (molecular wt. solute) = 10g / [(342.3 (g / mole)] = 2.921 × 10^-2 moles. You have 150 g of solvent, ethanol. Butmolalityis per 1 kg, so that you must multiply 150 g by (1 kg / 1000 g). Molality = (.02921mole) / [(150 / 1000 kg)] = .195 M Thus, the elevation of the boiling point is ∆T_b =K_b\eta= (1.20\textdegreeC / M)(.195 M) = .23\textdegreeC. Thus, the boiling point of the solution is 78.5\textdegreeC + .23\textdegreeC = 78.73\textdegreeC.

Question:

What wavelength of light is needed to excite an electron in a 0.2 nanometer (1 nm = 10^-9 m) box from the ground state to the second excited state? What wavelength of light is emitted when the same electron falls from the second excited state to the first excited state?

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Solution:

The solution to this problem requires the application of two equations, one to determine the energies of the various states, from which may be obtained the energy of a transition between states, and one to relate energy to wavelength. The first equation is the energy of a particle-in-a-box, E_n = [(n^2 h^2) / (8mL^2)] where n is the principal quantum number (n = 1, 2, 3, ...), h is Planck's constant (6.626 × 10^-27 erg-sec = 6.626× 10^-34 J-sec), m is the mass of the particle, and L is the length of the box. For an electron in a 0.2 nm = 0.2 × 10^-9 m = 2 x 10^-10 m box, E_n = (n^2 h^2) / (8mL^2)] = [{n^2 (6.626 × 10^-34 J-sec)^2} / {8 × 9.11 × 10^-31 kg × (2 × 10^-10 m)^2}] = 1.506 × 10^-18 j × n^2. The second equation relates the energy difference, ∆E, between two states and the wavelength \lambda: ∆E =hc/\lambda, or \lambda =hc/∆E, where c is the speed of light. In the following table, we calculate the energies corresponding to the ground state (n = 1), the first ex-cited state (n = 2) , and the second excited state (n = 3). n E_n = n^2 × 1.506 × 10^-18 J 1E_1 = (1)^2 × 1.506 × 10^-18 J = 1.506 × 10^-18 J 2E_2 = (2)^2 × 1.506 × 10^-18 J = 6.024 × 10^-18 J 3E_3 = (3)^2 × 1.506 × 10^-18 J = 1.355 × 10^-17 J In going from the ground state to the second excited state, the electron must absorb a quantum of light of energy ∆E = E_3 - E_1 = 1.355 × 10^-17 J - 1.506 × 10^-18 J = 1.204 × 10^-17 J. The corresponding wavelength is \lambda = [(hc) / (∆E)] = (6.626 × 10^-34 J × 3 × 10^8 m/sec) / (1.204 × 10^-17 J)] = 16.510 × 10^-9 m = 16.510 nm. In falling from the second excited state to the first excited state the electron will emit a quantum of light of energy ∆E = E_3 - E_2 = 1.355 × 10^-17 J - 6.024 × 10^-18 J = 7.526 × 10^-18 J. This corresponds to a wavelength of \lambda = [(hc) / (∆E)] = (6.626 × 10^-34 J × 3 × 10^8 m/sec) / (7.526 × 10^-18 J)] = 26.412 × 10^-9 m = 26.412 nm.

Question:

AssumingpD= - log [D_3O^+] in analogy to pH, what ispDof pure D_2O? K = 2 × 10^-15.

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Solution:

This question can be solved with the knowledge that D_2O undergoes dissociation in a similar manner to H_2O. The only difference is that D_2O is composed of deuterium instead of hydrogen. Therefore, one can write: D_2O + D_2O \rightleftarrows D_3O^+ + OD^- K = [D_3O^+] [OD^-] = 2 × 10^-15. [D_2O] is a constant and does not appear in the above equi-librium expression . From the reaction, it becomes apparent that the concentrations of D_3O^+ and OD^- are equal. There-fore, both concentrations can be represented by x, x \textbullet x = 2 × 10^-15. Solving for x, x = 4.47 × 10^-8. This, then, represents the concentration of D_3O^+ (and OD^-). pD = - log [D_3O^+] = - log 4.47 × 10^-8 = 7.35.

Question:

Differentiate between a placenta and an umbilical cord.

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Solution:

As the human embryo grows, the region on the ventral side from which the folds of the amnion, the yolk sac, and the allantois grow, becomes relatively smaller, and the edges of the amniotic folds come together to form a tube which encloses the other membranes. This tube is called the umbilical cord and contains, in addition to the yolk sac and allantois, the large umbilical blood vessels through which the fetus obtains nourishment from the wall of the uterus. The umbilical cord, about 1 cm in diameter and about 70 cm long at birth, is composed chiefly of a peculiar jelly- like material found nowhere else. The placenta is the area where a portion of the chorion and a portion of the uterine wall join. The placenta is thus a structure of double origin, partly embryonic (chorion) and partly maternal (uterus). It functions to exchange nutrients, wastes, and gases between the mother and the fetus. In mammals such as the human, the placenta is a disk about 7 inches in diameter and about 1 to 1(1/2) inches thick. Not all mammals have such disklike placentas. For example, cows have numerous isolated clusters of fetal-maternal associations, other animals have fingerlike projections of the chorion that provide a very diffuse type of association. In cats and dogs the placenta forms a ring around the fetus.

Question:

You have 100-proof (50 percent alcohol by volume) bonded Scotch whisky. Calculate itsmolarity, mole fraction, and molality. If the temperature were to drop to - 10\textdegreeC, could you still drink the Scotch? Assume density = .79 g/ml for ethyl alcohol (C_2H_5OH) andK_f= 1.86\textdegreeC/m for water.

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Solution:

To findmolarity, mole fraction, andmolality, you need to know how many moles of alcohol are present in the whiskey. You can obtain this from a calculation using the density. To find whether you can still drink the whisky at - 10\textdegreeC, requires you to know its freezing point, since you cannot drink something if it is frozen. You can deter mine the freezing point of the Scotch by determining the freezing point depression of the water containing the alcohol. This is found from\DeltaT_f=K_fm, where\DeltaT_f= the depression,K_f= freezing point depression constant, and m =molality. You now proceed as follows: You are told that 100 proof means 50 percent alcohol by volume. This means that 1 liter of components possess 500 cm^3 of ethyl alcohol (a liter = 1000 cm^3). Density = mass/volume. You know the density and volume, so that mass of alcohol = density × volume = .79(500) = 395 g. Molecular wt. of the alcohol = 46.07 g. Thus, you have 395/46.07 = 8.57 moles of alcohol. Hence, molarity= [(moles of solute)/(liters of solution)] = 8.57 / 1 = 8.57 M. Molarity= [(moles of solute)/(kilograms of solvent)]. To find the number of grams of solvent, you need to use the density again. In 1 liter of Scotch Whisky, you have 500 cm^3 of water. Density of water = 1 g/ cm^3. Thus, mass of water = 500 cm^3 (1 g/ cm^3) = 500 grams. Thus, Molality= [(8.57) / {(500/1000)}] = 17.14 m You divided 500 by 1000 sincemolalityis moles per 1000 g of solvent. Mole fraction: (MW of H_2O =18.) moles of H_2O = 500 g/18.0 g/mole. Mole fraction of ethyl alcohol = [(moles ethyl alcohol)/(moles ethyl alcohol + moles H_2O)] = [(8.57) / {8.57 + (500/18)}] = .2357. Now, you answer whether the scotch can be consumed at - 10\textdegreeC. You already calculated themolality. Thus,\DeltaT_f=k_fm = (1.86)(17.14) = 31.88\textdegreeC. Thus, the freezing point of whisky is 0\textdegreeC - 31.88\textdegreeC = - 31.88\textdegreeC. The temperature is only - 10\textdegreeC, however. This means whisky is still a liquid and can be consumed.

Question:

A positively charged ring of radius R lies in the Y-Z plane. Compute the electric intensity at point P on the axis of the ring.

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/Users/wenhuchen/Documents/Crawler/Physics/D17-0569.htm

Solution:

To calculate the electric field intensity at a point P from a continuous distribution, we calculate the electric field intensity due to one arbitrary element of charge, dq, at the point P, and then integrate this over the entire region under consideration. The electric field intensity, due to a differential charge element dq is given by dE^\ding{217} = (dq r^\ding{217})/(4\pi\epsilon_0 r^3)(1) where dE^\ding{217} is the field intensity due to dq, dq is an element of charge, r^\ding{217} is a vector from dq to P, and r is the magnitude of this vector. In this problem, however, we can simplify the integral in (1). Note from the figure that the Z com-ponents of the field intensities due to charge elements dq on opposite sides of the ring will cancel because they are of equal magnitude, but opposite in direction. Hence, the net electric field intensity will point along the x-axis and will be due to the x component of dE^\ding{217}, which is of magni-tude (dE) cos \alpha. Now we may neglect the vector nature of the integrand in (1), because the problem has been reduced to a calculation in one dimension. Therefore, E_net = \intdE cos \alpha. From (1), we obtain E_net = \int(dE cos \alpha)/(4\pi\epsilon_0r^2). Because cos \alpha, 4\pi\epsilon_0, and r are constants, we may write E_net = (1/4\pi\epsilon_0) [(cos \alpha)/r^2 ]\int dq(2) Integrating (2), we find E_net = (q cos \alpha)/(4\pi\epsilon_0 r^2)(3) where q is the total charge on the ring. Note that this field intensity is directed along the x-axis, and varies inversely with r^2. At the center of the ring, \alpha = 90\textdegree, cos \alpha = 0, and E = 0, as is evident by symmetry. At distances large compared to R, the angle \alpha is small, cos \alpha is approximately 1, and s^2(see figure) is nearly equal to r^2. Hence at large distances E = [1/(4\pi\epsilon0)] (\alpha/s^2) In other words, at large distances the ring can be con-sidered a point charge.

Question:

Devise a BASIC program to calculate the first N numbers of the FIBONACCI series.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0240.htm

Solution:

The FIBONACCI series is defined recursively by F(1) = 1, F(2) = 1, and for I greater or equal to 3, F(I) = F(I - 1) + F(I - 2). It is convenient to tabulate the functional values in an array dimensioned to be, say, 50. DIM (dimension) statement is used for this purpose. BASIC automatically provides for 10 places in its memory cells for any one dimensional array (11 places in those systems, which accept 0 subscript). If one desires N \not = 10 places to be reserved, he can use the statement of the form: DIM V(N) where V can be any valid variable. Thus the program will handle N up to 50. 1\O DIM F(5\O) 2\O INPUT N 3\O LET F(1) = 1 4\O LET F(2) = 1 45 PRINT F(1), F(2), 5\O FOR 1=3 TO N 6\O LET F(I) - F(I - 1) + F(I - 2) 7\O PRINT F(I), 8\O NEXT I 1\O\O END

Question:

You have been given an unordered table to search for a par-ticular entry. When this entry is found you are to replace it with an updated piece of data.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G09-0205.htm

Solution:

Since the table is not ordered it would be best to use a linear search in which all the elements of the table are inspected. To do this problem we will assume that the table has 100 entries and each entry is 14 bytes long. Program: RDCDREADCARDDATA, CRDEOF LA4, SYMTBLSTORE ADDRESS OF TABLE SEARCHCLC0(14, 4), SYMBOLCOMPARE ENTRY WITH ITEM TO BE SEARCHED FOR (SYMBOL) BESYMFOUNDIF EQUAL WE'VE FOUND SYMBOL C4, LAST'LAST' HOLDS THE ADDRESS OF THE LAST ENTRY IN THE TABLE BENOTFOUNDHERE, IF EQUAL AND THE SYM-BOL IS NOT FOUND, THEN IT IS NOT IN THE TABLE A4, = F '14'GO TO NEXT ENTRY IF SYMBOL IS NOT FOUND BSEARCHCONTINUE TO SEARCH NOTFOUND \textbullet \textbullet \textbullet \textbullet \bullet \bullet SYMFOUND MVC0 (14 , 4),REPLACE ENTRY WITH \textbulletNUDATAUPDATED DATA \textbullet CRDE OF- - - - - \textasteriskcenteredDATA DEFINITION FOR INPUT AREAS SYMTBLDS100CL14ALLOCATE SPACE FOR A HUN-DRED 14-BYTE ENTRIES. SYMBOLDSCL14HOLDS SYMBOL BEING CHECKED. LASTDCA (FINDAT)STORES ADDRESS OF FINAL DATA IN TABLE. \textasteriskcenteredDATA DEFINITION FOR INPUT AREA DATADC\varphiCL80 \varphi \varphi NUDATADSCL14 DSCL66 Comments: The 'NOTFOUND' segment of this program would usually involve printing a message which indicates the data was not in the table. This is always a good procedure, since when we do any table searching there is the possiblility that the given data won't be found. For the 'LAST DC A (FINDAT)' instruction, the 'A' indicates that the address of the symbol in the parentheses should be stored in the location represented by the symbol in the label field. In other words, this instruction tells the computer to store the address of 'FINDAT' in the location denoted by LAST. In coding the table, in our program, we would attach the label 'LAST' to the final piece of data in the table. On reaching the label 'CRDEOF', program execution would have generally been completed, and as such, we would restore the contents of the registers stored at the begin-ning of our program.

Question:

Small amounts of cyanide can kill a person. Why is cyanide such a deadly poison?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0085.htm

Solution:

Cyanide acts by inhibiting the final reaction of the respiratory chain, shutting it off and preventing the energy extraction process from taking place. Normally, reduced cytochrome a, the final element in the respiratory chain, would be reoxidized by oxygen, but since cyanide inhibits this from occuring, the whole chain is shut off as the intermediate acceptors are all converted to the reduced condition. Consequently, there is a severe decrease in the amount of ATP that can be produced fox life processes, leading to the death of the victim. Hydrogen sulfide and carbon monoxide have similar effects at this site. Other poisons block either of the first two sites. Looking at the electron transport chain it may not be obvious how the inhibition of the reoxidation of reduced cytochrome a can shut off the entire respiratory chain. Cyanide blocks ATP formation of the third site of ATP synthesis in the chain. However, this does not mean that ATP production still goes on as usual at the first and second site. Some ATP may be produced immedi-ately after cyanide enters the electron transport chain, but eventually production stops. ATP production ceases because of the inability of the intermediate acceptors to oxidize NADH. Inhibition by cyanide generates an accumulation of reduced cytochrome a by preventing its oxidation. Since cytochrome a is in a reduced state, it cannot oxidize its neighbor, reduced cytochrome c. This leads to an accumulation of reduced cytochrome c and then c_1, which, being unable to oxidize cytochrome b, prevents the formation of ATP at the second site. Of more importance is the "backing up" of reduced states towards the beginning of the chain leading even-tually to the reduction of all intermediate acceptors. These reduced intermediates cannot accept electrons; ATP formation in the respiratory chain requires inter-mediates to receive electrons which are liberated going from a reduced state to an oxidized state. Therefore, ATP formation cannot proceed normally and, in fact, finally stops if any step in the electron transport chain is blocked.

Question:

John is injured and badly needs a blood transfusion. He is type B. His girlfriend Jane volunteers, but she has type O blood. The physician, in view of the urgency of the situation, uses Jane's blood anyway. A few months later, Jane needs a transfusion. John volunteers but the physician turns him down. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0370.htm

Solution:

Before we explain this particular problem, it would be worthwhile to discuss the A\rule{1em}{1pt}B\rule{1em}{1pt}O blood type system. There are four types of blood determined geneti-cally: A, B, AB and O. The basis of these blood types is the presence or absence of certain antigens, calledagglutinogens, on the surface of the erythrocytes. The erythrocytes of type A blood carry agglutinogen A; those of type B blood carryagglutinogenB. Type O blood has neither of theseagglutinogens, while type AB has both A and B agglutinogens . Theseagglutinogensreact with certain antibodies, called agglutinins, that may be present in the plasma. An agglutinin-agglutinogen reaction causes the cells to adhere to each other. This clumping, or agglutination, would then block the small blood vessels in the body, causing death. Agglutination is simply the clumping of erythrocytes by agglutinins. The blood clotting mechanism, or co-agulation is not involved. A Type A person has the A antigen (agglutinogen) and the anti-B antibody (agglutinin) . A Type B person has the B antigen and anti-A antibody. A Type O person has no antigens but both anti-bodies. Conversely, a Type AB person has bothanitgensbut no antibodies. Basically, a person does not have the agglutinin in his plasma which would clump his own red blood cells. A summary of the antibodies and anti-gens that each blood type contains is outlined in the following table: Blood Type Agglutinogens (antigens) Agglutinins (antibodies) A A anti-B B B anti-A AB A and B none O none anti-A and anti-B Blood typing is a critical factor in blood trans-fusions. Since the agglutinins in the plasma of one type will react with theagglutinogenson the erythrocytes of the other types, transfusions are usually between people with the same blood type. However, one may use another blood type provided that the transfusion is small and the plasma of the recipient and the erythrocytes of the donor are compatible (does not cause a reaction) . This type of transfusion was performed in the given case. Type O blood from Jane was given to John, a blood-type B patient. The anti-A agglutinin in John's plasma finds noagglutinogensto attack on Jane's erythrocytes. We can usually ignore theagglunitins, specifically anti-B, in Jane's plasma, which would normally react with theagglutinogenB on John's erythrocytes. This is because Jane's plasma is sufficiently diluted during transfusion so that the con-centration of agglutinins is too low to cause appreciable agglutination. Jane's blood, type O, can be given to any-one since her erythrocytes contain noagglutinogens. Type O blood is therefore called the universal donor. Although type O blood can be donated to any type, it can only accept type O since it has the agglutinins, anti-A and anti-B, which react with theagglutinogensof every other blood type. In contrast, type AB blood can receive from any type since it has no agglutinins in its plasma to react with the donor's erythrocytes. Type AB blood is therefore called the universal recipient. But it can only donate to other AB patients since itsagglutinogens, A and B, react with the plasma of every other type. The reason John's offer was turned down is that his blood, type B, would agglutinate with Jane's blood, type O. The agglutinins in Jape's plasma, specifically anti-B, would react with the Bagglutinogenon John's erythrocytes and cause agglutination.

Question:

Three coordinate systems S, S', and S" have a common x-axis. With respect to S, S' moves in the direction of the x-axis with a speed v, and S" accelerates along the x-axis with acceleration a. At time t = 0, the origins of all three coordinate systems coincide and S" has zero velocity with respect to S. At that instant/ a man starts out running from the origin along the x-axis and an observer in S measures his speed as constant and of magnitude u(u > v). How do observers in S' and S" describe his motion, using Galilean relativity?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0217.htm

Solution:

The observer in S sees the man running along the x-axis with constant speed u. The motion is thus observed as one in which the x displacement increases linearly with time. On a displacement-time diagram, the motion appears as the dot-dashed straight line at an angle \texttheta to the t-axis, where tan \texttheta = x/t = u; and on a velocity -time diagram it appears as the dot-dashed straight line parallel to the t-axis. (see figs. (a) and (b)) . To an observer in s', the man has only the speed (u - v), the relative speed between the two. But the man is still seen as moving with constant speed and his motion is shown in the diagrams by the dashed straight lines. The angle \o is such that tan \o= u - v. An observer in S\textquotedblright, who is accelerating along the x-axis with acceleration a relative to S and therefore to the running man, considers himself at rest and therefore attributes to the runner an acceleration of -a. The runner appears to start off with velocity u but to decelerate gradually to rest and then go backward. His vel-ocity at any time is seen to be, using the kinematics equations for constant acceleration, V = u - at, and on the velocity-time diagram this is represented by the full line at an anglevto the t-axis, where tanv= -a. Also his x-displacement relative to the origin of s" is given by the constant acceleration kinematics equation x = ut - (1/2)at^2, which is represented by a parabola in the displacement- time diagram, tangent to the dot-dashed line at the origin and having its highest point at the time t where the corresponding velocity--time graph cuts the t-axis.

Question:

Find the ionic radius of I-, given that the unit-cell edge length of Lil is measured to be .6240 nm. Assume that the large negative ions are in actual contact with the diagonal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0227.htm

Solution:

To solve this problem, determine the length of the face diagonal in Lil space lattice, which is the pattern of the arrangement of atoms or molecules in a crystal. Because Lil has the same form of the cubic unit cell as NaCl, its space lattice will possess 3 I- ions on the face diagonal. Thus .if one knows the length of the face diagonal, one can divide by 4 to deduce the radius of I-. The face diagonal can be found (from the Pythagorean Theorem) by multiplying (\surd2) times (cubic- edge length) . Thus, the face diagonal = (\surd2) (.6240) = .8823 nm. However, face diagonal = 4I- radii. Thus, I- radius = .8823 / 4 = .2206 nm.

Question:

Using the pedigree below, fill in the genotype of each individual. Assume the trait is recessive and an individual who marries into the family and does not exhibit a trait does not carry the recessive gene for it,

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0686.htm

Solution:

For simplicity and conciseness, we shall use the following system to refer to any one of the members in the pedigree: Let the allele for left-handedness be a and the allele for right-handedness be A. Consider P_1 (that is, the first member of the Parental generation). Since she expresses the trait, her genotype is aa. To determine the genotype of P_2 , we look at the progeny, F_1 (first generation). From the diagram, we see that F_1,2 is left- handed; hence his genetic make-up is aa. This means that he must have received a copy of a from each of his parents. Therefore we know that P_2 carries a copy of the recessive gene. But since P_2 is phenotypically normal, his genotype must be Aa. F_1,1 and F_1,3 are right-handed. Yet they must have received an a allele from P, since she has only a alleles to transmit. Therefore their genotype of F_1,1 and F_1,3 is Aa. Consider the cross between P_3 and P_4. We know P_3 is aa (because he is left-handed) and P_4, can be AA or Aa (since she is left handed). Since all the offspring are right-handed, it seems probable that P_4 is AA; however, we cannot be entirely certain of this, since both genotypes AA and A_2 are compatible with the phenotypes of the offspring. We know, however, that F_1,4, F_1,5, and F_1,6 must all have the genotype Aa since they carry an a allele donated by P_3. In the F_2 generation, we see that F_2,2 is left- handed, and therefore her genotype is aa. This is compatible with what we have determined to be the genotypes of the parents (F_1,3 and F_1,4) namely Aa. F_2,1 and F_2,3 are both right-handed, and are either AA or Aa. Since both parents are Aa, either genotype is possible, though Aa is more probable.

Question:

500 g of alcohol at 75\textdegree C are poured into 500 g of water at 30\textdegree C in a 300 g glass container (c_glass = .14). The mixture displays a temperature of 46\textdegree C. What is the specific heat of alcohol?

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/Users/wenhuchen/Documents/Crawler/Physics/D13-0479.htm

Solution:

When the alcohol is poured into the glass- water system, the former loses heat energy, and its temperature drops. The latter gains heat energy and its temperature rises. Hence, by the principle of conservation of energy heat loss by alcohol = heat gained by H_20 + heat gained by glass(1) In general, if a sample of mass m, composed of a substance of specific heat c, is exposed to a temperature change ∆T, it will lose or gain heat energy Q = m c ∆T. Using this fact, we may write heat gained by H_20 = m(H)2 0c(H)2 0∆T(H)2 0 = (500 g)(1 cal/g\textdegreeC) (46\textdegreeC - 30\textdegreeC) = 8000 cal heat gained by glass = m_g c_g ∆T_g = (300 g) (.14 cal/g\textdegreeC) (46\textdegreeC - 30\textdegreeC) = 672 cal heat lost by alcohol = m_a c_a ∆T_a = (500 g)(c_a) (75\textdegreeC - 46\textdegreeC) = (14500 cal) c_a (Note that the heat gained by the alcohol is negative. Hence, the heat lost by the alcohol is positive. This is the reason why ∆T_a > 0). Using these facts in (1) (14500 g \textdegreeC)c_a = 8000 cal + 672 cal = 8672 cal orc_a = (8672 / 14500) (cal / g0C) = .598 (cal / g0C)

Question:

When ultraviolet light of frequency 1.3 × 10^15 sec^-1 is shined on a metal, photoelectrons are ejected with a maximum energy of 1.8 electron volts. Calculate the work function of the metal in ergs and electron volts. What is the threshold frequency of this metal?

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/Users/wenhuchen/Documents/Crawler/Physics/D33-0963.htm

Solution:

This problem is an example of the photo- electric effect. We want to relate the work function of a metal to the energy of the incident radiation. (The work function is the amount of energy needed to release an electron from the attraction of the rest of the metal, and thereby pull it out of the metal.) Let us examine the reaction occurring. A photon of energyhu(the ultraviolet radiation) collides with an electron of the metal, and, in the process, gives up all of its energy. Ifhuis large enough, the electron will travel through the metal and be released with kinetic energy (1/2) mv^2, where m and v are the electron mass and velocity, respectively. In leaving the metal, the electron loses energy due to collisions with atoms of the metal. The electron also loses energy, equal in amount to the work function, in order to overcome the attraction of the metal and escape. If we examine only those electrons which lie close to the surface of the metal, we observe that they do not go through any collisions and hence can-not lose energy by this means. These surface electrons lose less energy than the interior electrons and there-fore will have a maximum amount of kinetic energy when they leave the metal. Using the principle of conservation of energy, we obtain hu= (1/2) m v^2_max + \textphi(1) where \textphi is the work function of the metal. (Note that this relation is true only for surface electrons.) Solving (1) for \textphi, we obtain \textphi =hu- (1/2) mv^2_max(2) The energy of a photon of the ultraviolet light is hu= 6.625 × 10^-27 × 1.3 × 10^15 erg = 8.61 × 10^-12 erg Since 1 electron volt = 1.60 × 10^-12 erg, the maximum kinetic energy of the photoelectrons is (1/2) mv^2_max = 1.8 X 1.6 X 10^-12 erg = 2.88 X 10^-12 erg From equation 2, the work function is \textphi =hu- (1/2) mv^2_max = 8.61 X 10^-12 - 2.88 X 10^-12 erg = 5.73 X 10^-12 erg Since 1 erg = .625 × 10^12eV \textphi = (5.73 × 10^-12 erg) [.625 × 10^12 (eV/erg)] \textphi = 3.58eV The threshold frequency is the smallest frequency of indecent radiation for which the electrons will be released from the metal with no kinetic energy. Setting (1/2) mv^2_max = 0 in equation (1) we find that the threshold frequency, u_0, is u_0 = (\textphi/h)(3) Using the value of \textphi calculated in the first part of the problem, u_0 = [(5.73 X 10^-12 erg)/(6.625 X 10^-27erg\textbulletsec)] oru_0 = 8.65 X 10^14 sec^-1 This frequency is in the ultraviolet, and just beyond the blue end of the visible spectrum.

Question:

Suppose there are three typists in a typing pool. Each typist can type an average of 6 letters/hr. If letters arrive to be typed at the rate of 15 letters/hr., a) What fraction of the time are all three typists busy? b) What is the average number of letters waiting to be typed? c) What is the average time a letter spends in the system (waiting and being typed)? d) What is the probability of a letter taking longer than 20 min. waiting to be typed and being typed? e) Suppose each individual typist receives letters at the average rate of 5/hr. Assume each typist can type at the average rate of 6 letters/hr., what is the average time a letter would spend in the system (waiting and being typed)?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0456.htm

Solution:

a) The following information is given \lambda = 15 letters/hr. \mu = 6 letters/hr. S = 3 Thus, we want P(n \geq 3). To obtain this, we need Po, the probability of no customers in the system given by the equation: Substituting the given values into the above equation we have the following: Therefore, P(n \geq S) = probability of the arrived letter wait-ing for service = probability of at least S customers in the system P(n \geq 3)= {(15/3)3(0.044944)} / [6{1 - (15/18)}] = 0.70225 b) Lq=average # of customers in queue = {(\lambda/\mu)S+1Po} / [s. s! {1 - (\lambda / \mus)2}] = {(15/6)4(0.044944)} / [(3) (6) {1-(15/18)^2}] = 3.51124 c) We first need the average number of letters in the system. L = L_q + (\lambda/\mu) = 3.51124 + (15/6) = 6.01124 W = average time a customer spends in a system = (L/\lambda) = (6.01124) / 5 = 0.40075 hr. \approx 24 min. d) The probability of a letter taking longer than 20 min. waiting to be typed and being typed e) Each queue and server can be treated as a separate single queue, single-server system so that W = 1 / (\mu - \lambda) = 1 / (6 - 5) = 1 hr. \textasteriskcentered\textasteriskcentered QUEUING PROGRAM \textasteriskcentered\textasteriskcentered \textasteriskcentered\textasteriskcentered INFINITE SOURCE, INFINITE QUEUE, MULTIPLE SERVERS \textasteriskcentered\textasteriskcentered WE ASSUME ARRIVALS FORM A SINGLE QUEUE INFINITE SOURCE AND INFINITE QUEUE FIFO QUEUE DISCIPLINE THERE ARE MULTIPLE SERVERS WITH EXPONENTIAL SERVICE TIME DEPARTURES FROM SYSTEM OCCUR COMPLETELY AT RANDOM REAL\textasteriskcentered4 TITLE (20) , LAMBDA, MU, N, L, LQ, NFACT 5READ(5, 10, END = 2000) TITLE 10FORMAT (20A4) WRITE (6, 11) TITLE 11FORMAT ('1', 20A4, //) READ (5, 20) LAMBDA MU 20FORMAT (3F10.0) READ (5, 20) N, S, T IF (LAMBDA. GE. S\textasteriskcenteredMU) GO TO 700 C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PZERO\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered NS=S SUM = 0.0 DO 205 NK = 1, NS SK = NK-1 CALL FACT (SK, NFACT) SUM = SUM+ (1/NFACT) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredSK 205CONTINUE CALL FACT(S, SFACT) PZERO = 1/ (SUM+ (1/ (SFACT\textasteriskcentered (1-LAMBDA/ (MU\textasteriskcenteredS) ) )) \textasteriskcentered (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PN\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered IF(N.GE.S) GO TO 210 CALL FACT (N, NFACT) PN = (1/NFACT)\textasteriskcentered((LAMBDA/MU)\textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO GO TO 215 210CALL FACT(S, SFACT) PN = (1/ (SFACT\textasteriskcenteredS\textasteriskcentered\textasteriskcentered (N-S) ) ) \textasteriskcentered ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredN) \textasteriskcenteredPZERO C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredCALCULATE PS, PT, L, LQ, W, WQ\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered 215CALL FACT(S, SFACT) PS = ( ( (LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredPZERO) / (SFACT\textasteriskcentered(1-LAMBDA/ (MU\textasteriskcenteredS)) ) PTN = ((LAMBDA/MU) \textasteriskcentered\textasteriskcenteredS) \textasteriskcenteredPZERO\textasteriskcentered (1-EXP (-MU\textasteriskcenteredT\textasteriskcentered (S- 1-LAMBDA/ MU))) PTD = SFACT\textasteriskcentered (1-LAMBDA/MU\textasteriskcenteredS)) \textasteriskcentered (S- 1-LAMBDA/MU) PT = EXP(-MU\textasteriskcenteredT)\textasteriskcentered(1+PTN/PTD) LQ = (((LAMBDA/MU) \textasteriskcentered\textasteriskcentered (S+1)) \textasteriskcenteredPZERO) /(S\textasteriskcenteredSFACT\textasteriskcentered (1-LAMBDA / (MU\textasteriskcenteredS))\textasteriskcentered\textasteriskcentered2) L = LQ+LAMBDA/MU W = L/LAMBDA WQ = LQ/LAMBDA C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered C\textasteriskcenteredPRINT RESULTS\textasteriskcentered C\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered\textasteriskcentered WRITE (6,100) LAMBDA, MU, N, S, T, L, LQ, W, WQ, PT, PN, PZERO, PS 100FORMAT (' LAMBDA = ', F10.5/' MU = ', F10.5/' N = ', F10.5/ ' S \textasteriskcentered = ', F10.5/'T = ' , F10.5/' L = ', F10.5/1 LQ = ', F10.5 \textasteriskcentered/' W = ' , F10.5/' WQ = ', F10.5/' PT = ', F10.5/' PN = '\textasteriskcentered, F10.5/'PZERO = ', F10.5/'PS= ', F10.5) GO TO 5 700WRITE(6, 705) 705FORMAT('QUEUING SYSTEM NOT VALID BECAUSE LAMBDA \leq S\textasteriskcenteredMU') GO TO 5 2000STOP END SUBROUTINE FACT (P, PROD) NUM = P PROD = 1. IF (NUM.EQ.0) GO TO 20 DO 10 K = 1, NUM 10PROD = PROD\textasteriskcenteredK 20RETURN END /DATA SINGLE QUEUE - MULTIPLE SERVER MODEL 15.6. 1.3..3333 SINGLE QUEUE - MULTIPLE SERVER MODEL LAMBDA= 15.00000 MU= 6.00000 N= 1.00000 S= 3.00000 T= 0.33330 L= 6.01124 LQ= 3.51124 W= 0.40075 WQ= 0.23408 PT= 0.46198 PN= 0.11236 PZERO= 0.04494 PS= 0.70225

Question:

Convert the following logical expressions to reverse Polish string notation: (a) (Z \downarrow ((((X) \odot Y) / (X \odot (Z))) > ((Y) \equiv X))) (b) (X / Y) / ((X / X) / (Y / Y))

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G07-0135.htm

Solution:

Polish notation is a way of representing logical and algebraic expressions without using parentheses, unless parentheses represent some variables or operators in the particular expression. If you think about it, the purpose of parentheses is to override the natural hierarchy of the operators. Polish notation does away with that notion. All operators have the same precedence: their position in the string determines their order of evaluation. With reverse Polish notation, also called Polish post-fix, the operator is written after its two operands. For example, the usual logical equation A \oplus B would be written AB \oplus in postfix notation. It is possible to use trees to evaluate Polish strings, but we will not do so here. a) As we can see in this example, the abundance of parentheses makes this expression look formidable. We will proceed step-by-step, enclosing in brackets each portion of the expression that has been converted. Step 1. First, we will take care of the negations. We use the symbol \rceil for this operator. (Z \downarrow ((([X\rceil] \odot Y) / (X \odot [Z\rceil])) > ([Y\rceil] \equiv X))) (Z \downarrow ((([X\rceil] \odot Y) / (X \odot [Z\rceil])) > ([Y\rceil] \equiv X))) Step 2. We consider next the innermost partial expressions and convert them. (Z \downarrow (([X\rceilY \odot] / (X Z\rceil \odot]) > ([Y\rceilX\equiv])) Step 3. Now we include the NAND operator / to complete the innermost partial expression. (Z \downarrow ([X\rceilY \odot XZ\rceil \odot / ] > [Y\rceilX\equiv])) Step 4. We evaluate the operator >, inserting it at the end of the string. (Z \downarrow [X\rceilY \odot XZ\rceil \odot / Y\rceilX \equiv >]) (Z \downarrow [X\rceilY \odot XZ\rceil \odot / Y\rceilX \equiv >]) Step 5. Finally, we add the NOR operator \downarrow to the end of the string. The final reverse Polish string is ZX\rceilY \odot XZ\rceil \odot / Y\rceilX\equiv> ZX\rceilY \odot XZ\rceil \odot / Y\rceilX\equiv> \downarrow b) This problem uses solely the NAND operator/and several parentheses. We will proceed as before, evaluating the innermost parentheses first. Step 1. The two inner expressions on the right side get translated and bracketed first. (X / Y) / ([XX / ] / [YY / ]) (X / Y) / ([XX / ] / [YY / ]) Step 2. Next we merge the two bracketed expressions together over the NAND operator. (X / Y) / [XX / YY / / ] (X / Y) / [XX / YY / / ] Step 3. Now we convert the right side. [XY / ] / [XX / YY / / ] [XY / ] / [XX / YY / / ] Step 4. Finally, the entire expression, translated into a reverse Polish string looks like this. XY / XX / YY / / /

Question:

A copper bar laid perpendicularly across a pair of parallel metal tracks running north-south, one meter apart, is slid in a northerly direction. At one end, the tracks are connected through a galvanom-eter. At the given location the vertical component of the earth's magnetic field is .55 oersted. Is there any reason to suppose that the galvanometer will read a deflection as the bar is moved at the rate of 150 cm/sec? If so, what is it, which way does the current point, how much electromotive force is developed in the bar between the rails, and which end of the bar is at the higher potential?

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/Users/wenhuchen/Documents/Crawler/Physics/D22-0751.htm

Solution:

Note that as the bar moves along the track, as indicated in the diagram, it cuts the vertical component of the earth's field, which points downward, or into the paper. Hence, an electromotive force must be generated between the points A and B, and a current must be established in the' circuit, to be detected by the galvano-meter. This results from Faraday's Law because the magnetic flux is changing. The direction of this current follows from Lenz' Law, which states that the induced current will flow in a direction such as to set up an induced flux opposite in direction to the change in flux. The current points from B to A in the bar. This means that the point A is at a higher potential than point B. To determine the magnitude of the electromotive force Induced, use Faraday's law E = - (d\varphi/dt) where the negative sign indicates that the EMF is Induced in such a direction as to oppose the change in flux that created it. The flux \varphi through the loop formed by the bar and the galvanometer is \varphi = \int B^\ding{217} \bullet dA^\ding{217} = BLx for B is constant in the loop and it is parallel to the area vector (i.e. , \texttheta = 0 o ) . Then E = - BL (dx/dt) Sincev = (dx/dt) E = - BLv where B here is 5.5 × 10^-5 weber/meter^2 , L = 1.0 m, and v = 1.5 (m/sec) E = - {5.5 × 10^-5 (w/m^2)} (1.0 m) {1.5 (m/sec)} = -8.25 × 10^-5 volt .

Question:

A water pipe in which water pressure is 80 lb/in.^2 springs a leak, the size of which is .01 sq in. in area. How much force is required of a patch to "stop" the leak?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0395.htm

Solution:

This problem is quickly recognized as an application of the definition of fluid pressure. Commencing with the definition of pressure as the ratio of the force f on an area a, and the area a, we find p = f/a It follows that:f = pa Substituting values: f = 80 lb/in^2 × .01 in^2 = .8 lb.

Question:

(ClO^-_3) + Fe^2+ \rightarrow (Cl^-) + Fe^3+

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/Users/wenhuchen/Documents/Crawler/Chemistry/E16-0567.htm

Solution:

Reactions in which electrons are transferred from one atom to another are known as oxidation-reduction re-actions or as redox reactions. To balance this type of re-action , you want to conserve charge and matter, i.e., one side of the equation must not have an excess of charge or matter. To perform this balancing, you need to (1)Balance charge by adding H^+ (in acid) or OH^- (in base) . (2) Balance oxygen by adding water. (3) Balance atoms (of hydrogen) by adding hydrogen to appropriate side. These three rules will balance the redox equation. These rules apply to balancing only the half-reactions. The overall reaction; the sum of these, will be balanced by their addition. Proceed as follows: Fe^2+ goes to Fe^3+. It lost an electron thus, it's the oxidation half reaction. To balance charge, add H^+. Thus, Fe^2+ + H^+ \rightarrow Fe^3+. To conserve mass, add an H atom to right. You obtain Fe^2+ + H^+ \rightarrow Fe^3+ + (1 / 2)H_2. The reduction must be ClO^-_3 \rightarrow Cl^-. The charges are already balanced. To balance the 3 oxygen atoms on left side, add 3 water molecules on right side. You obtain ClO^-_3 \rightarrow Cl-+ 3H_2O. To balance hydrogens, add 6 hydrogens (or 3H_2molecules) to left side. 3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O. In summary, you have oxid:Fe^2+ + H^+ \rightarrow Fe^3+ + (1 / 2)H_2 red:3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O. So that no free H's appear in the overall equation, multiply the oxidation reaction by six. You obtain 6Fe^2+ + 6H^+ \rightarrow 6Fe^3+ + 3H_2 Thus,oxid:6Fe^2+ + 6H^+ \rightarrow 6Fe^3+ + 3H_2 red:3H_2 + ClO^-_3 \rightarrow Cl^- + 3H_2O overall:6Fe^2+ + ClO^-_3 + 6H^+ \rightarrow 6Fe^3+ + Cl^- + 3H_2O Notice: The H atoms dropped out. This is the balanced equation for charge and mass are equal on both sides of the equation. H^+ comes from the acid; recall, it's in an acidic solution.

Question:

A chemist found the value of the specific rate constant for the decomposition of nitrous oxide, 2NO + N_2 + O_2, at two separate temperatures: k, = .14 liter/mole-sec at 970\textdegreeK, k_2 =3.7 liter/mole-sec at 1085\textdegreeK. (a) Calculate the activation energy, Ea, for the reaction, (b) Calculate the A factor in the Arrhenius equation, (c) Calculate the specific rate constant at 800\textdegreeK.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0472.htm

Solution:

This problem deals with the effect of tem-perature on the kinetics of a reaction. (a) E_a is the activation energy, the difference between the heat content of (a) E_a is the activation energy, the difference between the heat content of the active molecules and that of the inert molecules. To find the value of E_a for the decomposition of nitrous oxide, use the Arrhenius equation: Ea= [(-2.303) / {(1/T_2) (1/T_1)}] log (k_2 / k_1) where T = temperature in degrees Kelvin (Celsius plus 273\textdegree) , k = specificrate constant and R = universal gas constant. Substitute known values to obtain (T_1 = 970\textdegree, T 2 = 1085\textdegree) E_a = [{(-2.303)(1.987)(1/1000)} / {(1/1085) (1/970)}] log (3.7 / .14) = 59.2k cal/mole [Note: 1/1000 is a conversion factor of 1 k cal/1000 cal.] (b) The Arrhenius equation can also be written in the from l_nk=l_nA- (E_A / RT) , where A is called the frequency factor. To find A, first solve the equation in term of A . l_nA=l_nk+(E_A / RT) or A = log k + (E_a /2.303 RT). l_nA =l_nk+(E_A / RT) or A = log k + (E_a /2.303 RT). Substitute for k and T and calculate log A = log 3.7 + [{(59.2) (1000cal / 1 k cal) } / {(2.303)(1.987)(1085)}] = 0.57 + 11.9 = 12.5. Since antilog 12.5 = A, A = 3 × 10^12 liters/mole-sec. (Notice: k2and T2 Since antilog 12.5 = A, A = 3 × 10^12 liters/mole-sec. (Notice: k 2 and T 2 valueswere used in the calculations but k_1 and T_1 could have been used instead.) (c) To find k at 800 solve the Arrhenius equation for k log k = log A - (E_a / 2.303RT) . Substitute in known values for A, Ea, R, and T to obtain: log (3 × 10^12) - [{(59.2)(1000cal / 1k cal)} / {2.303 (1.987)( 800 )}] = 12.5 - 16.2 = - 3.7 k. = antilog (- 3.7) = 2 × 10^-4 liters / mole-sec.

Question:

The mean proper lifetime (T) of \pi ^+ mesons is 2.5 × 10^-8 s. In a beam of \pi^+ mesons of speed 0.99c, what is the average distance a meson travels before it decays? What would this value be if the relativtistic time dilation did not exist?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0945.htm

Solution:

The proper lifetime, T, of a particle is the lifetime of the particle as measured by an observer traveling with the particle. In a coordinate system S' moving with the mesons the average lifetime of the mesons is 2.5 × 10^-8 s. A laboratory observer makes his measurements in a systemS fixed in the laboratory, with respect to which S' is moving at a speed of 0.99 c. According to the theory of special relativity, an event occurring in space-time will have coordinates (x, y, z, t) re-lative to S, and coordinates(x', y', z', t') relative to S'. If the relative velocity between the 2 frames is v, these 2 sets of coordinates are related by the Lorentz Transformation x = (x' +vt') / \surd{1 - (v/c)^2}(1a) y = y'(1b) z = z'(1c) t = {t' + (vx' / c^2)} / \surd{1 - (v/c)^2}(1d) where c is the speed of light. Now, note that the equations in (1) relate the co-ordinates of one event as recorded in 2 different reference frames. We need 2 events to describe the life-time of the meson: its "birth" and its "death". If we travel with the meson, we record the following coordi-nates for these 2 events: Birth:(x_1', y_1', z_1' , t_1') = (0, 0, 0, 0) Death: (x_2' , y_2', z_2' , t_2') = (0, 0, 0, T) Here, x_2' = 0 because, if we travel with the meson, we do not see it move. It appears stationary. The coordi-nates of these 2 events in S are found from equations using the above data: x_1 = {(0) + v(0)} / \surd{1 - (v/c)^2} = 0 y_1 = 0 z_1 = 0 t1= {(0) + (v (0) / c^2)} / \surd{1 - (v/c)^2} = 0 and x_2 =vT/ \surd{1 - (v/c)^2} = 0 y_2 = 0 z_2 = 0 t_2= T / \surd{1 - (v/c)^2} = 0 Relative to S, the meson travels a distance x_2 - x1=vT/ \surd{1 - (v/c)^2} in a time t2- t1= T / \surd{1 - (v/c)^2} = 0 Hence, t2- t1is the lifetime of the particle relative to S. (Notice that t2- t1> T) Using the given data x_2 - x_1 = (.99c) (2.5 × 10-8s) / \surd{1 - (.99)^2} x_2 - x_1 = {(2.5 × 10-8s) (.99c) (3 × 10^8 m/s)} / .141 x_2 - x1= 52.7 m. If time dilation did not exist, the distance traveled would be d_0, where d_0 = 2.5 × 10-8s × 0.99c × 3 × 10^8 m/s^-1 = 7.43 m.

Question:

A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar which moves the end up and down through a distance of 0.50cm. The motion is continuous and is repeated regularly 120 times per second. a) If the string has a linear density of 0.25(kg / m) and is kept under tension of 90 N, find the speed, amplitude, fre-quency, and wavelength of the wave motion. b) Assuming the wave moves in the +x-direction and that, at t = 0, the end of the string described by x = 0 is in its equilibrium position y = 0, write the equation of the wave. c) As this wave passes along the string, each particle of the string moves up and down at right angles to the direc-tion of the wave motion. Find the velocity and acceleration of a particle 2.0 ft. from the end.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1091.htm

Solution:

(a) The end moves 0.25cm away from the equilibrium position, first above it, then below it; therefore, the ampli-tudey_mis 0.25cm. The entire motion is repeated 120 times each second so that the frequency is 120 vibrations per second, or 120 Hz. The wave speed is given by v = \surd(F / \mu). But F = 90 N and \mu = 0.25(kg / m), so that v = \surd[(90N) / (0.25 kg/m)] = 19(m/s). The wavelength is given by \lambda = v/\lambda, so that The wavelength is given by \lambda = v/\lambda, so that \lambda = [(19 m/s) / (120vib/s)] = 16cm. (b) The general expression for a transverse sinusoidal wave moving in the +x-direction is y =y_msin(kx-\cyrchar\cyromegat- \varphi). ). Requiring that y = 0 for the conditions x = 0 and t = 0yeilds 0 =y_msin(- \varphi) ) which means that the phase constant \varphimay be taken to be zero. Hence may be taken to be zero. Hence for this wave y=y_msin(kx-\cyrchar\cyromegat), and with the values just found, y_m= 0.25cm, \lambda = 16cmork = (2\pi / \lambda) = (2\pi / 16cm) = 0.39cm^-1, v = 19 m/s = 1900 cm/s or \cyrchar\cyromega =vk= (1900 cm/s) (0.39cm^-1) = 740s^-1 = 740Hz, the equation for the wave is y = 0.25 sin(0.39x - 740t) where x and y are in centimeters and t is in seconds. c) The general form of this wave is y =y_msin(kx-\cyrchar\cyromegat) =y_msink (x -vt). The v in this equation is the constant horizontal velocity of thewavetrain. It is not the velocity of a particle in the string through which this wave moves; this particle velocity is neither horizontal nor constant. Each particle moves vertically, that is, in the y-direction. In order to deter-mine the particle velocity designated by the symbol u, con-centrate on a particle at a particular position x-that is, x is now a constant in this equation-and see how the particle displacement y changes with time. With x constant u = (\partialy / \partialt) = -y_m\cyrchar\cyromegacos(kx-\cyrchar\cyromegat), in which the partial derivative (\partialy / \partialt) shows that although in general y is a function of both x and t, assume that x remains constant so that t becomes the only variable. The accelera-tion a of the particle at this (constant) value of x is a = (\partial^2y / \partialt^2) = (\partialu / \partialt) = -y_m\cyrchar\cyromega^2 sin(kx-\cyrchar\cyromegat) = -\cyrchar\cyromega^2y. For a particle at x = 62cm with y_m= 0.25cm, k = 0.39cm^-1, \cyrchar\cyromega = 740 s^-1, u = -y_mvcos(kx-\cyrchar\cyromegat) oru = -0.25(740)cos[(0.39) (62) - (740)t] = -185cos(24 - 740t) anda = -\cyrchar\cyromega^2y ora = -(740)^20.25 sin[(0.39) (62)-(740)t] = -13.7 ×10^4sin(24 - 740t) where t is expressed in seconds u in (cm/s) and a in cm/s^2 .

Question:

How does thetypedefstatement increase portability?

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Solution:

In C new data types can be created usingtypedefstatement. Once a type is defined, it can be used in the same manner as the standard C types. The main advantage of usingtypedefstatement is for porta-bility of code, i.e. code which can be run without change on a variety of hardware. Some machines define the typeintto contain 16 bits while other machines provide 32 bits. If we create a type called integer based on the word-size of the machine, then we can de-fine all our integer variables in terms of this type. This is done asfollows: typedeflongintinteger; main ( ) { integer x; integer y; /\textasteriskcenteredstatements\textasteriskcentered/ } Two things are to be noted: 1)typedefstatement is defined outside the main. 2) Qualifier long tells the compiler to use two words (32 bits) for storage. Now if we want the same program to run on 16-bit machine instead of modifying variables x, y, and z, all we have to do is modify thetypedef i.e. typedeflongintinteger; changeto typedefintinteger; All the following integer type statements in the program will be changed automatically.

Question:

What are the functions of (a) the vascular cambium, (b)stomata, (c) heartwood, (d) lenticels (e) abscission layer and (f)cutin?

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Solution:

The vascular cambium is ameristemicregion found in the stem. Sometimes it is found in the roots of higher plants. It exists as a continuous ring of cells extending between the cluster of phloem cells and xylem cells. The cambium produces these two types of cells during periods of active mitotic cell division. The ones facing the center of the stem differentiate into xylem cells; and the ones facing the periphery become phloem cells. The stomata are small specialized pores scattered over the epidermal surfaces of leaves and some stems. These pores are found mainly on the lower surface of the leaves. The stomata are delineated by a pair of cells called the guard cells. The guard cells, by changing their shape (due to changes in theirturgorpressure), regulate the size of the stomatal aperture. An open stoma permits the escape of water and the exchange of gases. The heartwood is the inner layer of the xylem in a woody plant. They are composed of dead, thick-walled xylem cells which have lost the ability to conduct water and minerals. Instead it has become tough supporting fibers. The heartwood increases the strength of the stem, and accommodates the increasing load of foliage as the tree grows. Lenticels are masses of cells which rupture the epidermis and form swellings . These masses of cells are formed from cells of the cork cambium which divide repeatedly and rupture the epidermis lying above. Lenticels represent a continuation of the inner plant tissues with the external environment, and permits a direct diffusion of gases into and out of the stem or twig. Such direct passages are necessary because the cork cambium forms a complete sheath around the vascular bundles and effecti-vely obstructs the ventilation of the vascular bundles. Anabcissionlayer is a sheet of thin-walled cells, loosely joined together , which extends across the base of the petiole. The formation of the abscission layer separates and loosens the point of attachment of the petiole to the stem. After the abscission layer is formed, the petiole is held on only by the epidermis and the fragile vascular bundles, so that wind or other mechanical disturbances will cause the leaf to fall off. This is the mechanism that accounts for the fall of leaves during autumn. Cutin is a waxy organic substance secreted by the epidermal cells of the stems and leaves, but not the roots. Its waterproof property retards evaporative water loss to the atmosphere.

Question:

Two thirds of the atoms in a molecule of water (H_2O) are hydrogen. What percentage of the weight of a water molecule is the weight of the two hydrogen atoms? The atomic weight of hydrogen is 1.008 g/mole and of oxygen is 16.00 g/mole.

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Solution:

The most direct way to solve this composition problem is to consider the total weight of one molecule of water: mass H_2O = 2 mass_H + mass_O. = 2(1.008 g/mole) + 16.00 g/mole = 18.016 g/mole. The mass of two hydrogen atoms is 2 × 1.008 = 2.016 g/mole. Hence, the percentage mass of hydrogen in water is: % mass = (2 mass_H-) / (mass H_2O) ×100% = [(2.016 g/mole) / (18.016 g/mole)] × 100% = 11.19

Question:

What are the lower vascular plants? Why are they considered more advanced than the bryophytes but less advanced than the seed plants?

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Solution:

The vascular plants (phylumTracheophyta) are subdivided into five groups . These are thePsilophyta(fork ferns),Lycophyta(club mosses, quill worts)Sphenophyta(horsetails or the "scouring rushes"),Pterophyte (ferns) andSpermophyta(the seed plants gymnosperms, and angiosperms ). Vascular plants, in contrast to the algae, fungi, and bryophytes , posses a vascular system that serves for support and for the conduction of water, mineral salts, and foods. The lower vascular plants refer to all the non-seed bearing classes, namely all classes except the gymnosperms and angiosperms. Most of the lower vascular plants are land plants. As mentioned before , they contain xylem and phloem, a system missing in the bryophytes . They have also evolved a more complicated anatomy than the bryophytes - more efficient roots, more elaborate stem and leaves, more complicated embryonic structure and a larger body size. Thesporophyte of the lower vascular plants has become the independent, dominant generation , and corres-pondingly, the gametophyte has become reduced. These characteristics of the lower vascular plants place them higher on the evolutionary ladder than the bryophytes. Structurally,most lowervascular plants resemble the seed plants far more than the bryophytes in their roots, stems and leaves. However, they are not that much better adapted for a life on land than the bryophytes . Because their sperm retain thebiflagellatedstructure necessitating a film of moisture to be active, and because the young sporophyte develops directly from the zygote without passing through any stage where it is protected by a seed, these plants are restricted to moderately moist habitats for active growth and re-production. Most of the seed plants, on the other hand, have evolvednonflagellatedsperm, a mechanism of pol-lination for sperm transport andgameticunion, and a seed structure within which the embryonicsporophyteis protected and nourished . These three features make the seed plants extremely successful inhabitants of the land. In addition, while the lower vascular plants have a reduced gametophyte, the seed plants have reduced it even further in size and have simplified its structure to the point where it is completely dependent on thesporophyte. The seed plants have also evolved heterospory, that is, the production of two types of spores. The lower vascular plants arehomosporous, and produce only one type of spore . The seed plants are therefore most advanced in the evolutionary trend toward greater embryonic protection, flexible and efficient ways of fertilization , the production of two types of spores, and the reduction of the haploid generation. While the lower vascular plants, then, are more advanced than the bryophytes, they trail the seed plants in evolutionary development .

Question:

(a) 2^3 \textbullet 2^2(b) a^3 \bullet a^5(c) x^6 \textbullet x^4

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Solution:

If a is any number and n is any positive integer, the product of the n factors a\bulleta\bulleta ... a is denoted by a^n \bullet a is called the base and n is called the exponent. Also, for base a and exponents m and n, m and n being positive integers, we have the law: a^m \bullet a^n = a^m+n. Therefore, (a) 2^3 \bullet 2^2= (2 \bullet 2 \bullet 2) (2 \bullet 2) = 8 \bullet 4 = 32 or2^3 \bullet 2^2= 2^3+2 = 2^5 = 32 (b) a^3 \bullet a^5= (a \bullet a \bullet a) (a \bullet a \bullet a \bullet a \bullet a) = (a \bullet a \bullet a \bullet a \bullet a \bullet a \bullet a \bullet a) = a^8 ora^3 \bullet a^5= a^3+5 = a^8 (c) a^6 \bullet x^4 = x^6+4 = x^10.

Question:

A mixture of gaseous oxygen and nitrogen is stored at atmospheric pressure in a 3.7 l iron container maintained at constant temperature. After all the oxygen has re-acted with the iron walls of the container to form solid iron oxide of negligible volume, the pressure is measured at 450 torr. Determine the final volume of nitrogen and the initial and final partial pressures of nitrogen and oxygen.

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Solution:

The partial pressure of each component, which is independent of any other component in a gaseous mixture, at a defined volume, is equal to the pressure each com-ponent would exert if it were the only gas in that volume. The total pressure of a gaseous mixture is the sum of the partial pressures of the components. These two definitions are sufficient to solve this problem. Once all the oxygen has reacted, no oxygen is present in the gaseous phase, so that the final partial pressure of oxygen is zero. By the second definition, final total pressure = final partial pressure of N_2 + final partial pressure of O_2 450 torr = final partial pressure of N_2 + O torr or,final partial pressure of N_2 = 450 torr. But since, after all the oxygen has reacted, only nitrogen fills the entire volume, the final volume of nitrogen is 3.7 l, By the first definition, the initial partial pressure of nitrogen is the same as the final partial pressure of nitrogen, 450 torr. We now again employ the second definition to determine the last re-maining unknown quantity, the initial partial pressure of oxygen. Proceeding, we obtain initial total pressure = initial partial pressure of O_2 + initial partial pressure of N_2 1 atm = initial partial pressure of O_2 + 450 torr or,initial partial pressure of O_2 = 1 atm - 450 torr = 760 torr - 450 torr = 310 torr. Summarizing these results in tabular form we have: InitialFinal partial pressure of O_2310 torr0 torr partial pressure of N_2450 torr450 torr total pressure760 torr450 torr

Question:

A deuterium atom moving with kinetic energy of 0.81× 10^-13 J collides with a similar atom at rest. A nuclear inelastic reaction takes place and a neutron is observed to be emitted at right angles to the original direction of motion. Determine its kinetic energy, given that the other product of the reaction is an atom of the light isotope of helium and that the rest masses of a neutron, a deuterium atom, and a light helium atom are 1.6747, 3.3441, and 5.0076, respectively, in units of 10^-27 kg.

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Solution:

Before doing this problem, we must see if it is necessary to use relativistic considerations. The total relativistic energy of a particle is the sum of its rest mass energy and its relativistic kinetic energy, or E = m_0c^2 + T where m_0 is the particle's rest mass. Hence, T = E - m_0c^2 ButE = m_0c^2 / \surd{1 - (v2/ c^2)} where v is the particle's velocity. Therefore, T = m_0c^2 [ \surd{1 / (v2/ c^2)} - 1] Solving for v/c, T / m_0c^2 + 1 =1 / \surd{1 - (v2/ c^2)} 1 -(v2/ c^2) = [m_0c2/ (T + m_0c^2)]2 v/c =\surd {1 - [m_0c2/ (T + m_0c^2)]2} For the incident deuterium atom, v/c = \surd {1 - [(3.3441 × 10^-27 kg × 9 × 1016m^2 / s2)/ (.81 × 10^-13J +(3.3441 × 10^-27 kg × 9 × 1016m^2 / s^2 ))]2} or ,v/c =5.382 × 10^-4 Since this is very small, the problem is extremely non- relativistic. We now proceed with a classical analysis of the collision. Looking at the figure, we may write the following 2 equations from the principle of conservation of momentum. m_Dv = m_Hv_2 cos \texttheta(1) m_nv_1 = m_Hv_2 sin \texttheta(2) Since the collision is inelastic, we must account for the fact that energy, Q, is gained or lost in the reaction. Hence, the principle of conservation of energy yields. 1/2 (m_Dv^2) + Q = 1/2 (m_n v^2_1) + 1/2 (m_H v^2_2)(3) If Q > 0, energy is released in the reaction, and is absorbed as kinetic energy by the reaction products. Now we solve for v_1 in terms of Q. First, square (1) and (2), then add [m^2_Dv^2= m_H^2v^2_2 cos^2 \texttheta] + [m^2_nv^2_1= m_H^2v^2_2 sin^2 \texttheta] m^2_Dv^2 + m^2_nv^2_1 = m_H^2v^2_2 orm_Hv^2_2 = ( m^2_Dv2+ m^2_nv^2_1) / m_H(4) From (3)m_Hv^2_2 = 2Q +m_Dv2- m_nv^21(5) Combining (4) and (5) 2Q +m_Dv2- m_nv^2_1 = m_Hv^2_2 = ( m^2_Dv^2 + m^2_nv^2_1) / m_H or2QmH+ m_Dm_Hv^2 - m_nm_Hv^2_1 = m_D^2v^2 - m^2_nv^2_1 (m^2 n + m_nm_H ) v^2_1 = 2Qm_H + (m_DmH- m^2_D) v^2 Finally ,v^2_1 = {2Qm_H + m_D(mH- m_D) v^2} / {m_n(mn+ m_H)} and(1/2) m_n v^2_1 = {Qm_H + (1/2) m_D v^2 (mH- m_D) } / (mn+ m_H)(6) This is the neutron kinetic energy. Now, note that in (6), we have an unknown variable, namely, Q. But, we can obtain Q by noting that the only mechanism which can be responsible for Q is a difference in mass between the products and reactants of the collision. This follows from Einstein's mass-energy relation, since a mass difference \Deltam is equivalent to an energy E = (\Deltam)c^2(7) Now, the mass entering the reaction is the sum of the masses of the incident and target deuterons, or 2m_D(8) The sum of the masses of the products is m_n + m_H Hence,\Deltam = m_n + m_H - 2m_D \Deltam = [1.6747 + 5.0076 - 2(3.3441)] × 10^-27 kg \Deltam = 59 ×10^-31 kg Now,Q = E = (59 × 10^-31kg) (9 × 10^16 m^2/s^2) Q = 5.31 × 10^-13 J(9) Inserting (9) and the given data in (6), we obtain (1/2) m_nv^2_1 = [(5.31 × 10^-13 J) (5.0076 × 10^-27kg) + (.81 × 10^-11 J) (5.0076 - 3.3441) × 10^-27kg] / [(1.6747 + 5.0076) × 10^-27kg] = [(26.5904 × 10^-40 + 1.3474 ×10-40) / (6.6823 × 10^-27 )] J = 4.1809 × 10^-13J

Question:

A toroidal coil has 3000 turns. The inner and outer diameters are 22 cm and 26 cm, respectively. Calculate the flux density inside the coil when there is a current of 5.0 amp.

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0705.htm

Solution:

A toroid is a wire wound in a helix and bent into the shape of a doughnut with a current i running through it. The magnetic field B^\ding{217} forms concentric circles inside the toroid. Let r be the mean radius of the toroid. Apply Ampere's law around the circular path of radius r, \oint B^\ding{217} \bullet dl^\ding{217} = \mu_0I where the symbol \oint indicates that the integral is taken over a closed path and I is the total current enclosed by the path of integration. We get (B) (2\pir) = \mu_0 N i Therefore,B = [(\mu_0 N i)/(2\pir)] For the toroid given, mean diameter = [(22 cm + cm)/2] = 24 cm = 0.24 m mean radius = r = (0.24 m/2) = 0.12 m The magnetic field is B = [{( 4 \pi × 10^-7 weber/arap-m) (3000) (5.0 amp)}/{(2\pi) (0.12 m)}] = 0.025 weber/m^2 Note that the magnetic field is not constant over the cross section of The toroid, but is inversely proportion-al to the radius. The magnetic field outside the toroid is zero since around any closed path encircling the toroid the net current enclosed is zero, due to equal amounts of current travelling in opposite directions.

Question:

What type of flip-flop does the diode-transistor circuit of fig. 1 represent?

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Solution:

The circuit of fig. 1 is broken down into two identical "sub-circuits" in fig. 2(a) and (b). In sub--circuit (a) if either or both of the input diodes are con-nected to OV (200), current flows through R2. This makes V_be = OV and the transistor does not conduct, leaving its output Y_1 high (5V). If both diodes are connected to 25V (high), no current flows through R2. This makes V_be = 5V and the transistor conducts, pulling V_1 low (OV). Thus, each sub-circuit is a NAND gate and the output of each gate is connected to the input of the opposite gate. The logic diagram of fig. 1 is shown in fig. 3. The truth table for fig. 3 is shown in Fig. 4. InputsPresent StateNext State X_1X_2 Y_1Y_2 Y_1Y_2 00 01 not used 00 10 not used 01 01 10 01 10 10 10 01 01 10 10 01 11 01 01 11 10 10 Fig.4 This truth table is similar to that of an S-R flip-flop except in this case X_1 =S; X_2 =R. Hence, this is anS-Rflip-flop.

Question:

Define antibiotics, where are they derived from? What is meant by broad-spectrum antibiotics?

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Solution:

Compounds that can inhibit and/or destroy bacteria may be defined as antibiotics. These substances may be bacteriostatic (growth-inhibiting) or bactericidal (bacteria killing) in their action. Currently, antibiotics are derived from microorganisms and are, in effect, metabolic products of these organisms. Common classes include the penicillins and the tetracyclines. Tetracyclines have been isolated from "streptomyces" bacteria strains and all have a basic structure of four fused rings. These compounds are also called broad- spectrum antibiotics because of their effectiveness against a wide variety of microorganisms. Aureomycin, the first member of the tetracycline family, has its structure written above:

Question:

A chemist dissolves BaSO_4 , in pure water at 25\textdegree C. If its K_sp= 1 × 10-^10 , what is the solubility of the barium sulfate in the water?

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Solution:

The solubility of a compound is defined as the limiting concentration of the compound in a solution before precipitation occurs. To find the solubility of the barium sulfate, you need to know the concentration of its ions in solution. BaSO_4 will dissociate into ions because it is a salt. There will be an equilibrium between these ions and the BaSO_4 . The equilibrium can be measured in terms of a constant,K_sp, called the solubility constant. TheK_spis ex-pressed in terms of the concentrations of the ions. As such, to answer the question, you want to represent thisK_sp. For this reaction, the equation is BaSO_4\rightleftarrowsBa++ + SO-_4 K_sp= [Ba^2+ ] [ SO4^2-] = 1 × 10-^10 . Let x = [Ba^2+ ] . Thus, x = [SO4^2-], also, since both ions will be formed in equimolaramounts. Therefore, x \bullet x = 1 × 10-^10 . Solving, x = 1 × 10-^5 M = [Ba^2+ ] = [SO_4 ^2- ] .

Question:

A steel rod of length L = 20 cm and cross-sectional area A = 3 cm^2 is heated at one end to T_1 = 300\textdegreek while the other end rests in ice. Assuming that heat transmission occurs exclusively through the rod (without losses from the walls), calculate the mass m of the ice melting in time ∆t = 10 min. The thermal conductivity of steel is k = 0.16 cal deg^-1 sec^-1cm^-1.

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Solution:

Thermal conductivity k as given above, indicates the amount of heat transferred per second per square centi-meter per degree centigrade through 1 cm length of steel. The rate of heat transfer through the steel rod is given by heat/sec = ∆Q/∆t = k × [A × (T_1 - T_2)/L] where T_2 is the temperature of the colder end of the rod. Hence, ∆Q/∆t = [(0.16 cal K\textdegree^-1 sec^-1 cm^-1) × (3 cm^2) × (300 - 273)K\textdegree] / [20 cm] = 0.648 cal/sec In time ∆t = 10 min., the amount of heat transferred by the rod to the ice is ∆Q= (0.648 cal/sec) × ∆t = (0.648 cal/sec) × (10min×60sec/min) = 388.8 cal. The heat of fusion of ice is 80 cal/gr, which means in order to melt 1 gr. of ice, 80 cal. must be added to the ice. Therefore, 388.8 calories will melt m = (∆Q cal)/(80 cal/gr) = (388.8)/(80 gr) = 4.86 gr. of ice.

Question:

How much energy is emitted by Avogadro number of atoms, if they each emit a light wave of 400 nm wavelength?

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Solution:

Energy is related to wavelength in the following equation, E =hc/\lambda, where E is energy, h Planck's constant (6.626 × 10^-34 Js) , c the speed of light (3.0 × 10^8 m/sec) and \lambda the wavelength. Using this equation, one calculates the energy emitted by each atom. To find the total amount of energy produced byAvogrado'snumber of atoms, multiply the amount of energy one atom emits by 6.02 × 10^23. Solving for E:1 nm = 10^-9 m. E = (hc/\lambda) = [(6.626 × 10^-34 Js) (3.0 × 10^8 m/s)] / (400 × 10^-9 m) = 4.97 × 10^-19 J/atom Thus, the total amount of energy emitted is E = (4.97 × 10^-19 J/atom) (6.02 × 10^23 atoms/mole) = 2.98 × 10^5 J.

Question:

A rod of negligible mass with length 2 meters has a small 2kg mass mounted on each end. (a) Calculate the moment of inertia of this rod about an axis perpen-dicular to the rod and through its center. (b) Find its angular momentum if the rod rotates about this axis with an angular velocity of 10 radians per second.

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Solution:

(a) Rotational inertia I of a system is defined as the sum of the products of the masses m of the particles in the system and the squares of their respective distances r from the rotational axis. Then I = I = \sum_i m_i r^2 _i For the system shown in the figure, we have I = (2kg)(1m)^2 + (2kg)(1m)^2 = 4 kg-m^2. (b) The angular momentum is defined as the product of the rotational inertia and the angular velocity \cyrchar\cyromega. Therefore the angular momentum L in this problem is L = Iw = (4 kg-m^2)(10 rad/sec) = 40 kg-m^2/sec . Note that angular momentum is analogous to linear momentum which equals the product of inertial mass m and linear velocity v.

Question:

It was found that a magnesium oxide contained .833 g of oxygen and 1.266 g of magnesium. Calculate the gram- equivalent weight of magnesium.

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Solution:

You begin this problem by establishing what is meant by the term gram-equivalent weight. Gram-equivalent weight may be defined as the number of grams of an element that will involve a gain or loss of N electrons, i.e., the Avogadro's number (6.02 × 10^23) of electrons, when the element enters into chemical combination with another element. In this problem, you know that oxygen is present. Oxygen has an oxidation state of - 2. In a reaction, one mole of oxygen will gain two moles of electrons. The molecular weight of oxygen is 16 g/mole. Its equivalent weight becomes 16/2 = 8.00 g, since 2 moles of electrons will be gained and, by definition, equivalent weight is the amount of a substance that will gain or lose one mole of electrons. The gram-equivalent weight of magnesium is that amount of the element that combines with 8.00 g of oxygen. You are told that 1.266 g Mg combines with .833 g O_2. If you let x = grams of Mg that will combine with 8.00 of oxygen, you can set up the following proportion: (1.266 g/Mg) / (.833 g O_2) = (x g Mg) / (8.00 g O_2) Solving for x, you obtain x = 12.16 g of Mg, which is its gram-equivalent weight.

Question:

If the atomic weight of oxygen was 50, what would its equivalentweight be?

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Solution:

The relationship of the atomic weight and the equivalent weight is (Atomic weight)/ (Equivalent weight) = valence number The valence number is a measure of the number of atoms of hydrogenthat will combine with one atom of the element.Two hydrogen atomscombine with one oxygen. The valence number of oxygen is therefore2. One can now solve for the equivalent weight of oxygen when theatomic weight is taken as 50. 50 / (equivalent weight) = 2 equivalent weight = 50/2 = 25.

Question:

Draw a labelled phase diagram for a substance Z which has the following properties? normal boiling point = 220\textdegreeC, normal freezing point 80\textdegreeC, and triple point 60\textdegreeC and .20 atm. Predict the freezing and boiling, if the pressure were .80 atm?

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Solution:

To draw this diagram you want to understand all the terms involved. The relation between solid, liquid, and gaseous states as a function of the given temperature and pressure can be summarized on a graph known as a phase diagram. From the given experimental observations, you can draw the diagram. The lines which separate the states in a diagram represent an equilibrium between the phases. The intersection of the three lines is called the triple point, where all three phases are in equilibrium with each other. By normal boiling and melting points you mean those readings taken at 1 atm. Thus, the phase diagram can be written as shown in the accompanying figure. From the diagram you see that if the atm was .80, the b.p. and f.p. would drop, respectively, to 2150C and85\textdegreeC.

Question:

Starting from rest at the top, a small sphere rolls without slipping off a large fixed sphere. At what point will the small sphere leave the surface of the big sphere? (See figure).

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0207.htm

Solution:

The small sphere will leave the surface of the large sphere when the normal force (N^\ding{217}) of the latter on the former is zero, for this means that contact has ceased. Applying Newton's Second Law to the tangential and radial components of motion of the small sphere (see figure), we obtain F_radtal = Ma_radial = Mg cos \varphi - N F_tangential = Ma_tangential = mg sin \varphi where a_radial is positive in the direction of BA^\ding{217}, and a_tangential is positive in the direction of motion of the sphere. Furthermore, a_radial = v^2/(R + r), since the small sphere is traveling along a circular arc. (Here, v is the latter's speed.) Then (Mv^2)/(R + r) = Mg cos \varphi - N orN = Mg cos \varphi - [(Mv^2)/(R + r)] We require N = 0, or Mg cos \varphi = [(Mv^2)/(R + r)] Thereforecos \varphi = v^2/[(R + r) g](1) We are not yet finished, since we don't know v in (1). In order to find it, we may use the principle of conservation of energy to relate the energy of the small sphere at points O and S. Since the sphere starts from rest at O, it has only potential energy. Measuring poten-tial energy from A, the energy at O is E = Mg(R + r)(2) At S, the sphere has potential and kinetic energy equal in amount to the energy at O. Hence E = (1/2) Mv^2 + (1/2) Iw^2 + Mg(R + r) cos \varphi(3) where the first and second terms are the translational and rotational kinetic energies of the sphere respectively. Equating (2) and (3) (1/2) Mv^2 + (1/2) Iw^2 + Mg(R + r) cos \varphi = Mg(R + r)(4) Since the small sphere rolls without slipping v = wr, whence, using (4) (1/2) Mv^2 + (1/2) I(v^2/r^2) + Mg(R + r)(1 - cos \varphi) The moment of inertia of a sphere about its center is I = (2/5) Mr^2 Then(1/2) Mv^2 + (1/2)(2/5)(Mr^2)(v^2/r^2) + Mg(R + r)(1 - cos \varphi) (1/2) Mv^2 + (1/5) Mv^2 = Mg(R + r)(1 - cos \varphi) (7/10) v^2 = g(R + r)(1 - cos \varphi) v^2 = (10/7) g(R + r)(1 - cos \varphi)(5) Utilizing (5) in (1), cos \varphi = [10g (R + r)(1 - cos \varphi)]/[7 (R + r) g] cos \varphi = (10/7) (1 - cos \varphi) Solving for \varphi (10/7) cos \varphi + cos \varphi = 10/7 cos \varphi (17/7) = (10/7) cos \varphi = (10/17) and\varphi = cos^-1 (10/17) \approx 54\textdegree

Question:

When 10.0 g of silicon dust, Si, is exploded with 100.0 g of oxygen, O_2, forming silicon dioxide, SiO_2, how many grams of O_2 remain uncombined? The reaction equation is Si + O_2 \rightarrow SiO_2 .

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/Users/wenhuchen/Documents/Crawler/Chemistry/E05-0192.htm

Solution:

From the equation, it can be seen that Si and react in a 1 : 1 ratio. This means that 1 mole of Si will react with 1 mole of O_2. To determine the amount of left unreached after the reaction is performed, calculate the number of moles of O_2 and Si present and then subtract the number of moles of Si from the number of moles of O_2. (MW of Si = 28, MW of O_2 = 32.) number of moles = [(number of grams) / (MW)] number of moles of O_2 = (100.0 g) / (32 g/mole) = 3.12 moles number of moles of Si = [(10.0 g) / (28 g/mole) = 0.357 moles number of moles of excess O_2 = 3.12 - 0.357 = 2.763 moles weight of excess O_2 = number of moles × MW = 2.763 moles × 32 g/mole = 88.5 g.

Question:

What are the advantages and disadvantages of an exoskeleton? How have some of the disadvantages been overcome by themolluscsand by the arthropods?

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0304.htm

Solution:

An exoskeleton serves several functions: (1) it provides support for the soft body to counter-balance the force of gravity; (2) it serves as a point of attachment for muscles; (3) it protects the body against desiccation; and (4) it serves to protect the animal from predators. The disadvantages of an exo-skeleton are that movement is somewhat restricted and that growth within the exoskeleton is limited. Different species ofmolluscshave evolved different mechanisms for overcoming these limitations. Mostmolluscslead very sluggish lives; many are sedentary or burrow in sand or wood. The fast moving squids have a greatly reduced exoskeleton enclosed within the mantle. Octopuses, also fast moving, have lost the exoskeleton entirely. Movement of bivalves" is made possible by muscles attached to two hinged shells. Growth of mostmolluscsproceeds by additions to the shell, creating a larger exoskeleton. In gastropods, growth proceeds spirally, resulting in a spiral shell. Nautilus, a cephalopod, builds a new chamber as it grows, and lives within the latest and largest chamber of the shell. The animal secretes a gas into the other chambers, and is thus able to float. The arthropods have a segmented body. The rigid exoskeleton is thinner in certain regions, such as the leg joints and between body segments. Muscle attachments at these points offer arthropods a fairly wide range of movement. Most arthropods walk, jump, or crawl. Some insects are capable of flying and many aquatic arthropods swim actively. Growth in arthropods occurs by a series of molts. When the growing animal reaches a certain size, the exoskeleton is shed and a new, larger exo-skeleton is formed. During the process, the arthropod is left "naked", and is thus temporarily vulnerable.

Question:

A braking device in a rifle is designed to reduce recoil. It consists of a piston attached to the barrel. The piston moves in a cylinder filled with oil. When the shot is fired the barrel and the piston recoil at an initial velocity V_0 . The oil under pressure passes through tiny holes in the piston causing the piston and the barrel to decelerate at a rate proportional to their velocity a = \rule{1em}{1pt}kv. Express (a) v in terms of t, (b) x in terms of t, (c) v in terms of x. Draw the corresponding motion curves.

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/Users/wenhuchen/Documents/Crawler/Physics/D37-1063.htm

Solution:

(a) The basic definition of acceleration is a = dv / dt .(a) Substituting the expression given for the accelera-tion yields \rule{1em}{1pt}kv = dv / dt .(b) Rearranging to separate variables yields \rule{1em}{1pt}k dt = dv / v .(c) Now it is possible to integrate equation (c) to get \rule{1em}{1pt}k^t\int_0dt = \rule{1em}{1pt}k^t\int_0dt = ^v\int_(v)0 (dv / v) \rule{1em}{1pt}kt = In (v / v_0) .(d) If antilogs of equation (d) are taken, we arrive at the desired form v = v0e^\rule{1em}{1pt}kt(e) This curve is shown in Figure 2. (b) Again, starting with a basic definition, this time for v, we have v = dx / dt(f) Substitution of the expression (e) into (f) yields v_0 e^\rule{1em}{1pt}kt = dx / dt, v_0 e^\rule{1em}{1pt}ktdt = dx.(g) Integrating (g) v_0^t\int_0 e^\rule{1em}{1pt}ktdt =^x\int_0 dx, \rule{1em}{1pt}v_0 / ke^\rule{1em}{1pt}kt│t_0 = x, \rule{1em}{1pt}v_0 / k (e^\rule{1em}{1pt}kt \rule{1em}{1pt} e^0) = x, x = \rule{1em}{1pt}v_0 / k (1 \rule{1em}{1pt} e^\rule{1em}{1pt}kt).(h) This curve is shown in Figure 3. (c) Part c may be done in two ways: Using the chain rule for differentials, equation (a) may be rewritten a =dv / dt = (dx / dt)( dv / dx) = v(dv / dx).(i) Substituting a = \rule{1em}{1pt}kv in equation (i), we get \rule{1em}{1pt}kv = v(dv / dx) ordv / dx = \rule{1em}{1pt}k, dv = \rule{1em}{1pt} k dx. Integrating to solve yields v \rule{1em}{1pt} v_0 = \rule{1em}{1pt} kx. orv = v_0 \rule{1em}{1pt} kx. As an alternate method, equations (e) and (h) can be combined. From (e) we know e^\rule{1em}{1pt}kt = v / v_0 . Substituting for the exponential in (h) yields x = v_0 / k [(1 \rule{1em}{1pt} (v / v_0)] orkx = v_0 \rule{1em}{1pt} v orv = v_0 \rule{1em}{1pt} kx. This curve is illustrated in Figure 4.

Question:

If blood is carefully removed from a vessel and placed on a smooth plastic surface, will it clot? Explain why or why not.

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0347.htm

Solution:

Vertebrates have developed a mechanism for preventing the accidental loss of blood. Whenever a blood vessel is ruptured, one of the soluble plasma proteins, fibrinogen, is enzymatically converted into an insoluble protein, fibrin, which forms a semisolid clot. Many people think that blood clots when it becomes exposed to the air or when it stops flowing. However, if one were to carefully remove blood from a vessel- without allowing it to contact the damaged part of the vessel, and then place this blood on a smooth plastic dish or one lined with paraffin, it would not clot. However, if this blood were allowed to touch any damaged tissues or were placed on glass or some other relatively rough surface, the blood would clot. Either the damaged tissues or the blood itself must release some chemical which initiates the clotting mechanism. Actually, it is both damaged tissues and disintegrated platelets in the plasma that release substances responsible for the clotting reaction. Platelets are very small disc-shaped bodies found in mammalian blood and formed in the bone marrow from large cells called megakaryocytes. Platelets seem to disintegrate more readily upon contacting glass surfaces than on plastic surfaces. Platelets are also called thrombocytes. When a blood vessel is cut, the damaged tissues release a lipoprotein, called thromboplastin, which initiates the clotting mechanism. Calcium ions and certain protein factors in the plasma must be present in order for thrombo-plastin to be effective. Thromboplastin interacts with Ca^+2 and these proteins to produce prothrombinase, the enzyme that catalyzes the second step in the clotting mechanism. Prothrombinase can also be made by the interaction of a substance released from the disintegrated platelets (platelet factor # 3) and other factors in the plasma, including Ca^2+ and proteins. The prothrombinase made from either the tissues or the platelets catalyses the conversion of prothrombin, a plasma globulin, into thrombin. Finally, thrombin enzymatically converts fibrinogen into fibrin, an insoluble protein. Fibrin forms long fibers which mesh and trap red cells, white cells, and platelets, forming the clot. Usually the clot forms within 5 minutes of the rupturing of the vessel. The clot then begins to contract and squeeze out most of the plasma from itself within an hour. This process, called clot retraction, serves to increase the strength of the clot, and also pulls the vessel walls adhering to the clot closer together. The extruded plasma is now called serum, since all the fibrinogen and most other clotting factors have been removed. Because it lacks these constituents, serum cannot clot.

Question:

Discuss the sequence of reactions that constitute the "dark reactions" of photosynthesis. What are the products of these reactions?

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/Users/wenhuchen/Documents/Crawler/Biology/F09-0218.htm

Solution:

The dark reactions of photosynthesis, in which carbohydrates are synthesized, occur in a cyclic sequence of three phases - the carboxylative, reductive, and regenerative phases. These dark reactions do not require the presence of sunlight. In the carboxylative phase, a five - carbon sugar, ribulose- 5-phosphate, is phosphorylated by ATP to yield ribulose diphosphate (see Figure). Ribulose diphosphate is then carboxylated (that is, CO_2 is added), presumably yielding a six - carbon intermediate, which is split immedi-ately by the addition of water to give two molecules of phosphogiyceric acid. The three - carbon phosphogiyceric acid is reduced by NADPH in an enzymatic reaction in the reductive phase, utilizing energy from ATP. The product of this reduction is phosphoglyceraldehyde, another triose. Two molecules of phosphoglyceraldehyde can then condense in the regenerative phase to form one hexose molecule, fructose diphosphate. This is subsequently converted to a glucose-6-phosphate molecule and finally transformed into starch. Ribulose - 5 - phosphate, the cycle - initiating reactant, can be regenerated from fructose diphosphate via certain reactions of the pentose phosphate pathway. Some scientists tend to believe that the ribulose molecule is regenerated from phosphoglyceraldehyde by a complicated series of reactions. However, the cyclicity of the dark reactions is a well established and universally accepted fact. The materials consumed in the production of one hexose molecule are one molecule each of CO_2 and H_2O three ATP molecules, and four H atoms (from two molecules of NADPH_2). Ribulose - 5 - phosphate is not consumed but gene-rated at some point in the cycle. The glucose - 6 - phosphate produced can either be polymerized into starch and stored, or broken down to yield energy for work.

Question:

A certain solution has pH 3.89 at 0\textdegreeC. FindpOHand [OH-].

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0340.htm

Solution:

pH is a measure of the [H^+] andpOHis a measure of [OH^-]. Their product gives K_W, the ionization constant of water: [H^+][OH^-] = K_W. pH andpOHare related by the equation, pOH+ pH =pK_W. At 0\textdegreeC,pK_W= 14.94. Therefore: pOH=pK_W- pH = 14.94 - 3.89 = 11.05. To find [OH^-] use the equation, pOH= - log [OH^-] 11.05 = - log [OH^-] [OH^-] = 10^-11.05 = 10^-12+.95 = (10^-12)(10^.95) Find the antilog of 10^.95 . It is 8.9 which gives [OH^-] = 8.9 × 10-12.

Question:

A cube of copper metal has a mass of 1.46 × 10^-1 kg. If the length of each edge of the cube is 2.5 x 10^-2 m and a copper atom has a mass of 1.06 x 10^-25 kg, determine the number of atoms present in the sample and then estimate the size of a copper atom. Although the actual crystal structure is more complex, assume a simple cubic lattice.

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/Users/wenhuchen/Documents/Crawler/Physics/D31-0913.htm

Solution:

The number of atoms, N, is found in the follow-ing manner: Mass of N atoms = (Mass of 1 atom) (number of atoms, N) 1.46 × 10^-1 kg = [1.06 × 10^-25 (kg/atom)] (N atoms) N = [{1.46 × 10^-1 kg} / {1.06 × 10^-25 (kg/atom)}] = 1.38 × 10^24 atoms. Let us assume that each copper atom may be represented by a sphere whose diameter is d. If the atoms are touching one another, there will be the same number, say n, along each edge of the cube. The total number of atoms N is re-lated to n by the volume relationship n^3 = N = 1.38 × 10^24. The number of atoms along one edge is then approximately n= 1.1 × 10^8 atoms This number is approximate due to the inaccuracy of the initial assumption that each atom has the shape of a sphere. Also, Length of an edge of the cube = (diameter, d, of one copper atom) (number, n, of atoms along an edge) Thennd= 2.5 x 10^-2 m. Therefore, the diameter of a cop-per atom is approximately d = [(2.5 x 10^-2 m) / (1.1 x 10^8)] = 2.27 x 10^-10 This value agrees reasonably well with the value of 2.56 × 10^-10 m obtained using other methods.

Question:

Write a BASIC program which converts inches to feet and inches.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G10-0230.htm

Solution:

Use the INT function 1\O READ I 2\O LET F = INT(I/12) 3\O LET I1 = I - F\textasteriskcentered12 4\O PRINT I "INCHES = "; F;"FEET"; I1; "INCHES" 45 GO TO 1\O 5\O DATA 9, 86, 47, 37, 947, 48\O 6\O END

Question:

In hunting a submarine, two ships A and B, separated by 3000 ft, find a sonar echo on the line between them but 30\textdegree from vertical for A and 60\textdegree for B. However, 200 ft below the surface, the water temperature changes suddenly so that the velocity of sound in the depths is 0.9 that in the shallows, (a) For what depth will the depth charges be set, if the ship captains do not know of this? (b) What is the actual depth of the submarine? (c) What will the ship captains think the submarine's horizontal distance from ship A is? (d) What is the submarine's actual horizontal distance from ship A?

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/Users/wenhuchen/Documents/Crawler/Physics/D27-0868.htm

Solution:

Figure (a) shows the actual paths of the sonar waves represented by solid lines, and their apparent paths as broken lines. The submarine will appear to the ship captains to be at point 0' when in reality it is at point 0, as shown. AC + CB = 3000 (AC/CS) = tan 30 (CB/CS) = tan 60 AC + CB = CS (tan 30 + tan 60) CS = [(AC + CB)/(tan 30 + tan 60)] = [(3000 ft)/(0.577 + 1.732)] = 1310 ft since tan 30 = 0.577 tan 60 = 1.732 (b) In actual fact the sound waves are refracted at a depth of 200 ft. By definition, the refraction is such that: (sin i/sin r) = (v_s/v_d) = (v_s/0.9 v_s) = (1/0.9) where i is the angle that the path the sonar wave takes in the shallow water makes with an imaginary line normal to the interface between shallow and deep water. The letter r stands for a similar angle for the path in deep water, and v_s and v_d represent the velocities of sound in the shallow and deep water, respectively. If i = 30\textdegree, then r = 26.7\textdegree and if i = 60\textdegree, r = 51.3\textdegree We see from figure (c) that: AD + DC' + C'E + EB = 3000 ft AD = 200 tan 30 EB = 200 tan 60 (FH/HS) = tan 26.7(HG/HS) = tan 51.3 We can see from the diagram that: DC' = FH C'E = HG Thus, from the first equation in the above group: AD + FH + HG + EB = 3000 ft. 200 tan 30 + HS(tan 26.7 + tan 51.3) + 200 tan 60 = 3000 HS = [(3000 - 115 - 346) / (0.503 + 1.248)] = 1450 ft The actual depth of the submarine is 200 + 1450 = 1650 ft. The horizontal distance of the submarine from ship A will appear to be AC in figure (b): AC = 1310 tan 30 = 755 ft The submarine's actual horizontal distance is AC' in figure (c): AC' = 200 tan 30 + 1450 tan 26.7 = 115 + 730 = 845 ft Thus without knowledge of the water temperature the depth charges will be shallow by 340 ft and off in a horizontal direction by 90 ft. That is, the ex-plosion will be 350 ft away from the submarine, a near miss rather than a direct hit.

Question:

The circulatory system of insects does not function in gas exchange. What is its function? Describe the circulatory and respiratory systems in insects.

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/Users/wenhuchen/Documents/Crawler/Biology/F12-0312.htm

Solution:

Insects, which have high metabolic rates, need oxygen in large amounts. However, insects do not rely on the blood to supply oxygen to their tissues. This function is fulfilled by the tracheal system. The blood serves only to deliver nutrients and remove wastes. The insect heart is a muscular dorsal tube, usually located within the first nine abdominal segments. The heart lies within a pericardial sinus. The pericardial sinus is not derived from the coelom, but is instead a part of the hemocoel. It is separated by connective tissue from the perivisceral sinus which is the hemocoel surrounding the other internal structures. Usually, the only vessel besides the heart is an anterior aorta. Blood flow is normally posterior to anterior in the heart and anterior to posterior in the perivisceral sinus. Blood from the perivisceral sinus drains into the pericardial sinus. The heart is pierced by a series of openings or ostia, which are regulated by valves, so that blood only flows in one direction. When the heart contracts, the ostia close and blood is pumped forward. When the heart relaxes, the ostia open and blood from the pericardial sinus is drawn into the heart through the ostia. After leaving the heart and aorta, the blood fills the spaces between the internal organs, bathing them directly. The rate of blood flow is regulated by the motion of the muscles of the body wall or the gut. A respiratory system delivers oxygen directly to the tissues in the insect. A pair of openings called spiracles is present on the first seven or eight abdominal segments and on the last one or two thoracic segments. Usually, the spiracle is provided with a valve for closing and with a filtering apparatus(composed of bristles) to prevent entrance of dust and parasites. The organization of the internal tracheal system is quite variable but usually a pair of longitudinal trunks with cross connections is found. Larger tracheae are supported by thickened rings of cuticle, called taenidia. The tracheae are widened in various places to form internal air sacs. The air sacs have no taenidia and are sensitive to ventilation pressures (see below). The tracheae branch to form smaller and smaller subdivisions, the smallest be-ing the tracheoles. The smallest tracheoles are in direct contact with the tissues and are filled with fluid at their tips. This is where gas exchange takes place. Within the tracheae, gas transport is brought about by diffusion, ventilation pressures, or both. Ventilation pressure gradients result from body movements. Body move-ments causing compression of the air sacs and certain elastic tracheae force air out; those causing expansion of the body wall result in air rushing into the tracheal system. In some insects, the opening and closing of spiracles is coordinated with body movements. Grasshoppers, for example, draw air into the first four pairs of spiracles as the abdomen expands, and expel air through the last six pairs of spiracles as the abdomen contracts.

Question:

Assuming the same declarations as in Problem 14, write PROCEDURECONCAT(S1,S2:string;varS3:string) which concatenates strings S1 and S2 and puts the result in S3.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G16-0408.htm

Solution:

To concatenate strings S1 and S2 means to create a new string which consists of S1 and S2 appended to it. In the following program, the lengths of the two strings are first added to see if the resulting string could fit in the array "housing" it, which can hold only 80 characters. If this is the case , S3 is first set to S1, and then, starting with position S1.length+1, the significant characters of S2 are copied into S3. PROCEDUREC0NCAT(SI,S2:string:VAR S3:string); Var I : integer; BEGIN IF( S1.length + S2.length) >strsizeTHEN write1n ('concatenation impossible-strings too long') ELSE BEGIN FORI: =1 TO S1.length DO S3\bulletWord [I]: =SI\bulletWord[I]; FORI: =1 TO S2\bulletlength DO S3 \bullet Word [S1 \textbullet length + I] := S2\bulletWord [I] ; END END;

Question:

What Fahrenheit temperature corresponds to \rule{1em}{1pt}40\textdegree Centigrade?

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0440.htm

Solution:

\rule{1em}{1pt}40\textdegree C is 40 Centigrade degrees below the freezing point of water. Now 40\textdegreeC = 9/5(40) = 72\textdegreeF since 1 Centigrade degree is 9/5 of 1 Fahrenheit degree. But 72\textdegreeF below the freezing point, which is 32\textdegreeF, is 40\textdegreeF below 0\textdegreeF (72 \rule{1em}{1pt} 32 = 40). Thus \rule{1em}{1pt}40\textdegreeC = \rule{1em}{1pt}40\textdegreeF. This could have been obtained directly as follows. The formula for converting temperature in degrees Centigrade to degrees Fahrenheit is \textdegreeF = (9/5) \textdegreeC + 32 If \textdegreeC = \rule{1em}{1pt}40\textdegree then \textdegreeC = (9/5)(\rule{1em}{1pt}40) + 32 = \rule{1em}{1pt}40\textdegree

Question:

What is the wavelength of the sound wave emitted by a standard 440 cycles per second tuning fork?

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/Users/wenhuchen/Documents/Crawler/Physics/D26-0830.htm

Solution:

Noting that the velocity (v), frequency (n), and wavelength (\lambda) of sound are related by v =n\lambda, and assuming the velocity of sound to be 34,000 cm/sec, or approx. 1100 ft/sec, we find \lambda = v/n = [1100 ft / 440] = 2.5 ft (approx.)

Question:

Consider the array of three charges shown in the diagram. Find the force on charge 1 caused by the other two charges. Calculate the field at the position of 1 due to the other two charges.

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/Users/wenhuchen/Documents/Crawler/Physics/D18-0584.htm

Solution:

The force at 1 will be the vector sum of forces caused by charges 2 and 3. The magnitudes of these forces are F_12 = (K_E) [(2q)q / L^2] = K_E [(2q)^2 / (L)^2 ] F_13 = K_E [(2q)q / (d)^2 ] F_13 = K_E [(2q)q / (d)^2 ] But d^2 = 2L^2(figure 1). Therefore But d^2 = 2L^2(figure 1). Therefore F_13 = [(K_Eq^2) / (2L)^2 ] F_13 = [(K_Eq^2) / (2L)^2 ] F_12 is an attractive force, and F_13 repulsive. Their x- and y-components are F_12 is an attractive force, and F_13 repulsive. Their x- and y-components are (figure 3). (figure 3). (F_12)_x = 0 (F_12)_x = 0 (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_12)_y = \rule{1em}{1pt}(K_E)[(2q)2/ (L)^2 ] (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_x = \rule{1em}{1pt}(K_E)[(q)2/ (2L)^2 ] cos 45\textdegree (F_13)_y = (K_E)[(q)2/ (2L)^2 ] sin 45\textdegree (F_13)_y = (K_E)[(q)2/ (2L)^2 ] sin 45\textdegree Thus the resultant force F at 1 is Thus the resultant force F at 1 is F_x = (F_12)_x + (F_13)_x = \rule{1em}{1pt}(K_E)[(q)^2 / (2L)^2](1 / \surd2) for cos 45\textdegree = 1 / \surd2 F_x = (F_12)_x + (F_13)_x = \rule{1em}{1pt}(K_E)[(q)^2 / (2L)^2](1 / \surd2) for cos 45\textdegree = 1 / \surd2 F_y = (F_12)_y + (F_13)_y = \rule{1em}{1pt}(K_E)[(2q)^2 / (L)^2 ]+ (K_E)[(q)2/ (2L)^2](1 / \surd2) for sin F_y = (F_12)_y + (F_13)_y = \rule{1em}{1pt}(K_E)[(2q)^2 / (L)^2 ]+ (K_E)[(q)2/ (2L)^2](1 / \surd2) for sin 45\textdegree = 1 / \surd2 45\textdegree = 1 / \surd2 The magnitude of F is The magnitude of F is F = \surd( F^2_x + F^2_y) = 1.59 K_E (q2/ L^2) F = \surd( F^2_x + F^2_y) = 1.59 K_E (q2/ L^2) If q = e and L = 1 \AA = 10^\rule{1em}{1pt}10 m, then If q = e and L = 1 \AA = 10^\rule{1em}{1pt}10 m, then F = 1.59 × [9× 109{(nt\rule{1em}{1pt}m^2) / (coul)2}] × [{(1.6 × 10^\rule{1em}{1pt}19 coul)^2 } F = 1.59 × [9× 109{(nt\rule{1em}{1pt}m^2) / (coul)2}] × [{(1.6 × 10^\rule{1em}{1pt}19 coul)^2 } /(10^\rule{1em}{1pt}10m)^2 ] /(10^\rule{1em}{1pt}10m)^2 ] = 2.24 × 10^\rule{1em}{1pt}8N = 2.24 × 10^\rule{1em}{1pt}8N The electric field E at position 1 is given by the force at 1 divided by the The electric field E at position 1 is given by the force at 1 divided by the net charge at 1, 2q: net charge at 1, 2q: E = F / 2q =1.59 KE(q / 2L^2) E = F / 2q =1.59 KE(q / 2L^2) = (2.24 × 10^\rule{1em}{1pt}8N) / [2(1.6 × 10^\rule{1em}{1pt}19 coul)] = (2.24 × 10^\rule{1em}{1pt}8N) / [2(1.6 × 10^\rule{1em}{1pt}19 coul)] = 2.1 × 10^11 (N / coul) = 2.1 × 10^11 (N / coul)

Question:

Describe briefly the characteristics of the ontogeny of a humanbeing from a zygote to a mature fetus. What is the significanceof this?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0746.htm

Solution:

A human being, like any .other organism, starts out as a fertilized egg, or zygote. The fertilized egg may be compared to the single-celled flagellatezygote of some ancient organism. The zygote then under-goes cellulardivisions and gives rise to the blastula. The blastula may be comparedto some spherical multicellular embryonic form of a primitive species. The blastula next develops into a gastrula with two layers of cells, thenan embryo with three layers of cells. The early human embryo resemblesa fish embryo, with gill slits, pairs of aortic arches, a fishlike heartwith a single atrium and ventricle, a primitive fish kidney, and a tail thatcan wag. Later in its development/ the human embryo resembles a reptilianembryo, with closed gill slits, fused vertebra, a new kidney, and anatrium par-titioned into left and right chambers. Still later, the human embryodevelops a mammalian, four-chambered heart, and a newer kidney. The seven-month old human embryo, covered with hair, resemblesa baby ape more than an adult human. The ontogeny of the human embryo reveals to some extent the evolutionaryhistory of man. The one-celled, zygotic stage may reflect the earlyperiod of life on earth. Some unicellular flagellate organism is believedto be the ultimate origin of all living things. The fish-like embryonicstage may parallel the age of the fish, theretipilianstage the ageof the reptiles, the mammalian stage the age of the mammals, and the apelikestage the period of the early primates. The sequence of all these embryonicstages parallels the chronological development of the ages as wellas the evolution of animal forms. Thus, one can trace man's evolution backfrom some more primitive, apelike primate, through the forms of someancient mammal reptile, amphibian, fish, primitive ver-tebrate, invertebrate, andmulticellularstructure, to an ultimate one-celled ancestor. The significance of this is that ontogeny recapitulates phylogeny.

Question:

Under ordinary conditions of temperature and pressure, nitrogen gas exists as diatomic molecules (N_2). As a result of its electronic configuration, N_2 is very inert and requires extreme conditions before it will react with any other species. What is the electronic ground state configuration of a nitrogen atom (atomic number =7)?

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Solution:

The solution to this problem involves the application of the Pauli Exclusion Principle and of Hund's Rule. Since the atomic number of nitrogen is 7 and it is a neutral atom, there are 7 electrons to place in atomic orbitals. The two orbitals of lowest energy are the is and the 2s. These are filled according to the Pauli Exclusion Principle - 2 electrons of unpaired spin in each orbital. The orbitals with the lowest energy of those remaining are the three 2p orbitals, 2p_x, 2p_y, 2p_z. Hund's rule dictates that the remaining three electrons enter these orbitals singly and with parallel spin. The ground state electronic distribution for nitrogen can therefore be pictured as shown in the accompanying Figure.

Question:

The volume of the bulb of a mercury thermometer at 0\textdegreeC is V_0, and the cross section of the capillary is A_0. The coefficient of linear expansion of the glass \alpha_G is per C\textdegree, and the coefficient of cubical expansion of mercury is \beta_M per C\textdegree. If the mercury just fills the bulb at 0\textdegreeC, what is the length of the mercury column in the capillary at a temperature of t\textdegreeC?

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Solution:

An exaggerated view of the expansion is shown in figure (b). Figure (a) represents the initial situation. When exposed to a temperature change \DeltaT, the cross section of the capillary, the volume of the bulb, and the volume occupied by the mercury all change (see figures). The final volume occupied by the mercury is V_Hg = V_0 (1 + \beta_M \DeltaT)(1) where \beta_M is the coefficient of volume expansion of mercury. The new cross section of the capillary will be A = A_0 (1 + 2\alpha_G \DeltaT)(2) where \alpha_G is the coefficient of linear expansion of glass. Similarly, the new volume of the bulb is V_g = V_0 (1 + 3 \alpha_G \DeltaT)(3) (Note that, initially, the volume of Hg = the volume of the bulb). Now, the volume of mercury outside the bulb in figure (b) will be V_Hg - V_g = V_0 (1 + \beta_M \DeltaT - 1 - 3\alpha_G\DeltaT) V_Hg - V_g = V_0 (\beta_M - 3\alpha_G)\DeltaT(4) If the length of the mercury column in figure (b) is h, then V_Hg - V_g = hA = hA_0 (1 + 2 \alpha_G \DeltaT)(5) where we have used (2). Equating (5) and (4) hA_0 (1 + 2\alpha_G \DeltaT) = V_0 (\beta_M - 3\alpha_G)\DeltaT orh = (V_0/A_0) [(\beta_M - 3\alpha_G)/(1 + 2\alpha_G \DeltaT)] \DeltaT Since\DeltaT = t\textdegreeC - 0\textdegreeC h = (V_0/A_0) [(\beta_M - 3\alpha_G)/(1 + 2\alpha_Gt)]t This is the length of the mercury column in the capillary at temperature t.

Question:

Suppose a salesman gets $6 commission on each sale as longas the sale is under $150. However, if the sale is over $150 he gets a bonus of 2% of the amount over the $150. Write a PL/I program that will take as data the values of the amountof sale and print the commission the salesman gets.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0447.htm

Solution:

Formally, we can express the commission C as follows. If the amountA is less than $150, then C = $6. Otherwise, if A>150, then C = 6 + 2 x (A-150)/100 /\textasteriskcenteredCOMMISSION CALCULATION\textasteriskcentered/ L2\O:GET LIST (A); L3\O:IF A< = 15\O THEN GO TO L6\O; L4\O:C = 6 + 2\textasteriskcentered(A-15\O)/1\O\O; L5\O:GO TO L7\O; L6\O:C = 6; L7\O:PUT EDIT (A, C) (F (7, 2) F (7, 2)); L8\O:PUT SKIP; L9\O:GO TO L2\O;

Question:

Define the term "debugging."

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Solution:

The program, even after being tested and accepted as valid, may still have faults that will manifest themselves for certain input sequences. Such hidden faults that are not readily detectable are called "bugs." Debugging a program means the modification of programs, throughout its lifetime , in order to fix the faults that produce unacceptable or unexpected results . Thus, as opposed to "unit testing" that simply makes the program "acceptable" for the specified input-output sequences, the de-bugging process involves the entire life cycle of a computer program.

Question:

The moon revolves about the earth in a circle (very nearly) of radius R = 239,000 mi or 12.6 x 10^8 ft, and requires 27.3 days or 23.4 × 10^5 sec to make a complete revolution, (a) What is the acceleration of the moon toward the earth? (b) If the gravitational force exerted on a body by the earth is inversely proportional to the square of the distance from the earth's center, the acceleration produced by this force should vary in the same way. Therefore, if the acceleration of the moon is caused by the grav-itational attraction of the earth, the ratio of the moon's acceleration to that of a falling body at the earth's surface should equal the ratio of the square of the earth's radius (3950 mi or 2.09 × 10^6 ft) to the square of the radius of the moon's orbit. Is this true?

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Solution:

(a) The velocity of the moon is v = (distance)/(time) = (circumference)/(time for one orbit) = (2\pi R)/(T) = [(2\pi × 12.6 × 10^8 ft)/(23.4 × 10^5 sec)] = 3360 ft/sec . Its radial acceleration is therefore a = v^2/R = [(3360 ft/sec)^2]/[12.6 × 10^8 ft] = 0.00896 ft/sec^2 = 8.96 × 10^-3 ft/sec^2 (b) The ratio of the moon's acceleration to the acceleration of a falling body at the earth's surface is: a/g = (8.96 × 10^-3 ft/sec^2)/(32.2 ft/sec^2) = 2.78 × 10^-4 The ratio of the square of the earth's radius to the square of the moon's orbit is: (2.09 × 10^6 ft)^2/(12.6 ×10^8 ft)^2 = 2.75 × 10^-4 The agreement is very close, although not exact because we have used average values.

Question:

How would you explain the secondary response to antigens on the basis of the clonal selection theory?

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/Users/wenhuchen/Documents/Crawler/Biology/F14-0366.htm

Solution:

When an antigen is first injected into an animal, only a small number of lymphocytes become trans-formed into plasma cells. The amount of antibodies produced rises slowly to a low peak and then decreases. This is termed the primary response. If a second injec-tion of the same antigen occurs some weeks later, a much more vigorous response is induced. Not only is the amount of antibodies much greater than in the primary response, but their production is much more rapid (see Fig.). Further injections will result in the maximum amount of antibody that can be produced. Primary and secondary responses of antibody formation can be explained on the basis of the clonal selection theory. When an antigen first comes in contact with its corresponding lymphocyte, it stimulates the lymphocyte to divide and differentiate into plasma cells (or lymphoblasts). The antibodies Of these cells interact with the antigen to form complexes which are engulfed and destroyed by macrophages. Some days after the primary response, these plasma cells (or lymphoblasts) eventually die. However, the antigen caused production of many more copies of the originally stimulated small lymphocytes. It is thought that these lymphocytes, called memory cells, arise from division of the originally stimulated cells without concomitant dif-ferentiation into plasma cells (lymphoblasts). This immunological memory explains the process in which an animal "remembers" previous exposure to an antigen. When the animal is exposed to the antigen the second time, many more lymphocytes (specific for the antigen) are stimulated, resulting in a greater and more rapid production of anti-bodies. This explains the characteristics of the secon-dary response.

Question:

Mostmarine vertebrates have body fluids with osmotic marine vertebrates have body fluids with osmotic pressure lower than that of their saline environment. How do these organismsosmoregulatein face of aprepetualthreat of dehydration?

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Solution:

To survive their hostile saline environment, marine organisms show a variety of adaptations. The concentration of body fluids in marine fish is about one-third that of the sea water. Osmotic pressure, being higher in sea water, tends to drive water out of the body fluid of the fish.One site of severe water loss is the gill, which is exposed directly to the surrounding water for gaseous diffusion. To compensate for this loss, the fish must take in large amounts of water. How-ever, the only source of water available is sea water; if consumed, the sea water would cause further water loss from the body cells because of its higher osmotic pressure. Marine fish, however, overcome this problem. In the gills, excess salt from the consumed sea water is actively transported out of the blood and passed back into the sea. Marine birds and reptiles also face the same problem of losing water. Sea gulls and penguins take in much sea water along with the fish which they scoop from the sea. To remove the excess salt, there are glands in the bird's head which can secrete saltwater having double the osmolarityof sea water. Ducts from these glands lead into the nasal cavity and the saltwater drips out from the tip of the bill. The giant sea turtles, a marine reptile, has similar glands but the ducts open near the eyes. Tears from the eyes of turtles are not an emotional response, in contrast to popular belief. Rather, 'crying' is a physiological mechanism to get rid of excess salt.

Question:

Define a buffering system. What significance does it have in the living cell?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0027.htm

Solution:

A buffering system is one that will prevent significant changes in pH, upon addition of excess hydrogen or hydroxide ion to the system. Buffering systems are of great importance in the maintenance of the cell. For example, the enzymes within a cell have an optimal pH range, and outside this range, enzymatic activity will be sharply reduced. If the pH becomes too extreme, the enzymes and proteins within the cell may be denatured, which would cause cellular activity to drop to zero and the cell would die. Hence a buffering system is essential to the existence of the cell. The average pH of a cell is 7.2, which is slightly on the basic side. A buffer will prevent significant pH changes upon variation of the hydrogen ion concentration by abstracting or releasing a proton. Relatively weak diprotic acids (Ex: H_2CO_3, carbonic acid) are good buffers in living systems. When carbonic acid undergoes its first dissociation reaction, it forms a proton and a bicarbonate ion: H_2CO_3 \rightleftarrows H^+ + HCO_3^-. The pH of the system is due to the concentration of H^+ formed by the dissociation of H_2CO_3. Upon addition of H^+, the bicarbonate ion will become protonated so that the total H^+ concentration of the medium remains about the same. Similarly, upon addition of ^-OH, the bicarbonate ion releases a hydrogen to form water with the ^-OH. This maintains the H^+ concentration and hence keeps the pH constant. These reactions are illustrated below:

Question:

The sodium D-lines, two very distinct lines of yellow light emitted by heated sodium metal, are often used in the calibration of spectrometers. The wavelength of one of these lines is 5890 \AA. How much energy does an electron emit (or absorb) in undergoing the electronic transition associated with this line?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0689.htm

Solution:

To solve this problem we will combine two relationships: ѵ = c/\lambda, where ѵ is the frequency, \lambda the wavelength, and c the speed of light; and Planck's re-lationship, E = hѵ, where E is the energy and h is Planck's constant, 6.62 × 10^-27 erg-sec. Then, E = hѵ = hc/\lambda. Since 1 \AA = 10^-8 cm, 5890 \AA = 5890 \AA × 10^-8 cm/\AA = 5.890 × 10^-5 cm. The energy associated with the 5890 \AA sodium D-line is then E = (hc/\lambda) = [(6.62 × 10^-27 erg-sec × 3.0 × 10^10 cm/sec) / (5.89 × 10^-5 cm)] = 3.37 × 10^-12 erg.

Question:

Distinguish betweenpenetranceexpressivity.

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Solution:

A recessive gene produces a given trait when it is present in the homozygous state. A dominant gene produces its effect in both the homozygous and heterozygous states. Geneticists, however, have found that many genes do not always produce their phenotypes when they should. Genes that always produce the expected phenotype in individuals who carry the gene in an expressible combi-nation are said to have completepenetrance. If only 70 percent of such individuals express the characterphenotypically, then the gene is said to have 70 percent penetrance .Penetranceis thus defined as the percentage of individuals in a population who carry a gene in the correct combination for its expression (homozygous for recessive, homozygous or heterozygous for dominant) and who express the genephenotypically. Some genes that are expressed may show wide variations in the appearance of the character. Fruit flies homozygous for the recessive gene producing shortening of the Wings exhibit variations in the degree of shortening. Expres-sivity is defined as the degree of effect or the extent to which a gene expresses itself in different individuals. If it exhibits the expected trait fully, then the gene is said to be completely expressed. If the expected trait is not expressed fully, the gene shows incomplete expres-sivity. The difference between the two terms-penetranceand expressivity, lies in the fact that the former is a function of the gene at the population level, while the latter varies on an individual level. Thus a given gene having a certainpenetrancewithin a population may have varying expressivity in individuals of that population who express it. Bothpenetranceand expressivity are functions of theinterractionof a given genotype with the environment. Changes in environmental conditions can change both thepenetranceand expressivity of a gene. For example, a given gene may code for an enzyme required for the synthesis of a given metabolite in bacteria. If that metabo-lite is provided in the organisms nutrient environment, the organism might not produce the enzymes needed for its synthesis, and then the gene will not be expressed. If however, the nutrient is depleted from the media, the organism will begin to manufacture the enzyme, and thus express the gene. In humans, it is thought that allergy is caused by a single dominant gene; the different types of allergies are due to the varying expressivity of the gene, as a result of the interaction of the gene with both the environment and a given individual's genetic and physical make up.

Question:

A nail of mass M is driven into a board against a constant resistive force F by a hammer of mass m which is allowed to fall freely at each stroke through a height h. The hammer does not rebound after striking the nail. Find the distance the nail is driven in at each blow. Show that the total energy expended in raising the hammer during the operation of driving the nail fully in to a depth d is independent of the value of h, and can be decreased by making the hammer more massive.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0329.htm

Solution:

The hammer falls through a height h, losing potential energy and gaining kinetic energy. By the principle of conservation of energy, mgh = 1/2 mv^2,and that the hammer strikes the nail with a velocity v, where v = \surd(2gh). An inelastic collision takes place and momentum is conserved, so thatmv= (m + M)V. Therefore the kinetic energy of hammer and nail together, after the impact, is (1/2) (m + M)V^2 = (1/2) [m^2/(m + M)] v^2. When the hammer hits the nail, we assume it loses all its kinetic energy. This energy is converted into the work necessary to drive the nail a distance x against the re-sistive force F. If the hammer drives the nail this distance x at each stroke, then we have from the work- energy theorem, W = F^\ding{217} \bullet x^\ding{217} = -Fx0 - (1/2) [m^2/(m + M)] v^2 since the final kinetic energy is zero and F opposes the nail's displacement. Solving for x and substituting v = \surd(2gh), we have x = (m^2v^2) / [2F(m + M)] = [ghm^2] / [F(m + M)]. If it requires n strokes to drive the nail fully home a distance d, then d =nx= [nghm^2] / [F(m + M)].(1) But the energy E supplied to the system is that energy required to raise the hammer n times through a height h. Thus E =nmgh In equation (1), solve fornmgh. Then E = [Fd(m + M)] / m =Fd+Fd(M/m). Thus the energy does not depend on h and will be decreased if m is made larger.

Question:

Parturition is the separation of the fetus and its membranes from the mother's body at the time of birth. Describe this process.

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Solution:

As the period of gestation ends, a hormonal substance termed relaxin appears in the bloodstream. This substance seems to be a product of the corpusluteumof pregnancy and of the placenta and causes the cervix of the uterus to become readily dilatable.Oxytocin, one of the hormones released from the posterior pituitary, is an ex-tremely potent uterine-muscle stimulant.Oxytocinisreflexlyreleased into the bloodstream as a result of afferent input to the hypothalamus from receptors in the uterus, par-ticularly in the cervix.Oxytocinstimulates the uterine muscles to contract with greater force and frequency during labor. Relaxin andoxytocinare two important hormones in-volved in parturition. At the time of birth, rhythmic contractions of the smooth muscle of the uterus begin. Soon they are accompanied by reflex and voluntary activity of skeletal muscle. The result is that internal pressure forces the fetus, normally head first, into the cervix of the uterus. The fetal membranes then bulge through the cervix and burst. Amniotic fluid is discharged. The child then enters the vagina and is born, a process lasting from twenty minutes to one hour. Some fifteen minutes after birth, the uterus again goes into rhythmic contraction, and an "afterbirth" consisting of the fetal membranes is expelled. Surprisingly enough, there is only a minor amount of bleeding. Within the next few days, the lining of the uterus is restored and the uterus returns to nearly its original dimensions.

Question:

In certain abnormal conditions, the stomach does not secretehydrochloric acid. What effects might this haveon thedigestive process?

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Solution:

The highHClconcentration in the stomach serves several importantfunctions. Without the acid, the gastric enzymes would show no activity. For example, pepsin, an important protein-digesting enzyme, wouldremain inactive. Pepsin exists in the form of a precursor compound, calledpepsinogen, which has noproteolyticactivity in the absence of acidity. Thepepsinogenwill remain in the stomach as is, untilHClis secretedand acti-vates it. Thus the absence ofHClwould lead to some lossof protein digestion. This loss, however, can be made up by the intestinalproteases. HClis also important for its direct effect on the conformation of proteins. The high acidity ofHClde-natures proteins.Denaturationdoes notmean cleavage of a molecular bond, but rather an unwinding of the molecule. Proteins are normally tightly packed, wovenchains, thatare difficultfor enzymes to attack. The acid alters the protein to facilitate enzymeaction. Another function of stomach acid is to kill ingested bacteria. Many ofthe bacteria ingested are harmless, but some, if allowed to grow, could haveserious effects.HClis also important to pancreatic secretion. Pancreatic juice secretion is controlled, in part, by the hormonesecretin. TheHClsecreted by the stomach glands passes through the pyloricsphincter, along with the food, into the duodenum. Some of the acidstimulates special receptor cells in the intestinal lining. These cells, in turn, secretesecretininto the blood. When thesecretinreaches the pancreas, a pancreatic juice rich in bicarbonate travels down the pan-creaticduct into the intestine. If noHClis present, there will be less thanone quarter of the secretion, and less one quarter of the normal amountof digestion.

Question:

In acommensalisticrelationship between two species living symbioticallywhat are the consequences of their interactions?

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Solution:

Commensalism is a relationship between two species in which one speciesbenefits while the other receives neither benefit nor harm. The advantagederived by thecommensalspecies from its association with the hostfrequently involves shelter, support, transport, food or a combination ofthese. For example, in tropical forests, numerous small plants called epiphytesusually grow on the branches of the larger trees or in the forks oftheir trunks. They use the host trees only as a base of attachment and donot obtain nourishment from them. A similar type of commensalism is theuse of trees as nesting places by birds. Such relationships do not produceany apparent harm to the hosts. Commensalism is widespread throughout the animal kingdom, and isespecially common among the marine in-vertebrates. The host organismsare typically slow- moving or sessile and live in shells or burrows, which can be readily shared by the smallercommensalspecies. One example is a certain species of fish that regularly lives in association withsea anemones, deriving pro-tection and shelter from them and sometimessharing some of their food. These fish swim freely among the tenta-clesof the anemones even though these tentacles quickly paralyze otherspecies of fish that touch them. The anemones feed on fish, yet the particularspecies that live ascommensalswith them can actually enter thegastrovascularcavity of their host, emerging later with no ill effects. This implies a certain amount of physiological and behavioral adaptation betweenmembers making up acommensalrelationship. Still another example is a small tropical fish (Fierasfer) that lives in therectum of a particular species of sea cucumber. The fish periodically emergesto feed and then returns by first poking its host's rectal opening withits snout and then quickly turning so that it is drawn tail first into the rectal chamber.

Question:

A chemist reacts ferric sulfate with barium chloride and obtains barium sulfate and ferric chloride. He writes the following balanced equation to express this reaction: Fe_2 (SO_4)_3 + 3BaCl_2 \rightarrow 3BaSO_4\downarrow + 2FeCl_3 (A)How much BaC1_2 should be used to react with 10 grams of Fe_2 (SO_4)_3? (B) How much Fe_2 (SO_4)_3 will be necessary to produce 100 g of BaSO_4? (C) From a mixture of 50 g of Fe_2 (SO_4)3 and 100 g of BaC1_2, how much FeC1_3 can be produced?

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Solution:

To answer these questions, you must understand the mole concept, and how it is used to calculate the amount of material required in a chemical reaction. A mole is defined as the number of grams of a sub-stance divided by its molecular weight. In other words, a mole = [(amount in grams of substance) / (molecular weight of substance)] In the given equation: Fe_2 (SO_4)_3 + 3BaCl_2 3BaSO_4\downarrow + 2FeCl_3, the numbers before each compound are termed coefficients (the coefficient of Fe_2 (SO_4)_3 is one, and by convention is not written). The equation is balanced. This means that all of the elements are equal in number on both sides of the equation. For example, we have a total of 12 oxygen atoms on the left and 12 on the right. Before doing any problem involving a chemical equation, we must always balance it. All the atoms must be accounted for. The coefficients serve this purpose. After balancing, they also tell you the relative mole amounts that will react. In other words, in this reaction 1 mole of Fe_2 (SO_4)_3 reacts with 3 moles of BaCl_2. For the reaction to occur, we must have a mole ratio of 1 : 3 between Fe_2 (SO_4)_3 and BaCl_2. If such a condition exists, then the equation tells us that 3 moles of BaSO_4 will be produced per mole of Fe_2 (SO_4)_3. In addition, 2molesof FeCl_3 will also be produced. There-fore, the coefficients tell the relative number of moles of each substance that must be either present or produced. With this information, we can now answer the questions. (A)We have 10 grams of Fe_2 (SO_4)_3 and want the number of grams of BaCl_2 that are required to react with it. The molecular weight of Fe_2 (SO_4)_3 is 399.88 g/mole. This number is obtained by adding up all the atomic weights of the individual atoms in the formula. Recalling the definition of a mole, we have 10/399.88molesof Fe_2 (SO_4)_3. The equation tells us that 3 moles of BaCl_2 react with one mole of Fe_2 (SO_4)_3. Therefore, we must have 3 times the number of moles of Fe_2 (SO_4)_3or 3 × [(10)/(399.88)] = [(30)/(399.88)]. The number of BaCl_2 required is, therefore, [(30)/(399.88)]. The molecular weight of BaCl_2 is 208.24 g/mole. Recall, mole = number of grams/mole-cular weight. The number of grams of BaCl_2 required is, therefore, [(30) / (399.88)] moles × 208.24 g/mole = 15.62 grams (B) We want to produce 100 grams of BaSO_4. The molecular weight of BaSO_4 is 233.40 g/mole. Therefore, we want to produce 100/233.40molesof BaSO_4. How much Fe_2 (SO_4)_3 should be used? Again, we must go back to the equation and look at the coefficients to determine the mole requirements. We see that 3 moles of BaSO_4 are pro-duced for every mole of Fe_2 (SO_4)_3. Therefore, the required number of moles of Fe_2 (SO_4)_3 is 1/3 the number of moles of BaSO_4 or (1/3) × [(100) / (233.40)] = [(100)/(700.20)]. The molecular weight of Fe_2 (SO_4)_3 is 399.88 g/mole. Consequently, the number of grams required of Fe_2 (SO_4)_3 is [(100) / (700.20) mole × 399.88 g/mole = 57.11 grams (c) In this question, we are working with a mixture of two substances to produce a third. The first thing is to compute the moles: Moles Fe_2 (SO_4)_3= [(50.00 g) / (399.88)] = .125 moles Moles BaCl_2= [(100 g) / (208.24 g/mole)] = .48moles Because we are dealing with two substances to yield one, we must also consider their mole ratios as well as with the product FeC1_3. The equation calls for three moles of BaC1_2 to react with one mole of Fe_2 (SO_4)_3 We have .125 moles of Fe_2 (SO_4)_3. 3 times this amount is .375moles. However, you have .48 moles of BaCl_2. This means, there-fore, that we have .48 - .375 or .105. moles of BaCl_2 that will not react. In other words, BaCl_2 is in excess. We always take the limiting reagent, which is Fe_2 (SO_4)_3. The amount of FeCl_3 present will reflect the number of moles of Fe_2 (SO_4)_3, and not BaCl_2. Because 2 moles of FeCl_3 will be produced for every mole of Fe_2 (SO_4)_3, .125 × 2 = .25 moles of FeCl_3 will be produced. The molecular weight of FeC1_3 is 162.20. There-fore, the maximum weight of FeCl_3 produced is .25(162.20) = 40.56 grams.

Question:

How many liters ofphosphine(PH_3) gas at STP could be made from 30 g of calcium by use of the following sequence of reactions: 3Ca + 2P\rightarrowCa_3 P_2 Ca_3 P_2 + 6HCl\rightarrow2PH3 + 3CaCl_2 (Molecular weights: Ca = 40, PH_3 = 34.)

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Solution:

From thestoichiometryof these two equations 3 moles of Ca will yield 2 moles of PH_3 gas. Thus, if one knows the number of moles of Ca, then one can determine the number of moles of PH_3 produced. The number of moles of Ca given is its weight divided by its molecular weight. Therefore, number of moles Ca = [(30 g Ca) / (40 g/mole)] = 0.75 moles From 0.75 moles of Ca, one produces 2/3 as much or 2/3(0.75) = 0.50 moles of PH_3 gas. At STP, one mole of any gas occupies 22.4 liters, hence, the total volume of PH_3 produced at STP = (22.4 l/mole) (0.50 moles PH_3) = 11.2 l.

Question:

Outline briefly the events occurring in each stage of meiosis. Illustrate your discussion with diagrams if necessary.

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0643.htm

Solution:

Meiosis is the process by which diploid organisms (having two sets of chromosomes) produce haploid gametes (having only one set of chromosomes). When two gametes fuse in fertilization, the zygote formed will thus have the full diploid chromosomal complement. Meiosis consists of two cell divisions, the first (Meiosis I) called a reduction division, and the second (Meiosis II) a mitotic type division. Interphase I:This phase is similar to mitotic interphase The cell appears inactive in reference to cell division; but it is during interphase that chromosome duplication occurs. Prophase I:The chromosomes become thicker and more visible. While they are still long thin threads, an attractive force (as yet not identified) causes homologous chromosomes to come together in pairs, a process known as synapsis. This is the stage during which cross-over between homologous chromosomes will occur. After synapsis, the chromosomes continue to shorten and thicken; their double nature becomes visible, so that each homologous pair appears as a bundle of four chromatids called a tetrad. Each tetrad is composed of two doubled homologous chromosomes. The number of tetrads is thus equal to the haploid number. The cent-romeres of homologous chromosomes are connected, and there are thus two centromeres for the four chromatids. While these events are occurring, the centrioles migrate to opposite poles, the spindle begins to form between them, and the nucleolus and nuclear membrane dissolve. The tetrads move to the equatorial plane of the spindle. Metaphase I:Migration to the equatorial plane is complete, and the nuclear membrane and nucleolus have dissolved. Anaphase I:At this point the homologoues chromosomes that had paired in prophase separate and move to opposite poles of the cell. Each is still composed of two identical daughter chromatids joined at the centromere. Thus the number of chromosome types in each resultant cell is reduced to the haploid number. Telophase I:Cytoplasmic division occurs as in mitosis. Meiosis I concludes and meiosis II begins. There is no definable interphase between the two series of divisions. The chromosomes do not separate or duplicate, nor do they form chromatin threads. Prophase II:The centrioles that had migrated to each pole of the parental cell, now incorporated in each haploid daughter cell, divide, and a new spindle forms in each cell. The chromosomes move to the equator. Metaphase II:The chromosomes are lined up at the equator of the new spindle, which is at right angles to the old spindle. Anaphase II:The centromeres divide and the daughter chromatids, now chromosomes, separate and move to opposite poles. Telophase II:Cytoplasmic division occurs. The chromo-somes gradually return to the dispersed form and a nuclear membrane forms. The two meiotic divisions yield four cells, each carrying only one member of each homologous pair of chromosomes. These cells are for this reason called haploid cells.

Question:

Consider the following variant ofNim. A pile of N objects is given. Two players take turns removing 1, 2 or 3 objects fromthe pile. The person who has to take the last object, loses. Write a Basic program which enables you to play this gamewith a computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0566.htm

Solution:

Let us consider 2 players: you and the computer. The computer losesif it takes the last object. The question now is: How many objects mustyou leave to the computer at its penultimate move to ensure your win? The answer is:1 + the maximum number of objects always possible toremove in one move by each you and the computer. This number is 4, becauseif the computer removes the minimum possible number of objects (i.e., 1), and you - the maximum (i.e., 3) - that will give you the desired number. So, by leaving the compu-ter with 5 objects at its penultimate move, you ensure your win (providing you do not make a mistake). In fact, ifyou leave the computer with 1 + 4N (where N = 0,1,2,3...) objects before everymove, you guarantee yourself a win by removing (4 - A) objects, where"A" is the number of objects removed by the computer. Using this strategyand providing that you move first, you are guaranteed to win everytime, except when there are 4N objects in a pile. In that case, by usingthe same strategy, the computer can beat you. 10PRINT "31 OBJECTS GAME" 20PRINT "LET'S FLIP A COIN TO SEE WHO GOES FIRST." 25PRINT "IF IT COMES UP HEADS, I WIN THE TOSS." 30N = 31 40Q = INT (2\textasteriskcenteredRND (5)) 50IF Q = 1 THEN 80 60PRINT "TAILS! YOU GO FIRST" 70GOTO 140 80PRINT "HEADS! I GO FIRST" 90PRINT "I TAKE TWO OBJECTS" 100N = N - 2 110PRINT "THE NUMBER OF OBJECTS IS NOW" N 120PRINT "YOUR TURN. YOU MAY TAKE 1, 2 OR 1303 OBJECTS." 140PRINT "HOW MANY DO YOU WISH TO REMOVE?", 150INPUT K 160IF K > 3 THEN 360 170IF K < = 0 THEN 360 180N = N - K 190PRINT "THERE ARE NOW" ;N; "OBJECTS REMAINING." 200IF N = 4 THEN 260 210IF N = 3 THEN 280 220IF N = 2 THEN 300 230IF N < = 1 THEN 400 240Z = 4 - K 250GOTO 320 260Z = 3 270GOTO 320 280Z = 2 290GOTO 320 300Z = 1 320PRINT "MY TURN. I REMOVE" Z "OBJECTS" 330N = N - Z 340IF N < = 1 THEN 380 350GOTO 110 360PRINT "IMPOSSIBLE. HOW MANY; 1, 2 OR 3" 370GOTO 150 380PRINT "YOU LOST" 390GOTO 410 400PRINT "YOU WERE LUCKY, YOU WON" 410STOP 420END

Question:

Describe the life cycle of a typical fern plant.

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/Users/wenhuchen/Documents/Crawler/Biology/F07-0184.htm

Solution:

The fern plant, like other vascular plants, has a life cycle in which the sporophyte generation is dominant. The spprophyte generation is the relatively large, leafy green plant we recognize as the fern plant growing in both the tropics and temperate regions. The sporophyte consists of a horizontal stem, or rhizome, lying at or just under the surface of the soil and bearing fibrous roots. From the rhizome grows several leaves or fronds, usually finely divided into pinnae. On the under surfaces of certain pinnae develop small, brown sori. A sorus is a cluster of sporangia or spore cases in which haploid spores are formed. Most modern ferns are homosporous, that is, all their spores are alike. When mature and proper conditions, the spores are shed. After germination, each spore develops into a gametophyte which is typically tiny, thin, and often heartshaped. Small and obscure as it is, the fern gametophyte, also called a prothallus, is an independent photosynthetic organism. It grows in moist, shady places, especially on decaying logs and on moist soil and rocks. A number of rhizoids grow from the gametophyte into the substrate, anchoring it and absorbing water and salts. The male and female sex organs (antheridia and archegonia) develop on the under-surface of the game-tophyte. Each archegonium, usually located near the notch of the heart-shaped prothallus, contains a single egg. The antheridia, located at the other end of the gametophyte, each develop a number of flagellated sperm, which are ovoid in shape and have many flagellae on a spiral band at their anterior end. The sperm, released after a rain and attracted by a chemical substance released by the archegonium, swim through a film of moisture to the egg. Although the fern plant is usually monoecious (that is, it possesses both the male and female sex organs), the sperm of one plant usually fertilizes the egg of another, thus accomplishing cross--fertilization. The zygote begins to develop within the archegonium into a sporophyte embryo. At first the sporophyte develops as a parasite on the gametophyte, but it soon acquires its own roots, stems, and photosynthetic leaves and becomes an independent sporophyte, thus completing the life cycle.

Question:

What hormone, if any, acts antagonistically to the parathormone?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0546.htm

Solution:

In 1961 a hormone called calcitonin was discovered, and was found to be directly antagonistic to parathormone, the hormone secreted by the parathyroid glands. Calcitonin, also called thyrocalcitonin, is secreted by cells within the thyroid gland which surround but are completely distinct from the thyroxine secreting cells. It is the function of calcitonin to lower the plasma calcium concentration. This is achieved primarily by the deposition of calcium-phosphate in the bones. Parathormone has an exactly opposite effect in calcium- phosphate metabolism. The secretion of calcitonin is regulated by the calcium content of the blood supplying the thyroid gland. When blood calcium rises above a certain level, there is an increase in calcitonin secretion in order to restore the normal concentration of calcium. In lower invertebrates calcitonin is produced by separate glands, which in mammals become incorporated into the thyroid during embryonic development.

Question:

What is the flux density B^\ding{217} in the region between the plates of a circular capacitor that is being charged?

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0638.htm

Solution:

Figure 2 represents a plane midway between the plates of the capacitor in Fig. 1. During the process of charging, an effective current will pass across the plates as a result of charge transfer between the plates over a finite period of time. The electric field inside the capacitance will increase in time from zero to a maximum value, and give rise to the displacement current density J^\ding{217}D, which is given by Maxwell's equations density J^\ding{217}D, which is given by Maxwell's equations as \mu_0\epsilon_0 =\partialE^\ding{217} / \partialt = J^\ding{217}D The magnetic field due to J\ding{217}Dcan be obtained from Stoke' s theorem \ding{217} \int_C B^\ding{217} \bullet dr^\ding{217} = \mu0\int_S J^\ding{217}D\bullet dA^\ding{217}(1) where B^\ding{217} is the magnetic induction For the first integral, let the circle of Fig. 2 be the contour C about which we evaluate the left side of (1). The integral involving J^\ding{217}_D is then evaluated about the area S enclosed by C. Because of the circular symmetry of the charge distribution on the plates, i.e., the uni-formity of the J^\ding{217}_D lines, the induced magnetic field B^\ding{217} will be circular. Then, we find using (1), that at every radius r 2\pir B_r = \mu0ID,r,Br =(\mu_0 / 2\pi)[(I_D, r) / r)] where I_D, r is the displacement current passing through the indicated surface. If r is less than the radius of the plates, then I_D, ris smaller than the conduction current I_C, because not all the electric field lines go through S. In the idealized case in which fringing fields are neglected, and r equals the radius of the plates, the displacement current equals the conduction current and B_r = (\mu_0 / 2\pi)( I_C, / r) which is the same as the field at a distance r^\ding{217} from an infinitely long straight conductor.

Question:

Suppose the velocity of the particle in the diagram is given by the equationv\ding{217} = [m + nt^2] where m = 10 cm/s and n = 2 cm/s^3, (a) Find the change in velocity of the particle in the time interval between t_1 = 2 sec and t_2 = 5 sec. (b) Find the average accelera-tion in this time interval. (c) Find the instantaneous acceleration at t_1 = 2 sec.

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/Users/wenhuchen/Documents/Crawler/Physics/D03-0069.htm

Solution:

(a) At time t_1 = 2 sec v_1\ding{217}= [10(cm/s) + {(2cm)/(s^3)} (2s)^2] v_1\ding{217}= 18(cm/s) (a) Particle moving on the x-axis. (b) Velocity time graph of the motion. The average acceleration between t_1, and t_2 equals the slope of the chord pq. The instantaneous acceleration at p equals the slope of the tangent at p. whereis a unit vector in the positive x direction. At time t_2 = 5 sec v_2\ding{217} = [10(cm/s) + (2cm/s^3)(5s)^2 ] v_2\ding{217} = 60cm/s The change in velocity is therefore v_2\ding{217} - v_1\ding{217} = (60cm/s)- (18cm/s)= (42cm/s) (b) The average acceleration is defined as a_avg\ding{217} = (v_2\ding{217} - v_1\ding{217})/(t_2 - t_1) where v_2\ding{217} and v_1\ding{217} are the velocities at t_2 and t_1, respec-tively. Hence, in the given interval a_avg\ding{217} = [{(60cm/s) - (18cm/s)}]/(5s - 2s) = 14(cm/s^2) This corresponds to the slope of the chord pq in the diagram. (c) The instantaneous acceleration is a\ding{217} = dv\ding{217}/dt = 2nt At t_1 = 2 sec, a= (2)(2cm/s^3)(2s) = 8 cm/s^2 This corresponds to the slope of the tangent at point P in the figure.

Question:

We consider a particle with charge q which moves along the x axis with a high momentum p_0 and enters a region of length L in which there is a transverse electric field \epsilon\^{y}. Find the angle through which the particle is deflected by the electric field.

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/Users/wenhuchen/Documents/Crawler/Physics/D19-0631.htm

Solution:

Since we are given .the momentum and not the mass of the particle, we use a momentum-energy approach. We first note that we are given the electric field \epsilon. We can translate this into the force exerted on the particle by multiplying the field by the charge on the particle (q\epsilon) . But, force is the derivative of momentum. Since the field is only present in the y direction, the resulting force is exerted solely in the y direction. Thus, d(p)x/ dt = 0 ;d(p)y/ dt = q\epsilon, Integrating, we have p_x = p(o)x, py= q\epsilont + p_(o)y where p(o)xis the initial x momentum, and p_(o)y is the initial y momentum. Since the particle initially has no y velocity p(o)y= 0 andpx= p(o)xp_y = q\epsilont(1) We want to find the velocity v. If we can find the energy E, then we can find the velocity from the momentum by using the relation v = pc2/ E. The energy is given by E2= M^2c4+ p^2c^2 + M^2c4+ (p^2_(0)x + q^2 \epsilon^2 t^2)c2 from (1). But ^2E_0 = M^2c4+^2p_0 c^2and, then , E2= (M^2c4+ p^2_(0)x c^2) + (q\epsilontc)2 Since p_0 = p_(0)x E2= E20+ (q\epsilontc)2(2) where E_0 is the intial energy. Therefore from (1),(2) and the velocity- momentum relation we have v_x = p_(x)c^2 / [E20+ (q\epsilontc)^2]^1/2 =p_0(x)c^2 / [E^2_0 + (q\epsilontc)^2]^1/2 v_y = p_(y)c^2 / [E20+ (q\epsilontc)^2]^1/2 =q\epsilontc^2 / [E^2_0 + (q\epsilontc)^2]^1/2 Note that v_x decreases as t increases. At a time t, the angle \texttheta^X the trajectory makes with the x axis is given by tan \Phi (t) = v_y / v_x = q\epsilontc^2 / p_0 c^2 =q\epsilont / p_0 .

Question:

The gold leaves of the machine in the figure periodically diverge and then collapse. Describe how this perpetual- motion machine operates and explain why it is not a perpetual-motion machine.

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/Users/wenhuchen/Documents/Crawler/Physics/D34-1017.htm

Solution:

Radium atoms are radioactive and have a half- life of 1,622 years. For radium ^226 _88 Ra \ding{217}^222 _86 Rn +4_2 He(1) The alpha particle (the ^42He nucleus) with 2 quanta of positive charge, is ejected with a kinetic energy of 4.77 MeV and escapes from the surface of the metal. Therefore, if a piece of radium is placed on an electroscope stem, the electroscope becomes negatively charged. This is due to the fact that only a part of the radium atom is emitted as an alpha particle. The num-ber of electrons around the radium nucleus remains the same. Thus, for each alpha particle that escapes 2 quanta of ne-gative charge are left on the electroscope mount and gold leaves. As radium atoms decay, the leaves slowly diverge due to the mutual repulsion of the negative charge on the two foil strips. At the instant they touch the grounded metal surface of the container, the negative charge mi-grates from the leaves. The gravitational forces cause the leaves to collapse to their original position. The leaves immediately begin to diverge again, repeating the cycle over and over until there are no more radium nuclei present. Since the half-life of radium is 1,622 years, the machine would "run" for many years without any apparent slowing down. From today's perspective, though, we see that the energy source is derived from the nuclear energy released in the nuclear reaction described in (1). This energy is converted to mechanical motion of the gold leaves. There-fore, the device is not a perpetual-motion machine.

Question:

Explain what is meant by primary, secondary, and tertiary treatment of sewage.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E24-0855.htm

Solution:

Sewage treatment is carried out in three stages known as primary, secondary, and tertiary. Each uses a different process to eliminate different substances from the sewage. Primary treatment uses mechanical means to separate out solid objects such as sticks and rags. These solid objects are removed by means of a coarse screen. The liquid that passes through this screen flows into settling tanks, where insoluble material settles and forms sludge. This removes approximately one third of the pollutants. Whereas primary treatment is mechanical, secondary treatment is biological. The liquid or effluent from the primary facility filters slowly through a bed of rock. Bacteria present consume most of the organic material. In many plants, oxygen is bubbled in to accelerate the growth of bacteria and, as such, the decomposition of the organic waste. This process removes about 90 % of the biodegradable materials in the water. The last type of treatment, tertiary, is the most expensive. It is used to remove organic chemicals, nutrients, and excessive salts. To accomplish this, the treatment may include such processes as chemical coagulation, distillation and reverse osmosis.

Question:

A conductor of capacitance 10^-2 \muF to the ground is insulated from the ground by a silica plate 2.5 mm thick and 5 cm^2 in cross-sectional area. What is the minimum resistivity of the silica if the rate of decrease of potential is to be no greater than 0.1% per min of its instantaneous value?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0695.htm

Solution:

In this problem, the capacitor is initially charged to its maximum value, and then discharges to ground through the silica plate (see figure (a)). The latter may be considered to be an effective resistance (see figure (b)). Before solving the problem, we must know the voltage across the capacitor as a function of time. Kirchoff's Voltage Law states that the net voltage around a closed circuit loop is zero. Hence, V_cb + V_ba = 0(1) Starting at c, and traversing the loop in the given direction (see figure (b)), we note that c is at a higher potential than b. Hence, V_c > V_borV_b - V_c = V_cb < 0. But, by definition of capacitance C = [(\midQ\mid)/(V_bc )] orV_bc = (\midQ\mid/C) where \vertQ\mid is the absolute value of the charge on 1 plate of the capacitor. Therefore, V_cb = -V_bc = (-Q/C)(2) whereQ > 0.(2a) Furthermore,V_b > V_aorV_a - V_b = V_ab < 0.By Ohm's Law, V_ab = -iR(3) Using (3) and (2) in (1), (-Q/C) - iR = 0(4) But i = [(dQ)/(dt)], where Q is the net charge passing a point of the circuit per unit time. Equation (4) then becomes (1/C) Q + R [(dQ)/(dt)] = 0 orQ + RC [(dQ)/(dt)] = 0(5) Solving (5) (dQ/Q) = -(1/RC) dt \int (dQ/Q) = - (1/RC) \int dt ln Q = - (1/RC) t + F(6) where F is a constant. Taking the exponential of both sides of (6) Q = e- t/RC + F= e^F e^-t/RC Let A = e^F .Then Q(t) = Ae^- t/RC . We find A by noting that at t = 0, Q = Q_0 , the maximum charge C has. Then Q(0) = A = Q_0 whence Q(t) = Q_0 e- t/RC(7) The voltage across C is then, by definition of C V = (\midQ\mid/C) = (Q_0 /C) e- t/RC Note that Q_0 > 0 because Q(t) > 0. [See eq. (2a)]. Defining the initial voltage across C as V_0 = (Q_0 /C) we find V(t) = V_0 e^- t/RC(8) The time rate of change of (8) is [{dV(t)} / dt] = - (1/RC) V_0 e- t/RC= - (1/RC) V(t) But, the question states that [{dV(t)} / dt] \leq (.001 min^-1 ) V(t) or- (1/RC) \leq .001 min^-1 Hence, solving for R - 1 \leq (.001 min^-1 ) RC orR \geq [(-1)/{(.001 min^-1 ) (c)}](9) Now, R is the resistance of the silica, which may be written as R = (\rhol/A)(10) where \rho is silica's resistivity, and l and A are the thickness and cross-sectional area of the plate. Using (9) in (8), and solving for \rho \rho \geq - [(A)/{(.001 min^-1 ) (l) (c)}] Using the given value \rho \geq [(5 × 10^-4 m^2 )/{(.001/605) (2.5 × 10^-3 ) (10^-2 \muF)}] Since 1\muF = 10^-6 F \rho \geq [{(5 × 10^-4 m^2 )(605)} / {(10^-3)(2.5 × 10^-3 m)(10^-8 F)}] \rho \geq 1.2 × 10^12 \Omega \bullet m We have used the fact that1F = 1 C/V. Then \rho_min = 1.2 × 10^12 m \bullet \Omega

Question:

One of the two most important uses of ammonia is as a reagent in the first step of theOsfwaldprocess, a synthetic route for the production of nitric acid. This first step proceeds according to the equation 4NH_3(g) + 50_2(g) \rightleftarrows 4N0(g) + 6H_20(g) . What is the expression for the equilibrium constant of this reaction?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E09-0304.htm

Solution:

This problem is an exercise in writing the equilibrium constant of a reaction. In general, for a reaction in which re-actants A,B,C,... go to products W,X,Y,... according to the equation aA+bB+cC+ ... \rightleftarrowswW+xX+yY+ ... , wherea,b,c,w,x,y,... are thestoichiometriccoefficients, the equilibrium constant is given by K = { [W]^w [X]^x [Y]^y ...} / { [A]^a [B]^b [C]^c ...} Hence, for the reaction 4NH_3 + 5 O_2 \rightleftarrows 4N0 + 6H_2O , the equilibrium constant is given by K = {[NO]^4 [H_2O]^6 } / { [NH_3]^4 [O_2]^5}.

Question:

What is the fractional increase of mass for a 600─mi/hr jetliner?

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/Users/wenhuchen/Documents/Crawler/Physics/D32-0939.htm

Solution:

Fractional increase of mass is defined as change in mass divided by the original mass or \Deltam/m_0, The equation for the variation of mass with velocity is m = m_0 / (\surd1 ─ \beta^2) ,\beta^2 = v^2 / c^2 Therefore the change in mass is \Deltam = m ─ m0[ {1 / (\surd1 ─ \beta^2)} ─ 1] And \Deltam / m_0 =[ {1 / (\surd1 ─ \beta^2)} ─ 1] For velocities much less than light, \beta = v/c is very small and{1 / (\surd1 ─ \beta^2)} can be approximated by 1 + \beta2/ 2. Since 600 mi/hr is small compared to c, we can say that for this problem the fractional mass is \Deltam/ m_0 \allequal (1 + \beta2/ 2) ─ 1 = \beta^2 / 2 V = 600 mi/hr \allequal 2.7 × 104cm / sec \beta = v / c = ( 2.7 × 104cm / sec) /(3 × 1010cm/sec) \allequal 10^─6 Therefore , \Deltam/ m_0 \allequal 1/2 \beta^2 \allequal .5 × 10─12 so that the mass is increased by only a trivial amount .

Question:

An ancient Egyptian mummy is found. A count of the radiationemitted from the C^14 in the body is only one eighth thatfound in persons who have died recently. About how old isthe mummy? (The half-life of C^14 is 5568 years.)

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0733.htm

Solution:

Certain radioactive elements are transformed spontaneously into otherelements at rates which are gradual and essentially independent of thetemperatures and pressures to which the elements have been subjected. In addition, unstable isotopes of certain elements also undergo gradualtransformation to more stable forms. An example of this is the decayof radioactive isotope carbon-14 into the element nitrogen. This processof transformation is termed radioactive decay. It is established that carbon-14 has a half-life of 5568 years. A half- lifeis defined as the period of time (in units of seconds, minutes, hours, yearsand so forth) that it takes for one half of a sample to undergo radioactivedecay. Since one half of the original sample remains after one half-life, one fourth will remain after two half lives, and one eight after threehalf lives. Therefore, the fact that the mummy contains one- eighth the amountof C^14 of that in persons who have died recently indicates to us thatthree half lives have passed since the mummification. During this period, 7/8 of the total C -14 has decayed into the more stable element, nitrogen. The mummy is hence 3 × 5568 or 16,704 years old.

Question:

What are the differences between the growth patterns of dicotsthat use their cotyledons as an absorption organ and those that use them for both storage and absorption?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0203.htm

Solution:

In thedicots, the cotyledons can either serve as only an absorption organ or as both an absorption and storage organ. When the cotyledons function only in absorption, the embryo remains relatively small and is surrounded by endosperm. The cotyledons of these plants absorb the stored food of the endosperm. After the seed germinates, and the endosperm has been depleted, these cotyledons develop into leaf-like photosynthetic organs. In otherdicots, such as beans and peas, the cotyledons function in storage as well as absorption of nutrients. The embryo grows until all of the endosperm is absorbed by the cotyledons. Subsequently the cotyle-dons undertake the function of food reserve, resulting in their appearance as enlarged, thickened structures, which are not photosynthetic in function. In thesedicots, the endosperm is usually completely absorbed before the seed germinates. The cotyledons remain as a food supply until the seedling is capable of photosynthesis, at which point they shrivel and fall off.

Question:

Write a FORTRAN program to simulate the daily activity at a gasstation. Assuming that one customer is served every six minutes, what is the average waiting time per customer?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0571.htm

Solution:

This is a problem in discrete simulation. A random number generatoris used in the program to model the arrival of customers. Since thegas station attendant cannot know if the stream of customers will be continuous, randomness of arrivals is assumed. To calculate the average waiting time, the program should compute the waitingtime of each individual car, add the times, and divide that sum by thetotal number of cars. The first arrival incurs no waiting time, but all subsequentarrivals must wait six minutes for each car in front of them. To obtainan average the service time remaining for each car, the total waitingtime for all the cars, and the number of cars that are lined up must befound. For each simulated minute, the program must reduce the service time SERV by one and check to see if another car has arrived.If so, the waiting timefor that last car is given by SERV. WAIT - the total waiting time, is increasedby adding SERV to it, while the number of cars, CARTOT, is Increased by one, and SERV is increased by six minutes. The parameter N is the number of minutes to be simulated (N cannot contain more than 6 digits). Another assumption made is that customers arrive each minute with a probabilityof 0.1. To simplify the problem, it is assumed that there is only onegas pump at this station. Note, that several values of N can be entered in the DATA section. The programprocedure will be done for each of those time intervals. In order toend the program, enter N \leq 0 as the last value. CGASOLINE LINE SIMULATION GASOLINE LINE SIMULATION INTEGER WAIT, SERV, CARTOT, TIME 5READ (5,1)N 1 FORMATE (I6) CDO WHILE N GREATER THAN ZERO IF (N. LE. 0) GO TO 99 SERV = 0 WAIT = 0 CARTOT = 0 DO 10 TIME = 1,N SERV = MAX (SERV - 1,0) 222 CALL RAND (X) IF (X.GT.0.1) GO TO 22 CARTOT = CARTOT + 1 WAIT = WAIT + SERV SERV = SERV + 6 10CONTINUE AVWAIT = FLOAT (WAIT)/FLOAT (CARTOT) WRITE (6,101) CARTOT,AVWAIT 101FORMAT (IX,'AVERAGE WAIT FOR EACH OF THE', 16,'CUSTOMERS IS', F6.2,'MINUTES.') GO TO 5 CEND DO-WHILE 99STOP END

Question:

Outline the development of the reproductive organs in the human embryo.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0583.htm

Solution:

The reproductive system develops relatively late in the history of the embryo. In fact, its full development is not complete until after puberty. More-over, the reproductive system remains remarkably labile and is subject to the influence of the sex hormones. At first, male and female embryos cannot be distinguished. This stage is known as the indifferent stage of develop-ment; the primordial gonads have the potential to become either ovaries or testes. If the gonad is to different-iate into an ovary, the outer layer of cells, or cortex, progresses, and the inner core, or medulla, regresses (see Figure 1). If it is to be a testis, the opposite process occurs. In addition, the indifferent embryo has two sets of primordial cells that will give rise to the reproductive ducts. The indifferent germ glands, or gonads, first appear as thickenings of the coelomic epithelium on the side of the embryonic kidney, termed the mesonephros (see Figure 2) and contain the germ cells. There has been much dispute as to the origin of the germ cells themselves. The bulk of the evidence indicates that they are not derived from the initial germinal epithelium but migrate into it via the blood from some other part of the embryo. Soon each gonad protrudes into the body cavity as a longi-tudinal ridge, called the germinal ridge. Cord-like clumps of cells, called the medullary cords, from the germinal epithelium push into the gonad. In the male, the medullary cords push into the core of the gonad and become the seminiferous tubules. Their walls consist of immature germ cells (the spermatogonia) and accompanying nurse or Sertoli cells which serve a nourishing function. Ultimately, the spermatogonia will transform into mature sperm. The seminiferous tubules become connected to the rate testes, a system of thin tubules that develop from the dorsal part of the gonad, and form connections with the adjoining tubules of the mesonephros, (which later become the epididymis). The epididymis is attached to the Diagram showing development of gonads in higher vertebrates. epithelium of the ridge and located partly in the adjacent mesenchyme. (future seminiferous tubules). cortex contains germ cells. vas deferens, whose embryonic origin is a duct in the mesonephros called the Wolffian duct, which opens to the outside. The development of the male system proceeds under the stimulation of testosterone secreted by the interstitial cells of the developing testes. In female embryos, the Wolffian ducts degenerate when the mesonephros stops functioning as an excretory organ. The mesonephros is replaced by the metanephros, which remains as the functioning kidney all throughout life. New formations called Mullerian ducts develop lateral to the Wolffian ducts. These open at the anterior end by a funnel-shaped ostium into the coelomic cavity, and at the posterior end to the outside. The anterior portions become the Fallopian tubes. Farther back, the Mullerian ducts fuse and give rise to the uterus. The posterior parts of the Mullerian ducts unite to produce a single vagina. In females, the mesonephric and Wolffian ducts degenerate while the Mullerian ducts proceed to develop. In the male, the testis appears to secrete, in addition to testosterone, a second hormone, that inhibits the development of the Mullerian ducts so that no oviduct or uterus appears. The external genitals of the male and female are derived from the same embryonic source, and at the in-different stage they cannot be distinguished (see Figure 3). In mammals the genital opening of the embryo is flanked on both sides by elongated thickenings, the genital folds. A genital tubercle (phallus) is located on the ventral end where the genital folds meet. Prom this sexually indifferent condition, under the stimulation of the sex hormones, development proceeds toward the male or female condition. In the male, the phallus grows large and elongates to form the penis. A groove on its ventral surface is bounded by the genital folds. These close in beneath and form a tube, extending from the urethra to the tip of the penis. In the female, the phallus is smaller and is known as the clitoris. The genital folds do not fuse but remain as the labia minora. An outer fold bordering each labia minor develops into the labia majora, which is comparable to the skin of the scrotum.

Question:

Name an organism which produces spores in its life cycle and explain how this aids in its proliferation.

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Solution:

While spores are primarily used as a means of reproduction in plants, there is one class of protozoans (the Sporozoa) which produces spores. These spores are highly resistant forms of the organism which can withstand extreme environmental conditions and are specialized for asexual reproduction. The sporozoanPlasmodiumis a spore-forming protozoan that causes the disease malaria.Plasmodiumspores are produced initially by sexual reproduction in the body of a mosquito (see figure 1) which has sucked the blood of an individual infected with the organism. These spores (sporo-zoites) may now be injected into another individual's bloodstream when the mosquito bites the new host. After invading the red blood cells, the spores reproduce asexually, producing many spores (merozoites) within each invaded cell. This causes the cell to burst, releasing the spores. Each is able to invade another red blood cell. It is thus through repeated sporulation that the organism proliferates in the bloodstream of the infected individual. Some merozoites develop into male and female gametocytes. It is these gameto-cytes which a mosquito sucks up when she bites. In the mos-quito the gametocytes develop into gametes. By sexual repro-duction, the gametes fuse into zygotes which become sporozoites and the cycle repeats.

Question:

A metal sphere of radius 5 cm has an initial charge of 10^-6 coul. Another metal sphere of radius 15 cm has an initial charge of 10^-5 coul. If the two spheres touch each other what charge will remain on each?

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Solution:

First we find the field of the charged sphere by using Gauss's law, \int E^\ding{217} \bullet dS^\ding{217} = q/\epsilon_0 Here, q is the total charge enclosed by the Gaussian surface and the integral is a surface integral. Consider a spherical Gaussian sur-face of radius R centered at the center of the sphere. The magnitude of E^\ding{217} is the same at all points on the Gaussian surface by symmetry considerations, E^\ding{217} and dA^\ding{217} are in the same directions, as well (i.e., both point radially outward). Gauss' law then reduced to E \int dA = E 4\piR^2 = q/\epsilon_0 For 4\piR^2 is the surface area of the sphere. Thus, For 4\piR^2 is the surface area of the sphere. Thus, E = (1/4\pi\epsilon_0) (q/R^2) The field external to the sphere is as if all the charge were con-centrated at the center; the field internal to the sphere must be zero. Since the sphere is a conductor. Similarly the potential ex-ternal to the sphere will be as if all the charge were concentrated at the center. The potential inside the sphere must be constant since E, the potential gradient, [i.e.,E = \rule{1em}{1pt}(dV/dr)], is zero (see figure). The potential, then, at any point on the sphere or external to the sphere, is, by definition V(R) - V(\infty) = \rule{1em}{1pt} ^R\int_\infty E dR V(R) - V(\infty) = \rule{1em}{1pt} ^R\int_\infty E dR = \rule{1em}{1pt} (q/4\pi\epsilon_0) ^R\int_\infty dR/R^2 = \rule{1em}{1pt} (q/4\pi\epsilon_0) ^R\int_\infty dR/R^2 = (1/4\pi\epsilon_0)(q/R) ^R\vert_\infty = (1/4\pi\epsilon_0)(q/R) ^R\vert_\infty But V(\infty) = 0, V(R) = (1/4\pi\epsilon_0) q/R. If we set K_E = 1/4\pi\epsilon_0, V(R) = K_Eq/R. When the two spheres are touched, their potentials must become equal. This is accomplished by a movement of charge from one sphere to the other, until the potentials equalize. After touching, V_1 = V_2 or K_E (q_1/R_1)= K_E(q_2/R_2) where q_1 and q_2 are the charged on spheres 1 and 2 respect-ively. Thus q_1/q_2 = R_1/R_2 = 15/5 = 3 q_1 = 3q2 But the total charge is q_1 + q_2 = 10^-5 coul +10^06 coul = 10^-5 (1 + 0.1) coul(1) = 1.1 × 10^-5 coul Also,3q_2 = q_1, and using (1) 4q_2 = 1.1 × 10^-5 coul q2= 2.75 × 10^-6 coul q_1 = 8.25 × 10^-6 coul Initially we had q_1 = 10^-5 coul and q_2 = 10^-5 coul.

Question:

Suppose that two particles are traveling opposite to each other with velocity v'_x = \pm 0.9 c as observed in the S' system. What is the velocity of one particle with respect to the other, that is as measured by the other?

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Solution:

In the reference frame S', the two particles move with velocities - v'(1)x= v'_(2)x = 0.9 c (see Fig. 1). If we observe from the reference frame S in which the first particle is at rest, then S' will appear to move with velocity + 0.9 c with respect to this particle, i.e. S' has velocity v = + 0.9 c relative to S (see Fig. 2). The velocity of the second particle in S, as observed by the first particle, will be given by the formula for the relativistic addition of v'_(2)x and v: v_(2)x =(v'(2)x+ v ) / {1 + (v'_(2)xv / c^2)} = 1.8c / {1 + (0.9)^2 } = 0.994c , (i.e. we transformed the velocity of the second particle from S' to S). Notice that the relative velocity of the two particles is less than c. If a photon is travelling at velocity + c in S', and S' is traveling relative to S at velocity + c, the photon as viewed from S is traveling only at velocity + c, and not at + 2c. If v'_s is the velocity of the photon in S' and v the relative velocity of S' with respect to S, then in S we have v_s = (v's+ v ) / {1 + (v'_sv / c^2)} = (c + c) / {1 + (c2/ c^2)} = c , for the photon. The fact of an ultimate speed is a consequence of the structure of the velocity-addition equations which we have derived from the Lorentz transformation. Note further that there is no frame in which a photon (light quantum) is at rest.

Question:

Derive the mirror formula for rays incident on a mirror of radius of curvature R, if the rays make a small angle with the mirror's axis. mirror

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Solution:

Our proof will be totally geometrical. (See figure.) Ray OD is incident on mirror DG, at an angle s with the normal to the mirror CD. (C is the center of curvature of the arc forming the mirror.) It is re-flected at an angle r, and intersects the mirror axis at point I. By the law of reflection, angle i is equal to angle r. Hence s = r(1) Furthermore, \beta = (DG/CG)(2) But CG is the mirror radius R, whence \beta = (DG/R) Alsotan \alpha = (DE/OE) andtan \Upsilon = (DE/IE)(3) However, if \alpha, \beta and \Upsilon are small angles (that is, ray OD is close to the mirror axis) we may write tan \alpha \approx \alpha(4) tan \Upsilon \approx \Upsilon Furthermore, DE \approx DG(5) Using (5) and (4) in (3) \alpha \approx (DG/OE) \Upsilon \approx (DG/IE)(6) We also note that, since \Upsilon is an exterior angle of triangle ICD \Upsilon = \beta + r(7) Similarly,\beta = \alpha + s(8) Using (1) in (7) and (8) \Upsilon = \beta + s(9) \beta = \alpha + s(10) Eliminating s in (9) and (10), \Upsilon - \beta = \beta - \alpha or2\beta - \Upsilon - \alpha = O(11) Substituting (6) and (2) in (11) [(2DG)/R] - (DG/IE) - (DG/OE) = O or(1/IE) + (1/OE) = (2/R)(12) However, if \alpha, \beta and \Upsilon are small, IE \approx IG = i OE \approx OG = O(13) where i and O are the distances of the image from the mirror (image distance) and the object from the mirror (object distance). Hence, using (13) in (12) we have the mirror formula (1/i) + (1/O) = (2/R)

Question:

Besides their actions, the sympathetic and parasympathetic systems also differ in the neurotransmitter they release. Explain.

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/Users/wenhuchen/Documents/Crawler/Biology/F20-0515.htm

Solution:

Nerves of the parasympathetic system secrete a neurotransmitter called acetylcholine. For this reason they are usually referred to as cholinergic neurons. Acetylcholine is also the transmitter at the neuromuscular junction. Nerves of the sympathetic system release noradrenaline, also called norepinephrine, and are thus noradrenergic. Acetylcholine is a strong base, containing a choline moiety. [-CH_2CH_2-N_+-(CH_3)_3] In exists as a cation (positive ion) at physiological pH (about 7.4). Because of its ability to attach to a membrane and create a reversible change in the membrane's permeability to different ions, acetylcholine released by the presynaptic neuron acts to bring about depola-rization and generate an impulse in the postsynaptic neuron. The molecular structure of acetylcholine is: Noradrenaline has a molecular structure containing a ring moiety: The closely related compound adrenaline (epinephrine) - is a hormone released by the adrenal glands, and although its action is similar to that of noradrenaline, it is not a neurotransmitter. The adrenal medulla (inner part of the adrenal gland) acts like a modified post- ganglionic sympa-thetic "neuron": it is derived embryologically from neural tissue; it is innervated by a sympathetic preganglionic fi-ber; and it has become specialized to secrete noradrenaline and adrenaline and thus prolongs sympathetic activation.

Question:

What are the three laws of thermodynamics? Discuss their biological significance.

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0007.htm

Solution:

The first law of thermodynamics states that energy can be converted from one form into another, but it cannot be created or destroyed. In other words, the energy of a closed system is constant. Thus, the first law is simply a statement of the law of conservation of energy. The second law of thermodynamics states that the total entropy (a measure of the disorder or randomness of a system) of the universe is increasing. This is characterized by a decrease in the free energy, which is the energy available to do work. Thus, any spontaneous change that occurs (chemical, physical, or biological) will tend to increase the entropy of the universe. The third law of thermodynamics refers to a com-pletely ordered system, particularly, a perfect crystal. It states that a perfect crystal at absolute zero (O Kelvin) would have perfect order, and therefore its entropy would be zero. These three laws affect the biological as well as the chemical and physical world. Living cells do their work by using the energy stored in chemical bonds. The first law of thermodynamics states that every chemical bond in a given molecule contains an amount of energy equal to the amount that was necessary to link the atoms together. Thus, living cells are both transducers that turn other forms of energy into chemical bond energy and liberators that free this energy by utilizing the chemical bond energy to do work. Considering that a living organism is a storehouse of potential chemical energy due to the many millions of atoms bonded together in each cell, it might appear that the same energy could be passed continuously from organism to organism with no required extracellular energy source. However, deeper consideration shows this to be false. The second law of thermodynamics tells us that every energy transformation results in a reduction in the usable or free energy of the system. Consequently, there is a steady increase in the amount of energy that is unavailable to do work (an increase in entropy). In addition, energy is constantly being passed from living organisms to non-living matter (e.g., when you write you expend energy to move the pencil, when it is cold out your body loses heat to warm the air, etc.). The system of living organisms thus cannot be a static energy system, and must be replenished by energy derived from the non-living world. The second law of thermodynamics is also helpful in explaining the loss of energy from the system at each successive trophic level in a food pyramid. In the following food pyramid, The energy at the producer level is greater than the energy at the consumer I level which is greater than the energy of the consumer II level. Every energy transformation between the members of the successive levels involves the loss of usable energy, and this loss serves to increase the entropy. Thus, this unavoidable loss causes the total amount of energy at each trophic level to be lower than at the preceding level .

Question:

How does energy release in combustion differ fromenergy releasein aerobic respiration?

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/Users/wenhuchen/Documents/Crawler/Biology/F03-0080.htm

Solution:

Since glucose serves as the primary energy source, we will considerthe oxidation of glucose. Com-bustion refers to the rapid oxidation ofa substance with the liberation of heat. When a mole of glucose is burnedoutside of a living cell, approximately 690 kilocalories are released asheat. This great amount of heat would destroy a cell if released in one burst. Instead, the cell oxidizes glucose by a series of reactions (in aerobic respiration), each catalyzed by a specific enzyme. The re-lease of energy issequential, and in small packets thus preventing damage to the cell. During oxidativephosphor-ylation, the energy is transformed and stored in theform of ATP. The complete degradation of one mole of glucose yields 36 ATPs.Since each high energyphosphate bond (\simP) holds about 7 kilocaloriesof energy, 252 (36 × 7) kilocal-ories are biologically useful, whilethe other 438 (690 252) kilocalories are released as heat. Aerobic respira-tion, although only 36.5% efficient, is able to harness the energy fromthe oxidation of glucose to provide an en-ergy source for later needs, withoutresulting in trauma to the organism.

Question:

(a)Ag_2 O \rightarrow Ag + O_2 ; (b)Zn +HCl+ ZnCl_2 + H_2 ; (c)NaOH + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O.

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Solution:

When balancing chemical equations, one must make sure that there are the same number of atoms of each element on both the left and right side of the arrow. For example, H_2 + O_2 \rightarrow H_2 O is not a balanced equation because there are 2 O' s on the left side and only one on the right. 2H_2 + O_2 \rightarrow 2H_2 O is the balanced equation for water be-cause there are the same number of H and 0 atoms on each side of the equation. (a)Ag_2 O \rightarrow Ag + O_2 is not a balanced equation because there are 2 Ag on the left and only one on the right, and because there is only one O on the left and two O on the right. To balance this equation one must first multiply the left side by 2 to have 2 O' s on each side. 2Ag_2 O \rightarrow Ag + O_2 There are now 4 Ag on the left and only one on the right, thus the Ag, on the right must be multiplied by 4. 2Ag_2 O \rightarrow 4Ag + O_2 The equation is now balanced. (b)Zn + HC1 \rightarrow ZnCl_2 + H_2 In this equation, there are 2 H and 2Clon the right and only one of each on the left, therefore, the equation can be balanced by multiplying the HC1 on the left by 2. Zn + 2HCl \rightarrow ZnCl_2 + H_2 Because there are the same number of Zn, C1, and H on both sides of the equation, it is balanced. (c)NaOH + H_2 SO_4 \rightarrow Na_2 SO_4, + H_2 O Here, there are 1 Na, five O, 3 H and 1 S on the left and 2 Na, 1 S, five O, and 2 H on the right. To balance this equation, one can first adjust the Na by multiplying the NaOH by 2. 2NaOH + H_2 SO_4 \rightarrow Na_2 SO_4 + H_2 O There are now 2 Na, six O, 4 H, and 1 S on the left and Na, five O, 2 H, and 1 S on the right. Because there are two more H and one more O on the left than on the right, you can balance this equation by multiplying the H_2 O by 2. 2NaOH + H_2 SO_4 Na_2 SO_4 + 2H_2 O The equation is now balanced.

Question:

A reacts with B in a one-step reaction to give C. The rate constant for the reaction is 2.0 × 10^-3 M^-1 sec^-1. If 0.50 mole of A and 0.30 mole of B are placed in a 0.50 liter-box, what is the initial rate of the reaction?

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Solution:

The equation for this reaction can be written: A + B \rightarrow C From this, one can write the rate law assuming that the reaction is first order in both A and B. When a re-action is first order in a particular reactant, it means that the rate is proportional to the concentration of the reactant. Thus, the rate law is written Rate = k[A] [B], where k is the rate constant and [ ] indicates concentration. In this problem, one is given k, [A] and [B]. Thus, Rate= 2.0 × 10^-3 M^-1 sec^-1 × (1 M) (.60M) = 1.20× 10^-3 M/sec k = 2.0 × 10^-3 M^-1 sec^-1 [A] = 0.50 mole / 0.5 liter = 1 M [B] = 0.30 mole / 0.5 liter = 0.60 M .

Question:

Show that the average distance, l, a molecule travels between collisions in a gas is related to the number density n and the molecular diameter d by l = [1/(n\pid^2)].

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0492.htm

Solution:

Consider a molecule moving through a region of stationary molecules. It can collide with those molecules whose centers are at a distance less than or equal to d from its center. If the speed of the molecule is v, it moves a distance vt in time t and collides with every molecule in the cylindrical volume vt\pid^2 (Fig,A). The molecules in this cylinder cannot avoid the incident molecule because the diameter of the cylinder is 2d, twice that of a molecule. If there are n molecules per unit volume, the number of molecules in the cylinder is N = nvt\pid^2. Actually, after each collision, the molecule changes its direction and the cylinder mentioned before makes zigzags as shown in Fig. B. The number of collisions is just the number of molecules in the cylinder, hence the average distance per collision is l \approx (vt)/(nvt\pid^2) = 1/(n\pid^2). l is called the mean free path.

Question:

Insulin controls the glucose level in the blood. In 1969, Hodgkin announced the secondary structure of insulin, the first protein hormone to be resolved by X-ray diffraction, (a) Chemial analysis shows insulin is 2.15% zinc by weight. Calculate the minimum molecular weight, (b) Osmotic pressure studies show the molecular weight of insulin is about 6000. How many zinc atoms does the molecule contain?

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Solution:

The minimum molecular weight of insulin can be determined by assuming that there is one zinc atom per molecule. This means that in 1 mole of insulin there exists 1 mole of zinc. From the knowledge that zinc is 2.15% by weight of insulin, and that the MW of zinc is 65.3, one can determine the minimum MW of insulin by using, the following relationship. (.0215)(MW of insulin) = (MW of zinc) MW of insulin = (MW of Zn) / (.0215) = (65.3 g/mole) / (.0215) = 3037 (g/mole) (b) If the true molecular weight of insulin is about 6000, then two of the units weighing 3037 (g/mole) make up each molecule of insulin. Thus, there are 2 atoms of Zn per molecule.

Question:

Follow the flow chart and the group of items given to write a program called DATA-ALTERATION. You may omit the Configuration Section, but use the required division header. Use two independent data items, LAST-NAMB, which is fifteen characters long, and SOCIAL-SEC, which is nine characters long. FIRST-NAME STUDENT-RECORD INIT 1 char-acter REST-OF-NAME 8 characters STUDENT-NO. 9 char-acters STUDENT-NAME 12 characters S-INIT 1 char-acter In the above, note that FIRST-NAME, STUDENT-RECORD, INIT, REST-OF-NAME, STUDENT-NO., STUDENT-NAME, and S-INIT are variables used in the program to be written.

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Solution:

In following the flow chart, the START statement covers the IDENTIFICATION and ENVIRONMENT DIVISION. In reality a programmer who works for a company almost always uses the same IDENTIFICATION and ENVIRONMENT divisions, keeping in mind that changes such as DATE-WRITTEN and DATE- COMPILED are essential. In DATA-DIVISION the programmer is required to use two independent data items. In COBOL independent data items are identified by level number 77; therefore the requireditems may be written as: 77 LAST-NAME PIC X(15). 77 SOCIAL-SEC PIC X(9). Observe that the first item has 15, and the second item has 9 characters from the standard character set. Examining the group of items given in the question, it is clear that there are two group items each with their own subdivisions, namely, FIRST-NAME, and STUDENT-RECORD. In format, these group items get identified by level numbers 01 through 49, the first one being the highest level of the data item, and starting from Margin A. The rest of the level numbers are called secondary and start from margin B. Following these rules, the given group of items may be formatted as: AB 01FIRST-NAME. 02 INIT PIC X(1). 02 REST-OF-NAME PIC X(8). 01STUDENT-RECORD 02 STUDENT-NAME X(12) 02 STUDENT-NO PIC X(9). 02 S-INIT PIC X(1) Now having finished the IDENTIFICATION, ENVIRONMENT, and DATA DIVISIONS the flow chart may be followed to write the PROCEDURE DIVISION. Step by step examination of the given flow-chart will enable the programmer to write the required statements. There are two paragraph names given in the flow-chart; FIRST- PARA, and LAST-PARA. The first paragraph consists of the following opera-tions : 1) Display the sentence "ENTER LAST NAME" on console (screen or typewriter); after the name is entered by the user, ac-cept it from typewriter or screen. 2, 3) Do the same for FIRST NAME, and SOCIAL SECURITY Number. 4) Take these accepted items and move them to their appro-priate places in Student-RECORD 5) Print the STUDENT-RECORD on the standard output device. In the last paragraph the computer is told to display the sentence "THANK YOU THATS ALL" on screen or on the type-writer and stop the execution. Following these rules and explanations the complete program is given in fig. 1.

Question:

The figure shows a battery and a resistor in series. What is the power P supplied by the battery and the power dissipated as heat?

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Solution:

The total resistance of the circuit is R_t = r + R = (5.5 + 0.5) \Omega = 6 \Omega Therefore the current in the circuit is, by Ohm's Law, i = (\epsilon/R_t) = (12 v/6 amp) = 2 amp. Point b is at a higher potential than point a, and V_ba = iR = 2 amp (5.5\Omega) = 11 volts or, by Kirchoff's Law V_ba = \epsilon \rule{1em}{1pt} ir = 12 \rule{1em}{1pt} (2 × 0.5) = 11 volts and V_ab = \rule{1em}{1pt} V_ba = \rule{1em}{1pt} 11 volts. In the lower rectangle, the direction of the current is from b to a, hence the power dissipated in R is P_R = V_ba × i = (2 amp) × (11 v) = + 22 watts. The energy supplied to R is converted to heat at a rate energy/time = i^2R = (2 amp)2 x 5.5 \Omega = 22 watts which equals the power input to R. The energy loss in the battery is due to the internal resistance r; P_r = i^2r = (2 amp)^2 × 0.5 \Omega = 2 watts. The rate of conversion of nonelectrical to electrical energy in the seat of emf is given by P_\epsilon = i\epsilon = (2 amp) × (12 v) = 24 watts. The total rate of energy dissipation in the resistors should add up to the rate at which energy is supplied by the emf, orP_\epsilon = P_R + P_r = (22 + 2) watts = 24 watts.

Question:

Snails can usually be seen crawling around during the night and on dark, damp days. What happens to them on bright, sunny days and why?

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Solution:

The soft body of the snail is protected by a hard calcareous shell secreted by the mantle. The mantle is a fold of tissue that covers the visceral mass and projects laterally over the edges of the foot. During the daytime, the snail usually draws up inside the shell. In order to move and to feed, the foot and head must emerge from the shell. This usually occurs at night. In bright sunlight, the soft body of the snail would dry out. Slugs aremolluscssimilar to snails, but without hard shells. Slugs are generally found in damp forest environments, and remain in the moist earth during bright daylight. Mostmolluscsare aquatic animals, and hence are in no danger of drying out. Land snails are one of the few groups of fully terrestrial invertebrates. In most land snails, the gills have disappeared, but the mantle cavity has become highlyvascularizedand functions as a lung.

Question:

Describe the composition and structure of the plant cell wall.

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Solution:

Because plant cells must be able to withstand high osmotic pressure differences, they require rigid cell walls to prevent bursting. This rigidity is provided primarily by cellulose, the most abundant cell wall component in plants. Cellulose, a polysaccharide, is a long, linear, unbranched polymer of glucose molecules. In the cell wall cellulose molecules are organized in bundles of parallel chains to form fibrils. The fibrils are often arranged in criss-cross layers reinforced and held together by three other polymeric materials: pectin and hemicellulose (complex polysaccharides) and extensin (a complex glycoprotein). Lignin is an addi-tional polymeric substance found in the wood of trees. Together these substances provide a matrix capable of withstanding enormous stress. A layer known as the middle lamella lies between and is shared by adjacent cells. The lamella is composed primarily of pectin. By binding the cells together, it provides additional stiffness to the plant. The cell wall is usually interrupted at various locations by plasmodesmata. These, are tiny holes in the cell walls through which run protoplasmic connections between adjacent cells. This provides for intercellular exchange of such materials as water and dissolved sub- stances. Being extremely small, the plasmodesmata do not prevent the cell wall from exerting pressure on a swollen cell.

Question:

(a) If the x direction is normal to the surface of the earth and directed upward, the gravitational force is F_G = -Mgx˄, where g is the acceleration of gravity and has the approximate value 980 cm/sec^2. Calculate the work done by gravity when a mass of 100 gm falls through 10 cm. (b)If the particle in (a) was initially at rest, what is its kinetic energy and its velocity at the end of its 10-cm fall?

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Solution:

(a) Work done by a force F^\ding{217} is calculated by \intF^\ding{217} \textbulletds^\ding{217}. Here F^\ding{217} =mgx˄ which is a constant, and since the object falls in a straight line, the work is given by W =mgx This is a scalar quantity since we are taking the dot product of F^\ding{217} \textbulletds^\ding{217}. Since F^\ding{217} and s^\ding{217} are parallel, we take the simple arithmetic product: W = (100 gm) (980 cm/s)(10 cm) = 980,000 ergs (b) The initial value K_A of the kinetic energy is zero; the terminal value K_B is equal to the work done by gravity on the particle, so that K_B = W \approx 10^6 ergs = (1/2) mv^2_B, whence v^2_B \approx 2(10^6 ergs)/(100 gm) \approx 2 × 10^4 cm^2/sec^2. Therefore,v_B\cong 1.41 × 10^2 cm/sec We may obtain the same result from the equations of motion. We have v =gtand h = (1/2) gt^2. (These equations are adaptations of the more complete equations of motion, d = (1/2) at^2 + v_0t + d_0 and v = at + v_0, to the initial conditions of our problem where v_0 = 0, d_0 is assumed to be 0, and a = g.) Eliminating t, we have v^2 = 2gh. Therefore, v^2 = 2(980 cm/s^2) (10 cm) v^2 \cong 2 × 10^4 cm^2/s^2 v \cong 1.41 × 10^2 cm/sec.

Question:

If the coefficient of sliding friction for steel on ice is 0.05, what force is required to keep a man weighing 150 pounds moving at constant speed along the ice?

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Solution:

To keep the man moving at constant velocity, we must oppose the force of friction tending to retard his motion with an equal but opposite force (see diagram). The force of friction is given by: F = \mu_kineticN By Newton's Third Law F_forward = F_friction Therefore F_forward = \mu_kineticN F_forward = (.05)(150 lb) = 7.5 lb.

Question:

The standard E\textdegree for(1/2)F_2(g) + e^- \rightarrow F^-(aq), which is +2.87V, applies when the flouride ion concentration is 1M . What would the corresponding E be in 1M H_3O^+ , i.e., for the electrode reaction (1/2)F_2(g) + e^- + H_3O^+ \rightarrow HF(aq) +H_2O? (The K_diss of HF is 6.7 × 10^-4).

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Solution:

One can solve for E for the electrode reaction by using the Nernst equation. This is stated E = E\textdegree - (0.591 / n)logQ , where E is the actual electrode potential for the half-reaction, E\textdegree is the standard potential for the half-reaction, n is the number of electrons transferred in the reaction, Q is the dissociation con-stant. Thus, one must find the concentration of F^-, in order to eval-uate Q, before solving for E\textdegree . The dissociation constant is equal to. [H_3O^+][F^- ] / [HF] . One is given that the dissociation constant (K_diss)for 1M HF is 6.7 × 10^-4. Thus, K_diss = [H_3O^+][F^- ] / [HF] = 6.7 × 10^-4 = Q . Solving for the E of the electrode reaction: For the half-reaction (1/2)F_2(g) + e-\rightarrow F^-(aq), one electron is trans-ferred, thus n = 1. One is given that E\textdegree = 2.87V and one has de-termined Q. Thus, E = 2.87V -(0.0591 / 1) log (6.7 × 10^-4) E = 2.87V - 0.0591 × (-3.17) E = 2.87V + .188 = (+3.06V) .

Question:

Find the minimum radius for a planet of mean density 5500 kg \bullet m^-3 and temperature 400\textdegreeC which has retained oxygen in its atmosphere.

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Solution:

The escape velocity from a planet is given by the relation V = \surd(2GM/r) = \surd[2\surd{(G × (4/3)\pir^3\rho)/r}] = \surd[(8/3)G\pi\rhor^2](1) where r is the planet radius, M is its mass, and \rho is its density If most oxygen molecules have velocities greater than this, then, when they are traveling upward near the top of the atmosphere, they will escape into space and never re-turn. A slow loss of oxygen from the atmosphere will there-fore take place. In this case, however, we are told that the planet has retained its oxygen and we can assume that escape velocity from the planet is greater than thermsvelocity of the oxygen molecules. When the two are equated, the minimum radius for the planet results. We need thermsvelocity of oxygen molecules. This speed V is so defined that the internal energy U would be the same if all the atoms had this speed. For a gas consisting of N atoms, V is defined by U = N(1/2)mv^2(2) where m is the mass of one atom. If the gas is ideal and monatomic, then we also know that U = N(3/2)kT(3) where k is Boltzmann's constant and T is the temperature of the gas in \textdegreeK. Oxygen is neither ideal or monatomic but equation (3) is still a good approximation since the gas is not very dense and thereforeinteratomic forces can be ignored. Equating equations (2) and (3), we get mv^2 = 3kT. Multiplying both sides by Avogadro's number N_A (the number of molecules in one mole of the gas), N_Amv^2 = 3N_AkT(4) ButN_Amequals the mass, M', of one mole of the gas. Also, by definition N_Ak = R where R is the universal gas constant. Substituting these two expressions in equation (4) yields M'v^2 = 3RT. Solving for the velocity, we have v = \surd(3RT/M')(5) Set equations (1) and (5) equal to each other so as to find the minimum radius. Then \surd[(8/3)G\pi\rhor^2_min] = \surd(3RT/M'). where T is the absolute temperature in the atmosphere, and M' is the mass per mole of O_2. The temperature of the oxygen is 400\textdegreeC + 273\textdegree = 673\textdegreeK. Oxygen gas is diatomic, and its effective molecular mass is therefore twice the atomic mass of monatomic oxygen M' = (2 × 16)g \bullet mole^-1 = 32 × 10^-3 kg \bullet mole^-1 r_min= \surd[(9RT)/(8G\pi\rhoM')] = \surd[(9 × 8.315 J\bulletmole^-1\bulletK deg^-1 × 673 K deg)/(8 × 6.67 × 10^-11 N\bulletm^2\bulletkg-2 × \pi × 5500 kg\bulletm^-3 × 32 × 10^-3 kg\bulletmole^-1)] \thereforer_min= \surd(1.708 × 10^11 m^2) = 4.131 × 10^5m = 413.1 km.

Question:

Define top-down structured programming. Also define bottom-up structured programming. What importance does each of these terms have in relation to the computer implementation of an algorithm?

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Solution:

The top-down design technique is the offspring of the idea of refinement. This technique requires that the programmer define the algorithm coarsely, in plain English or with some sort of code. Next, a structured language or a flow chart should be used to explain the parts of the algorithm that need greater clarification. This process of writing more and more detailed descriptions continues until the algorithm is sufficiently documented. Thereafter, the translation of the algorithm into a standard computer language, such as COBOL or FORTRAN, becomes a simple mechanical procedure. This technique is especially useful when a team of programmers is working on a specific problem. If each programmer is assigned a small part of a large task, the parts can be combined easily by assembling the codes from each part. If the modular programming rule of one-entry- one- exit per module is followed as well, the sections of code become like interchangeable parts of a machine. Any other programmer can pick up your section of code and substitute it into a program that needs what you have written. Bottom-up design is just the opposite. It refers to the creation of the program in a highly detailed fashion before coding or documentation is begun. This is sometimes bad practice when writing programs because the programmer is prone to miss important details. Also, it is both time-saving and economical to iron out errors in the coding stage before typing out a program for computer execution. This idea of program- first, code-second may be used if one wants to debug a program. By looking at a program first, you might be able to find syntactical errors that cause problems. After this is done, then a look at the code allows you to debug any logical errors.

Question:

Profound changes in uterine morphology occur during the menstrual cycle and are completely attributable to the effects of estrogen and progesterone. Describe these changes.

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0559.htm

Solution:

In the uterus estrogen stimulates growth of the smooth muscle layer called the myometrium, and the glandular epithelium called the endometrium, lining its inner surface. Progesterone acts on the endometrium and converts it to an actively secreting tissue. The glands become coiled and filled with glycogen; the blood vessels become more numerous; various enzymes accumulate in the glands and connective tissue of the lining. These changes are ideally suited to provide a favorable environment for implantation of a fertilized ovum. Progesterone also causes the mucus secreted by the cervix (a muscular ring of tissue at the mouth of the uterus which projects into the vagina) to become thick and sticky. This forms a "plug" which may constitute an important blockage against the entry of bacteria from the vagina. This is a protec-tive measure for the fetus should conception occur. As a result of the regression of the corpus luteum when no fertilization occurs, there is a fall in blood progesterone and estrogen levels. Consequently, the highly developed endometrial lining becomes deprived of its hormonal support. The most immediate result of this deprivation is the constriction of the uterine blood vessels, resulting in a decreased flow of oxygen and nutrients to the tissue. Subsequently, the lining disintegrates and gets sloughed off as the mentrual flow begins. This phase is called the menstrual phase and usually lasts for about 5 days. The flow ceases as the endometrium repairs itself, and then grows under the influence of the rising blood estrogen concentration. This phase is called the proli-ferative phase, and lasts for 9 days. This phase covers the time from menstruation up until ovulation. After ovulation and formation of the corpus luteum, progesterone acts synergistically with estrogen to induce formation of the glandular endometrium. This period, called the secretory phase (lasting about 14 days) is terminated by disintegration of the corpus luteum. Thus the cycle is completed and a new one begins. The phases of the menstrual cycle can be named either in terms of the ovarian or uterine events. The cycle is di-vided into follicular and luteal phases with respect to the ovary; proliferative and secretory phases are the terms re-ferring to the uterine state. It is important to realize that the uterine changes simply reflect the effects of vary-ing blood concentrations of estrogen and progesterone throughout the cycle.

Question:

When a Kundt's tube contains air, the distance between several nodes is 25 cm. When the air is pumped out and replaced by a gas, the distance between the same number of nodes is 35 cm. The velocity of sound in air is 340 m \bullet s^-1. What is the velocity of sound in the gas?

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Solution:

A Kundt's tube (see the figure) is a glass tube with sawdust spread over its interior. A piston P is set into longitudinal vibration and the wall W moved until a standing wave pattern of nodes and antinodes is set up within the tube. The sawdust will accumulate at all the nodal points, for these are the points of the wave for which there is no vibration of the gas. By measuring the distance d between the nodes, we may find the velocity of sound of the gas in the tube. Let the number of nodes involved be 2n + 1. There are 2n intervals between these 2n + 1 nodes, and the total distance thus corresponds to 2n half-wavelengths, or to n full wavelengths. The frequency of the emitted sound is the same in both cases, for it depends only on the frequency of the vibra-ting membrane which is causing the standing wave within the Kundt's tube. Thus c_A = f\lambda_A and c_G = f\lambda_G. Since there are, in the case of the air filled tube, n wavelengths in 25 cm of space, then the number of centi-meters of space one wavelength will contain is found by the following proportion. (25 cm / n wavelength) = (x cm / 1 wavelength) = \lambda_A Similarly for the case of the gas filled tube, (35 cm / n wavelength) = (x cm / 1 wavelength) = \lambda_G Thus c_G / c_A = \lambda_G / \lambda_A = [(35/n) cm] / [(25/n) cm] = 7/5. \therefore c_G = (7/5) × 340 m \bullet s^-1 = 476 m \bullet s^-1. If the number of nodes involved is 2n, then there are 2n - 1 intervals between these 2 nodes. The total dis-tance thus corresponds to 2n - 1 half-wavelengths, or to n - 1/2 full wavelengths. The same analysis as before holds, and c_G / c_A = \lambda_G / \lambda_A = [{(35/n) - (1/2)}cm] / [{(25/n) - (1/2)}cm] = 7/5 The same result is obtained, as it must, if an odd number of nodes are counted as if an even number of nodes are counted.

Question:

Calculate the pH of (a) a 0.5 M solution with respect to CH_3COOH and CH_3COONa; (b) the same solution after 0.1 mole HCl per liter has been added to it. Assume that the volume is unchanged. K_a = 1.75 × 10^-5.

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Solution:

Part (a) involves the determination of the concentration of acid that has dissociated. Since one is adding a strong acid, part (b) is a buffer type problem. For part (a), CH_3COOH is the acid and for each x moles of it that dissociates, x moles of H_3O^+ and x moles of CH_3COO^- are formed. This is seen more clearly below. CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- After reaction 0.5 - xxx There is also CH_3COONa present in this solution. Its effect on the equilibrium must be also taken into account. Since it is a salt it will completely dissociate, according to the following equation, in this dilute solution. CH_3COONa \rightleftarrows CH_3COO^- + Na^+ It is 0.05 M, thus, in the solution there will be 0.05 M CH_3COO^- and Na^+ contributions from it. The con-centrations of the various species found in this solution are shown in the table below. CH_3COOH(0.05 - x)M H_3O^+x M CH_3COO^-(0.05 + x)M (contributions are from both the CH_3COOH and CH_3COONa equilibria.) Na^+0.05 M pH is defined as being equal to - log [H_3O^+].Thus, to solve for pH, one must first determine [H_3O^+]. This can be done by setting up the equation for the equilibrium of the reaction CH_3COOH + H_2O \rightleftarrows CH_3COO^- + H_3O+ K_a = {[H_3O^+] [CH_3COO^-]} / [CH_3COOH] = 1.75 × 10^-5. 1.75 = {(x) (0.05 + x)} / (0.05 - x) Since CH_3COOH is a weak acid, there will be little dissociation and x will be very small. Thus one can approximate 0.05 + x \cong 0.05 and 0.05 - x \cong 0.05. Solving: {x (0.5)} / (0.5) = 1.75 × 10^-5 x = [H_3O^+] = 1.75 × 10^-5 pH = - log [H_3O^+] = - log (1.75 × 10^-5) = 4.76. Part (b): A buffer solution contains an acid and its conjugate base, both are moderately weak. What this type of solution does is to prevent a large change in pH when more acid or base is added. In a nonbuffered solution, the pH will change drastically. If a strong acid is added to a buffered solution, it reacts with the base A^- to produce a weak acid. H_3O^+ + A^- \rightleftarrows HA + H_2O. When a strong base is added the same effect occurs. It reacts with the acid ro produce a weak base. OH^- + HA \rightleftarrows A^- + H_2O. The acid-base ratio is altered somewhat but not to any great extent, thus allowing the pH to stay relatively stable. However, if the number of moles of acid or base added is greater than the number of moles of base or acid present, then the pH will change, thereby destroying the buffer. It is given that 0.1 M HCl is added. Since HCl is a strong acid, and dissociates completely, the H_3O^+ and Cl^- formed each have an initial concentration of 0.1 M. How-ever, the H_3O^+ will react with the CH_3COO^- and drive the equilibrium to the left by Le Chatelier's principle and produce a weak acid. Therefore, the buffer has converted the acid ions so that they cannot change the pH. Le Chatelier's principle states that when a stress is brought to bear on a system at equilibrium, the system tends to change so as to relieve the stress. The stress is relieved by driving the reaction to the other side of the equilibrium sign to prevent excesses of ions from building up. The net effect can be seen below. CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- Before:0.5 - xx0.5 + x After add of0.5 + 0.1 - xx0.5 - 0.1 + x 0.1 M H_3O^+(0.6 - x)(0.4 + x) At the new equilibrium, there will be x moles of H_3O^+ still in solution, which were formed from the redissociation of CH_3COOH. x moles of CH_3COO^- are also formed resulting in the new concentrations of CH_3COOH + H_2O \rightleftarrows H_3O^+ + CH_3COO^- 0.6 - xx0.4 + x Upon setting up the equilibrium constant equation, one obtains [H_3O^+]. K_a = {[H_3O^+] [CH_3COO^-]} / [CH_3COOH] = 1.75 × 10^-5 1.75 × 10^-5 = {(x)(0.4 + x)} / (0.6 - x) Using the assumption outlined above, 0.6 - x \cong 0.6, and 0.4 + x \cong 0.4. Solving: 1.75 × 10^-5 = {(x)(0.4)} / (0.6) x = 2.63 × 10^-5 = [H_3O^+] Since, pH = - log [H_3O^+], one finds by substitution, pH = - log 2.63 × 10^-5 = 4.58.

Question:

A flask containing H_2 at 0\textdegreeC was sealed off at a pressure of 1 atm and the gas was found to weigh, 4512 g. Calculate the number of moles and the number of molecules of H_2 present.

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Solution:

A mole is defined as 6.023 × 10^23 molecules of a substance. Hydrogen has a molecular weight of 1.008, therefore, 1 mole of H_2 weighs 2.016 grams. The number of moles present is therefore moles = (gram/M.W.) = [(.4512 g)/(2.016 g/mole)] = .2238 mole of H_2 To calculate the number of molecules, recall that 1 mole of any gas at STP has 6.023 × 10^23 molecules/mole. no. of molecules = 6.023 × 10^23 (molecules/mole) × no. of moles = 6.023 × 10^23 (molecules/mole) × 0.2238 moles = 1.348 × 10^23 molecules of H_2 .

Question:

Given 25 equally spaced data points x_i (i= 1,...,25) and 25 functionalvaluesy_i= f(x_i ) [arranged in ascending order ], writea program to evaluate^(X)25\int_(X)1 f(x)dxusing the trapezoidalrule.

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Solution:

According to the trapezoidal rule of integration, (X)25\int_(X)1 f (x)dx= Area \approx (h/2) (y1 + 2y2 + 2y3 + ...+ 2y24 + y25). We adopt the following computational strategy: accumulate 24\sum_i=2y_iin a DO loop, double the result, add it to (y1+ y25), and multiply by h =X(2) - X(1). The program (which assumes values of X and Y have alreadybeen read in) is given below. You should also remember that the error margin for the trapezoid rule is givenby E \leq (1/12 )( )( b - a)h2 \vertf"(\xi)\vertX_n\leq \xi \leq X1. b - a)h2 \vertf"(\xi)\vertX_n\leq \xi \leq X1. Knowing this, you can use a suitable N to calculate the integration to a closeapproximation. REAL H SUM = 0.0 DO 50 I = 2,24 50SUM = SUM +Y(I) H =X(2) - X(1) AREA = 0.5\textasteriskcenteredH\textasteriskcentered(Y(1) + 2.0\textasteriskcenteredSUM + Y(25))

Question:

Can complete equilibrium in a gene pool exist in real situations?

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Solution:

Of the four conditions necessary for the genetic equilibrium described by the Hardy-Weinberg Law, the first, large population size is met reasonably often; the second, absence of mutations, is never met; the third, no migration, is met sometimes; and the fourth, random reproduction, is rarely met in real situations. Therefore it follows that complete equilibrium in a gene pool is not expected. With regard to the first condition, many natural populations are large enough so that chance alone is not likely to cause any appreciable alteration in gene frequen-cies in their gene pools. Any breeding population with more than 10,000 members of breeding age is probably not significantly affected by random changes. The second condition for genetic equilibrium the absence of mutations, is never met in any population be-cause spontaneous mutations are always occurring. Most genes probably undergo mutation once in every 50,000 to 1,000,000 duplications, with the rate of mutation for different genes varying greatly. However, since the rate of spontaneous mutation is usually low, it is usually insignificant in altering the gene frequencies in a large population. The third condition for genetic equilibrium, implies that a gene pool cannot exchange its genes with the out-side. Immigration or emigration of individuals would change the gene frequencies in the gene pool. A high percentage of natural populations, however, experience some amount of migration. This factor, which enhances variation, tends to upset. Hardy Weinberg equilibrium. The fourth condition, random reproduction, refers not only to the indiscriminate selection of a mate but also to a host of other requirements that contribute to success in propagating the viable offspring. Such factors include the fertility of the mating pair, and the survival of the young to reproductive age. An organism's genotype actually influences its selection of a mate, the physical efficiency and frequency of its mating, its fertility, and so on. Thus entirely random reproduction in reality is not possible. We can conclude, therefore, that if any of the conditions of the Hardy-Weinberg Law are not met, then the gene pool of a population will not be in equilibrium and there will be an accompanying change in gene frequency for that population. Since it is virtually impossible to have a population existing in genetic equilibrium, even with animals under laboratory conditions, there must then be a continuous process of changing genetic constitutions in all populations. This is ultimately related to evolu-tion in that evolutionary change is not usually automatic, but occurs only when something disturbs the genetic equilibrium.

Question:

Discuss the responses of stomata to environmental factors.

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Solution:

A number of environmental factors affectstomatalopening and closing. Water loss is the most important among them. When theturgorof a leaf drops below a certain critical point because of rapid evapo-ration to the atmosphere, thestomatalaperture responds to this loss by becoming smaller or closed. In this situation, the need for water conservation overrides the need for the intake of carbon dioxide for photosynthesis. With the stomata closed, the leaf cells can use the carbon dioxide produced by their respiration, and thus be able to sustain a very low level of photosynthesis. Factors other than water loss that affect the stomata include light, carbon dioxide concentration, and temperature. It has been shown that when a turgid leaf which has been kept in darkness for a few hours is exposed to light, the stomata open. When the light is switched off, the stomata close. Light apparently plays a role in the opening and closing of the stomata. Light is now believed to initiate a series of enzymatic reactions which break down insoluble starch to many smaller, soluble glucose molecules in the guard cells of a stoma. As the solute concentration increases, water enters the guard cells by osmosis, increasing theirturgor, thereby opening the stomata. The role of carbon dioxide is similar to that of light, in the opening and closing of the stomata. It is found experimentally that a decrease in carbon dioxide increases the pH in a cell and an increase in pH facilitates the conversion of starch to glucose. This is explained by the fact that the enzymephosphorylaseis stimulated by the increased pH to convert starch to glucose-l-phosphate, which is in turn converted to glucose. As discussed earlier, water now enters the leaf cells,turgorpressure increases and as the guard cells bend, the stomata open. Temperature also fits into this scheme because of its influence on carbon dioxide concentration. An increase in temperature increases the rate of respiration, and so increases in turn the cellular carbon dioxide level. A decrease in temperature, on the other hand, retards respiration, and thus decreases the cellular carbon dioxide levels. It is clear that the stomata are responsive to a complex interplay of factors and are able to respond efficiently to different combinations of these factors.

Question:

Given the coordinates of three points (X_1 ,Y_1 ), (X_2 ,Y_2 ), and (X_3 ,Y_3), developea program which outputs a message regarding theircollinearity. (Three points are collinear if they all lie on the same line.) Include statements that will take into account round-off errors.

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Solution:

The solution to the problem rests on the fact that three points P_1 , P_2 , and P_3 are collinear if and only if the slopes of the segments P_1 P_2 and P_2 P_3 are equal. This implies (Y_2 -Y_1)/ (X_2 - X_1) = (Y3- Y_2)/ (X3- X_2) where (X_1,Y_1), (X_2 ,Y_2 ) and (X_3 ,Y_3 ) are the coordinates of three points. Instead of calculating slopes, we can test the expressions (Y2- Y_1) × (X_3 - X_2) and (Y_3 - Y_2) × (X_2 - Y_1) , for equality. If they are equal, the three points are collinear, and an appropriate message is output. This expression may also be used to establish the degree of round-off error. One way of doing this is with the statement ABS((Y2 - Y1){_\ast}(X3 - X2) - (Y3 - Y2){_\ast}(X2 - X1)) If this quantity is very small, say, less than .0001, we can accept the accuracy ofcollinearity. If this condition is not met, a message indicates that there is an unacceptable accuracy in the determination. We use a format-free READ statement in the program for simplicity's sake, a feature available on some machines. CCOLLINEARITY PROGRAM READ (5,{_\ast}) X1, Y1, X2, Y2, X3, Y3 ,ACCUR CARE THEY COLLINEAR? IF((Y2 - Y1){_\ast}(X3 - X2). NE. (Y3 - Y2){_\ast}(X2 - X1)) GO TO 30 WRITE (5,100) X1, Y1, X2, Y2, X3, Y3 100FORMAT (1X, 'THE PAIRS OF POINTS ', 3(1X,2(1X,F7.5)), 'ARE 1COLLINEAR') TEST = ABS((Y2 - Y1){_\ast}(X3 - X2) - (Y3 - Y2){_\ast}(X2 - X1)) IF (TEST. GE. ACCUR) GO TO 25 WRITE (5,101) ACCUR 101FORMAT(1X, 'WITH DESIRED ACCURACY TO ', F7.5) GO TO 99 25WRITE (5,101) ACCUR 101FORMAT (1X 'WITHOUT ACCURACY OF', F7.5) 30WRITE (5,102) 102FORMAT (1X,'POINTS ARE NOT COLL INEAR') 99STOP END

Question:

Give the approximate boiling points at sea level for the following: (a) 2molalHBr; (b) Suspension of 100 g of powdered glass in one liter of water; (c) 1.2 × 10^24 (sucrose molecules / liter); (d) 0.5molalBaCl_2.

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Solution:

The boiling point is acolligativeproperty and, therefore, depends upon the number of particles present in 1 kg of solvent. The boiling point constant for water is 0.52\textdegreeC. This means that for each mole of particles dis-solved in 1 kg of H_2O the boiling point will be elevated by 0.52\textdegreeC. boiling pt. elevation = 0.52\textdegreec × (molalityof solute). Themolalityis defined as the number of moles in one kg. of solvent. (a) 2molalHBr.HBris a strong acid and will thus ionize completely when diluted with H_2O to form H^+ and Br^- ions. HBr \rightleftarrows H^+ + Br^- Therefore, 2 particles will be formed by eachHBrthat ionizes. The effectivemolalityof the solution is then twice themolalityof theHBr. effectivemolality= 2 × 2molal= 4molal. One can now find the boiling point elevation, boiling pt. elevation = 0.52\textdegree × 4molal= 2.1\textdegree The normal boiling point of H2O is 100\textdegreeC the new boiling point is (100\textdegree + 2.1\textdegree or 102.1\textdegreeC). (b) Suspension of 100 g of powdered glass. These particles will not dissolve to their molecular components in H_2O, thus the boiling point of the water will not be changed. (c) 1.2 × 10^24 (sucrose molecules / liter). One liter of H_2O weighs 1 kg, therefore, 1.2 × 10^24 sucrose molecules are dissolved in 1 kg of H_2O. The number of moles present is equal to the number of molecules present divided by Avogadro's number, the number of particles in one mole. no. of moles = (1.2 × 10^24 molecules) / (6.02 × 10^23 (molecules /moles) = 2 moles Hence, there are 2 moles of sucrose in 1 kg of water, thus the solution is 2 molal . Sucrose does not ionize in water, therefore, the truemolalityis equal to the effectivemolality. Solving for the boiling point elevations: boiling pt. elevation = 0.52\textdegree × 2molal= 1.04\textdegree The boiling point of this solution is 100\textdegree + 1.04\textdegree or 101.04\textdegree. (d) 0.5molalBaCl_2. BaCl_2 is a strong electrolyte and will completely ionize in H_2O. BaCl_2\rightleftarrowsBa^++ + 2Cl^- 3 ions will be formed for each BaCl_2 present. The effectivemolality will therefore be 3 times themolalityof unionized BaCl_2. effectivemolality= 3 × 0.5molal= 1.5molal One can now solve for the boiling point elevation. boiling pt. elevation = 1.5molal× 0.52\textdegree = 0.78\textdegree The boiling point of H_2O, 100\textdegree, will be raised 0.78\textdegree to be 100.78\textdegreeC.

Question:

A string is wrapped around a cylinder of mass M and radius R (see figure (a)). The string is pulled vertically upward to prevent the center of mass from falling as the cylinder unwinds the string, (a) What is the tension in the string? (b) How much work has been done on the cylinder once it has reached an angular speed w? (c) What is the length of string unwound in this time?

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Solution:

We must first find the moment of inertia I of a cylinder of mass M and radius R. To start, we first calculate the moment of inertia of a disk of mass density \rho and radius R, whose thickness is negligible (see figure(b)). The differential mass element dm, shown as the dotted ring in the figure, has area 2\pir dr. Thus: dm = 2\pi \rhor dr Therefore: I = \int r^2 dm = \int r^2 (2\pi\rhor dr) = 2\pi\rho ^R\int_0 r dr = 2\pi\rho (r^4/4) ]^R_0 = (\pi\rhoR^4)/2 = (1/2)(\piR^2\rho)R^2 = (1/2) mR^2 We have used the fact that \piR^2\rho is the mass m of the disk, since \rho is the mass density, and \piR^2 is the total area. We can think of a cylinder as many disks squeezed together, and we know that the total moment of inertia about a common axis, of many objects, is the sum of the individual moments of inertia: I_t = I_1 + I_2 + I_3 + ..... In the case of a cylinder: I = (1/2) m_1R^2 + (1/2) m_2R^2 + (1/2) m_3R^2 + ... = (1/2) MR^2 where M = m_1 + m_2 + m_3 + ... (a) The tension T of the string must exactly balance the cylinder's weight if the cylinder is not allowed to fall. Thus: T = Mg (b) The amount of work done on the cylinder equals its gain in (rotational) kinetic energy. Since the initial angular velocity is assumed to be zero: W = KE_r = (1/2) I \omega^2 = (1/2) [(1/2) MR^2]\omega^2 1/4MR^2\omega^2 (c) To calculate the length of string unwound we place ourselves in the reference frame in which the string is at rest. In this frame, the cylinder rolls forward at a linear acceleration: a = R\alpha leaving a trail of string behind. Here \alpha is the angular acceleration of the rotating cylinder. From the laws of rotational dynamics: \cyrchar\cyrt = I\alpha\alpha = \cyrchar\cyrt/I where \cyrchar\cyrt = MgR is the torque that the string exerts about the cylinder's center of mass. The length of string, S, that the cylinder unwinds after time t equals the distance it travels in this reference frames: S= v_0t + (1/2) at^2 ,v_0 = initial velocity = 0 S= (1/2) at^2 = (1/2) \alphaRt^2 = (1/2) (\cyrchar\cyrt/I) Rt^2 = (1/2) [(MgR)/(1/2 MR^2)] Rt^2 = gt^2 (These are the kinematics equations for constant acceleration.) We know that the time t it takes for the cylinder's angular velocity to each w is given by the angular kinematics equations for constant \alpha: w= w_0 + \alphat,w_0 = initial velocity = 0 w= \alphat t= w/\alpha = [w / {MgR/(1/2) MR2} ] = (wR)/(2g) Thus:S= g [(wR)/(2g)]^2 = (w^2R^2)/(4g)

Question:

A 0.240 g sample of a compound of oxygen and element X was found by analysis to contain 0.096 g of X and 0.144 g of oxygen, (a) Calculate the percentage compo-sition by weight, (b) Calculate from the above data, three possible atomic weights for X relative to oxygen (at. wt. = 16). (c) What additional information is needed to calculate the true atomic weight of X?

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Solution:

(a) The percent composition by weight of each element in the compound is found by dividing the weight of that element by the weight of the compound and then multiplying the quotient by 100. For O, there-fore, you obtain the following percentage: {(.144) / (0.240)} × 100 = 60 % O For X: {(.096) / (0.240)} × 100 = 40 % X (b) If you assume that the compound is XO, you can solve for the molecularweight of X by using the following ratio. Let x = MW of X. {(.144g O) / (16.0 g/mole O)} = {(.096 g) / (X g/mole x)} MW X = 10.7 If the compound is taken to be X_2O, there are twice as many X atoms present as O atoms. In solving for the molecular weight of X in this case, use the following ratio (.144) / (weight of 1 mole of O) = (.096) / (weight of 2 moles of x) Let x = MW of X .144g / 16.0 = 0.96g / x x = (16.0 × .096g) / .144g x = 5.35. Assume also that the formula for the compound might be XO_2. Then, the following ratio should be used: Let x = MW of X. .144g / (weight of 2 moles of O) = .096 / (weight of 1 moles of X) .144 / 32.0 = .096g / X X = (.096g × 32.0) / .144 g = 21.4 . (c) To solve for the true atomic weight of X, you must know the actual number of X atoms present per each O atom present.

Question:

What is the difference between a kinesis and a taxis?

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Solution:

Bothkinesesand taxes are types of animal orientation. Orientation is the process by which the animal directs theconsummatoryact at an appropriate object (the goal). It is involved in those forms of appetitive behavior in which the animal moves in a constant direction in search of a goal. Highly developed forms of orientation are involved in the migration and homing behavior of birds.Kinesesand taxes are two simpler forms of orientation. A kinesis is a form of orientation in which the animal does not necessarily direct its body toward the stimulus. The stimulus merely causes the animal to speed or slow down its movements, resulting in the eventual displacement of the animal toward or away from the stimulus. It is an undirected type of orientation. The movement of the woodlouse is a good example of a kinesis. A woodlouse is a crustacean that lives in damp environments beneath stones or pieces of wood. When the woodlouse finds itself dehydrating, it begins to move around. This increase in locomotoractivity may cause the woodlouse to chance upon a moist spot. Here, the dampness causes the woodlouse to slow down and it now establishes a new home. The woodlouse did not direct itself toward the moist spot, but simply encountered it by random movement. In contrast, a taxis is a type of orientation in which the animal directs its body toward or away from the stimulus. The type of stimulus is usually denoted by a prefix to the word taxis: a geotaxis is movement regulated by gravity; aphototaxisis movement guided by light; a chemotaxisis movement guided by chemical substances. Movement toward the stimulus is a positive taxis whereas movement away from the stimulus is a negative taxis. For example, the directed movement of cockroaches away from light is termed negativephototaxis. Another common example of a taxis is the directed movement of the bee flying to its hive or source of pollen. The bee maintains a straight course by moving at a constant angle to the sun, therefore exhibiting aphototaxis. The term taxis should be distinguished from tropism: Tropism indicates growth of an organism in a particular direction. The phototropism of plants is a typical tropism.

Question:

We know that if we drop an object of mass m while giving it a horizontal velocity component, the object will fall toward the surface of the Earth with the horizontal velocity remaining constant. With what velocity must an object be projected so that the curvature of its path is just equal to the curvature of the Earth?

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Solution:

In such a situation, the object would fall toward the Earth at the same rate that the surface of the Earth curves away from the instantaneous velocity vector; that is, the object would fall around the Earth. The height of the object above the surface of the Earth would therefore never decrease and the object would become a satellite of the Earth. Suppose that we start with the object at a distance h above the surface of the Earth. The radius of the Earth is R so that the radius of the desired circular path of the object is R + h. The centripetal force required to maintain the circular motion is F =ma_c= (mv^2)/r = (mv^2)/(R + h) The centripetal acceleration is furnished by gravity, so we can substitute g [=GM_e/(R + h)^2] for a_c where G is the gravitational constant and M_e is the mass of the Earth. Thus, m [v^2/(R + h)] = m [ GMe GMe /(R + h)^2] , /(R + h)^2] , hencev = \surd[(GMe)/(R + h)] hencev = \surd[(GMe)/(R + h)] The period of the motion is \cyrchar\cyrt = 2\pi/\omega = 2\pir/v forv = (length of one orbit)/(time to make one orbit) = (circumference)/(period)

Question:

Given that log_102 = 0.3010 and log_103 = 0.4771 find log_10\surd6.

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Solution:

\surd6 = 6^1/2, thus log_10\surd6 = log_106^1/2. Since log_bx^y = y log_bx, log_106^1/2 = (1/2) log_106. Therefore log_10\surd6 = (1/2) log_106. 6 = 3 \bullet 2, hence (1/2) log_106 = (1/2) log_10(3 \bullet 2). Recall log_10(a \bullet b) = log_10a + log_10b. Thus (1/2) log_10(3 \bullet 2) = (1/2) (log_103 + log_102). Replace our values for log_103 and log_102.

Question:

The maximum permissible dosage for scientific workers using \Upsilon- radiation is 6.25milliroentgensper hour. What is the safe working distance from a 20-Ci source of cobalt-60 which has a dose rate of 27 roentgens (R) per hour at a distance of lm? If the source is used in a lead container which reduces the \Upsilon-radiation emitted to 1%, how much closer can the scientists work?

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Solution:

The radiation from the source is emitted uniformly in all directions. Hence at any distance r the radiation is passing through an area 4\pir^2 . The intensity of radiation per unit volume thus de-creases as r increases according to an inverse-square-power law. The diminution of the intensity of the \Upsilon-radiation due to absorption in the air may be neglected in comparison with this inverse-square-law effect. Hence if r is the safe working distance, [I (r)] / [I(r_0 )] = (r^2 _0) / (r^2 ) where I(r) is the intensity at a distance r from the source, and I(r_0 ) is the value of the intensity atr_0. Hence r^2 = [I(r_0 )(r^2 _0)] / [I (r)] r = r_0 \surd[ {I(r_0 ) } / { I (r) } ] But we know that when r_0 = lm, l(r_0 ) is 27 roentgens/hr. Hence, if I = 6.25 milliroentgens/hr at r, r = (Im) \surd[(27 r/hr) / (6.25 ×10\Elzbar3r/hr)] or r = [\surd{(27 ×10^3 ) / (6.25 )} ] ×Im= 65.7 m. If the lead container cuts the radiation to 1% of its former value, then by the same arguments, the new safe working distance r_1 . is given by r_1 = r_0 \surd[ {I(r_0 ) } / { I (r) } ] Note that only (I r_0 ) is changed. I(r_0 ) is now 27 × 10\Elzbar2roentgens/hr, and r_1 = (Im)\surd{(.27 ×10^3 ) / (6.25 )} = 6.57 m \surd{(.27 ×10^3 ) / (6.25 )} = 6.57 m

Question:

Krypton crystallizes in the face-centered cubic system, with the edge of the unit cell 5.59 \AA. Calculate the density of solid krypton. Assume 1 \AA = 10^-8 cm.

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Solution:

To determine the density of krypton in g/cc, one can use the given information in the following way; 1) determine the number of atoms in one cube 2) determine the number of cubes in one cubic centimeter 3) determine the weight of one cube 4) determine the weight of one cubic centimeter (i.e. the density) . Determining the density: 1) In a face centered cube, there are 14 atoms contributing to the volume. 8 are corner atoms and 6 are face atoms. Corner atoms contribute 1/8 of their volume to the cube and face atoms contribute 1/2 of their volume. Thus, the number of atoms making up the cube can be found. no. of atoms = (1/8 atom/corner × 8 corners) + (1/2 atom/face × 6 faces) = 4 atoms. 2) 1 \AA = 10^-8 cm. Thus, the side of the cube is equal to 5.59 \AA or 5.59 × 10^-8 cm. The volume of a cube is equal to the length of the side cubed. volume of one cube = (5.59 × 10^-8 cm)^3 = 1.75 × 10^-22 cc The number of cubes in 1 cc is found by dividing 1 cc by the volume of one cube. no. of cubes in 1 cc = 1 cc / [1.75 × 10^-22 cc] = 5.71 × 10^21 3) The weight of one mole of krypton is 84. [6.02 × 10^23] atoms (Avogadro's number) of it must weigh 84 g. Therefore, the weight of one atom is found by dividing 84 g by 6.02 × 10^23 atoms. There are 4 atoms in one cube, thus the weight of one cube is equal to the weight of one atom times 4. weight of 1 atom = 84 g / [6.02 × 10^23 atoms] weight of 4 atoms = 4 atoms × {84 g / [6.02 × 10^23 atoms]} = 5.58 × 10^-22 g. 4) The density is now found by multiplying the weight of 1 cube by the number of cubes in 1 cc, density = no. of cubes in 1 cc × weight of 1 cube density = 5.71 × 10^21 /cc × 5.58 × 10^-22 g = 3.19 g/cc.

Question:

A 10 kg mass resting on a table is attached to a steel wire 5 m longwhich hangs from a hook in the ceiling.If the table is removed, the wire is momentarily stretched and the mass commences to oscillate up and down. Calculate the frequency of the oscillation if the cross-sectional area of the wire is 1 sq mm and the value of Young's Modulus is taken to be 19 × 10^11 dynes/cm^2.

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Solution:

Young's modulus is defined as Y = [(F/A)/(∆x/x)] where∆x/x is the fractional change in length of the wire, A is its cross-section and F is the force normal to A. \therefore F =Y(∆x/x)A But F = ma (Newton's second law, where a = acceleration). \therefore a = F/m = [(Y/m)(A/x)]∆x(1) Since a is proportional to the extension of the wire from its equilibrium length (∆x), the motion is simple harmonic. For this type of motion , a = \omega^2 _0 x(2) where \omega_0 is the angular frequency of the oscillation. Comparing (2) and (1) \omega^2 _0 = YA/mx and w_0 = \surd(YA/mx) If f is the frequency of the oscillation \omega_0 = 2\pif =\surd(YA/mx) where f = 1/2\pi \surd(YA/mx) Using the data supplied f = 1/2\pi\surd[{(19 × 10^11 dynes/cm^2)(1 mm)^2}/(10 kg)(5 m)] Transforming all quantities toc.g.s. values f = 1/2\pi\surd[{(19 × 10^11 dynes/cm^2)(10\Elzbar2mm)^2}/{(10000 g)(500 cm)}] f\approx 10vib/s.

Question:

What is the centripetal force required to cause a 3200 pound car to go around a curve of 880-ft radius at a speed of 60 mph?

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Solution:

When a body travels in uniform circular motion, it experiences an acceleration towards the center of the circle. Since the object has a mass, a force towards the center of the circle is produced. In circular motion, the acceleration of the body is given by a = v^2/r where v is the linear velocity of the object and r is the radius of the circle. In our case (using 60 mph = 88 ft/sec): a= (88 ft/sec)^2/(880 ft) a= (88 ft/sec)^2/(880 ft) = 808 ft/sec^2 = 808 ft/sec^2 The mass of the car is m= Wt/a = (3200 lb)/(32 ft/sec^2) =100[(lb - sec^2)/(ft)] [slugs] AndF=ma F= 100[(lb - sec^2)/(ft)] × 8.8 ft F= 100[(lb - sec^2)/(ft)] × 8.8 ft /sec^2 = 880 lb.

Question:

An electron with velocity v = 10^9 cm/s enters a region of length L = 1 cm in which there exists a transverse deflecting field E = .1 statvolt/cm (see figure). What angle with the x axis does the electron make on leaving the deflecting region?

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Solution:

From the figure, we see that the angle made by v_f (the exit velocity) with the x axis is tan \Phi = v_(f)y / v_(f)x(1) where v_(f)y / v_(f)x are the y and x components of the-electron's final velocity. We note that since there is no acceleration in the x- direction, the x component of the electron's velocity must remain constant. Hence v_(f)x = v(0)x= v = 10^9 cm / s(2) We must now calculate v_(f)y. In order to find v_(f)y, we must apply Newton's Second Law (F = ma) to the vertical (y) component of motion of the electron. By definition of electric field intensity E^\ding{217} = F^\ding{217} / q where F^\ding{217} is the force acting on charge q. For the electron, q = -e where e is the electron charge. Therefore F^\ding{217} = -e E^\ding{217} Furthermore, from the figure E^\ding{217} = E \^{\j} where \^{\j} is a unit vector in the positive y direction. Then F^\ding{217} = (- e)(- E\^{\j}) = eE \^{\j} ButF^\ding{217} = ma^\ding{217} where m, a^\ding{217} are the mass and acceleration of the electron. Hence ma^\ding{217} = eE\^{\j} anda^\ding{217} = eE\^{\j} / m We note that E^\ding{217} is constant, and, therefore, a^\ding{217} is constant. We may then use the kinematics equations for constant acceleration to find v_(f)y . Also , a^\ding{217} is one dimensional and we may use a scalar relation for v_(f)y v_(f)y= v_(0)y+ at = v_(0)y+ (eE / m) t where t is the time it takes for the electron to travel the distance L, and v_(0)y = 0, as shown in the figure. Then v_(f)y= (eE / m) t But by definition of v_x , (and since v_x is constant) t = L / v_(0)x Thereforev_(f)y = (eE / m) (L / v_(0)x ) = eEL / mv_(0)x Substituting in (1) and using (2) tan \Phi = v_(f)y/ v(f)x= (eEL / m v_(0)x ) (1 / v_(0)x ) tan \Phi = eEL / mv^2_(0)x tan \Phi = [(4.8 × 10^\rule{1em}{1pt}10 esu) (10^\rule{1em}{1pt}1 statvolt / cm)(1cm)] / [(9.11 × 10\rule{1em}{1pt}28gm)(1018cm^2 / s^2) But1 statvolt = 1 esu, and tan \Phi = [(4.8 × 10^\rule{1em}{1pt}11) esu^2 ] / [(9.11 × 10\rule{1em}{1pt}10)gm\bullet cm^2 / s^2] Also,1 (esu^2 / cm) = 1 dyne Hence1 dyne \bullet cm = 1 esu^2 and1gm \bullet cm^2 / s2= 1 dyne \bullet cm Thereforetan \Phi = .0527 \Phi \approx 3\textdegree

Question:

Design an experiment which would select for a nutritional mutant .

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Solution:

Organisms possess structures called genes which determine their characteristics . A gene is capable of changing, or mutating, to a different form so that it determines an altered characteristic. Any organism that has a specific mutation is called a mutant. Among the large variety of bacterial mutants are those which exhibit an increased tolerance to inhibitory agents such as antibiotics. There are also mutants that exhibit an altered ability to produce some end product, and mutants that are nutritionally deficient (unable to synthesize or utilize a particular nutrient.) These nutritional mutants are calledauxotrophsbecause they require some nutrient not required by the original cell type or prototroph. The first step in isolating a nutritional mutant is to increase the spontaneousmutation rate of the bacteria, which usually ranges from 10^\rule{1em}{1pt}6 to 10^\rule{1em}{1pt}10 bacterium per generation.(This means that only 1 bacterium in 1 million to 1 in 10 billion is likely to undergo a mutational change.) The mutationrate can be significantly increased by exposing a bacterial culture (e.g. Escherichia coli), to ultraviolet radiation or x-rays. A portion of the irradiatedE.coliculture is placed on the surface of a Petri dish, and spread over the surface to ensure the isolation of colonies. This Petri dish must contain a "complete" medium, such as nutrient agar, so that all bacteria will grow, including the nutritional mutants . Nutrient media contain all the essential nutrients needed for growth . After incubation of this "complete" medium plate, the exact position of the colonies on the plate is noted. A replica plating device is then gently pressed to the surface of the complete plate, raised, and then pressed to the surface of a "minimal" media plate. A replica plating device consists of a sterile velveteen (a cotton fabric woven like velvet) cloth. Cells from each colony adhere to the cloth in specific locations so that each colony remains isolated. These cells serve as aninoculumwhen the cloth is pressed to another plate. The replica plating devicetranfersthe exact pattern of colonies from plate to plate. The positioning of the cloth on the "minimal" agar plate must be identical to its positioning when originally pressed to the complete plate. Colony locations will then be comparable on each of the two plates, which are termed "replicas." The "minimal" medium consists only of glucose and inorganic salts, which are nutrients which normally permit the growth ofE.coli. From these basic nutrients , normalE. colican synthesize all required amino acids, vitamins, and other essential components. After incubation, colonies appear on the minimal plate at most of the positions corresponding to those on the complete plate. Those missing colonies on the minimal plate are assumed to be nutritional mutants, becausethey cannot grow on a glucose inorganic salts medium. These missingcolonies can be located on the complete media plate by comparing the location of colonies on the replicas. If they had not been irradiated , and if mutations did not occur, all the colonies would have been able to grow on this minimal medium. The colonies that did grow were not affected by the irradiation, and were non-mutants. To determine the exact nutritional deficiency of theauxotrophs, one can plate them on media which contain specific vitamins or amino acids in addition to glucose and inorganic salts. These compounds are normally synthesized byprototrophicE. coli. The mutation might have affected a gene which controls the formation of an enzyme which, in turn, regulates a stepin the biosynthesis of one of these nutrients. By plating the mutant on several plates with each plate containing an additional specific nutrient, wecould determine which nutrient the mutant cannot synthesize. The mutant will grow on that plate containing the specific nutrient which the mutant is unable to synthesize.

Question:

A car travels at the constant speed of 30 mph for 20 miles, at a speed of 40 mph for the next 20 miles, and then travels the final 20 miles at 50 mph. What was the average speed for the trip?

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Solution:

For situations in which the speed is variable, the rate at which distance d is traveled as a function of time, t, can be described by the average speed. The average speedvis equal to that constant speed which would be required for an object to travel the same distance d in the same time t. Therefore v= d/t. The total time the car travels is the sum of the times for each segment of the trip. t= t_1 + t_2 + t_3 = (d_1/v_1) + (d_2/v_2) + (d_3/v_3) t= [(20 mi)/(30 mph)] + [(20 mi)/(40 mph)] + [(20 mi)/(50 mph)] = (0.67 + 0.50 + 0.40)hr = 1.57 hr The total distance is d= d_1 + d_2 + d_3 = (20 + 20 + 20)mi = 60 mi Therefore, the average speed is v= d/t = (60 mi)/(1.57 hr) = 38.2 mph

Question:

Suppose a rain cloud has a water volume of 10^5 liters. If the cloud forms over a chemical factory which puts SO_3 into the air and each SO_3 molecule bonds with one out of 10^4 H_2O molecules (to form H_2SO_4), what is the molarity of the H_2SO_4 in the cloud?

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Solution:

To solve this problem, the number of moles of H_2SO_4 must be determined. This is done by applying the stoichiometric rules to the equation of this particular reaction: H_2O + SO_3 \rightarrow H_2SO_4 According to this equation for every mole of SO_3 consumed, one mole of H_2SO_4 is produced or, in other words, if the number of moles of SO_3 consumed is known, then the number of moles of H_2SO_3 produced is also known. The number of moles of SO_3 can be found since it is the same number as H_2O (from the stoichiometry of the equation) . The number of moles of H_2O consumed can be found by dividing its weight by its molecular weight (MW = 18.02) or number of moles of H_2O = [{(10^5 liter) (1000 g / liter)} / {18.02 g / mole}] = 5.5 × 10^6 moles Note: 1000 g/liter is the density of water (rain). However, the number of water molecules is needed and thus, number of H_2O molecules = (5.5 × 10^6 moles) (6.02 × 10^23 molecules/mole) = 3.3 × 10^30 molecules. The problem states that SO_3 molecules bond with one out of 10^4 H_2O molecules. From this, the number of SO_3 molecules is determined, {[3.3 × 10^30 H_2O molecules] / [10^4(H_2O molecules/SO_3 molecules)]} = 3.3 × 10^26 SO_3 molecules. The number of moles of SO_3 can now be obtained. number of moles of SO_3 = {[3.3 × 10^26 SO_3 molecules] / [6.02 × 10^23 (molecules / mole)]} = 5.5 × 10^2 moles or 5.5 × 10^2 moles of H_2SO_4 are formed. The molarity of the H_2SO_4 produced is thus (5.5 × 10^2 moles) / (10^5 liter) = 5.5 × 10^-3 M.

Question:

Prove that when light goes from one point to another via a plane mirror, the path chosen is the one which takes the least time.

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Solution:

Let the points be A and B, and let C be any general point on the mirror. Orient the diagram so that the x- and y-axes are as shown. Draw the normals to the mirror surface passing through A, B and C. Now in specular reflection, the reflected ray lies in the plane determined by the incident ray and the normal to the mirror at the point of reflection. Hence A, B and C must be in the same plane. The coordinates of the three points are A(x_1, 0); B(x_2, y_0); C(0, y). The length of the path ACB is, by the Pythagorean theorem, p = \surd(x^2_1 + y^2) + \surd[x^2_2 + (y_0 - y)^2] But the time of travel of light by this path, the velocity of light being, c, is t = p/c. For the path to be traveled in minimum time, we must have dt/dy = 0, where y is the variable which changes with path. Thus t = (1/c) [(x^2_1 + y^2)^1/2 + {x^2_2 + (y_0 - y)^2}^1/2] (dt/dy) = (1/c) [(1/2) (x^2_1 + y^2)^-1/2 (2y) + (1/2) { x^2_2 + (y_0 - y)^2}^-1/2 × 2(y_0 - y) (- 1)] (dt/dy) = (1/c) [{y / \surd(x^2_1 + y^2)} - {(y_0 - y) / \surd(x^2_2 + (y_0 - y)^2)}] To find the values of y which make t a minimum, we set dt/dy = 0. Hence, 0 = (1/c) = {y/\surd(x^2_1 + y^2)} - {(y_0 - y)/\surd[x^2_2 + (y_0 - y)^2]} But since c \not = \infty, the quantity in the braces must be zero. Therefore, [y/{\surd(x^2_1 + y^2)}] = (y_0 - y)/[\surd{x^2_2 + (y_0 - y)^2}](1) sin \texttheta_1 = [y/(\surdx^2_1 + y^2)]sin \texttheta_2 = (y_0 - y)/[\surd{x^2_2 + (y_0 - y)^2}] Hence, (1) becomes sin \texttheta_1 = sin \texttheta_2 and consulting the diagram we see that, \texttheta_1 = \texttheta_2 which is the law of reflection. Since light reflected specularly always satisfies this condition, the light ray follows the path which .takes the least time.

Question:

A 12-V automobile battery supplies 20C of charge in 5 s to a stereo tape player. What is the electrical current flowing into the device and how much work is done by the battery.

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Solution:

The charge supplied by the battery, q = 20 C, and the time interval, t = 5 s, are given. The current is, by definition, I = (q/t) = [(20C)/(5 s)] = 4 A The potential of the battery, V = 12 V, is also given. However, the potential difference between two points is the work required to move a unit charge from the point at lower potential to that of higher potential. Or V = (W/q ) ThenW = Vq = (12 V) (20 C) = 2.4 × 10^2 J The work done by the automobile battery is 240 J.

Question:

High levels of progesterone are essential for the maintenance of the uterine lining during pregnancy. But high levels of progesterone exerts a negative feedback action on LH secretion, with a consequent decrease in LH secretion and eventual atrophy of the corpus luteum. Atrophy of the corpus luteum cuts the progesterone supply resulting in menstruation. Clearly, these events cannot take place after conception, or the embryo would get sloughed off. When conception occurs, the corpus luteum is seen to last through most of pregnancy. How can this be explained?

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/Users/wenhuchen/Documents/Crawler/Biology/F21-0562.htm

Solution:

It has been shown that the portion of the placenta called the chorion secretes a hormone, called chorionic gonadotrophin, that is functionally very similar to LH. The function of this hormone is to take the place of LH in preserving the corpus luteum, LH production being inhibited by a high progesterone level. The corpus luteum is then able to secrete progesterone at high levels without being shut off. The amount of this chorionic hormone produced in a pregnant woman is very great. A pregnant woman produces so much of this hormone that most of it is excreted in the urine. Many commonly used tests for pregnancy are based on this fact. One test involves the injection of the patient's urine into a test animal such as a rabbit. If chorionic hormone is in the urine, development of a corpus luteum within 24 hours will take place in the rabbit, and pregnancy can be confirmed. Although the corpus luteum is essential during early pregnancy and is present during most of pregnancy, it has been shown that it is no longer necessary after about the first two months. It seems that the placenta begins to secrete progesterone (and estrogen) early in pregnancy, and once this secretion has reached a sufficiently high level the placenta itself can maintain the pregnancy in the absence of the corpus luteum.

Question:

The barred pattern of chicken feathers is inherited by a pair of sex-linked genes, B for barred and b for no bars. If a barred female is mated to a non-barred male, what will be the appearance of the progeny?

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/Users/wenhuchen/Documents/Crawler/Biology/F25-0668.htm

Solution:

This is an example of a sex-linked cross. Using the notations for the sex-linked traits: let X^B be the chromosome carrying the gene for the barred pattern, and X^b be the chromosome carrying the gene for the non-barred pattern. Since B is a dominant gene, a barred female could have one of the 2 genotypes: X^BX^B or X^BX^b. The only possible genotype for a non-barred male is X^bY. Therefore there are two possible crosses between a barred female and a non-barred male: Phenotypically, all the progeny of this cross have the barred feather pattern. F_1 X^b Y X^B X^BX^b X^BY X^b X^bX^b X^bY Phenotypically:X^B X^bisabarred female, X^b X^bisanon-barred female, X^B Xisabarred male,and X^b Xisanon-barred male.

Question:

Write a FORTRAN program to evaluate the polynomial expression a1x^n-1 + a2x^n-2 + ... + a_n forgiven values of n, a1, a2,...,a_n using various values of x whichare read in. Let the program terminate when a zero valueof x is read in.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0521.htm

Solution:

The most efficient polynomial evaluation procedure is based onthe nesting (...((a1x+ a2)x + a3)x + ... + a_n-1)x + a_n . We assume n \leq 25. The data deck consists of a value for n followed by the ncoefficientsa_ion one or more data cards. These are fol-lowed by successivedata cards, each with a value for x; the final card contains the zerovalue. This nesting procedure is known as Horner's method for evaluating poly-nomials. We present the program segment below. CPOLYNOMIAL EVALUATION DIMENSIONA(25) READ, N,(A(J), J = 1,N) 16READ, X IF (X.EQ.O.) STOP POLY =A(1) DO 12 I = 2,N 12POLY = PGLY\textasteriskcenteredX +A(1) GO TO 16 STOP END

Question:

A differential hoist is used to lift a load of 450 pounds. If the larger wheel of the hoist has 9 slots and the smaller wheel 8 slots and if the efficiency of the hoist is 33 (1/3)%, what is the effort required to raise the load?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0292.htm

Solution:

The IMA = 2 × 9 = 18. This is to since there are 9 strands supporting the load. Since efficiency = AMA/IMA , 0.33 (1/3) = AMA/18, whence AMA = 6. Since AMA = R/E , 6 =(450 pounds)/(E), whence E = 75 pounds.

Question:

A force of 0.20newtonacts on a mass of 100 grams. What is the acceleration?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0098.htm

Solution:

From Newton's Second Law we have F = ma a = F/m Also, 100 grams = 0.10 kg. Therefore a = (0.20nt)/(0.10 kg) = [(0.20 kg-m/g^2)/(0.10 kg)] = 2.0 m/sec^2.

Question:

A magnetic south pole of 15 units strength, when placed 10 cm away from another pole in air, experiences a repulsion of 300 dynes. What is the nature and strength of the second pole?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0721.htm

Solution:

Since the force is one of repulsion, the second pole is a like pole, i.e. , a south pole. Until the advent of quantum theory, the nature of the magnetic force of a magnet was not well understood. The concept of magnetic pole strength due to a magnetic pole was then de-veloped in a manner analogous to the concept of electric field strength and charge. The analogue to Coulomb's law for the electrostatic force between two charges is Coulomb's law for the magnetic force between two poles of field strength m and m' separated by a distance r. It is F = (1/\mu) (mm'/r^2 ) where \mu = 1 for the CGS system being used. We are given that F = 300 dynes, m is 15 poles, \mu = 1 for air, and r = 10 cm. Therefore, m = (\mur^2 F/m) = [{(1) (10 cm)^2 (300 dynes)}/(15 poles)] = 2000 poles Although this model of magnetic poles correctly predicts experimentally observed magnetic forces between poles, magnetic monopoles have never been observed.

Question:

Write a structural formula for each of the following molecular formulas. Name each structure that you draw, (a) CH_2O; (b) C_2H_6O_2; (c) C_2H_6O; (d) C_2H_4O_2; (e) C_2H_4O.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E20-0745.htm

Solution:

In order to do this problem, one must know how carbon, hydrogen, and oxygen bond. Carbon can share its 4 valence electrons to form 4 covalent bonds, hydrogen can share its 1 valence electron to form 1 covalent bond, and oxygen can share its 6 covalent electrons to form only 2 covalent bonds. Thus, with this knowledge in mind, one can construct several types of molecular structures. A few of these structures have the following functional groups: Alcohols contain the hydroxyl group, - OH, acids contain the carboxyl group, (represented as - COOH), aldehydes and ketones contain the carbonyl group, Structure (a), CH_2O, is an aldehyde, and has the name formaldehyde. Structure (b), C_2H_6O_2, is an alcohol and has the name 1,2 dihydroxyethane or ethylene glycol. Structure (c), C_2H_6O, is either an alcohol or an ether and is named ethanol or dimethyl ether, respectively. Structure (d) C_2H_4O_2, is an acid and has the name acetic acid. Structure (e), C_2H_4O, is an aldeyde and has the name acetaldehyde.

Question:

In what distance can a 3000-lb automobile be stopped from a speed of 30 mi/hr (44 ft/sec) if the coefficient of friction between tires and roadway is 0.70?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0103.htm

Solution:

The retarding force furnished by the roadway can be no greater than F_s =\muN= (0.70)(3000 lb) = 2100 lb. Since the work done by this force is equal to the kinetic energy of the car, the stopping distance can be found from W = Fs = 1/2mv^2. We must divide the weight by the acceleration due to gravity, g, to obtain the mass m = W/g = (3000 lb) / (32 ft/sec^2) = 94 slugs s = (1/2mv^2)/F = [{94 slugs (44 ft/sec)^2}/{2 × 2100 lb}] = 43 ft.

Question:

An elevator is descending with an acceleration of 5 ft\bullets^-2. The shade on the ceiling light falls to the floor of the elevator, 9 ft below. At the instant that it falls, one of the passengers sees it, and realizes that the shade will hit his foot. How long does he have to get his foot out of the way?

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Solution:

At the instant that the light shade becomes detached, it has a downward speed of v_0, and it will drop freely under the action of gravity. Suppose that there is an observer outside the elevator. By the time the shade strikes the (moving) floor of the elevator, it will have been seen by the observer to have dropped a distance s in a time t. This distance is given by the free fall equation s = v_0t + (1/2)gt^2. In the same time t, the elevator floor will have traveled s - 9 ft (9 ft less than the shade, since the elevator floor doesn't have to traverse the length of the elevator) with the acceleration of 5 ft \textbullet s^-2, having started with the same downward speed v_0. \therefore s - 9 ft = v_0t + (1/2) × 5 ft\bullets^-2 × t^2. Subtract one equation from the other. Thus 9 ft = (1/2)(g - 5 ft\bullets^-2)t^2 = (1/2) × 27 ft\bullets^-2 × t^2. This equation could have been obtained more easily by considering the motion of the shade relative to the elevator in an accelerated frame of reference. For, relative to the elevator, the light shade starts off with zero velocity and has an acceleration of g - 5 ft\bullets^-2. Thus, applying the same equation of motion as before, we find that the shade drops 9 ft relative to the elevator in a time t, and 9 ft = 0 × t + (1/2)(g - 5 ft\bullets^-2)t^2. t = \surd[(2 × 9 ft)/(27 ft\bullets^-2)] = \surd(2 s^2/3) = 0.82 s. The passenger has therefore less than 1 s to get his foot out of the way, and must react rapidly.

Question:

The speed of a certaincompressionalwave in air at standard temperature and pressure is 330 m/sec. A point source of frequency 600/sec radiates energy uniformly in all directions at the rate of 5.00 watts What is the intensity of the wave at a distance of 20.0 m from the source? What is the amplitude of the wave there?

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/Users/wenhuchen/Documents/Crawler/Physics/D25-0803.htm

Solution:

At any concentric spherical surface the energy from a point source is spread over an area 4\pir^2. At a distance of r = 20.0 m, the intensity I is I = (E/tA) = (P/A) = (5.00 watts/[4\pi × (20.0 m)^2] = 0.99 × 10^\rule{1em}{1pt}3 watt/m^2 The rate of transfer of energy depends on the square of the wave amplitude and the square of the wave frequency for all types of waves, with I the average intensity and the density of air \rho = 1.29 gm/liter = 1.29 kg/m^3, we have, A^2 = I/(2\pi^2vpf^2) = [(0.99 × 10^\rule{1em}{1pt}3 watt/m^2)/{2\pi^2 (330 m/sec) (1.29 kg/m^3) (300/sec)^2}] = 1.32 × 10^\rule{1em}{1pt}12 m^2 A = 1.15 × 10^\rule{1em}{1pt}6 m = 1.15 × 10^\rule{1em}{1pt}4 cm

Question:

Using the data from the accompanying table find the enthalpy change for the combustion of a mole of C_2H_4 (g) to form CO_2 (g) and H_2O(g) at 298\textdegreeK and 1 atm standard state conditions.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E14-0497.htm

Solution:

When a chemical reaction occurs heat is usually absorbed or given off to the environment. This means that the enthalpy or heat content of the system will also change. If one knows the enthalpy change of the combustion of each reactant and product, one can find the change in enthalpy of a reaction. The reaction in this problem can be written as C_2H_4 (g) + 3O_2 (g) \rightarrow 2CO_2 (g) + 2H_2O(g) For the reaction, ∆H_combustion = \sum∆H_products - \sum∆H_reactants. Remembering that the enthalpy change for the formation of an element (i.e. O_2) in its standard state is zero, one then has, ∆H_combustion= 2∆H_CO2 + 2∆H_H2O -∆HC2 H4- 3∆H_O2 = 2(-394) + 2(-242) - (52) - 3(0) = -1324 KJ Enthalpies of formation in kilojoules per mole from the elements for various compounds at 298\textdegreek and 1 atm pressure Component ∆H C_2H_4 (g) + 52 CO_2 (g) - 394 H_2O (g) - 242 O_2 (g) 0

Question:

Calculate the equilibrium constant for the following reaction at 25\textdegreeC or 298\textdegreeK C _(graphite) + 2H_2 (g) \rightarrow CH_4 (g) ∆H\textdegree for this reaction is - 17, 889 cal.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E15-0539.htm

Solution:

At equilibrium the equilibrium constant of any reaction is independent of the amount of pure solid (or liquid) phase. To solve this problem (1) determine ∆S\textdegree, the entropy change or randomness of the system. (2) Determine ∆G\textdegree, the free energy or the energy available to do work, by using the formulas:∆G\textdegree = ∆H\textdegree - T∆S\textdegree, where ∆H\textdegree = the change in enthalpy or heat content, and T = absolute tem-perature (Celsius plus 273\textdegree) . (3) DetermineK_p, the equilibrium constant, by using ∆G\textdegree = - RT InK_p. The entropy change (from any entropy table) for this reaction is ∆S\textdegree of reaction = ∆S\textdegree _products - ∆S\textdegree _reactants = ∆S\textdegree_CH4 - 2∆S\textdegree_H2 - ∆S\textdegree_C = 44.50 - 2 (31.211) - 1.3609 = - 19.28 cal/k ∆G\textdegree = ∆H\textdegree - T∆S\textdegree = - 17, 889 - (298) (- 19.28) = - 12, 143 cal. The equilibrium constant is thus ∆G\textdegree = - RT InK_p= - (1.987 cal\textdegreeK^-1 mole^-1) (298\textdegreeK) × (2.303) logK_p - 12,143 cal = - 1364 logK_p logK_p= 8.90 K_p= 7.94 x 10^8 .

Question:

Write a FORTRAN program segment to compute values of the POISSON distribution given by P(X) = e^-mm^x/x! Assume (1) that X and m are integer variables; (2) that X and m are small enough that no term will exceed the size permitted for a real variable in your computer; and (3) that the function FACT (n) for computing n! is available.

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Solution:

Remember that the library functione^yis defined for real y, so we set y = FLOAT(m) to convert the integer m to a real variable when evaluating P. The rest of the Integer expressions, need not be FLOATED since the FORTRAN compiler allows mixing of modes in a product expression of real \textasteriskcentered integer; the result being converted to a real number. Bearing the above in mind, we can write out the program below: FUNCTION P(X,M) INTEGER X Y = FLOAT (M) P = EXP(-Y) \textasteriskcentered M \textasteriskcentered\textasteriskcentered X / FACT (X) RETURN END The factorial function, FACT (X), is defined in chapter 19.

Question:

One end of a horizontal wire is fixed and the other passes over a smooth pulley and has a heavy body attached to it. The frequency of the fundamental note emitted when the wire is plucked is 392 cycles\textbullets^\rule{1em}{1pt}1. When the body is totally immersed in water, the frequency drops to 343 cycles\textbullets^\rule{1em}{1pt}1. Calculate the density of the body.

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Solution:

Let the density of the body be \rho and its volume be V. The density of water is 1 g\textbulletcm^\rule{1em}{1pt}3 = \rho_0. In the first case, the weight of the body is balanced by the tension S^\ding{217} in the wire. In the second case, a third force, the buoyancy, enters into the calcula-tion. (See the figure.) The weight of the body is balanced partly by the new tension in the wire, S_0, and partly by the buoyancy, U^\ding{217}, acting on it according to Archimedes' principle. That is, U is equal to the weight of water displaced by the body. Since the volume of the displaced water is equal to the volume of the body, then U = (V\rho_0)g where V\rho_0 is the mass of the displaced water. Thus S = V \rhog where V\rho is the mass of the body and S_0 + U = S_0 + V \rho_0g = V \rhog, the buoyancy due to the air in the first case being ignored. The frequencies of the fundamental notes emitted in the two cases are f_1 = (1/2L) \surd(S/\mu)andf_0 = (1/2L) \surd(S_0/\mu) where \mu is the mass of the wire per unit length (the mass density). Thus (f^2_0)/(f^2_1) = (S_0/S) = [(V \rhog - V \rho_0g)/(V \rhog)] = [(\rho - \rho_0)/\rho] \therefore\rho = [(f^2_1 \rho_0)/(f^2_1 - f^2_0)] = [(392^2 s ^\rule{1em}{1pt}2 × 1 g\textbulletcm^\rule{1em}{1pt}3)/(392^2 - 343^2) s^\rule{1em}{1pt}2] = 4.27 g\textbulletcm^\rule{1em}{1pt}3.

Question:

Use the Euler method to solve the equation of motion of a dampedhar-monic oscillator. Let the initial conditions be that att = 0, x(0) = 10 cm anddx/dt= v = 0.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G21-0531.htm

Solution:

The equation of motion is given by (md2x / dt2) = - ( cdx cdx /dt) -kx /dt) -kx where(-cdx/dt) is the damping force and -kxis the restoring force. Since Euler's method is based on first order differential equations, we rewrite the equationof motion as two first order equations: dx/dt= v(1) dv/dt= -(c/m) v - (k/m)x(2) According to Euler's method, the solution is given by dx\approx ∆x =x_new-x_old;dx=vdt) x_new=x_old+v_old∆t(1') v_new=v_old+ [(-c/m)v_old-(k/m)x_old)∆t(2') Here we have used the general result that ifdy/dx= f(x,y), then y_i+_1 =y_i+y'_i∆x, wherey'_i_-= f(x_i ,y_i). For the sake of concreteness we use the following data: m = k = 10, c = 2, x_0 = 10, v_0 =0, D \equiv ∆t = 0.1 andwe follow the motion for 2 seconds. 1\O\O REM DAMPED HARMONIC OSCILLATOR 1\O1 REM EULER'S METHOD 1\O5 PRINT "TIME\textquotedblright, "VELOCITY", "POSITION" 1\O6 PRINT 11\O READ M, 1 K, C, X\O, V\O, D 12\O LET X1 = X\O 121 LET V1 = V\O 13\O FOR T = \O TO 2 STEP D 14\O PRINT T, V1, X1 15\O LET X2 =X1 + V1 \textasteriskcenteredD 151 LET V2 = V1 + (-C\textasteriskcenteredV1/M - K\textasteriskcenteredX1/M)\textasteriskcenteredD 16\O LET X1 = X2 161 LET V1 =V2 162 NEXT T 8\O\O DATA 1\O,1\O, 2, 1\O, \O, 0.1 999 END Note that the exact solution of t = 2 gives x = -2.5807, compared with the Euler method result of -2.96507. More sophisticated numeri-cal methods (suchas the improved Euler method orRunge-Kuttatype methods) are necessaryfor better accuracy. Finally,v_newcan also be solved with the quadratic equation. If you rearrangethe equation of motion, you can obtain m(d2x / dt2) + c (dx/dt)kx= 0, whichis the quadratic form. This equation has a solution of the form x =e^\alphat.If we substitute x into the first equation and divide bye^\alphat,we get thesimpler form m\alpha2 +c\alpha+ k = 0. By using thediscrim-inantc2 - 4mk, we candetermine what type of motion is present: If c2 - 4mk is the motion is positive over-damped zero critically-damped negative oscillatory-damped

Question:

Consider an elementary inventory control system where goods are received at a rate R (units of good per time period) , where R is directly proportional to the number of goods on order, G(t), with proportionality constant 1/L, where L is the delay in ordering. Suppose the order rate is directly proportional to the difference between desired inventory and actual inventory, with proportionality con-stant 1/A where A is the time that would be required to correct the inventory if the order rate were constant. Write a FORTRAN program to simulate this system, using modified Euler's method (the method is fully explained, and the subroutine for it is given in the SIMULATION chap-ter) from time t=0 to t=t_f if I(t) is the inventory level at time t, I(o)=Io and G(o) = Go.

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Solution:

Note that this system's goal is maintaining the desired inventory. A first picture of this system might be as shown in Fig. 1. Where the solid arrows represent the flow of goods and the broken arrows represent the flow of information. If \.{I} is the time rate of change of I(t): \.{I} = R = G/L(1) If G is the net time rate of change of G(t), W the order rate, and D the desired inventory, then \.{G} = W - R \.{G} = (D-I)/A - G/L(2) Converting equations (1) and (2) into block diagram form details the relationships in this coupled negative feed-back system: The program applies the modified Euler method first to equation (1) and then to equation (2). The variable GOODS is used for G(t). I and GOODS are kept in a COMMON block so that the new computed value of I will be used in determining GOODS. A logical flag, ORDER, is used to determine which equation is being integrated: REAL T/0, 0/, TFIN, DT, REALN, ACCUR, I, GOODS, I\O, G\O, L, D, A INTEGER J, N LOGICAL ORDER/. FALSE. / COMMON ORDER, I, GOODS, L, D, A INPUT N, TFIN, ACCUR, I\O, G\O, L, D, A OUTPUT T, I\O, G\O REALN = N DT = TFIN/REALN I = I\O GOODS = G\O DO 1\O I = 1, N T = T + DT CALL MEULER (T, I, ACCUR, DT) ORDER = .TRUE. CALL MEULER (T, GOODS, ACCUR, DT) ORDER = .FALSE. OUTPUT T, I, GOODS 1\OCONTINUE STOP END FUNCTION G(W) REAL G, W, I, GOODS, D, L, A LOGICAL ORDER COMMON ORDER, I, GOODS, L, D, A G = GOODS/L IF (ORDER)G = (D - I)/A - W/L RETURN END

Question:

In. the figure, suppose that the block weighs 20 lb., that the tension T can be increased to 8 lb. before the block starts to slide, and that a force of 4 lb. will keep the block moving at constant speed once it has been set in motion, (a) Find the coefficients of static and kinetic friction, (b) What is the frictional force if the block is at rest on the surface and a horizontal force of 5 lb. is exerted on it?

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Solution:

a) There are 2 forces of friction which can act on a body. These are the forces of kinetic and static friction. If a body, such as that in the figure, is initially at rest and we begin pulling on it with a variable force T, the block will remain at rest. This means that no matter what force T we apply to the body, the frictional force f always balances it. However, at some value of T, the frictional force no longer balances it, and the block begins translating, We may describe this static frictional force by f_s \leq\mu_sN(1) where \mu_s is the coefficient of static friction and N is the normal force of the table on the block. (The equality holds when the block begins translating). Once the block begins translating, the static frictional force stops acting and the kinetic frictional force takes over. This force is f_k = \mu_kN(2) It is also found that f_k < fs max, and that once the block starts moving, we may reduce T and the block will still move. Applying the Second Law to the block of mass m N - mg = ma_y T - f = ma_x Here a_x and a_y are the x and y components of the block's acceleration. Because the block coesn't leave the surface of the table, a_y = 0. Hence N = mg(3) T - f = ma_x(4) The block is initially at rest, and just begins to slip when T = 8 lb. If we examine the block just before it moves, a_x = 0, and f is the maximum force of static friction. Then, using (1) T - fs max= 0 T= \mu_sN 8 lb= \mu_sN Using (3)\mu_s= (8 lb)/mg = (8 lb)/(20 lb) = .4 Once translation at constant velocity begins, T = 4 lb, a_x = 0 and f in (4) is f_k = \mu_kN. Hence T - f_k= 0 T= \mu_kN \mu_k= (4 lb)/(20 lb) = .20 b) Note that, if the block is initially at rest, a force of 8 lb is needed to start the motion of the block. Hence, if we pull the block with a force of 5 lb., a_x = 0 and, the force of static friction is acting. Then, from (4) T - f_s = 0 T = f_s = 5 lb.

Question:

Find the change in volume of an aluminum sphere of 5.0-cm radius when it is heated from 0 to 300\textdegree C.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0454.htm

Solution:

In the case of one dimensional thermal ex-pansion, we may relate the change in length of a sample to the temperature change which it experiences by \DeltaL =\alphaL\DeltaT(1) where \alpha is the coefficient of linear thermal expansion. Dividing both sides of (1) by \DeltaT, and taking the limit as \DeltaT \rightarrow 0, we obtain the exact relation: (dL/dT) =\alphaL(2) We are specifically concerned with the change in the volume of a sphere due to a change in temperature, or (dV/dT)(3) where V is the volume of the sphere. Using the chain rule. (dV/dT) = (dV/dR) (dR/dT) assuming that V changes only as a result of a change in radius of the sphere (R). ( dV dV /dR) = (d/dR) [(4/3) \piR^3] = 4\piR^2 /dR) = (d/dR) [(4/3) \piR^3] = 4\piR^2 (dV/dT) = 4\piR^2 (dR/dT) Also, from (2), (dR/dT) =\alphaRhence, (dV/dT) = [(4\piR^3)/3] (3\alpha) = 3\alphaV(4) where V is the original volume of the sphere. If \DeltaT is small, we may write \DeltaV = 3\alphaV \DeltaT(5) Substituting the information provided into (5), we obtain: \DeltaV = 3(2.2 × 10^-5/C\textdegree) (4/3) (\pi) (5 × 10^-2 m)^3 (300\textdegreeC) \DeltaV = 10 cm^3

Question:

The most abundant element in sea water is chlorine (Cl, atomic weight = 35.5 g/mole), which is present as chloride ion, Cl^-, in a concentration of 19 g Cl^- per 1 kg of sea water. If the volume of the earth's oceans is 1.4 × 10^21 liters, how many moles of Cl^- are present in the oceans? Assume that the density of sea water is 1.0 g/cm3.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E04-0113.htm

Solution:

The total number of moles of Cl^-is equal to the total mass of Cl^- divided by its atomic weight. Hence we must find the total mass of Cl^-. This will be done by determining the total mass of the oceans and using this value in conjunction with the concentration of Cl^-to find the total mass of Cl^-. The volume of the oceans is 1.4 × 10^21 liters = 1.4 × 10^21 liters × 1000 cm^3/liter= 1.4 × 10^24 cm^3. Multi-plying this volume by the density of sea water gives the total mass of the oceans, or 1.4 × 10^24 cm3× 1.0 g/cm^3 = 1.4 × 10^24 g. Expressing this total mass as kilograms, we obtain 1.4 × 10^21g = 1.4 × 10^24 g × 10^-3 kg/g = 1.4 × 10^21 kg. For every kg of sea water, there are 19 g of Cl^-. Hence, for 1.4 × 10^21 kg there are 1.4 × 10^21 kg × 19 g Cl^-/kg of sea water = 2.7 × 10^22 g Cl^- . Therefore, the number of moles of Cl^- is massCl-= (massCl-) / (atomic weightCl-) = (2.7 × 10^22 g) / (35.5 g/mole) = 7.6 × 10^20 moles.

Question:

Two straight parallel wires each 90 cm long are 1.0 mm apart. There are currents of 5.0 amp in opposite directions in the wires. What are the magnitude and sense of the force between these currents?

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/Users/wenhuchen/Documents/Crawler/Physics/D21-0724.htm

Solution:

There are 2 steps in the solution of problems like these. First we find the magnetic field due to current #1 at any point on wire #2. (See figure). This field (B^\ding{217}_21) exerts a force on each moving charge that constitutes current #2. It is the sum of these forces which constitute the net force on wire #2 due to wire #1. (By Newton's Third Law, the force on wire #1 due to wire #2 is equal and opposite to the force on wire #2 due to wire #1). The force on an element dl_2^\ding{217} of wire #2, which is immersed in the field of wire #1 (B_21^\ding{217}) is, by definition, dF_21^\ding{217} = I_2 dl_2^\ding{217} × B_21^\ding{217} andF_21^\ding{217} = I_2\intdl_2^\ding{217} × B_21^\ding{217}(1) where the integral in (1) is evaluated over the length of wire #2 and dl_2 has the direction of I_2 . In order to calculate F_21^\ding{217} , we must find B_21^\ding{217}. The lines of magnetic induction of a long wire are circles centered on the axis of the wire. Hence, to calculate B_21^\ding{217} , we may use Ampere's Law, which is \oint B^\ding{217} \textbullet dl^\ding{217} = \mu_0 I_enc(2) In (2), the integral is evaluated over a closed path (this is the meaning of the circle on the integral sign) and I_enc is the net current enclosed by the closed path. In evaluating B_21^\ding{217}, we take the integral about a circular line of magnetic induction, of radius r, centered on wire #1.Therefore, 2\pirB_21 = \mu_0 I_1 since I_1 is enclosed by the path. Then B_21 = [\mu_0 I_1 /(2\pir)](3) The direction of this field is shown in the figure by the "X" symbol, which indicates that the field is perpen-dicular to the plane of the figure and directed into the figure. Now, we may write the magnitude of F_21^\ding{217} as F_21 = I_2\intdl_2 B_21 sin \O where \O is the angle between dl_2^\ding{217} and B_21^\ding{217} . Looking at the figure, \O is 90\textdegree and F_21 = I_1\intdl_2B_21 Using(3)F_21 = I_2\int[(\mu_0 I_1)/(2\pir)] dl_2 F_21 = [(\mu_0 I_1 I_2 l_2 )/(2\pir)] F_21 = [{4\pi × 10^-7 (Weber/amp.m)} (25 amp^2 ) (90 × 10^-2 m)}/{(2\pi) (10^-3 m)}] F_21 = 4.5 × 10^-3nt By Newton's Third Law, \vert F_21^\ding{217} \vert = \vert F_21^\ding{217} \vert

Question:

A boy can throw a baseball horizontally with a speed of 20 m/sec. If he performs this feat in a convertible that is moving at 30 m/sec in a direction perpendicular to the direction in which he is throwing (see figure), what will be the actual speed and direction of motion of the base-ball?

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/Users/wenhuchen/Documents/Crawler/Physics/D01-0018.htm

Solution:

Since the baseball is originally travelling with the convertible, it has the speed of 30 m/sec in the direction the car is travelling. When the boy throws the ball perpendicular to the car's path, he imparts an additional velocity of 20 m/sec in that direction. The ball's velocity is then 30 m/sec in the direction the convertible is moving and 20 m/sec perpendicular to this movement. Its resultant velocity can be found through adding vectors as shown in the diagram. If the resultant velocity is R m/sec at an angle \texttheta to the direction in which the convertible is moving, then R^2 = (20)^2 + (30)^2 = 1300 R = \surd1300 = 36.06 m/sec Also, tan \texttheta = (20/30) = 0.666 From tables of tangents, \texttheta = 33.69\textdegree. Therefore, the ball has a speed of 36.06 m/sec in a direction at an angle of 33.69\textdegree to the direction in which the convertible is travelling.

Question:

A surface probe on Mars transmitsradiowavesat a frequency of 6.0 × 10^5 sec-^1 to an earth station 8.0 × 10^7 km away. How long does it takeradiowavesto traverse this distance?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E30-0918.htm

Solution:

Sinceradiowavesare a form of electromagnetic radiation, they move at the speed of light and we can use the relationship time = distance / speed where the speed is that of light. We do not need to know the frequency. Since 1 km = 10^5 cm, 8.0 × 10^7 km = 8.0 × 10^7 km × 10^5 cm/km = 8.0 × 10^12 cm and we have time = distance / speed = (8.0 × 10^12 cm) / (3.0 × 10^10 cm/sec) = 2.7 × 10^2 sec, or4 (1/2) minutes.

Question:

Express 20\textdegreeC and - 5\textdegreeC on the Kelvin scale.

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/Users/wenhuchen/Documents/Crawler/Physics/D12-0443.htm

Solution:

The relationship between the Celsius and Kelvin scale is K\textdegree = 273\textdegree + C\textdegree Therefore 20\textdegreeC is T = 273\textdegree + 20\textdegree = 293\textdegreeK and \rule{1em}{1pt} 5\textdegreeC is T = 273\textdegree+ (\rule{1em}{1pt} 5\textdegree) = 268\textdegreeK

Question:

Ribonuclease has a partial specific volume of (0.707 cm^3 g^-1) and diffusion coefficient of (13.1 × 10^-7 cm^2 sec^-1) cor-rected to water at 20\textdegreeC. The density of water at 20\textdegreeC is 0.998 (g/cm^3). Calculate the molecular weight of ribonuclease using the Svedberg equation. Assume that the sedimentation coefficient for ribonuclease is 2.02 × 10^-13 sec.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E22-0819.htm

Solution:

The molecular weight M of a protein can be calculated from the sedimentation coefficient by means of the Svedberg equation, which is derived by equating the centrifugal force with the opposing frictional force, the condition existing when the rate of sedimentation is constant: M = [RT_s] / [D(1 - ѵ\rho)], where R is the gas constant in ergs per mole per degree (8.314 × 10^7 ergs/\textdegreeK mole), T the absolute temperature, s the sedimentation coefficient, ѵ ̅ the partial specific volume of the protein, \rho the density of the solvent, and D the diffusion coefficient. All of the values necessary to solve for M are given. 20\textdegreeC = 293\textdegreeK. Solving for M: M = [(8.314 × 10^7 ergs/\textdegreeK mole) (293\textdegreeK) (2.02 ×10^-13 sec)] / [(13.1 × 10^-7 cm^2 sec^-1) {1-(0.707 cm^3 g^-2)(0.998 g/cm^3)}] = 13,000 g/mole.

Question:

Which are the first two organs to appear in the human embryo, and how are they formed?

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/Users/wenhuchen/Documents/Crawler/Biology/F23-0602.htm

Solution:

The first two organs to appear in the embryo are the brain and spinal cord, which form by a process termed neurulation. During the third week of development, the ectoderm in front of the primitive streak develops a thickened plate of cells called the neural plate. At the center of this plate there appears a depression, known as the neural groove. At the same time, the outer edges of the plate rise in two longitudinal neural folds that meet at the anterior end. These appear, when viewed from above, like a horseshoe. These folds gradually come together at the top, forming a hollow neural tube. The cavity of the anterior part of this neural tube becomes the ventricles of the brain. At the same time, the cavity of the posterior part becomes the neural canal, extending the length of the spinal cord. The brain region is the first to form, and the long spinal cord develops slightly later. The anterior part of the neural tube, which gives rise to the brain, is much larger than the posterior part and continues to grow so rapidly that the head region comes to bend down at the anterior end of the embryo. By the fifth week of development, all the regions of the brain, i.e., the forebrain, midbrain, and hindbrain are established. A short time later, the outgrowths that will form the large cerebral hemispheres begin to grow. While the various motor nerves grow out of the brain and spinal cord, the. sensory nerves do not, having a separate origin. During the formation of the neural tube, bits of nervous tissue, known as neural crest cells, are left over on each side of the tube. These cells migrate downward from their original position and form the dorsal root ganglia of the spinal nerves and the postganglionic sympathetic neurons. From sensory cells in the dorsal root ganglia, dendrites grow out to the sense organs and axons grow in to the spinal cord. Other neural crest cells migrate and form the medullary cells of the adrenal glands, the neurilemma sheath cells of the peripheral neurons, and certain other structures.

Question:

Let the acceleration of a non-inertial frame, a freely falling elevator, be a_O^\ding{217} = -g\^{y} where \^{y} is measured upward from the surface of the earth and g is the acceleration of gravity. Under these conditions, what is the net force acting on a mass M at rest in the elevator?

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Solution:

Note that in this problem, we are doing an experiment in a non-inertial reference frame. It is very important to realize that Newton's Laws only hold in inertial frames. If we want to analyze this experiment from the point of view of an observer in a non inertial frame, we must use a modified form of Newton's Second Law, Ma"^\ding{217} = F^\ding{217} - F'^\ding{217}(1) where a"^\ding{217} is the acceleration of M as observed in the non-inertial frame (the elevator) ,F^\ding{217} is the sum of all real forces acting on M (i.e., tensions, gravity, etc.) and -F'^\ding{217} is the sum of all fictitious forces. These latter forces arise because we are doing our experiment in a non--inertial frame. (Fictitious forces include Carioles forces, centrifugal forces, etc.) In this particular problem, a"^\ding{217} = 0 because the mass M is at rest in the elevator. F^\ding{217}, the sum of the real forces acting on M, includes Mg, the weight of M, and N, the normal force exerted by the floor of the elevator on the mass. Hence, F = N - Mg(2) where the minus sign appears because N and Mg are in opposite directions. We may now substitute equation (2) into equation (1) and solve the latter for F'^\ding{217} : 0 = F - F' F' = F F' = N - Mg(3) Now, - F', the fictitious force, is defined as - Ma'^\ding{217}, where a'^\ding{217} is the acceleration of the non-inertial frame with respect to the earth (-g\^{y}in our case). Hence, - F' = Mg(4) Inserting this in (3), we find - Mg = N - Mg(5) N = 0(6) Hence, the floor exerts no force on M. If this is the case, then M must be floating inside the elevator, i.e., M is weightless.

Question:

What is reverse transcriptase and what is its function? How might it be used in cancer research?

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/Users/wenhuchen/Documents/Crawler/Biology/F24-0639.htm

Solution:

We have seen in protein synthesis that biological information flows from DNA to RNA to protein. This flow had become the rule until 1964, when Temin found that infection with certain RNA tumor viruses (can- cer- causing substances) is blocked by inhibitors of DNA synthesis and by inhibitors of DNA transcription. This suggested that DNA synthesis and transcription are re-quired for the multiplication of RNA tumor viruses. This would mean that the information carried by the RNA of the virus is first transferred to DNA, whereupon it is trans-cribed and translated, and consequently, that information flows in the reverse direction, that is, from RNA to DNA. Temin proposes that the RNA of these tumor viruses, in their replication, are able to form DNA. His hypothesis requires a new kind of enzyme - one that would catalyze the synthesis of DNA using RNA as a template. Such an enzyme was discovered by Temin and by Baltimore in 1970. This RNA - directed DNA polymerase, also known as re-verse transcriptase, has been found to be present in all RNA tumor viruses. An infecting RNA virus binds to and enters the host cell (the cell which the virus attacks). Once in the cytoplasm, the RNA genome is separated from its protein coat (see figure). Then, through an as yet unknown mechanism, the viral reverse transcriptase is used to form a DNA molecule using the viral RNA as a template. This DNA molecule is integrated into the host's chromoso-me, with a number of possible consequences. The viral DNA may now be duplicated along with the host's DNA, and its information thus propogated to all offspring of the infected cell. Its presence in the genome of the host may "transform" the cell and its offspring (cause them to become cancerous). In addition, the viral DNA may be used as a template for the synthesis of new viral RNA, and the virus thus multiplies itself and continues its infectious process, often without killing the cell. Because the viral genome is RNA, it cannot as such be integrated into the host's genome. Reverse transcriptase enables the virus to convert its genetic material to DNA, whereupon it is capable of inserting itself into the host chromosome. There are three reasons why reverse transcriptase is so important in cancer research. First, human leukemia sarcomas (different forms of cancer) have been shown to contain large RNA molecules similar to those of the tumor viruses that cause cancer in mice. Second, these human cancer cells contain particles with reverse transcriptase activity. Third, the DNA of some human cancer cells have virus-like sequences of nucleotides, sequences not found in the DNA of comparable normal cells. Since the action of RNA from tumor viruses depends on the presence of the enzyme reverse transcriptase, reseach into its chemical composition (amino acid' sequence) may bring better understanding of its function. The ultimate objective of such reseach would of course be the prevention of cancer, perhaps through the inhibition of reverse transcriptase activity.

Question:

Metallic gold crystallizes in the face-centered cubic lattice. The length of the cubic unit cell, a, is 4.070 \AA. (a) What is the closest distance between gold atoms? (b) What is the density of gold?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E06-0224.htm

Solution:

An ordered array of atoms, ions, or molecules is called a lattice. Every lattice, is a three-dimensional stacking of identical building blocks called unit cells. The most symmetrical crystals have cubic lattices. There are three kinds of elementary cubic lattices: the simple cubic, the face- centered cubic, and the body-centered cubic. The length of the cube edge is designated by the symbol a. In a face-centered lattice, an atom would be located at each corner of the unit cell and at the center of each of the six faces. (a) In the gold crystal above, the closest distance from one corner atom to another corner atom is a, 4.070 \AA. The distance from any corner atom to an atom at the center of a face is one-half the diagonal of that face. Using the fact that the diagonal of any square is \surd2 × length of one side, the diagonal is then a\surd2. Half the diagonal is 1/2 a\surd2 or 1/2 (4.070 \AA) \surd2 = 2.878 \AA This distance is the closest distance between atoms, since this distance is shorter than the distance between adjacent corners. (b) The density is computed by first counting the number of gold atoms that can occupy one unit cell. This number is 1/8 times the number of occupied corners in the unit cell plus 1/2 times the number of occupied face- centers. Since a cube has 8 corners and 6 faces, the number is . 1/8 \textbullet 8 + 1/2 \textbullet 6 = 4 The volume of the cubic unit cell is a^3. The density if the mass of 4 gold atoms (4 atomic Wt / Avogadro's #) divided by the volume of the unit cell, a^3. Thus, the formula for the density of a gold crystal lattice is Density = {(4 atoms)(Atomic Wt. of gold)} / {a^3(Avogadro's number)} Substituting into the equation the Atomic Weight of gold, 197.0 g/mole, Avogadro's number, 6.023 × 10^23 atoms/mole, and converting 4.070 \AA to 4,070 × 10^-8 cm, then Density = {(4 atoms) (197.0 g/mole)} / {(4.070 × 10^-8 cm)^3 (6.023 × 10^23 atoms/mole)} = 19.4 g/cm^3.

Question:

What is dormancy? What factors play a role in the termi-nation of dormancy and the germination of a seed?

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/Users/wenhuchen/Documents/Crawler/Biology/F08-0201.htm

Solution:

Dormancy is a special condition of rest which enables an embryo to survive long periods of un-favorable environmental conditions, such as water scarcity or low temperature. During this period of rest, the embryo ceases or limits its growth, and metabolizes at very low levels, although ordinary plant rest can be termi-nated and normal growth resumed by the onset of warmer temperatures or the availability of water. Germination or the resumption of normal growth by a dormant embryo requires certain, very precise combinations of environ-mental cues. This is of great survival importance to the plant in that it prevents the dormant seed from germinating in response to conditions such as a warm spell in winter, which, although apparently favorable, are only temporary. Should the seed germinate during such a period, the seedling would begin to grow, only to be killed by the ensuing cold and frost. Apparently the dormant seed contains certain endogenous inhibiting factors which must be overcome before germination can begin, and thus requires more complex triggers then just a favorable temperature or water supply. Some seeds are required to pass through a period of cold before they are able to germinate. This is seen in the seeds of almost all plants growing in areas with marked seasonal variations. Many seeds require drying before they can germinate. Such a requirement has adaptive significance in that it prevents germination within the moist fruit of the parent plant. Some seeds such as those of the lettuce, require a light exposure, while the germination of others is inhibited by light. Certain seeds cannot germinate, even in highly favorable conditions of water, light, oxygen, and temperature, until they have been abraded by factors such as soil action. This abrasion wears away the seed coat, allowing water and oxygen to enter the seed and signal the initia-tion of germination. The seeds of some desert species germinate only when sufficient rainfall has drained inhibitory chemicals from their seed coats. Most seeds "force" the embryo into dormancy by the fact that the seed coat is impermeable toundissolvedoxygen and the seed itself has a low water content. The seed is protected from freezing and desiccation by the seed coat and provided with nourishment by the endosperm. During germination, the embryo and endosperm absorb water and swell; then the seed coat ruptures, freeing the embryo and enabling it to resume development. It is important to note that most seeds do not need soil nutrients in order to germinate-they germinate equally well on moist paper as in soil. Dormancy plays a similar role in plant buds.

Question:

(a)Write a FORTRAN subroutine to find approximate solutionsto the initial-value problem: dy/dx = f(x,y), y(x_0) = y_0 using modified Euler's (predictor-corrector) method. (b)Write a FORTRAN program to find an approximate solution to the initial-value problem: dy/dx = g(y), y(x_0) = y_0 using modified Euler's method.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G24-0576.htm

Solution:

The modified Euler's method uses Euler's formula: y_n+1 = y_n + \Deltaxf(x ,y )(1) to obtain a predicted valuey_n+1,_p \cyrchar\cyro\pm y_i+1 and then iterates the following formula to obtain (k+1) corrected valuesyi+1,c: y^(k+1)_i+1,c = y_i + 1/2 [f(x_1 , y_1) + f(x_i+1,y^(k)_i+1)] (x_i+1 - x_i)(2) The iterative process is stopped as soon as y^(k+1)_i+1 = y^(k)_i+1to the required degree of accuracy. The following flowchart details the programming of this method : (a)The subroutine closely follows the flowchart, but the indices i and k have been omitted: SUBROUTINE EULER (N,XN,X,YCOR,ACCUR) REAL N CUSE A REAL VALUE FOR N TO AVOID MIXED CMODE ARITHMETIC DELTAX = (XN - X)/N 50Y = YCOR FOLD = F(X,Y) PRINT, X,Y X = X + DELTAX IF(X.GT.XN) STOP YPRED = Y + FOLD \textasteriskcentered DELTAX 100YCOR = Y + \O.5\textasteriskcentered (FOLD + F(X,YPRED))\textasteriskcenteredDELTAX YDIF = ABS(YCOR - YPRED) IF(YDIF.LE.ACCUR) GO TO 50 YPRED = YCOR GO TO 100 RETURN END (b)A subroutine to perform the loop that computes the corrected value, leaving 1/0 tasks to the main program will be used here. The advantage of this approach is the availability of each y(x). (An alternative approach would be to store the y's in an array, but for large N this might require a large storage area). The dis-advantage is the increased execution time required to pass arguments back and forth. The program looks as follows: CMAIN PROGRAM READ, N,XN,X,YCOR,ACCUR PRINT, X, XCOR REALN = FLOAT (N) DELTAX = (XN - X)/REALN DO 10 I = 1,N X = X + DELTAX CALL MEULER(X,YCOR,ACCUR,DELTAX) PRINT, X,YC0R 10CONTINUE STOP END SUBROUTINE MEULER(X,YCOR,ACCUR,DELTAX) Y = YCOR GOLD = G(Y) YPRED = Y + GOLD \textasteriskcentered DELTAX 100YCOR = Y + 0.5\textasteriskcentered(G0LD + G(YPRED))\textasteriskcenteredDELTAX YDIF a ABS(YCOR - YPRED) IF (YDIF.LE.ACCUR) GO TO 11 YPRED = YCOR GO TO 100 11RETURN END

Question:

A solid cylinder 30 cm in diameter at the top of an incline 2.0 m high is released and rolls down the incline without loss of energy due to friction. Find the linear and angular speeds a the bottom.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0199.htm

Solution:

This problem can be solved using the conser-vation of energy principle. The cylinder initially at rest at the top of the incline has only gravitational (potential) energy. Taking the bottom of the incline as the zero level of the potential energy (see the figure above), we get E_P = mgh where m is the mass of the cylinder, g is the acceler-ation of gravity, and h = 2.0 m is its height above ground level. When the cylinder reaches the bottom of the incline, all of its energy will be kinetic: E_k = (1/2)mv^2 + (1//2) I\omega^2 where v is the cylinder's linear velocity, I its moment of inertia about the central axis, and \omega its angular momentum. (l = (1/2)mR^2 for cylinders, where R is the radius.) In the process of rolling down the incline the cylinder's potential energy turns to kinetic, the total change in each being equal to: ∆E_p = ∆E_k mgh= (1/2)mv^2 + (1/2) I\omega^2 = (1/2)mv^2 + (1/2)[(1/2) mR^2] [v/R]^2 = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2 Using \omega = v/R Thus, gh = (3/4)v^2 v= [(2\surd3)/3](gh)^1/2 = 1.15 [(9.8 m/sec^2)(2.0 m)]^1/2 = 5.09 m/sec. Note that the linear speed does not depend upon the size or mass of the cylinder. To find \omega, we use the formula: w= v/R = (5.09 m/sec)/(0.15 m) = 34 rad/sec.

Question:

Mesogleaand mesoderm are two terms used to describe the middle tissue layer of an organism. What distinction can be made between the two and in what organisms can each be found?

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0281.htm

Solution:

Mesogleais the term used to describe the layer between the outer epidermis and the inner layer of cells lining thegastrovascularcavity. It is found in adult Coelenterates (Cnidarians) such as hydra,Obelia, and Portuguese man-of-war (Physalia). Themesogleaconsists of either a thin non-cellular membrane or thick, fibrous, jellylike,mucoidmaterial with or without cells. The term mesoderm is used to describe the middle of the three embryonic tissue layers first delineated during an early developmental stage of the embryo. This layer gives rise to the skeleton, circulatory system, muscles, excretory system, and most of the reproductive system. It is found in higher invertebrates, the insects, and all vertebrate groups. The mesoderm is always a cellular layer and always refers to an embryonic layer.Mesoglea, on the other hand, is usuallyacellularand merely denotes the middle layer of the body wall of an organism.

Question:

Predict the final hydrogenation products in each of the compounds in figure A.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E21-0769.htm

Solution:

Hydrogenation refers to the reduction of unsaturated hydrocarbons, by addition of the H_2 across the bond. Frequently a catalyst is used to increase the rate of the reaction. After complete hydrogenation, the product contains only single bonds. A triple bond requires two moles of hydrogen gas, while a double bond requires only 1 mole for complete hydrogenation. With this in mind, you have the reactions shown in figure B.

Question:

A space vehicle ejects fuel at a velocity u relative to the vehicle. Its mass at some instant of time is m. Fuel is expelled at the constant rate ∆m/∆t . Set up and solve the equation of motion of the space vehicle, neglecting gravity.

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/Users/wenhuchen/Documents/Crawler/Physics/D11-0438.htm

Solution:

At any given time t, the momentum is mv. At time t + ∆t the mass of the rocket is m + ∆m where ∆m is negative since the total mass of the vehicle is de-creasing, while the velocity is v + ∆v and the momentum is (m + ∆m) (v + ∆v). However, after time ∆t, since ∆m in mass leaves the rocket, it introduces its own momentum. In the inertial frame of reference, the velocity of the fuel is v - u which is the velocity of the rocket minus the velocity of the fuel with respect to the rocket. The mass of the fuel ejected during time ∆t is ∆m, and so its momentum is ∆m(v - u). The law of conservation of momentum tells us that, in an isolated system the momentum at time t equals the momentum at t + ∆t. Therefore, mv = (m + ∆m) (v + ∆v) - ∆m (v - u) Simplifying, we have m∆v = -∆m (u + ∆v) If we let ∆v \rightarrow 0 to get instantaneous velocity and ∆m \rightarrow 0 we have mdv = - dm(u) dv = - dm(u/m) Integrating from v_0 to v (initial and instantane-ous velocities) and from M to m (where M is the initial mass and m is the instantaneous mass) we have: ^v\int_v0 dv = - ^m\int_M u(dm/m) Integrating, we have v = u In [1 + (m/M)] + v_0

Question:

A flywheel has a mass of 30 kg and a radius of gyration of 2 m. If it turns at a rate of 2.4 rev/sec, what isits rotational kinetic energy?

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/Users/wenhuchen/Documents/Crawler/Physics/D05-0248.htm

Solution:

The rotational kinetic energy of a body about a given axis is T = (1/2) I \omega^2(1) where I and \omega are the rotational inertia and angular velocity of the body about the given axis. By definition of the radius of gyration, \rho, we may write m\rho^2 = I(2) where m is the body's mass. Using (2) in (1) T = (1/2) m \rho^2 \omega^2 Using the given data T = (1/2)(30 kg) (4 m^2) (2.4 rev/s)^2 T = 345.6 kg\bulletm^2\bulletrev^2/s^2 To put this answer in conventional energy units, note that 1 rev/s = 2\pirad/s whenceT = 1382.4\pi^2 Joules T = 13643.74 Joules

Question:

Find the mistakes in the following FORTRAN expressions.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G12-0273.htm

Solution:

a) In FORTRAN any constant or variable in an expression, unless it is the first constant or variable of that expression, must be preceded immediately by one of the following: a left parenthesis one of the operators +, -, {_\ast},/, or {_\ast}{_\ast} . Any constant or variable in an expression, unless it is the last constant or variable of that expression, must be followed by one of the following: a right parenthesis one of the operatorsor +, -, {_\ast},/, or {_\ast}{_\ast} . The number of opened and closed parenthesis must be equal. Therefore the correct expression is (A - (B - (C + D(4.7)))). b) For the same argument as was used in part (a), the correct expres-sions are (A/B) or (A)/(B). c) The Statement is incorrectly formed, and has no meaning. Except for the unary minus, each operator +, {_\ast}, /, or {_\ast}{_\ast} must have a term or factor both to its left and to its right, in order for an expression to be correctly formed.

Question:

A PL/I program uses the EDIT form of input. The declare statementof the program is as follows: DCL(A,B,C,D,E,F) FIXED(5,2) ; GCHAR(15)VAR, H CHAR(4) ; A data card used by the program has the following informationpunched on it (Note: The symbolbdenotes a blankspace). \downarrowcol.8\downarrowcol.34 AB-123 \textbullet 45b123 \textbullet 459123 \textbullet 456bb12345b18 \textbullet b 345912345B2GbCARS CARS CARS Show what will be stored in the memory locations correspond-ingto A, B, C, D, E, F, G and H when the followingstatement is executed: GETEDIT(A, B, C, D, E, F, G, H) [COL(10), F(7, 2), F(7, 2), X(2), F(4, 1), COL(34), F(5, 2), X(1), F(6, 3), F(1, 1), X(5), A(3), X(1), A(4)];

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G13-0324.htm

Solution:

As a result of the GET EDIT statement, the data card will be read in thefollowing manner: All entries on the data card up to col. 10 are ignored. Starting from col. 10, a numerical fieldF(7,2) is read off, for A. Thus, a value -123\textbullet45 is stored for A. This brings the computer to col. 17 ofthe card. Next, a numeric fieldF(7,2) for B is read off. Thus, a value 123\textbullet45 is stored for B. This brings the computer to col. 24 of the card. The nextinstruction is X(2), i.e., skip 2 columns. Hence the computer ignores the91, and goes to col. 26 of the card. Now, it must read a numeric field ofF(4,1) corresponding to C. Thus, a value 23\textbullet4 is read off, and the computercomes to col. 30 of the card. However, C is declared as FIXED(5,2). Therefore, the value 23\textbullet4 which is read from the card is storedas 023\textbullet40 in the memory location corresponding to C. The next 4 columns of the card are to be skipped be-cause the programmust jump to col. 34 of the card according to the next instruction. Starting from col. 34, a numeric field ofF(5,2) is to be read. Thus, 12345 are read off from the card and assigned to D. And, as D is specified aformat of F(5,2), therefore 12345 is stored for D as 123\textbullet45. The program comesto col. 39 Column 39 is skipped due toX(1), and the program reaches col. 40. The next instruction isF(6,3) corresponding to E. So, 1\textbullet8345 are read fromthe card and assigned to E. However, E is specified a format of F(6,3). The format specifications wants E to be of the form XX\textbulletXXX. In caseof a conflict between the format specification and the value read from thecard, the value read from the card gets pre-ference. Hence, E becomes 1\textbullet8345. Now, E is declared asFIXED(5,2). Thus, the memory al-located for E on the basis of the declare statement (DCL) expects E to be of the form XXX\textbulletXX. Hence, the value stored in memory for E is 001\textbullet83. Note that the extra two decimal places to the right of 3 are truncated. Next to be read is a fieldF(1,1) corresponding to F which is declaredas FIXED(5,2). Thus,F(1,1) reads off a single digit, 9, from the card. The value of F stored is 000\textbullet90. The next 5 card columns are skipped due toX(5). The next instructionspecifies A(3). This means a character field of 3 characters is tobe read off from the card cor-responding to G. Thus, B2G is stored for G. (2 is treated as a character). The next card column is to be skipped be- causeof X(1). Finally, a value CARS is read off for H, corresponding to thespecification A(4). The contents of the memory locations are shown below: Name of the memory location Content of the memory location A -123\textbullet45 B 123\textbullet45 C 023\textbullet40 D 123\textbullet45 E 011\textbullet83 F 000\textbullet90 G B2G H CARS

Question:

With reference to fluid pressures, explain the mechanism underlyingglomerularfiltration.

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/Users/wenhuchen/Documents/Crawler/Biology/F18-0445.htm

Solution:

The mean blood pressure in the large arteries of the body is approximately 100 mm Hg. However, as the blood passes through the arterioles connecting the renal artery to theglomeruli. the blood pressure decreases, so that the pressure in theglomerularcapillaries is usually only about half the mean arterial pressure or 50 mm Hg. This level is considerably higher than in other capillaries of the body, due to the fact that the arterioles leading to theglomeruliare wider than most other arterioles and therefore offer less resistance to flow. The force driving fluid out of theglomerulusand into Bowman's capsule is the pressure of the blood in theglomerularcapillaries. But this hydrostatic pressure favoring filtration is not completely unopposed. There is fluid within Bowman's cap-sule, resulting in a capsular hydrostatic pressure of about 10 mm Hg, which resists the flow into the capsule. The presence of protein in the plasma and its absence in Bowman's capsule, results in a second force opposing fil-tration. This difference in protein concentration between the capsule and the blood induces an osmotic flow of water into the blood. This osmotic pressure, also known as colloidal osmotic pressure, equals roughly 30 mm Hg. Summing the forces, there is aglomerularcapillary blood pressure of 50 mm Hg favoring filtration, a 10 mm Hg opposition to filtration due to fluid in Bowman's capsule, and a 30 mm Hg. pressure working against filtration due to a protein concentration difference between the blood and the capsule fluid, resulting in a netglomerularfiltration pressure of 10 mm Hg. This net pressure forces fluid from the blood into Bowman's capsule.

Question:

Liquid nitrogen is an excellent bath for keeping temperatures around 77\textdegreeK, its normal boiling point. What pressure would you need to maintain over the liquid nitrogen if you wanted to set the bath temperature at 85\textdegreeK? Heat of vaporization is about 5560 (J / mole).

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/Users/wenhuchen/Documents/Crawler/Chemistry/E07-0266.htm

Solution:

One uses the Clausius-Clapeyron equation to solve for the final pressure. This equation is written In (P_1)(P_2) = (∆H / R) [(1 / T_2) - (1 / T_1)] where P_1 is the initial pressure, P_2 is the final pressure at 85\textdegreeK, ∆H is the heat of vaporization, R is the gas constant, (8.314 J / mole-\textdegreeK), T_1 is the initial temperature and T_2 is the final temperature. The pressure of a liquid at its boiling point is equal to the atmospheric pressure, which is 1 atm. Solving for the final pressure: P_1 = 1 atmT_1 = 77\textdegreeK P_2 = ?T_2 = 85\textdegreeK ∆H = 5560(J / mole) R = 8.314 (J / mole \textdegreeK) 1n (1 / P_2) = [(5560 J / mole)] / [(8.314 J /mole)] [(1 / 85\textdegreeK) - (1 / 77\textdegreeK)] -1nP_2 = 6.687 × 10^2\textdegreeK (1.176 × 10-^2\textdegreeK) - (1.298) × (10^-2\textdegreeK) -1nP_2 = 6.687 × 10^2\textdegreeK (-1.22) × (10^-3 / \textdegreeK) -1nP2= -8.1587 × 10^-1 P_2 = 2.26 atm.

Question:

Show that the energy eigenvalues E_n, of a particle moving between impenetrable walls is given by the relationship E_n = (h^2/8mL^2)n^2 where h is Planck's constant, m the particle mass, L the distance between the walls, and n = 1, 2, 3, ...

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/Users/wenhuchen/Documents/Crawler/Physics/D33-1011.htm

Solution:

In order to find the energy eigenvalues, we must, in general, solve Schroedinger's Equation. Since our problem is one dimensional, this equation reduces to (-h^2/2m) (მ2\psi/მx^2) + V\psi = ih(მ\psi/მt)(1) where m is the mass of the particle we are studying i = \surd- 1,his Planck's constant over 2\pi, and V is the potential that m is subject to. \psi is the wave function which describes the motion of the particle and it is generally a function of x and t. In order to solve (1), we use the method of separation of variables, and express \psi as the product of 2 functions, one with only a time dependence, and the other with only a spatial depend-ence . Hence, \psi = \psi (x) \textphi (t)(2) Substituting (2) in (1) (-h^2/2m) (მ2\psi/მx^2) [\psi (x) \textphi (t)] + V \psi (x) \textphi (t) = ih(მ/მt) [\psi (x) \textphi (t)] (-h^2/2m) \textphi (t) {[d^2 \psi (x)]/dx^2} + V \textphi (t) \psi (x) = ih\psi (x) [{d \textphi (t)}/dt] Dividing both sides by \psi = \psi (x) \textphi (t) (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] + V = ih[1/{\textphi (t)}] [{d \textphi (t)}/dt](3) Note that if v is a function of x alone, the left side of (3) is also a function of x alone, and the right side is a function only of t. The only way a function of x can equal a function of t is if both functions equal the same constant K. Hence, we separate (3) into (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2}] + V = K(4) ih[1/{\textphi (t)}] [{d \textphi (t)}/dt] = K(5) Remember that this is true only if V is time in-dependent. We first solve the differential equation in (5). Rewriting this equation [{d \textphi (t)}/dt] = (K/ih) \textphi (t) or[{d \textphi (t)}/{\textphi (t)}] = (K/ih) dt \int [{d \textphi (t)}/{\textphi (t)}] = (K/ih) \int dt In \textphi (t) = (K/ih) t + C where C is a constant. Taking the exponential of both sides \textphi (t) = e ^[(Kt/ih) + C]= e^c e^k ^t/ih(6) But e^c is a new constant. Letting e^c = A we may write (6) as \textphi (t) = A c^ Kt/ih= A e - ^i ^k ^t/h(7) Equation (7) may be written in terms of sines and cosines since e^-ix = cos x - i sin x Hence, \textphi (t) = A [cos Kt/h- i sin Kt/h] and \textphi (t) is an oscillatory function of time with fre-quency \omega_0 = K/h. The energy of particle is E = hv and\omega = (2\piE/h) = (E/h) But the only physical interpretation \omega_0 can have is that it is the particle's frequency. Hence \omega = \omega_0 or(E/h) = (E/h) andE = K.(8) Putting (8) in (5) [ih/{\textphi (t)}] [{d \textphi (t)}/dt] = E orih(d/dt) \textphi (t) = E \textphi (t)(9) Equation (9) is an eigenvalue equation, since an operator [(ihd)/dt] acts on a function [\textphi (t)] and gives back the function multiplied by a constant (E). The constant E, is called the eigenvalue of \textphi (t). Now that we have found K, we may rewrite (4) as (-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] + V = E or(-h^2/2m) [1/{\psi (x)}] [{d^2 \psi (x)}/dx^2] = E - V (+h^2/2m) [{d^2 \psi (x)}/dx^2] = (V - E) \psi (x)(10) The real problem is to solve (10) for the given V. A particle moving between impenetrable walls is a picturesque way of describing a particle in an infinite potential well. (See figure.) We may describe this V by \infty, x \leq 0 V =0, 0 \leq x \leq L \infty, x \geq L Hence, within the well, V = 0 and (10) becomes [{d^2 \psi (x)}/dx^2] + (2mE/h^2) \psi (x) = 0 This is the equation of simple harmonic motion, and has the general solution \psi (x) = B cos [\surd(2mE)/h] x + C sin [\surd(2mE)/h] x(11) In order to "taper" (11) to the problem at hand, note that \psi (0) = 0 \psi (L) = 0(12) since, if the potential is infinite at these points, we cannot expect to find the particle there. Using (12) in (11) \psi (0) = B = 0 \psi (L) = B cos [\surd(2mE)/h] L + C sin [\surd(2mE)/h] L = 0 HenceC sin [\surd(2mE)/h] L = 0 C cannot be 0, otherwise \psi (x) = 0 for all x, and this is a trivial solution. Therefore, it must be that sin [\surd(2mE)/h] L = 0 or[{L \surd(2mE)}/h] = n\pin = 1, 2, ...(13) Note that if n = 0 in (13), this implies that E = 0, always. Again, (11) would be a trivial solution for this case, since B = 0. Hence, E = [(n^2 \pi^2h^2)/(L^2 2m)] = [n^2 h^2/8mL]n = 1, 2, ... These are the energy eigenvalues of a particle in an infinite square well.

Question:

A cylinder is rolled up an incline by a tape arranged as shown. What minimum force T is required if the angle \texttheta = 30\textdegree? The weight of the cylinder is 2 lb.

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0205.htm

Solution:

The force, T, needed to pull the cylinder up the incline can be found by applying Newton's Second Law, F = ma, to the x and y directions of the cylinder's motion. Hence, using the figure, T - f - Mg sin \texttheta = ma_x(1) N - Mg cos \texttheta = ma_y(2) where f is the frictional force, N is the normal force of the plane, and a_x and a_y are the x and y directed accelerations of the cylinder. Since the cylinder does not leave the incline, a_y = 0, and, from (2) N = Mg cos \texttheta(3) Now, f is an unknown in (1) since we do not know the coefficient of rolling friction for the plane. In order to eliminate this unknown, we calculate the net torque, \cyrchar\cyrt^\ding{217}, on the cylinder. By definition, \cyrchar\cyrt^\ding{217} = r^\ding{217} × F^\ding{217} \cyrchar\cyrt= \vert\cyrchar\cyrt^\ding{217}\vert = rF sin \texttheta where \texttheta is the angle between r^\ding{217} and F^\ding{217}, and r^\ding{217} is a vector locating the point of application of F^\ding{217}. In our case, taking torques about the center of mass ButT=Rf + RT T=I\alpha where I and \alpha are the cylinder's moment of inertia and angular acceleration, respectively. Then I\alpha = fR + TR(4) Because the cylinder rotates without slipping v= wR(5) where v is the velocity of the cylinder's center of mass, and w is its angular velocity. Hence, differentiating (5) (dv/dt) = (dw/dt)R a = \alphaR Since v is directed along the x-axis, a is also, and a = a_x, whence a_x = \alphaR(6) Substituting (6) in (4) I(a_x/R) = (f + T)R f = (Ia_x/R^2 ) - T(7) Inserting (7) in (1) T - (Ia_x/R^2) + T - Mg sin\texttheta = ma_x 2T - Mg sin \texttheta = [m + (I/R^2)]a_x a_x = (2T - Mg sin \texttheta)/[m + (I/R^2)](8) The minimum applied force T will not accelerate the cylinder, but move it with constant velocity (a_x = 0). Hence, from (8) 2T= Mg sin \texttheta T= [(2 lb)(1/2)]/2 = 1/2 lb.

Question:

What is the value of log 0.0148?

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/Users/wenhuchen/Documents/Crawler/Chemistry/E33-0965.htm

Solution:

0.0148 = 1.48 × 10^\rule{1em}{1pt}2. The characteristic is the exponent of 10. Hence, the characteristic is \rule{1em}{1pt}2. The mantissa for 148 can be found in a table of logarithms. The mantissa is 0.1703. There-fore, log 0.0148 = \rule{1em}{1pt} 2 + 0.1703 = \rule{1em}{1pt} 2.0000 + 0.1703 = \rule{1em}{1pt}1.8297. Notice that the number 0.0148 is less than 1. Therefore, the value of log 0.0148 must be negative, as it was found to be.

Question:

A biologist deals with things on a microscopic level. To A biologist deals with things on a microscopic level. To describe cellular dimensions and the amount of materials present at the cellular level, units of an appropriately small size are needed. What are these units of measurements?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0001.htm

Solution:

In biology, the units of length commonly employed include the micron (abbreviated by \mu) and the \AAngstrom (abbreviated by \AA). A micron is equivalent to 10^-3 millimeters (mm). An \AAngstrom is equivalent to 10^-4 \mu or 10^-7 mm. Weights are expressed in milligrams (10^-3 grams) micrograms (10^-6 grams), and nanograms (10^-9 grams) . The unit of molecular weight employed is the dalton. A dalton is defined as the weight of a hydrogen atom. For example, one molecule of water (H_2O) weighs about 18 daltons. One dalton weighs 1.674 × 10^-24 grams.

Question:

A rigid rod whose own weight is negligible (see figure) is pivoted at point 0 and carries a body of weight w_1 at end A. Find the weight w_2 of a second body which must be attached at end B if the rod is to be in equilibrium, and find the force exerted on the rod by the pivot at 0.

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/Users/wenhuchen/Documents/Crawler/Physics/D02-0029.htm

Solution:

The question states that the rod is in equilibrium. In this case, the net force on the rod must be zero, and the net torque on the rod about the pivot must also be zero. The forces on the rod are T_1\ding{217} and T_2\ding{217}, the weights of masses 1 and 2, respectively, and P\ding{217}, the force of the pivot on the rod. Hence T_1 + T_2 - P = 0 orT_1 + T_2 = P(1) The torque about a point 0 is Շ\ding{217} = r × F where r\ding{217} is the vector from 0 locating the point of applica-tion of F\ding{217}. The net torque about 0 is T_2l_2 - T_1l_1 = 0 since the torque due to T_2 is opposite in direction to the torque due to T_1. Then T_2l_2 = T_1l_1(2) Substitution (2) in (1) T_1 + [(T_1l_1)/(l_2) = P P = T_1 [1+ (l_1-)/(l_2-)] But T_1 = w_1, and P = w_1 [1+ (l_1-)/(l_2-)] If l_2- = 3 ft, l_2= 4 ft and w_1 = 4 lb P= 4 lb[ 1+(3/4)]= 7 lb Furthermore, T_2 =[T_1l_1]/(l_2). Since T_2 = w_2, and T_1 = w_1 w_2 = (w_1l_1)/(l_2) = (4 lb)(3/4) = 3 lb

Question:

A microscope has an objective lens of 10.0 mm focal length and an eyepiece of 25.0 mm focal length. What is the distance between the lenses, and what is the magnifica-tion if the object is in sharp focus when it is 10.5 mm from the objective?

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/Users/wenhuchen/Documents/Crawler/Physics/D28-0881.htm

Solution:

This is a compound microscope which uses converging lenses to produce large magnification. A short-focus lens, the objective, is placed near the object. It produces a real, inverted and magnified image. The eyepiece further magnifies this inverted image and produces a virtual image 250 mm from the eye-piece. This is the distance of most distinct vision. Considering the image produced by the objective, we use the optics equation 1/p + 1/q = 1/f p represents the distance of the object from the ob-jective lens, q the distance of image 1 from the lens, and f is the focal length. Substituting, (1/10.5mm) + 1/q = (1/10mm) q = 210 mm Therefore image 1 is 210 mm from the objective lens, and is real. The eyepiece magnifies image 1 and produces a second virtual image 250 mm from the lens so as to provide most distinct vision. Since image 2 is virtual, on the same side of the lens as image 1, the distance q' of image 2 from the eyepiece is negative. Using the optics equation again, 1/p'^ + 1/q'= 1/f ' 1/p'^ + 1/-250mm = 1/25.0mm p' = 22.7 mm Therefore the eyepiece is 22.7 mm from image 1. The distance between lenses is q + p' = 210 mm + 22.7 mm = 233 mm = 23.3 cm To find the magnification, first find the magnifica-tion produced by each lens using the equation M = q/p Magnification by objective: M_0 = (210 mm)/(10.5mm) = 20.0 Magnification by eyepiece: M_E = (-250mm)/(22.7mm) = -11.0 Total magnification: M = MeM0= -11.0 × 20.0 = -220

Question:

Define and compare asexual reproduction with sexual reproduction in animals. Give examples.

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/Users/wenhuchen/Documents/Crawler/Biology/F22-0566.htm

Solution:

The fundamental difference between asexual reproduction and sexual reproduction is the number of parents involved in the production of offspring. In asexual reproduction only a single parent is needed. Offspring identical to the parent are produced when the parent splits, buds, or fragments. Asexual reproduction typically occurs in animals such as Euglena,Paramecium,Amoeba,Hydra, flatworms, and starfish. Two parents are usually involved in sexual repro-duction. A fertilized egg is produced through the union of specialized sex cells (egg and sperm) from each parent. Instead of having traits identical to a parent, the off-spring possess a variety of recombined traits inherited from both parents. In higher animals such as man, sexual reproduction is the most common form. It must be noted that in some cases, sexual repro-duction can take place with a single parent. In the hermaphroditic worms, such as the earthworm and the fluke, both male and female sex organs occur in the same individual. Whereas self-fertilization is avoided in the earthworm by certain adaptive mechanisms, it does take place in the parasitic fluke. This is an exception to the general observation that sexual repro-duction involves two parents.

Question:

The dew point (the temperature at which the water vapor in a given sample of air becomes saturated and condensation be-comes possible) of a mass of air at 15\textdegreeC is 10\textdegreeC. Calculate the relative humidity and the mass of 1 liter of the moist air if the barometric height is 76 cm of mercury. The gram- molecular weight of air is 28.9 g and the saturated aqueous vapor pressures at 10\textdegreeC and 15\textdegreeC are 9.2 mm of mercury and 12.8 mm of mercury, respectively.

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/Users/wenhuchen/Documents/Crawler/Physics/D14-0520.htm

Solution:

The relative humidity is defined as 100 × [(saturated vapor pressure at the dew point) / (saturated vapor pressure at the given temperature)] Therefore the relative humidity is RH = (9.2/12.8) × 100% = 71.9% Using the gas law, one can calculate the number of moles present independently for (a) the dry air and (b) the water vapor. Measuring the pressure p in atmospheres and the volume V in liters gives (a)n =pV/RT = (1atm× 1 liter)/(0.082litre-atm/mole - \textdegreeK × 288\textdegreeK) = 0.424 mole and (b)n' =p'V/RT = [(9.2 mm Hg/760 mm Hg/atm) × 1 liter]/[0.082 liter- atm/mole-\textdegreeK × 288\textdegreeK] = 0.00051 mole. The mass of moist air consists of dry air and water vapor. In one liter of moist air, there are 0.0424 moles of dry air and 0.00051 mole of water vapor, hence its mass is M = 0.0424 mole × 28.9 g/mole + 0.00051 mole × 18.0 g/mole = (1.2254 + 0.0092)g = 1.2346 g.

Question:

What is the resistance of a piece ofnichromewire 225 centimeters long with a cross-sectional area of 0.015 square centimeter?

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/Users/wenhuchen/Documents/Crawler/Physics/D20-0641.htm

Solution:

To solve this problem we use the relation R = \rho(L/A) Where R = resistanceL = wire length \rho = resistivityA = cross sectional area This basic relationship tells us that resistance is directly proportional to resistivity and length and in-versely proportional to cross sectional area. In the case of a wire this means that the resistance depends on the nature of the substance (which appears in the equation as the resistivity), that the resistance increases as the wire gets longer and decreases as the wire gets thicker. The resistivity (\rho) fornichromeis 100 × 10^-6 ohm-centimeter. The length is 225 centimeters, and the area is 0.015 square centimeter . ThenR = [(10^-4 ohm-cm × 225 cm)/(0.015 cm^2 )] = 1.5 ohms .

Question:

What properties of water make it an essential component of living matter?

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/Users/wenhuchen/Documents/Crawler/Biology/F01-0015.htm

Solution:

The chemistry of life on this planet is essentially the chemistry of water and carbon. Water is the most abundant molecule in the cell as well as on the earth. In fact, it makes up between 70 and 90 percent of the weight of most forms of life. Life began in the sea and the properties of water shape the chemistry of all living organisms. Life developed as a liquid phase phenomenon because reactions in solution are much more rapid than reactions between solids, and complex and highly structural molecules can behave in solution in a way that they cannot behave in a gas. Water is an excellent solvent for living systems. It can stay in the liquid stage throughout a very wide range of temperature variation. Almost all chemicals present in living matter are soluble in it. Water serves many functions in the living organism. It dissolves waste products of metabolism and assists in their removal from the cell and the organism. Water functions in heat regulation almost as an insulator would. It has a high heat capacity or specific heat in that it has a great capacity for absorbing heat with only minimal changes in its own temperature. This is because water molecules bond to one another by hydrogen bonds. Excess heat energy is dissipated by breaking these bonds, thus the living material is protected against sudden thermal changes. In addition, plants and animals utilize water loss to cool their bodies. When water changes from a liquid to a gas, it absorbs a great deal of heat. This enables the body to dissipate excess heat by the evapo-ration of water. In animals this process is sweating. Also, the good conductivity properties of water makes it possible for heat to be distributed evenly throughout the body tissues. Water serves as a lubricant, and is present in body fluids wherever one organ rubs against another, and in the joints where one bone moves on another. Water serves in the transport of nutrients and other materials within the organism. In plants, minerals dissolved in water are taken up by the roots and are transported up the stem to the leaves. Water is also very efficient in dissolving ionic salts and other polar compounds because of the polar physical properties of water molecules. The proper con-centration of these salts is necessary for life processes and it is important to keep them at extremely constant concentrations under normal conditions. These salts are important in maintaining osmotic relationships.

Question:

Describe locomotion in the amoeba.

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/Users/wenhuchen/Documents/Crawler/Biology/F11-0269.htm

Solution:

The locomotion of amoeba is considered to be the simplest type of animal locomotion. A moving amoeba sends out a projection, termed a pseudopodium. Following this, the organism advances as the inner, granular, gel--like endoplasm flows into the pseudopodium. Two or three pseudopodia may be formed simultaneously but ultimately one will become dominant for a time. As new pseudopodia are formed, the old ones withdraw into the general body region. In its locomotion the amoeba often changes its course in response to environmental stimulation, by forming a new dominant pseudopodium on the opposite side, thus moving in a very irregular fashion. Currently there is no fully complete explanation for the changes, both physical and chemical, which are involved in amoeboid movement. The theory accepted by zoologists at the present time is based on changes in the texture of the cytoplasm. As a result of some initial stimulus, ectoplasm, the outer clear, thin layer of the organism, undergoes a liquefaction and becomes endoplasm, which is gel-like. As a result of this change, internal pressure builds up and causes the endoplasm to flow out at this point, forming a pseudopodium. In the interior of the pseudopodium, the endoplasm flows forward along the line of progression; around the periphery, endoplasm is con-verted to ectoplasm, there by building up and extending the sides of the pseudopodium like a sleeve. At the posterior of the body, ectoplasm is assumed to be undergoing con-version to endoplasm. During this entire process, energy consumption is known to have taken place.

Question:

Verify that the fusion of four protons releases approximately 25 MeV for the proton-proton fusion cycle .

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/Users/wenhuchen/Documents/Crawler/Physics/D36-1061.htm

Solution:

The fusion of two hydrogen atoms produces one deuterium atom ^1_1H + ^1_1H^\ding{217} ^2_1H + e^+ + v_e .(1) The fusion of deuterium with hydrogen produces the helium atom ^1_1H + ^2_1H^\ding{217} ^3_2He + \Upsilon.(2) Finally, the fusion of helium with helium gives ^3_2He + ^3_2He^\ding{217} ^4_3He + ^1_1H + ^1_1H .(3) The chain of events (1), (2) and (3) is known as the proton-proton cycle, which is one of the schemes offered to describe the conversion of hydrogen into helium in stars. If we multiply equations (1) and (2) by two and sum the resulting equations and (3), we get (2^1_1H + 2^1_1H \ding{217} 2^2_1H +2e^+ + 2v_e) (2^1_1H + 2^2_1H \ding{217} 2^3_2He + 2\Upsilon) (3_2He + ^3_2He \ding{217} ^4_2He + ^1_1H + ^1_1H.) 6^1_1H + 2^2_1H+ 2^3_2He \ding{217}2^1_1H + 2^2_1H+ 2^3_2He + ^4_2He + 2e^+ + 2ve+ 2\Upsilon or4^1_1H \ding{217}^4_2He +2e^+ + 2ve+ 2\Upsilon Before the reaction the total nuclear mass is: 4 × (Atomic mass of hydrogen) = 4 × 1.0073 amu = 4.0292amu After reaction the total mass is Atomic mass of ^4He + 2 × (mass of positron) = (4.0015 + 2 × 0.0011) amu = 4.0026 amu The energy released in the fusion reaction is equivalent to the difference in mass: 4.0292 - 4.0026 = 0.0266 amu = {931 (MeV/amu)} (0.0266amu) = 25MeV.

Question:

The primary of a transformer, consisting of 20 turns, is connected to a varying voltage source which a volt-meter indicates to have a value of 110 volts. If the secondary circuit has 1000 turns and a resistance of 20,000 ohms, what voltage would a meter read if con-nected across it? What is the current in the primary if 100% efficiency is assumed?

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/Users/wenhuchen/Documents/Crawler/Physics/D23-0776.htm

Solution:

The voltmeter reads the root mean square (rms) value of the actual voltage. That is, if the actual instantaneous voltage is given by e = E_max sin \omegat where \omega is the angular frequency of the voltage variation, then E_rms = E = \surd[(1/2\pi) ^2\pi\int_0 E^2_max sin^2 \omegatd(\omegat)] If E_p and E_s are the rms values of the voltages on the primary (i.e. the side where the voltage source is located) and on the secondary sides, respectively, of the transformer, then according to the principle of the transformer (E_s/E_p) = (N_s/N_p) \therefore E_s = (N_s/N_p)E_p = (1000/20) (110) = 5500 volts The efficiency of a transformer is defined as E_ff = (Power out/Power in) × 100 If 100% efficiency is assumed, the power output is equal to the power input, where P = El. By Ohm's Law, I_s = (E_s/R_s) = (5500/20000) = .275 ampere \therefore P = E_s I_s = E_p I_p \therefore I_p = [(E_s I_s)/E_p] = (5500/110) (.275) = 50 (.275) = 13.75 amperes.

Question:

What would be the period of rotation of the earth about its axis if its rotation speed increased until an object at the equator became weightless?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0146.htm

Solution:

The two forces acting on a body at the equator are the force exerted on it due to the gravitational attraction of the earth, mg\ding{217}_0, where g\ding{217}_0^ is the free-fall acceleration at the equator and acts toward the center, and the normal force exerted by the surface of the earth on the body, N\ding{217}. This latter force acts upward. On a non--rotating earth, or at the poles (which by definition do not rotate), these forces are equal since a body would be in equilibrium. At the equator, where a body does ex-perience rotation and therefore a centripetal accelera-tion, the forces are unequal so that their resultant provides the centripetal force necessary to keep the body traveling in a circle. Therefore, Newton's Second Law yields: mg_0 - N = mv^2/R where v is the speed of the body and R is the radius of the earth. But the distance traveled in one period of rotation, T, is 2\piR. Therefore T = 2\piR/v, and v^2 = 4\pi^2R^2/T^2. Substituting this into the first expression, we get N = mg_0 - mv^2/R = mg_0 - (4\pi^2mR/T^2 ) = m[g_0 - (4\pi^2R/T^2)] = mg, where g is the acceleration as measured at the earth's surface, and N is a measure of the apparent weight of the body, which is thus less than the gravitational force exerted on the body by the earth. If the speed of revolution of the earth increases, the body becomes weightless when the normal force exerted on it by the surface be-comes zero. Thus weightlessness occurs when N = 0 = m[g_0 - (4\pi^2R/T^2)] g_0 = 4\pir^2R/T^2 or when the period of rotation is T= 2\pi\surd(R/g_0) = 2\pi\surd[(4 × 10^3mi × 5280 ft/mi)/(32.4 ft/s^2)] = (2\pi/3600 s/hr) × \surd[(4 × 5280 × 10^3)/32.4]s = 1.41 hr.

Question:

What exactly do the terms `'primitive" and "advanced" mean inevolution? Are "generalized" organisms necessarilymore advancedthan "specialized" ones?

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/Users/wenhuchen/Documents/Crawler/Biology/F28-0727.htm

Solution:

"Primitive" and "advanced" are two contras-ting terms. "Primitive" meansolder, more like the ancestral condition. "Advanced" means newer, lesslike the ancestral form. "Primitive" and "advanced" are just two terms referringto relative sequence in evolutionary time, and do not imply that oneis necessarily superior to or more complex than the other. The terms "specialized" and "generalized" refer to the relationships oforganisms, or particular characteristics of organisms, to their environment. "Specialized" organisms or structures are adapted to a special, usually rather narrow, mode of life. "Generalized" organisms or structuresare broadly adapted to a greater variety of environments and waysof life. The terms "specialized" and "generalized" are sometimes confusedwith "advanced" and "primitive." Generalized characteristics are likelyto be more primitive, and specialized ones more advanced, but this isnot an absolute rule.

Question:

The gene for PKU is found in the heterozygous state in about one person in fifty. About how frequently would you expect babies to be born with PKU? Assume here that the homozygous state for PKU is either lethal or severely restricts the affected person's ability to produce offspring.

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/Users/wenhuchen/Documents/Crawler/Biology/F26-0692.htm

Solution:

The trait for PKU disease is recessive, and thus the genotype of babies born with the disease must be homozygous recessive. This genotype could be produced through the mating of two phenotypically normal people who are heterozygous for the PKU gene. The chance that two such heterozygotes will marry can be determined by using simple probability procedures. We know that the probability of two separate events occurring together is the product of their separate probabilities. If the heterozygous genotype occurs in one person out of fifty, then the chance that any two people who marry will both be carriers is 1/50 × 1/50 or 1/2,500 and only one marriage out of 2,500 will be capable of producing children with PKU. We must now determine the probability within such a marriage that a child will be born with PKU. Let us use the small letter p to represent the recessive PKU allele, and the capital letter P, to represent the dominant, normal allele. The cross between two Pp individuals is illustrated below: P The offspring are obtained in the ratio: 1/4 PP : 1/2 Pp : 1/4 pp We see that only one child in four will be homozygous recessive for the PKU allele, and consequently have the disease. Now, if only one of every four offspring produced has the PKU disease, and only one of every 2,500 marriages has the genetic makeup to produce a PKU child, then the frequency of children born with PKU would be [1/(2,000)] × [1/4] or 1/(10,000) Therefore, one out of every 10,000 children born will have PKU disease.

Question:

When a block of wood of mass 1 kg is held in a vise, a bullet of mass 10 g fired into it penetrates to a depth of 10 cm. If the block is now suspended so that it can move freely and a second bullet is fired into it, to what depth will the bullet penetrate? (The retarding force in both cases is assumed constant.)

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0331.htm

Solution:

The wood exerts a constant retarding force F^\ding{217} on the bullet. The work this force does on the bullet is equal to its change in kinetic energy in accordance with the work-energy theorem. This work is given by W = F^\ding{217}\bullet S^\ding{217}(1) where s^\ding{217} is the distance the bullet penetrates the block. Since the force acts in a direction opposite to the path of motion of the bullet, equation (1) reduces to W = - Fs In the first case, all of the bullet's kinetic energy is converted to heat due to the work done on it by the re-tarding force. If the distance the bullet penetrates the block is d_1, then the work done on it is W_1 = - Fd_1 = (1/2)mv2_1 - (1/2)mv^2 = 0 - (1/2) mv^2 = - (1/2) mv^2 Fd_1 = (1/2)mv^2(2) where v is the bullet's initial velocity and v_1 is the bullet's final velocity, which equals zero since the bullet comes to rest in the block. In the second case, the bullet's final velocity, v_2,is not zero. Even after it stops inside the block, it has a speed since the block is now free to move and the bullet moves with it. Since there are no external forces acting on the system, the principle of conservation of momentum can be applied. This is an inelastic collision where the block and bullet move together with the same final velocity, v_2.If the bullet's mass is represented by m and that of the block by M, then we have mv= (m + M)v_2(3) where v has the same significance as before. Further, the work done by the retarding force in stopping the bullet over a distance d_2, must equal the total change in kinetic energy of the system. W_2 = - Fd_2 = (1/2) (m + M)v^2 _2 - (1/2) mv^2(4) Solving for v_2 in equation (3) and substituting into equation (4) - Fd_2 = (1/2) (m + M) [(mv) / (m + M)]^2 - (1/2) mv^2 Fd_2 = (1/2) mv^2 - (1/2) [m2/ (m + M)] v^2(5) Dividing equation (5) by (2), We obtain Fd2/ Fd_1 = [(1/2) mv^2 - (1/2) {m^2/(m + M)} v^2] / [ (1/2) mv^2] (d2/ d_1) = [m - {m^2/(m + M)}] / m = [{(m^2 +mM- m^2)/(m + M)} / m] = M/(m + M) Substituting the known values, we have for the distance d_2 that the second bullet penetrates d_2 = (m d_1)/(m + M) = [(1kg)(0.1 m)] / [(0.01 kg + 1kg)] = (0.1)/(1.01) m = 0.099m = 9.9cm

Question:

Calculate the characteristicvibrationalfrequency of the lowest vibration state for H_2. Assume k = 5.1newtons/cm and h = 6.63 × 10^-34 J-sec.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E18-0702.htm

Solution:

Picture a diatomic molecule, such as H_2, as two balls connected by a spring. The spring represents the chemical bond. When the two balls (atoms) are pulled apart the string acts as a restoring force to bring them back together to the equilibrium separation. When they are com-pressed, the string tends to force them apart. Such a system, in which the restoring force is proportional to the amount of distortion, is called a harmonic oscillator. Thevibrationalfrequency of the spring going back and forth is given by the formula ѵ = (1/2) \pi \surd(K/M_eff) where ѵ = characteristicvibrationalfrequency of the particular bond, k = force constant andM_eff= effective mass for the vibrating atoms: M_eff= [(M_A M_B) / (M_A + M_B)] If the masses of the two atoms are equal,M_effbecomes (M^2 /2M) = (1/2)M. To find the characteristic vibration frequency, use this formula. To find theM_efffind the mass of one hydrogen atom. The number of grams in one mole (molecular weight) is one. There are 6.02 × 10^23 atoms of H in one mole. Thus, the mass of one atom is [(1 g/mole) / (6.02 × 10^23 /mole)] = 1.66 × 10^-24 g.Thus, M_eff= [(1.66 × 10^-24 g) / (2)] = 8.31 × 10^-25 g. Solving for ѵ : k = 5.1 [(newtons) / (cm)] = 5.1 [(kg M) / (cm \bullet sec^2)] ѵ = (1/2\pi) \surd[{(5.1 (kg M/cm sec^2) × 1000 (g/kg) × 100 (cm/m)} /(8.31 × 10^-25 g)] = 1.25 × 10^14 sec^-1. Note: 1000 g/kg and 100 cm/m are conversion factors used to obtain the correct units.

Question:

Consider a molecule suspended in a liquid in the test chamber of an ultracentrifuge. Suppose that the molecule lies 10 cm from the axis of rotation and that the ultracentrifuge rotates at 1000 revolutions per second (60,000 rpm). What is the magnitude of the acceleration associated with the circular motion?

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/Users/wenhuchen/Documents/Crawler/Physics/D04-0162.htm

Solution:

The angular velocity of the molecule is \omega = 2\pin where n is the number of revolutions per second that the molecule is executing. Or \omega = 2\pi × 1 × 10^3 (rev/sec) \cong 6 × 10^3 (rad/sec) The linear velocity is v = \omegar \cong [6 × 10^3 (rad/sec) ](10 cm) \cong 6 × 10^4 cm/sec The magnitude of the acceleration associated with circular motion is a = v^2/r = \omega^2r since the particle is going around in a circle. a \cong [6 × 10^3 (rad/sec)]^2 (10 cm) \cong 4 × 10^8(cm/sec^2) Now the acceleration g due to gravity is only 980 cm/s^2 at the surface of the earth , so that the ratio of the rotational acceleration to the gravitational acceleration is a/g \approx (4 × 10^8)/(10^3) \approx 4 × 10^5

Question:

A capillary tube of length 50 cm is closed at both ends. It contains dry air at each end separated by a mercury column 10 cm long. With the tube horizontal, the air columns are both 20 cm long, but with the tube vertical the columns are 15 cm and 25 cm long. What is the pressure in the capillary tube when it is horizontal?

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/Users/wenhuchen/Documents/Crawler/Physics/D10-0412.htm

Solution:

When the mercury column is vertical, the pressure on the gas in the two parts is as shown in the diagram, where p_1 is the pressure at the foot of the mercury column and p_2 the pressure at the top. But the difference in pressure at two levels in a vertical column of liquid is known from the laws of hydrostatic pressure. Thus measuring distance from the top of the mercury column, p_2 = p_2 andp_1 = \rhog(h) + p_2 whencep_2 - p_1 = - \rhogh Here, \rho is the density of mercury. Applying Boyle's law to section C of the gas when the capillary is in its 2 positions, we obtain p_1Al_1 = p_0Al_0 Similarly, for section B p_2Al_2 = p_0Al_0 where p_0 and l_0 refer to conditions when the tube is horizontal. Thus p_1 = p_0 (l_0/l_1)andp_2 = p_0 (l_0/l_2). \thereforep_1 - p_2 = \rhogh = p_0l_0 [(1/l_1) - (1/l_2)] = \thereforep_1 - p_2 = \rhogh = p_0l_0 [(1/l_1) - (1/l_2)] = p_0l_0 [(l_2 - l_1)/(l_1l_2)] \thereforep_0 = [\rhogh l_1l_2]/[l_0(l_2 - l_1)] \therefore(p_0 / \rhog) = [(l_1l_2 ) / { l0(l2-l_1 )}] h = [ {15 cm × 25 cm} / {20 cm(25 - 15)cm} ] × 10 cm = 18.75 cm of mercury.

Question:

Write a Basic program that allows a player to play Tic-Tac-Toe with a computer. Tic-Tac-Toe with a computer.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G23-0558.htm

Solution:

The game board is numbered

Question:

Write apseudocodedprogram to input a sequence of numbers(representing kilowatt-hours, KWH) and output electricbills according to the following schedule: Each of the first 100 KWHs costs 10 cents Each of the next 100 costs 6 cents Each of the next 200 costs 8 cents Each KWH over 400 costs 10 cents. The program should terminate when a negative number is inputted. Also, how could you improve this programto handlethe situation in which the rate schedule changed rapidly?

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G18-0448.htm

Solution:

The program logic is straightforward. A DO-WHILE construct is set up, so that each entry of kilowatt-hours is processed separately until a negativevalue is encoun-tered . realKWH, COST inputKWH (1) through KWH (N) dowhile KWH > 0 if0 \leq KWH < 100 thenCOST\leftarrow10.0\textasteriskcenteredKWH if100 \leq KWH < 200 thenCOST\leftarrow(10. 0\textasteriskcentered100.0) + (6. 0\textasteriskcentered (KWH-100.0)) if200 \leq WH \leq 400 thenCOST\leftarrow(10.0\textasteriskcentered100.0) + (6.0\textasteriskcentered100.0) + (8.0\textasteriskcentered(KWH-200.0)) elseCOST\leftarrow(10.0\textasteriskcentered100.0) + (6.0\textasteriskcentered100.0) + (8.0\textasteriskcentered200.0) + (10.0\textasteriskcentered(KWH-400.0)) outputKWH, COST enddo while endprogram If the rates of the electric bills climb in price often, the program wouldhave to be rewritten just as often. To avoid this nuisance, we could writesubprograms to do the job. After each IF-THEN construct, a statementcalling a particular subprogram with a certain billing schedule couldbe inserted. In this way, the main program is left untouched, and you wouldneed to rewrite only a few lines of code in a subprogram. This is veryoften the case in programming for business applications. An alternative way of handling the situation of rapid rate schedule changeswould be to insert 2 READ statements in the program, one for inputtingthe KWH segments, and another for inputting the price of each segment.

Question:

Write a program which counts the number of times each letter is used in input text.

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/Users/wenhuchen/Documents/Crawler/ComputerScience/G15-0387.htm

Solution:

1&TRTM = 1 2CHAR = LEN(1) .CH 3LETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' 4COUNT = TABLE (30) 5READOUTPUT = INPUT:F(DISPLAY) 6TEXT = OUTPUT 7NEXTTEXT CHAR =:F(READ) 8COUNT = COUNT + 1:(NEXT) 9DISPLAYOUTPUT = 10LOOPLETTERS CHAR =: F (END) 11OUTPUT = NE(COUNT) CH 'OCCURS' +COUNT 'TIMES'(LOOP) 12END The first statement includes a keyword '&TRIM'. Keywords in SNOBGL IV allow access to several parameters and switches internal to the SNOBOL IV system. Keywords begin with ampersand (&) followed by certain identifiers. For example, keyword &TRIM controls the handling of trailing blanks on input of data. If the value of &TRIM is nonzero, trailing blanks are deleted. Other keywords will be explained upon occurrence in the programs. The second statement, CHAR = LEN (1). CH, includes two important features. First, LEN (1), is an example of primitive (Built-in) functions. This function returns a pattern that matches any string of the length specified by the integer in parentheses. INPUT, PUNCH, and OUTPUT are some other primitive functions. The second important feature of the second statement is that it is an example of conditional value assignment (in this case .CH). The conditional value assignment operator (.) is separated from its operands by blanks. In the state-ment CHAR = LEN (1) . CH, variable CHAR is assigned the very first found string of length one, and, in case of success, assigns the same string to CH, which is used as an array subscript later in the program. The fourth statement, COUNT = TABLE (30), introduces the use of tables. Tables of variables can be created by using the primitive function TABLE, for example: T = TABLE ( ) In a very similar way arrays of variables are constructed by using the primitive function ARRAY, for example: A = ARRAY ( ) The arguments of an array, written in parentheses, describe the number of dimensions, the bounds of each dimension, and the initial value of each variable in the array. Thus A = ARRAY (10,1.0) creates and assigns to Aaone-dimensional array of 10 variables, each initialized to the real value 1.0. A table is similar to a one-dimensional array. However, instead of referencing an element with an integer, any data object can be used. In this program, letters of the alphabet are used as subscripts for table elements. The fifth statement of the program introduces the most important feature of the flow of control - a transfer to alabelledstatement. It is specified in the GOTO field which appears at the end of a statement and is separated from the rest of the statement by a colon. Two types of transfers can be specified in the GOTO field: conditional and unconditional, A conditional transfer consists of letters F or S corresponding to failure or success respectively, followed by a label enclosed within parentheses. Thus OUTPUT = INPUT: F(DISPLAY) makes the computerprint outwhatever appears on the data card. If, however, there is no data in the input file, i.e., an end of file is encountered, the transfer is made to the statement labeled display. An unconditional transfer is indicated by the absence of an F or S before the enclosing parentheses. For example, consider state-ment #8 of this program: COUNT = COUNT + 1: (NEXT) Each time this statement is executed, the transfer is made to the statementlabelledNEXT. Statement #11 uses one of the primitive functions of a specific group called NUMERICAL PREDICATES. The main ones include: LT - less than... LE - less than or equal to... EQ - equal to... NE - not equal to... GE - greater or equal to... GT - greater than... Statement OUTPUT = NE (COUNT).... prints out whatever is following the parentheses if the value of COUNT is not equal to zero. Finally, a statement that is longer than one line can be continued onto successive lines by starting the continuation lines with a period or plus sign. This program employs a plus sign for continua-tion of statement #11. Also note, that the statements in this program are numbered only for easy reference. All labels in SNOBCL IV start with a letter, as was mentioned earlier. If a statement does not start with a label, it must start with at least one blank. The general flow of this program is as follows: statement 1 - deletes trailing blanks; statement 2 - indicates that CHAR and CH will be assigned the first one-character string matched; statement 3 - assigns a string of the 26 letters of the alphabet to variable LETTERS; statement 4 - sets a 30-element table and assigns it to COUNT; statement 5 - prints out the first card of the input text; statement 6 - assigns it to variable TEXT; statement 7 - tests the text for presence of the first one-character string (which is the first letter of the first word in the text), deletes them all one at a time, adding 1 to COUNT in statement 8 each time around; statement 8 - goes back to statement 7 every time after adding 1 to COUNT; statement 7 - repeats the loop until all the appearances of each letter are counted, and the table COUNT is set up. Then COUNT fails when no more characters are left; the transfer is made back to statement #5; statement 5 - repeats the whole procedure for every following input card, adding missing members (if there are any) to the table COUNT. Finally it fails when there is no input left and transfers the execution to statement #9; statement 9 - prints a blank line; statement 10 - tests string LETTERS for the presence of a one- character string, finds 26 of them, assigns the first one (i.e., A) to CHAR and deletes it from LETTERS; statement 11 - since CH takes the same value as CHAR, looks in the table for C0UNT which represents the number of times letter A appeared in the text. If finds it, outputs the letter A and the number of its occurrences. In any case it goes back to statement #10; statement 10 - repeats the same procedure for every letter. It fails when no more letters are left and finishes the program.

Question:

Differentiate clearly between diffusion, dialysis and osmosis.

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/Users/wenhuchen/Documents/Crawler/Biology/F02-0058.htm

Solution:

Diffusion is the general term for the net movement of the particles ofa substance from a region where the substance is at a high concentrationto regions where the substance is at a low concentration. The particles are in constant random motion with their speed being directly relatedto their size and the temperature. When the movements of all the particlesare considered jointly, there is a net movement away from the regionof high concentration towards regions of low concentration. This resultsin the particles of a given substance distri-buting themselves with relativelyuniform density or con-centration within any available space. Diffusion tends to be faster in gases than in liquids, and much slower in solids. The movement or diffusion of water or solvent molecules through a semipermeablemembrane is called osmosis. The osmotic pressure of a solutionis a measure of the tendency of the solvent to enter the solu-tion byosmosis. The more concentrated a solution, the more water will tend to moveinto it, and the higher is its osmotic pressure. The diffusion of a dissolved substance through asemipermeable membraneis termed dialysis. Dialysis is the movement of the solute, while osmosisis the movement of the solvent through asemipermeable membrane. Dialysis and osmosis are just two special forms of diffusion.

Question:

Find the pH of a solution in which [H^+] = 6.38 × 10^-6 mole/liter.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E10-0342.htm

Solution:

pH indicates the acidity orbasicityof a solution. It runs on a scale of1 to 14. 1 is most acidic, 7isneutral, and 14 is most basic.pHis a measureof [H^+]. The higher the [H^+], the lower the pH, and the stronger theacid. To calculatepH usethe equation: pH= - log [H^+] = - log [6.38 × 10^-6] = - [(- 6) + (log 6.38)] = 6 + (- .805) pH= 5.195.

Question:

Find the pH of 0.15 M H_2SO_4, solution, assuming K_2 =1.26 × 10^-2.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E12-0425.htm

Solution:

Begin this problem by noting that pH = - log [H^+]. Thus, to find pH, calculate the [H^+] in 0.15 M H_2SO_4. H_2SO_4 is an acid and dissociates into H^+ (= H_3O^+) ions. The general reaction for acid dissociation can be written as HA + H_2O \rightleftarrows H_3O^+ + A^-. H_2SO_4 undergoes this reaction, as H_2SO_4 + H_2O \rightleftarrows H_3O^+ + HSO_4^-. But note, HSO_4^- is also an acid, it still possesses a hydrogen that can dissociate. Thus, one also has HSO_4^- + H_2O \rightleftarrows H_3O^+ + SO_4^-. In other words, there exist two dissociation reactions. As given, the dissociation constant for the second reaction, K_2, measures the ratio of the concentrations of products to reactants. It is from this expression that [H^+] can be determined. H_2SO_4 is a very strong acid in its first dissociation reaction. This means it is completely ionized; i.e. 100% of the H^+ comes off (the first H^+ only). Thus, if there are 0.15 M of H_2SO_4, then, there are also 0.15 = [H^+] for the first dissociation. [H^+] = [HSO_4^-], as can be seen from the first dissociation reaction and are formed inequimolaramounts. Also, [HSO_4^-] = 0.15 (initially). For the second ionization, one is given K_2 = 1.26 × 10^-2, indicating dissociation is not complete. From the prior explanation of what K_2 indicates, one can write K_2 = 1.26 × 10^-2 = {[H_3O^+] [SO_4-]} / [HSO_4^-]. Note: Water is not included since it is assumed to be a constant. From the first dissociation, it is known that [H_3O^+] = 0.15. If x moles/liter of H^+ is produced in the second dissociation, the total [H_3O^+] = 0.15 + x. [SO_4 =] = [H_3O^+] from the second dissociation, since here they are formed in equimolar amounts. Thus, [SO_4-] = x. If one started with [HSO_4^-] = 0.15M (from the first dissociation) and x moles/liter dissociate in the second, one has [HSO_4^-] = 0.15 - x left. Substituting these values, 1.26 × 10^-2 = {[H_3O^+] [SO_4-]} / [HSO_4^-] = {(0.15 + x) (x)} / (0.15 - x) Solving for x (using the quadratic formula), x = .011. Thus, [H^+] = 0.15 + 0.011 = 0.161. pH = - log [H^+], so that pH = - log [.161] = 0.79.

Question:

You have the reaction HCrO^-4+ 3Fe^2+ + 7H^+ \rightarrow Cr3++ 3Fe^3 + 4H_2O with the rate law expression:rate = k [HCrO^-_4] [Fe^2+] [H^+].Explain why the rate of reaction is not proportional to the number of each species reacting according to the stoichiometryof the balanced equation.

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/Users/wenhuchen/Documents/Crawler/Chemistry/E13-0476.htm

Solution:

Stoichiometrypredicts only the amount of products expected; it cannot predict when they will be produced. The rate of production is dependent only on the number of molecules in the activated complex and their concentrations in the system. A reaction can occur only when the species come in contact with each other. Thus, the concentration reflects the probability of a collision among the species.Stoichiometrycannot provide any in-formation in these areas, and, as such, the rate of the reaction is not necessarily reflective of thestoichio-metryof the balanced equation.

Question:

A 3.60 x 10^4 -kg rocket rises vertically from rest. It ejects gas at an exhaust velocity of 1800 m/sec at a mass rate of 580 kg/sec for 40 sec before the fuel is expended. Determine the upward acceleration of the rocket at times t = 0,20, and 40 sec.

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/Users/wenhuchen/Documents/Crawler/Physics/D06-0333.htm

Solution:

Suppose a rocket is travelling with a velocity v^\ding{217} relative to a stationary coordinate system (S), and emits fuel at velocity u^\ding{217} with respect to the rocket (S') (see figure). The exhaust velocity with respect to S, w^\ding{217}, is the exhaust velocity with respect to S' plus the velocity of S' with respect to S or w^\ding{217} = u^\ding{217} + v^\ding{217}. Now that we know all the velocities with respect to a stationary frame, we may use the law of conservation of momentum in this frame to find the velocity of the rocket after a mass \Deltam has been emitted. Att = 0, no fuel has been emitted and the initial momentum of the fuel-rocket system is P_0^\ding{217} = Mv^\ding{217}. Att = \Deltat, a mass \DeltaM (where \DeltaM > 0) of fuel has been emitted and travels with velocity w^\ding{217}. The rocket now has mass M - \DeltaM and travels with a velocity v^\ding{217} + \Deltav^\ding{217}. The momentum is then P_f^\ding{217} = (M - \DeltaM)(v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(w^\ding{217}) orP_f^\ding{217} = (M - \DeltaM)(v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(u^\ding{217} + v^\ding{217}) Hence, the change in momentum is \DeltaP^\ding{217} = P_f^\ding{217} - P_0^\ding{217} = (M - \DeltaM) (v^\ding{217} + \Deltav^\ding{217}) + \DeltaM(u^\ding{217} + v^\ding{217}) - Mv^\ding{217} = Mv^\ding{217} - \DeltaMv^\ding{217} + M\Deltav^\ding{217} - \DeltaM\Deltav^\ding{217} + u^\ding{217}\DeltaM + v^\ding{217}\DeltaM - Mv^\ding{217} = M\Deltav^\ding{217} + u^\ding{217}\DeltaM - \DeltaM\Deltav^\ding{217} Then\DeltaP^\ding{217}/\Deltat = M (\Deltav^\ding{217}/\Deltat) + u^\ding{217} (\DeltaM/\Deltat) - (\DeltaM\Deltav^\ding{217})/(\Deltat) Taking the limit as \Deltat\rightarrow0 dP^\ding{217}/dt = lim_\Deltat\rightarrow0 \DeltaP^\ding{217}/\Deltat = M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) because \DeltaM∆v^\ding{217}/\Deltat\rightarrow0 as \Deltat\rightarrow0, whence dP^\ding{217}/dt = M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) But, for a constant mass system, which our's is, (see figure) the time rate of change of momentum of the system equals the net external force on the system and M (dv^\ding{217}/dt) + u^\ding{217} (dM/dt) = F^\ding{217}_ext(1) In this equation, dv^\ding{217}/dt is the rocket's acceleration, M is the instantaneous mass of the rocket, F^\ding{217}_ext is the netforce on the fuel-rocket system, and dM/dt is the rate of change of the rocket's mass. Note that in (1), dM/dt > 0 due to our derivation. Hence, we may replace dM/dt by- dM/dt if we redefine dM/dt to be less than zero. Then M (dv^\ding{217}/dt) - u^\ding{217} (dM/dt) = F^\ding{217}ext orM (dv^\ding{217}/dt) = orM (dv^\ding{217}/dt) = F^\ding{217}_ext + u^\ding{217} (dM/dt)(2) We define u^\ding{217} (dM/dt) as the rocket's thrust. Solving for (dv^\ding{217}/dt) dv^\ding{217}/dt = [(F^\ding{217}_ext)/M] + (u^\ding{217}/M) (dM/dt)(3) For our problem, F^\ding{217}_ext = - Mg\^{J}, where \^{J} is a unit vector in the positive y direction. Furthermore, if the rocket is propelled straight up, u^\ding{217} = - u\^{J}. Hence, (dv^\ding{217}/dt) = -g\^{J} - (u/M) \^{J} (dM/dt) = -\^{J} [g + (u/m) (dM/dt)](4) Note that since M is a function of time, dv^\ding{217}/dt will also be time dependent. Att = 0, M = 3.60 × 10^4 kg anddv^\ding{217}/dt = -\^{J} [9.8m/S^2 + {(1800 m/s)/(3.6 × 10^4 kg)} {(- 580 kg)/S}] dv^\ding{217}/dt = -\^{J} (9.8m/s^2 - 29m/s^2) dv^\ding{217}/dt = 19.2 m/s^2 \^{J} Att = 20 secs, M = 3.6 × 10^4 kg - (580 kg/s)(20s) = 3.6 × 10^3 kg - 11.6 × 10^3 kg = 24.4 × 10^3 kg anddv^\ding{217}/dt = -\^{J} [9.8 m/S^2 + {(1800 m/s)/(2.44 x 10^4)} {- (580 kg/S)} ] dv^\ding{217}/dt= -\^{J} (9.8m/s^2 - 42.79m/s^2) = 32.99 m/s2\^{J} Att = 40 secs, M = 3.6 × 10^4 kg - (580 kg/s) (40s) = 3.6 × 10^4 kg - 2.32 × 10^4 kg = 1.28 × 10^4 kg anddv^\ding{217}/dt = -\^{J} [9.8 m/S^2 + {(1800 m/s)/(1.28 x 104kg)} {- (580 kg/S)}] dv^\ding{217}/dt = - \^{J} (9.8 m/S^2 - 81.56 m/s^2) dv^\ding{217}/dt = 71.76 m/s^2 \^{J} In this example we have neglected air friction and the variation of g^\ding{217} with altitude.